Or Questions

1

4

Decision Analysis

MULTIPLE CHOICE

1. The options from which a decision maker chooses a course of action are

a. called the decision alternatives. b. under the control of the decision maker. c. not the same as the states of nature. d. All of the alternatives are true.

ANSWER: d TOPIC: Structuring the decision problem 2. States of nature

a. can describe uncontrollable natural events such as floods or freezing temperatures. b. can be selected by the decision maker. c. cannot be enumerated by the decision maker. d. All of the alternatives are true.

ANSWER: a TOPIC: Structuring the decision problem 3. A payoff

a. is always measured in profit. b. is always measured in cost. c. exists for each pair of decision alternative and state of nature. d. exists for each state of nature.

ANSWER: c TOPIC: Payoff tables 4. Making a good decision a. requires probabilities for all states of nature. b. requires a clear understanding of decision alternatives, states of nature, and payoffs. c. implies that a desirable outcome will occur. d. All of the alternatives are true. ANSWER: b TOPIC: Decision making without probabilities

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2 Chapter 4 Decision Analysis

5. A decision tree a. presents all decision alternatives first and follows them with all states of nature. b. presents all states of nature first and follows them with all decision alternatives. c. alternates the decision alternatives and states of nature. d. arranges decision alternatives and states of nature in their natural chronological order.

ANSWER: d TOPIC: Decision trees 6. Which of the methods for decision making without probabilities best protects the decision maker from

undesirable results? a. the optimistic approach b. the conservative approach c. minimum regret d. minimax regret

ANSWER: b TOPIC: Conservative approach 7. Sensitivity analysis considers a. how sensitive the decision maker is to risk. b. changes in the number of states of nature. c. changes in the values of the payoffs. d. changes in the available alternatives. ANSWER: c TOPIC: Sensitivity analysis 8. To find the EVSI, a. use the EVPI to calculate sample information probabilities. b. use indicator probabilities to calculate prior probabilities. c. use prior and sample information probabilities to calculate revised probabilities. d. use sample information to revise the sample information probabilities. ANSWER: c TOPIC: Expected value of sample information 9. If P(high) = .3, P(low) = .7, P(favorable | high) = .9, and P(unfavorable | low) = .6, then P(favorable) =

a. .10 b. .27 c. .30 d. .55

ANSWER: d TOPIC: Conditional probability 10. The efficiency of sample information is a. EVSI*(100%) b. EVSI/EVPI*(100%) c. EVwoSI/EVwoPI*(100%) d. EVwSI/EVwoSI*(100%) ANSWER: b TOPIC: Efficiency of sample information 11. Decision tree probabilities refer to a. the probability of finding the optimal strategy b. the probability of the decision being made c. the probability of overlooked choices d. the probability of an uncertain event occurring ANSWER: d TOPIC: Decision tree


Chapter 4 Decision Analysis 3

12. For a maximization problem, the conservative approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: b TOPIC: Decision making without probabilities 13. For a minimization problem, the optimistic approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: d TOPIC: Decision making without probabilities 14. For a maximization problem, the optimistic approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: c TOPIC: Decision making without probabilities 15. For a minimization problem, the conservative approach is often referred to as the a. minimax approach b. maximin approach c. maximax approach d. minimin approach ANSWER: a TOPIC: Decision making without probabilities

TRUE/FALSE

1. Sample information with an efficiency rating of 100% is perfect information. ANSWER: True TOPIC: Efficiency of sample information 2. States of nature should be defined so that one and only one will actually occur. ANSWER: True TOPIC: Structuring the decision process 3. Decision alternatives are structured so that several could occur simultaneously. ANSWER: False TOPIC: Structuring the decision problem 4. Square nodes in a decision tree indicate that a decision must be made. ANSWER: True TOPIC: Decision trees 5. Circular nodes in a decision tree indicate that it would be incorrect to choose a path from the node.



ANSWER: True TOPIC: Decision trees 6. Risk analysis helps the decision maker recognize the difference between the expected value of a decision

alternative and the payoff that may actually occur. ANSWER: True TOPIC: Risk analysis 7. The expected value of an alternative can never be negative. ANSWER: False TOPIC: Decision making with probabilities 8. Expected value is the sum of the weighted payoff possibilities at a circular node in a decision tree. ANSWER: True TOPIC: Decision making with probabilities 9. EVPI is always greater than or equal to EVSI. ANSWER: True TOPIC: Expected value of sample information 10. After all probabilities and payoffs are placed on a decision tree, the decision maker calculates expected

values at state of nature nodes and makes selections at decision nodes. ANSWER: True TOPIC: Developing a decision strategy 11. A decision strategy is a sequence of decisions and chance outcomes, where the decisions chosen depend on

the yet to be determined outcomes of chance events. ANSWER: True TOPIC: Decision strategy 12. EVPI equals the expected regret associated with the minimax decision. ANSWER: True TOPIC: Minimax regret approach 13. The expected value approach is more appropriate for a one-time decision than a repetitive decision. ANSWER: False TOPIC: Decision making with probabilities 14. Maximizing the expected payoff and minimizing the expected opportunity loss result in the same

recommended decision. ANSWER: True TOPIC: Minimax regret approach 15. The expected value of sample information can never be less than the expected value of perfect information. ANSWER: False TOPIC: Expected value of sample information

SHORT ANSWER

1. Explain why the decision maker might feel uncomfortable with the expected value approach, and decide to

use a non-probabilistic approach even when probabilities are available.



TOPIC: Decision making with probabilities 2. Why perform sensitivity analysis? Of what use is sensitivity analysis where good probability estimates are

difficult to obtain? TOPIC: Sensitivity analysis 3. How can a good decision maker “improve” luck? TOPIC: Introduction 4. Use a diagram to compare EVwPI, EVwoPI, EVPI, EVwSI, EVwoSI, and EVSI. TOPIC: Expected value of sample information. 5. Show how you would design a spreadsheet to calculate revised probabilities for two states of nature and

two indicators. TOPIC: Decision analysis and spreadsheets

PROBLEMS

1. Jim has been employed at Gold Key Realty at a salary of $2,000 per month during the past year. Because

Jim is considered to be a top salesman, the manager of Gold Key is offering him one of three salary plans for the next year: (1) a 25% raise to $2,500 per month; (2) a base salary of $1,000 plus $600 per house sold; or, (3) a straight commission of $1,000 per house sold. Over the past year, Jim has sold up to 6 homes in a month.

a. Compute the monthly salary payoff table for Jim. b. For this payoff table find Jim's optimal decision using: (1) the conservative approach, (2)

minimax regret approach. c. Suppose that during the past year the following is Jim's distribution of home sales. If one

assumes that this a typical distribution for Jim's monthly sales, which salary plan should Jim select?

Home Sales Number of Months 0 1 1 2 2 1 3 2 4 1 5 3 6 2 TOPIC: Decision making with and without probabilities 2. East West Distributing is in the process of trying to determine where they should schedule next year's

production of a popular line of kitchen utensils that they distribute. Manufacturers in four different countries have submitted bids to East West. However, a pending trade bill in Congress will greatly affect the cost to East West due to proposed tariffs, favorable trading status, etc.

After careful analysis, East West has determined the following cost breakdown for the four manufacturers (in $1,000's) based on whether or not the trade bill passes:

Bill Passes Bill Fails Country A 260 210 Country B 320 160 Country C 240 240



Country D 275 210

a. If East West estimates that there is a 40% chance of the bill passing, which country should they choose for manufacturing?

b. Over what range of values for the "bill passing" will the solution in part (a) remain optimal? TOPIC: Decision making with probabilities 3. Transrail is bidding on a project that it figures will cost $400,000 to perform. Using a 25% markup, it will

charge $500,000, netting a profit of $100,000. However, it has been learned that another company, Rail Freight, is also considering bidding on the project. If Rail Freight does submit a bid, it figures to be a bid of about $470,000. Transrail really wants this project and is considering a bid with only a 15% markup to $460,000 to ensure winning regardless of whether or not Rail Freight submits a bid. a. Prepare a profit payoff table from Transrail's point of view. b. What decision would be made if Transrail were conservative? c. If Rail Freight is known to submit bids on only 25% of the projects it considers, what decision

should Transrail make? d. Given the information in (c), how much would a corporate spy be worth to Transrail to find out if

Rail Freight will bid? TOPIC: Decision making with and without probabilities 4. The Super Cola Company must decide whether or not to introduce a new diet soft drink. Management feels

that if it does introduce the diet soda it will yield a profit of $1 million if sales are around 100 million, a profit of $200,000 if sales are around 50 million, or it will lose $2 million if sales are only around 1 million bottles. If Super Cola does not market the new diet soda, it will suffer a loss of $400,000.

a. Construct a payoff table for this problem. b. Construct a regret table for this problem. c. Should Super Cola introduce the soda if the company: (1) is conservative; (2) is optimistic; (3)

wants to minimize its maximum disappointment? d. An internal marketing research study has found P(100 million in sales) = 1/3; P(50 million in sales) = 1/2; P(1 million in sales) = 1/6. Should Super Cola introduce the new diet

soda? e. A consulting firm can perform a more thorough study for $275,000. Should management have

this study performed? TOPIC: Decision making with and without probabilities 5. Super Cola is also considering the introduction of a root beer drink. The company feels that the probability

that the product will be a success is .6. The payoff table is as follows:

Success (s1) Failure (s2) Produce (d1) $250,000 -$300,000 Do Not Produce (d2) -$ 50,000 -$ 20,000

The company has a choice of two research firms to obtain information for this product. Stanton Marketing has market indicators, I1 and I2 for which P(I1!s1) = .7 and P(I1!s2) = .4. New World Marketing has indicators J1 and J2 for which P(J1!s1) = .6 and P(J1!s2) = .3.

a. What is the optimal decision if neither firm is used? Over what probability of success range is this

decision optimal? b. What is the EVPI? c. Find the EVSIs and efficiencies for Stanton and New World. d. If both firms charge $5,000, which firm should be hired? e. If Stanton charges $10,000 and New World charges $4,000, which firm should Super Cola hire?

Why?



TOPIC: Computing branch probabilities 6. Dollar Department Stores has just acquired the chain of Wenthrope and Sons Custom Jewelers. Dollar has

received an offer from Harris Diamonds to purchase the Wenthrope store on Grove Street for $120,000. Dollar has determined probability estimates of the store's future profitability, based on economic outcomes, as: P($80,000) = .2, P($100,000) = .3, P($120,000) = .1, and P($140,000) = .4. a. Should Dollar sell the store on Grove Street? b. What is the EVPI? c. Dollar can have an economic forecast performed, costing $10,000, that produces indicators I1 and

I2, for which P(I1!80,000) = .1; P(I1!100,000) = .2; P(I1!120,000) = .6; P(I1!140,000) = .3. Should Dollar purchase the forecast?

TOPIC: Computing branch probabilities 7. An appliance dealer must decide how many (if any) new microwave ovens to order for next month. The

ovens cost $220 and sell for $300. Because the oven company is coming out with a new product line in two months, any ovens not sold next month will have to be sold at the dealer's half price clearance sale. Additionally, the appliance dealer feels he suffers a loss of $25 for every oven demanded when he is out of stock. On the basis of past months' sales data, the dealer estimates the probabilities of monthly demand (D) for 0, 1, 2, or 3 ovens to be .3, .4, .2, and .1, respectively.

The dealer is considering conducting a telephone survey on the customers' attitudes towards microwave ovens. The results of the survey will either be favorable (F), unfavorable (U) or no opinion (N). The dealer's probability estimates for the survey results based on the number of units demanded are:

P(F!D = 0) = .1 P(F!D = 2) .3 P(U!D = 0) = .8 P(U!D = 2) = .1 P(F!D = 1) = .2 P(F!D = 3) .9 P(U!D = 1) = .3 P(U!D = 3) = .1

a. What is the dealer's optimal decision without conducting the survey? b. What is the EVPI? c. Based on the survey results what is the optimal decision strategy for the dealer? d. What is the maximum amount he should pay for this survey?

TOPIC: Computing branch probabilities 8. Lakewood Fashions must decide how many lots of assorted ski wear to order for its three stores.

Information on pricing, sales, and inventory costs has led to the following payoff table, in thousands.

Demand Order Size Low Medium High

1 lot 12 15 15 2 lots 9 25 35 3 lots 6 35 60

a. What decision should be made by the optimist? b. What decision should be made by the conservative? c. What decision should be made using minimax regret? TOPIC: Decision making without probabilities 9. The table shows both prospective profits and losses for a company, depending on what decision is made

and what state of nature occurs. Use the information to determine what the company should do.

State of Nature Decision s1 s2 s3

d1 30 80 -30



d2 100 30 -40 d3 -80 -10 120 d4 20 20 20

a. if an optimistic strategy is used. b. if a conservative strategy is used. c. if minimax regret is the strategy. TOPIC: Decision making without probabilities 10. A payoff table is given as


d1 10 8 6 d2 14 15 2 d3 7 8 9

a. What decision should be made by the optimistic decision maker? b. What decision should be made by the conservative decision maker? c. What decision should be made under minimax regret? d. If the probabilities of s1, s2, and s3 are .2, .4, and .4, respectively, then what decision should be

made under expected value? e. What is the EVPI? TOPIC: Decision making with and without probabilities 11. A payoff table is given as

State of Nature

Decision s1 s2 s3 d1 250 750 500 d2 300 -250 1200 d3 500 500 600

a. What choice should be made by the optimistic decision maker? b. What choice should be made by the conservative decision maker? c. What decision should be made under minimax regret? d. If the probabilities of d1, d2, and d3 are .2, .5, and .3, respectively, then what choice should be

made under expected value? e. What is the EVPI? TOPIC: Decision making with and without probabilities 12. A decision maker has developed the following decision tree. How sensitive is the choice between N and P

to the probabilities of states of nature U and V? Topic: Sensitivity analysis

N

P

U

V

50

5



13. If p is the probability of Event 1 and (1-p) is the probability of Event 2, for what values of p would you choose A? B? C? Values in the table are payoffs.

Choice/Event Event 1 Event 2

A 0 20 B 4 16 C 8 0

TOPIC: Sensitivity analysis 14. Fold back the decision tree and state what strategy should be followed.

A

B

C .25

D .35

E .40

F .2

G .8

H

J

K

M

N

Q

R

S .55

T .45

U .25

V .75

W .5

X .5

Y .4

Z .6AA .3

BB .4

CC .3

300

100

10

80

-30

120

500

-200150

100

50

18020

-50

50

TOPIC: Expected value and decision trees



15. Fold back this decision tree. Clearly state the decision strategy you determine.

A

C

D .42

E .24

F .34

G .2

B

H .8

I .82

J .18

K

L

M

N

P .63

Q .37

R .3S .3T .4

100

-4030

6035

2550

-100120150

10

-50

TOPIC: Expected value and decision trees 16. If sample information is obtained, the result of the sample information will be either positive or negative.

No matter which result occurs, the choice to select option A or option B exists. And no matter which option is chosen, the eventual outcome will be good or poor. Complete the table.

Sample Result

States of Nature

Prior Probabilities

Conditional Probabilities

Joint Probabilities

Posterior Probabilities

Positive good .7 P(positive | good) = .8 poor .3 P(positive | poor) = .1

Negative good .7 P(negative | good) = poor .3 P(negative | poor) =

TOPIC: Posterior probabilities 17. Use graphical sensitivity analysis to determine the range of values of the probability of state of nature s1

over which each of the decision alternatives has its largest expected value.

State of Nature Decision s1 s2

d1 8 10 d2 4 16 d3 10 0

TOPIC: Graphical sensitivity analysis 18. Dollar Department Stores has received an offer from Harris Diamonds to purchase Dollar’s store on Grove

Street for $120,000. Dollar has determined probability estimates of the store's future profitability, based on economic outcomes, as: P($80,000) = .2, P($100,000) = .3, P($120,000) = .1, and P($140,000) = .4.

a. Should Dollar sell the store on Grove Street? b. What is the EVPI?



c. Dollar can have an economic forecast performed, costing $10,000, that produces indicators I1 and I2, for which P(I1!80,000) = .1; P(I1!100,000) = .2; P(I1!120,000) = .6; P(I1!140,000) = .3. Should Dollar purchase the forecast?

TOPIC: Posterior probabilities SOLUTIONS TO PROBLEMS 1. a. There are three decision alternatives (salary plans) and seven states of nature (the number of

houses sold monthly).

PAYOFF TABLE Number of Homes Sold 0 1 2 3 4 5 6 Salary Plan I 2500 2500 2500 2500 2500 2500 2500 Salary Plan II 1000 1600 2200 2800 3400 4000 4600 Salary Plan III 0 1000 2000 3000 4000 5000 6000

b. (1) Conservative Approach (Maximin): Plan I

REGRET TABLE Number of Homes Sold 0 1 2 3 4 5 6 Salary Plan I 0 0 0 500 1500 2500 3500 Salary Plan II 1500 900 300 200 600 1000 1400 Salary Plan III 2500 1500 500 0 0 0 0

(2) Minimax Regret Approach: Plan II

c. Use the relative frequency method for determining the probabilities. Using the EV approach: EV(Plan I) = 2500, EV(Plan II) = 3050, EV(Plan III) = 3417; Choose Plan III.

2. a. Using EV approach: EV(A) = 230, EV(B) = 224, EV(C) = 240; Choose Country B (lowest EV)

b. As long as the probability of the bill passing is less than .455, East West should choose Country B.



300

250

200

150

100

50

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

| | | | | | | | | | | | | | | | | |

Expected Value

.455

Probability

C B

A

0

3. a.

Rail Freight Transrail Bid $470,000 Doesn't Bid Bid $500,000 $0 $100,000 Bid $460,000 $60,000 $ 60,000

b. Bid $460,000 c. Bid $500,000 d. $15,000

4. a.

PAYOFF TABLE Sales ($millions) 100 50 1 Introduce $1,000,000 $200,000 -$2,000,000 Do Not Introduce -$400,000 -$400,000 -$400,000

b. REGRET TABLE

Sales ($millions)

100 50 1 Introduce $0 $0 $1,600,000 Do Not Introduce $1,400,000 $600,000 $0

c. (1) do not introduce; (2) introduce; (3) do not introduce d. Yes e. No

5. a. Introduce root beer; p < .483

b. EVPI = $112,000 NOTE: The answers to (c)-(e) are very sensitive to roundoff error.

Figures in parentheses are for two decimal places only. c. Stanton: EVSI = $13,200 ($11,862)

Efficiency = .118 (.106)



New World: EVSI = $6,400 ($6,424) Efficiency = .057 (.057)

d. Hire Stanton (Stanton) e. Hire New World (Stanton)

6. a. Yes, Dollar should sell store b. EVPI = $8,000 c. No; survey cost exceeds EVPI

7.

Demand For Ovens Ovens Ordered 0 1 2 3

0 0 -25 -50 -75 1 -70 80 55 30 2 -140 10 160 135 3 -210 -60 90 240

a. Order one oven: EV = $25.00 b. EVPI = $63.00 c. Favorable: order 2; Unfavorable: order 0; No opinion: order 1 d. EVSI = $9.10

8. a. 3 lots

b. 1 lot c. 3 lots

Regret table: Order Size

Demand Maximum Regret Low Medium High

1 lot 0 20 45 45 2 lots 3 10 25 25 3 lots 6 0 0 6

9. a. d3 b. d1 c. d4

Regret table: State of Nature Maximum

Regret Decision s1 s2 s3 d1 70 0 150 150 d2 0 50 160 160 d3 180 90 0 180 d4 80 60 100 100

10. a. d2 b. d3 c. a three way tie d. EV(d1) = 7.6 EV(d2) = 9.6 (the best) EV(d3) = 8.2 e. EVPI = 12.4 - 9.6 = 2.8



11. a. d2 b. d3 c. d1 d. EV(d1) = 695 (the best) EV(d2) = 385 EV(d3) = 500 e. EVPI = 925 - 695 = 230 12. Choose N if p < .78. 13. Choose A if p ≤ .5, choose B is .5 ≤ p ≤ .8, and choose C if p ≥ .8. 14.

100

62.5

A

B

C .25

D .35

E .40

F .2

G .8

H

J

K

M

N

Q

R

S .55

T .45

U .25

V .75

W .5

X .5

Y .4

Z .6AA .3

BB .4

CC .3

300

100

10

80

-30

120

500

-200

150

100

50

18020

-50

50

210

210

84.5

70

62.5

45

80

100

Strategy: Select A. If C happens, select H. If D happens, you are done. If E happens, select K.



15.

A

C

D .42

E .24

F .34

G .2

B

H .8

I .82

J .18

K

L

M

N

P .63

Q .37

R .3S .3T .4

100

-4030

6035

2550

-100120150

10

-50

74.8

41.133.39

6621.2

Choose A. If F happens, choose K.

16.

Sample Result

States of Nature

Prior Probabilities

Conditional Probabilities

Joint Probabilities

Posterior Probabilities

Positive good .7 P(positive | good) = .8 .56 .9492 poor .3 P(positive | poor) = .1 .03 .0508

Negative good .7 P(negative | good) = .2 .14 .3415 poor .3 P(negative | poor) = .9 .27 .6585

17. EV(d1) and EV(d2) intersect at p = .6. EV(d1) and EV(d3) intersect at p = .8333. Therefore when 0 ≤ p ≤

.6, choose d2. When .6 ≤ p ≤ .8333, choose d1. When p ≥ .8333, choose d3. 16

12

8

4

16

12

8

4

p = 0 p = 1

d2

d1d3

.6 .8333 18. a. Yes, Dollar should sell store.

b. EVPI = $8,000 c. No; survey cost exceeds EVPI.


1

5

Utility and Game Theory

MULTIPLE CHOICE

1. When consequences are measured on a scale that reflects a decision maker's attitude toward profit, loss, and

risk, payoffs are replaced by a. utility values. b. multicriteria measures. c. sample information. d. opportunity loss. ANSWER: a TOPIC: Meaning of utility 2. The purchase of insurance and lottery tickets shows that people make decisions based on a. expected value. b. sample information. c. utility. d. maximum likelihood. ANSWER: c TOPIC: Introduction 3. The expected utility approach a. does not require probabilities. b. leads to the same decision as the expected value approach. c. is most useful when excessively large or small payoffs are possible. d. requires a decision tree. ANSWER: c TOPIC: Expected utility approach 4. Utility reflects the decision maker’s attitude toward

a. probability and profit b. profit, loss, and risk c. risk and regret d. probability and regret

ANSWER: b TOPIC: Meaning of utility


2 Chapter 5 Utility and Game Theory

5. Values of utility a. must be between 0 and 1. b. must be between 0 and 10. c. must be nonnegative. d. must increase as the payoff improves. ANSWER: d TOPIC: Developing utilities for monetary payoffs 6. If the payoff from outcome A is twice the payoff from outcome B, then the ratio of these utilities will be

a. 2 to 1. b. less than 2 to 1. c. more than 2 to 1. d. unknown without further information.

ANSWER: d TOPIC: Meaning of utility 7. The probability for which a decision maker cannot choose between a certain amount and a lottery based on

that probability is a. the indifference probability. b. the lottery probability. c. the uncertain probability. d. the utility probability. ANSWER: a TOPIC: Developing utilities for monetary payoffs 8. A decision maker has chosen .4 as the probability for which he cannot choose between a certain loss of

10,000 and the lottery p(-25000) + (1-p)(5000). If the utility of -25,000 is 0 and of 5000 is 1, then the utility of -10,000 is

a. .5 b. .6 c. .4 d. 4 ANSWER: b TOPIC: Developing utilities for monetary payoffs 9. When the decision maker prefers a guaranteed payoff value that is smaller than the expected value of the

lottery, the decision maker is a. a risk avoider. b. a risk taker. c. an optimist. d. an optimizer.

ANSWER: a TOPIC: Risk avoiders versus risk takers 10. A decision maker whose utility function graphs as a straight line is a. conservative. b. risk neutral. c. a risk taker. d. a risk avoider. ANSWER: b TOPIC: Risk avoiders versus risk takers 11. For a game with an optimal pure strategy, which of the following statements is false? a. The maximin equals the minimax. b. The value of the game cannot be improved by either player changing strategies. c. A saddle point exists.


Chapter 5 Utility and Game Theory 3

d. Dominated strategies cannot exist. ANSWER: d TOPIC: Identifying a pure strategy 12. Which of the following statements about a dominated strategy is false? a. A dominated strategy will never be selected by a player.

b. A dominated strategy exists if another strategy is at least as good regardless of what the opponent does.

c. A dominated strategy is superior to a mixed strategy. d. A dominated strategy can be eliminated from the game. ANSWER: c TOPIC: Dominated strategies 13. A 3 x 3 two-person zero-sum game that has no optimal pure strategy and no dominated strategies a. can be solved using a linear programming model. b. can be solved algebraically. c. can be solved by identifying the minimax and maximin values. d. cannot be solved. ANSWER: a TOPIC: Larger mixed strategy games 14. For a two-person zero-sum game, which one of the following is false? a. The gain for one player is equal to the loss for the other player. b. A payoff of 2 for one player has a corresponding payoff of -2 for the other player. c. The sum of the payoffs in the payoff table is zero. d. What one player wins, the other player loses. ANSWER: c TOPIC: Introduction to game theory 15. If the maximin and minimax values are not equal in a two-person zero-sum game, a. a mixed strategy is optimal. b. a pure strategy is optimal. c. a dominated strategy is optimal. d. one player should use a pure strategy and the other should use a mixed strategy. ANSWER: a TOPIC: Mixed strategy games

TRUE/FALSE

1. The decision alternative with the best expected monetary value will always be the most desirable decision. ANSWER: True TOPIC: Introduction 2. When monetary value is not the sole measure of the true worth of the outcome to the decision maker,

monetary value should be replaced by utility. ANSWER: True TOPIC: Introduction 3. The outcome with the highest payoff will also have the highest utility.



ANSWER: True TOPIC: Developing utilities for monetary payoffs 4. Expected utility is a particularly useful tool when payoffs stay in a range considered reasonable by the

decision maker. ANSWER: False TOPIC: Meaning of utility 5. To assign utilities, consider the best and worst payoffs in the entire decision situation. ANSWER: True TOPIC: Developing utilities for monetary payoffs 6. A risk avoider will have a concave utility function. ANSWER: True TOPIC: Developing utilities for monetary payoffs 7. The expected utility is the utility of the expected monetary value. ANSWER: False TOPIC: Expected utility approach 8. The risk premium is never negative for a conservative decision maker. ANSWER: True TOPIC: Developing utilities for monetary payoffs 9. The risk neutral decision maker will have the same indications from the expected value and expected utility

approaches. ANSWER: True TOPIC: Expected monetary value versus expected utility 10. The utility function for a risk avoider typically shows a diminishing marginal return for money. ANSWER: True TOPIC: Developing utilities for monetary payoffs 11. A game has a pure strategy solution when both players’ single-best strategies are the same. ANSWER: False TOPIC: Identifying a pure strategy 12. A game has a saddle point when pure strategies are optimal for both players. ANSWER: True TOPIC: Identifying a pure strategy 13. A game has a saddle point when the maximin payoff value equals the minimax payoff value. ANSWER: True TOPIC: Identifying a pure strategy 14. The logic of game theory assumes that each player has different information. ANSWER: False TOPIC: Introduction to game theory 15. With a mixed strategy, the optimal solution for each player is to randomly select among two or more of the

alternative strategies. ANSWER: True TOPIC: Mixed strategy games 16. The expected monetary value approach and the expected utility approach to decision making usually result

in the same decision choice unless extreme payoffs are involved.



ANSWER: True TOPIC: Utility and decision making 17. A risk neutral decision maker will have a linear utility function. ANSWER: True TOPIC: Developing utilities for monetary payoffs 18. Given two decision makers, one risk neutral and the other a risk avoider, the risk avoider will always give a

lower utility value for a given outcome. ANSWER: False TOPIC: Developing utilities for monetary payoffs

SHORT ANSWER

1. When and why should a utility approach be followed? TOPIC: Expected value versus utility 2. Give two examples of situations where you have decided on a course of action that did not have the highest

expected monetary value. TOPIC: Introduction 3. Explain how utility could be used in a decision where performance is not measured by monetary value. TOPIC: Expected value versus expected utility 4. Explain the relationship between expected utility, probability, payoff, and utility. TOPIC: Expected value versus expected utility 5. Draw the utility curves for three types of decision makers, label carefully, and explain the concepts of

increasing and decreasing marginal returns for money. TOPIC: Risk avoiders versus risk takers

PROBLEMS

1. For the payoff table below, the decision maker will use P(s1) = .15, P(s2) = .5, and P(s3) = .35.


d1 -5000 1000 10,000 d2 -15,000 -2000 40,000

a. What alternative would be chosen according to expected value? b. For a lottery having a payoff of 40,000 with probability p and -15,000 with probability (1-p), the

decision maker expressed the following indifference probabilities.



Payoff Probability

10,000 .85 1000 .60 -2000 .53 -5000 .50

Let U(40,000) = 10 and U(-15,000) = 0 and find the utility value for each payoff. c. What alternative would be chosen according to expected utility? TOPIC: Expected utility approach 2. A decision maker who is considered to be a risk taker is faced with this set of probabilities and payoffs

State of Nature

Decision s1 s2 s3

d1 5 10 20 d2 -25 0 50 d3 -50 -10 80

Probability .30 .35 .35 For the lottery p(80) + (1 - p)(-50), this decision maker has assessed the following indifference probabilities Payoff Probability 50 .60 20 .35 10 .25 5 .22 0 .20 -10 .18 -25 .10 Rank the decision alternatives on the basis of expected value and on the basis of expected utility. TOPIC: Expected utility approach 3. Three decision makers have assessed utilities for the problem whose payoff table appears below.


d1 500 100 -400 d2 200 150 100 d3 -100 200 300

Probability .2 .6 .2

Indifference Probability for Person Payoff A B C 300 .95 .68 .45 200 .94 .64 .32 150 .91 .62 .28 100 .89 .60 .22 -100 .75 .45 .10



a. Plot the utility function for each decision maker. b. Characterize each decision maker's attitude toward risk. c. Which decision will each person prefer? TOPIC: Risk avoiders versus risk takers 4. A decision maker has the following utility function

Payoff Indifference Probability 200 1.00 150 .95 50 .75 0 .60 -50 0

What is the risk premium for the payoff of 50? TOPIC: Developing utilities for monetary payoffs 5. Determine decision strategies based on expected value and on expected utility for this decision tree. Use

the utility function

Payoff Indifference Probability 500 1.00 350 .89 300 .84 180 .60 100 .43 40 .20 20 .13 0 0

TOPIC: Expected utility approach 6. Burger Prince Restaurant is considering the purchase of a $100,000 fire insurance policy. The fire statistics

indicate that in a given year the probability of property damage in a fire is as follows:

Fire Damage $100,000 $75,000 $50,000 $25,000 $10,000 $0 Probability .006 .002 .004 .003 .005 .980



a. If Burger Prince was risk neutral, how much would they be willing to pay for fire insurance? b. If Burger Prince has the utility values given below, approximately how much would they be

willing to pay for fire insurance?

Loss $100,000 $75,000 $50,000 $25,000 $10,000 $5,000 $0 Utility 0 30 60 85 95 99 100

TOPIC: Decision making using utility 7. Super Cola is considering the introduction of a new 8 oz. root beer. The probability that the root beer will

be a success is believed to equal .6. The payoff table is as follows: Success (s1) Failure (s2) Produce $250,000 -$300,000 Do Not Produce -$50,000 -$20,000

Company management has determined the following utility values:

Amount $250,000 -$20,000 -$50,000 -$300,000 Utility 100 60 55 0

a. Is the company a risk taker, risk averse, or risk neutral? b. What is Super Cola's optimal decision?

TOPIC: Decision making using utility 8. Chez Paul is contemplating either opening another restaurant or expanding its existing location. The payoff

table for these two decisions is:


New Restaurant -$80,000 $20,000 $160,000 Expand -$40,000 $20,000 $100,000

Paul has calculated the indifference probability for the lottery having a payoff of $160,000 with probability p and -$80,000 with probability (1-p) as follows:

Amount Indifference Probability (p) -$40,000 .4 $20,000 .7 $100,000 .9

a. Is Paul a risk avoider, a risk taker, or risk neutral? b. Suppose Paul has defined the utility of -$80,000 to be 0 and the utility of $160,000 to be 80. What

would be the utility values for -$40,000, $20,000, and $100,000 based on the indifference probabilities?

c. Suppose P(s1) = .4, P(s2) = .3, and P(s3) = .3. Which decision should Paul make? Compare with the decision using the expected value approach.

TOPIC: Decision making using utility 9. The Dollar Department Store chain has the opportunity of acquiring either 3, 5, or 10 leases from the

bankrupt Granite Variety Store chain. Dollar estimates the profit potential of the leases depends on the state of the economy over the next five years. There are four possible states of the economy as modeled by



Dollar Department Stores and its president estimates P(s1) = .4, P(s2) = .3, P(s3) = .1, and P(s4) = .2. The utility has also been estimated. Given the payoffs (in $1,000,000's) and utility values below, which decision should Dollar make?

Payoff Table State Of The Economy Over The Next 5 Years Decision s1 s2 s3 s4 d1 -- buy 10 leases 10 5 0 -20 d2 -- buy 5 leases 5 0 -1 -10 d3 -- buy 3 leases 2 1 0 - 1 d4 -- do not buy 0 0 0 0

Utility Table Payoff (in $1,000,000's) +10 +5 +2 0 -1 -10 -20 Utility +10 +5 +2 0 -1 -20 -50 TOPIC: Decision making using utility 10. Consider the following two-person zero-sum game. Assume the two players have the same two strategy

options. The payoff table shows the gains for Player A.

Player B Player A Strategy b1 Strategy b2

Strategy a1 3 9 Strategy a2 6 2

Determine the optimal strategy for each player. What is the value of the game?

TOPIC: Mixed strategy games 11. Consider the following two-person zero-sum game. Assume the two players have the same three strategy

options. The payoff table below shows the gains for Player A.

Player B

Player A Strategy b1 Strategy b2 Strategy b3 Strategy a1 3 5 -2 Strategy a2 -2 -1 2 Strategy a3 2 1 -5

Is there an optimal pure strategy for this game? If so, what is it? If not, can the mixed-strategy probabilities be found algebraically? What is the value of the game?

TOPIC: Mixed strategy games 12. Suppose that there are only two vehicle dealerships (A and B) in a small city. Each dealership is

considering three strategies that are designed to take sales of new vehicles from the other dealership over a period of four months. The strategies, assumed to be the same for both dealerships, are:

Strategy 1: Offer a cash rebate on a new vehicle. Strategy 2: Offer free optional equipment on a new vehicle. Strategy 3: Offer a 0% loan on a new vehicle.

The payoff table (in number of new vehicle sales gained per week by Dealership A (or lost by Dealership B) is shown below.



Dealership B

Cash Rebate Free Options 0% Loan

Dealership A b1 b2 b3

Cash Rebate a1 2 2 1 Free Options a2 -3 3 -1 0% Loan a3 3 -2 0

Identify the pure strategy for this two-person zero-sum game. What is the value of the game?

TOPIC: Identifying a pure strategy 13. Consider the following two-person zero-sum game. Assume the two players have the same two strategy


Player B

Player A Strategy b1 Strategy b2 Strategy a1 4 8 Strategy a2 11 5

Determine the optimal strategy for each player. What is the value of the game?



Player B

Player A Strategy b1 Strategy b2 Strategy b3 Strategy a1 6 5 -2 Strategy a2 1 0 3 Strategy a3 3 4 -3

Is there an optimal pure strategy for this game? If so, what is it? If not, can the mixed-strategy probabilities be found algebraically?


options. The payoff table below shows the gains for Player A.

Player B

Player A Strategy b1 Strategy b2 Strategy b3 Strategy a1 3 2 -4 Strategy a2 -1 0 2 Strategy a3 4 5 -3

Is there an optimal pure strategy for this game? If so, what is it? If not, can the mixed-strategy probabilities be found algebraically? What is the value of the game?

TOPIC: Mixed strategy games



16. Two banks (Franklin and Lincoln) compete for customers in the growing city of Logantown. Both banks are considering opening a branch office in one of three new neighborhoods: Hillsboro, Fremont, or Oakdale. The strategies, assumed to be the same for both banks, are:

Strategy 1: Open a branch office in the Hillsboro neighborhood. Strategy 2: Open a branch office in the Fremont neighborhood. Strategy 3: Open a branch office in the Oakdale neighborhood.

Values in the payoff table below indicate the gain (or loss) of customers (in thousands) for Franklin Bank based on the strategies selected by the two banks.

Lincoln Bank

Hillsboro Fremont Oakdale

Franklin Bank b1 b2 b3

Hillsboro a1 4 2 3 Fremont a2 6 -2 -3 Oakdale a3 -1 0 5

Identify the neighborhood in which each bank should locate a new branch office. What is the value of the game?

TOPIC: Mixed strategy games

SOLUTIONS TO PROBLEMS

1. a. EV(d1) = 3250 and EV(d2) = 10750, so choose d2. b. Payoff Probability Utility 10,000 .85 8.5 1000 .60 6.0 -2000 .53 5.3 -5000 .50 5.0 c. EU(d1) = 6.725 and EU(d2) = 6.15, so choose d1. 2. EV(d1) = 12 EV(d2) = 10 EV(d3) = 9.5 EU(d1) = 2.76 EU(d3) = 3.1 EU(d3) = 4.13



3. a.

5 0 0 4 0 0 3 0 0 2 0 0 1 0 0 0 - 1 0 0 - 2 0 0 - 3 0 0 - 4 0 0

1 . 0

0 . 5

0 . 0

C 1

A

b. Person A is a risk avoider, Person B is fairly risk neutral, and Person C is a risk avoider. c. For person A, EU(d1) = .734 EU(d2) = .912 EU(d3) = .904 For person B, EU(d1) = .56 EU(d2) = .62 EU(d3) = .61 For person C, EU(d1) = .332 EU(d2) = .276 EU(d3) = .302 Decision 1 would be chosen by person C. Decision 2 would be chosen by persons A and B. 4. EV = .75(200) + .25(-50) = 137.50 Risk premium is 137.50 - 50 = 87.50 5. Let U(500) = 1 and U(0) = 0. Then After branch Expected value Expected utility

A 120 .336 J 316 .680 K 150 .522 B 127.2 .381 C 100 .430 Based on expected value, the decision strategy is to select B. If G happens, select J. Based on expected

utility, it is best to choose C. 6. a. $1,075

b. $5,000 7. a. Risk averse b. Produce root beer as long as p > 60/105 = .571 8. a. A risk avoider b. Amount Utility - $40,000 32 $20,000 56 $100,000 72 c. Decision is d2; EV criterion decision would be d1 9. Buy 3 leases. 10. Mixed strategy:



Player A: .4 for a1, .6 for a2 Player B: .7 for b1, .3 for b2

Value of game = 4.8

11. There is not an optimal pure strategy. However, there are dominated strategies. Strategy a3 is dominated (by strategy a1) and can be eliminated. Then strategy b1 is dominated (by strategy b2) and can be eliminated. Now it is a 2 x 2 game. Mixed-strategy probabilities are found algebraically: p = .3, (1 – p) = .7, q = .4, (1 – q) = .6 Value of game = 0.8 12. An optimal pure strategy exists for this game: Dealership A should offer a cash rebate on new vehicles.

Dealership A can expect to gain a minimum of 1 new vehicle sale per week. Dealership B should offer a 0% loan on new vehicles.

Dealership B can expect to lose a maximum of 1 new vehicle sale per week. Value of the game is 1 new vehicle.

13. The optimal mixed strategy solution for this game: Player A should select Strategy a1 with a .6 probability and Strategy a2 with a .4 probability.

Player B should select Strategy b1 with a .3 probability and Strategy b2 with a .7 probability. Value of the game is:

Player A: 6.8 = expected gain Player B: 6.8 = expected loss

14. There is not an optimal pure strategy. The optimal mixed-strategy probabilities can be found algebraically.

Player A should select Strategy a1 with a .2 probability and Strategy a2 with a .8 probability. Player B should select Strategy b1 with a .5 probability and Strategy b3 with a .5 probability.

Value of the game: For Player A: 2 = expected gain For Player B: 2 = expected loss

15. There is not an optimal pure strategy. Strategy a1 is dominated by Strategy a3, and then Strategy b1 is

dominated by Strategy b2. The optimal mixed-strategy probabilities can be found algebraically. Player A should select Strategy a2 with a .8 probability and Strategy a3 with a .2 probability. Player B should select Strategy b2 with a .5 probability and Strategy b3 with a .5 probability.

Value of the game = 1.

16. Franklin should select Hillsboro; Lincoln should select Fremont. Value of game = 2,000 customers


1

6

Forecasting

MULTIPLE CHOICE

1. Time series methods a. discover a pattern in historical data and project it into the future. b. include cause-effect relationships. c. are useful when historical information is not available. d. All of the alternatives are true. ANSWER: a TOPIC: Introduction 2. Gradual shifting of a time series over a long period of time is called a. periodicity. b. cycle. c. regression. d. trend. ANSWER: d TOPIC: Trend component 3. Seasonal components

a. cannot be predicted. b. are regular repeated patterns. c. are long runs of observations above or below the trend line. d. reflect a shift in the series over time.

ANSWER: b TOPIC: Seasonal component 4. Short-term, unanticipated, and nonrecurring factors in a time series provide the random variability known

as a. uncertainty. b. the forecast error. c. the residuals. d. the irregular component. ANSWER: d TOPIC: Irregular component


2 Chapter 6 Forecasting

5. The focus of smoothing methods is to smooth a. the irregular component. b. wide seasonal variations. c. significant trend effects. d. long range forecasts.

ANSWER: a TOPIC: Smoothing methods 6. Forecast errors

a. are the difference in successive values of a time series b. are the differences between actual and forecast values c. should all be nonnegative d. should be summed to judge the goodness of a forecasting model

ANSWER: b TOPIC: Smoothing methods 7. To select a value for α for exponential smoothing

a. use a small α when the series varies substantially. b. use a large α when the series has little random variability. c. use any value between 0 and 1 d. All of the alternatives are true.

ANSWER: d TOPIC: Exponential smoothing 8. Linear trend is calculated as Tt = 28.5 + .75t. The trend projection for period 15 is a. 11.25 b. 28.50 c. 39.75 d. 44.25 ANSWER: c TOPIC: Trend projection 9. The multiplicative model

a. uses centered moving averages to smooth the trend fluctuations. b. removes trend before isolating the seasonal components. c. deseasonalizes a time series by dividing the values by the appropriate seasonal index. d. provides a unique seasonal index for each observation of the time series.

ANSWER: c TOPIC: Multiplicative model 10. Causal models

a. should avoid the use of regression analysis. b. attempt to explain a time series’ behavior. c. do not use time series data. d. All of the alternatives are true.

ANSWER: b TOPIC: Regression analysis 11. A qualitative forecasting method that obtains forecasts through "group consensus" is known as the a. Autoregressive model b. Delphi approach c. mean absolute deviation d. None of these alternatives is correct. ANSWER: b TOPIC: Qualitative approaches in forecasting


Chapter 6 Forecasting 3

12. The trend component is easy to identify by using a. moving averages b. exponential smoothing c. regression analysis d. the Delphi approach ANSWER: c TOPIC: Trend component 13. The forecasting method that is appropriate when the time series has no significant trend, cyclical, or

seasonal effect is a. moving averages b. mean squared error c. mean average deviation d. qualitative forecasting methods ANSWER: a TOPIC: Moving averages 14. If data for a time series analysis is collected on an annual basis only, which component may be ignored? a. trend b. seasonal c. cyclical d. irregular ANSWER: b TOPIC: Seasonal component 15. One measure of the accuracy of a forecasting model is the a. smoothing constant b. trend component c. mean absolute deviation d. seasonal index ANSWER: c TOPIC: Forecast accuracy 16. Which of the following is a qualitative forecasting method? a. trend projection b. time series method c. smoothing method d. Delphi method ANSWER: d TOPIC: Introduction 17. Which of the following forecasting methods puts the least weight on the most recent time series value? a. exponential smoothing with α = .3 b. exponential smoothing with α = .2 c. moving average using the most recent 4 periods d. moving average using the most recent 3 periods ANSWER: b TOPIC: Smoothing methods 18. Using exponential smoothing, the demand forecast for time period 10 equals the demand forecast for time

period 9 plus a. α times (the demand forecast for time period 8) b. α times (the error in the demand forecast for time period 9) c. α times (the observed demand in time period 9) d. α times (the demand forecast for time period 9)



ANSWER: b TOPIC: Exponential smoothing 19. Which of the following exponential smoothing constant values puts the same weight on the most recent

time series value as does a 5-period moving average? a. α = .2 b. α = .25 c. α = .75 d. α = .8 ANSWER: a TOPIC: Smoothing methods 20. The time series component that is analogous to the seasonal component but over a longer period of time is

the a. irregular component b. trend component c. causal component d. cyclical component ANSWER: d TOPIC: Cyclical component

TRUE/FALSE

1. Time series methods base forecasts only on past values of the variables. ANSWER: True TOPIC: Introduction 2. Quantitative forecasting methods do not require that patterns from the past will necessarily continue in the

future. ANSWER: False TOPIC: Introduction 3. Trend in a time series must be linear. ANSWER: False TOPIC: Trend component 4. All quarterly time series contain seasonality. ANSWER: False TOPIC: Seasonal component 5. A four-period moving average forecast for period 10 would be found by averaging the values from periods

10, 9, 8, and 7. ANSWER: False TOPIC: Moving averages 6. If the random variability in a time series is great, a high α value should be used to exponentially smooth out

the fluctuations. ANSWER: False TOPIC: Exponential smoothing 7. Seasonal components with values above 1.00 indicate actual values below the trend line.



ANSWER: False TOPIC: Multiplicative model 8. To make period-to-period comparisons more meaningful and identify trend, the time series should be

deseasonalized. ANSWER: True TOPIC: Trend and seasonal components 9. With fewer periods in a moving average, it will take longer to adjust to a new level of demand. ANSWER: False TOPIC: Moving averages 10. Qualitative forecasting techniques should be applied in situations where time series data exists, but where

conditions are expected to change. ANSWER: True TOPIC: Qualitative approaches 11. For a multiplicative time series model, the sum of the seasonal indexes should equal the number of seasons. ANSWER: True TOPIC: Multiplicative model 12. A time series model with a seasonal component will always involve quarterly data. ANSWER: False TOPIC: Seasonal component 13. Any recurring sequence of points above and below the trend line lasting less than one year can be attributed

to the cyclical component of the time series. ANSWER: False TOPIC: Cyclical component 14. Smoothing methods are more appropriate for a stable time series than when significant trend and/or

seasonal variation are present. ANSWER: True TOPIC: Using smoothing methods in forecasting 15. The exponential smoothing forecast for any period is a weighted average of all the previous actual values

for the time series. ANSWER: True TOPIC: Exponential smoothing 16. The mean squared error is influenced much more by large forecast errors than by small errors. ANSWER: True TOPIC: Forecast accuracy 17. If a time series has a significant trend component, then one should not use a moving average to forecast. ANSWER: True TOPIC: Using smoothing methods in forecasting 18. If the random variability in a time series is great and exponential smoothing is being used to forecast, then a

high alpha (α) value should be used. ANSWER: False TOPIC: Exponential smoothing 19. An alpha (α) value of .2 will cause an exponential smoothing forecast to react more quickly to a sudden

drop in demand than will an α equal to .4. ANSWER: False



TOPIC: Exponential smoothing 20. Exponential smoothing with α = .2 and a moving average with n = 5 put the same weight on the actual

value for the current period. ANSWER: True TOPIC: Using smoothing methods in forecasting

SHORT ANSWER

1. Explain what conditions make quantitative forecasting methods appropriate. TOPIC: Introduction 2. What is a stable time series, and what forecasting methods are appropriate for one? TOPIC: Smoothing methods 3. How can error measures be used to determine the number of periods to use in a moving average? What are

you assuming about the future when you make this choice? TOPIC: Moving averages 4. Explain how to use seasonal index values to create a forecast. TOPIC: Smoothing methods 5. Explain how qualitative methods frequently incorporate the opinions of multiple analysts. TOPIC: Introduction

PROBLEMS

1. The number of cans of soft drinks sold in a machine each week is recorded below. Develop forecasts using

a three period moving average. 338, 219, 278, 265, 314, 323, 299, 259, 287, 302 TOPIC: Moving averages 2. Use a four period moving average to forecast attendance at baseball games. Historical records show 5346, 7812, 6513, 5783, 5982, 6519, 6283, 5577, 6712, 7345 TOPIC: Moving averages 3. A hospital records the number of floral deliveries its patients receive each day. For a two week period, the

records show 15, 27, 26, 24, 18, 21, 26, 19, 15, 28, 25, 26, 17, 23 Use exponential smoothing with a smoothing constant of .4 to forecast the number of deliveries. TOPIC: Exponential smoothing 4. The number of girls who attend a summer basketball camp has been recorded for the seven years the camp

has been offered. Use exponential smoothing with a smoothing constant of .8 to forecast attendance for the eighth year.

47, 68, 65, 92, 98, 121, 146



TOPIC: Exponential smoothing 5. The number of pizzas ordered on Friday evenings between 5:30 and 6:30 at a pizza delivery location for the

last 10 weeks is shown below. Use exponential smoothing with smoothing constants of .2 and .8 to forecast a value for week 11. Compare your forecasts using MSE. Which smoothing constant would you prefer?

58, 46, 55, 39, 42, 63, 54, 55, 61, 52 TOPIC: Exponential smoothing 6. A trend line for the attendance at a restaurant’s Sunday brunch is given by Number = 264 + .72(t) How many guests would you expect in week 20? TOPIC: Trend projection 7. The number of new contributors to a public radio station’s annual fund drive over the last ten years is 63, 58, 61, 72, 98, 103, 121, 147, 163, 198 Develop a trend equation for this information, and use it to predict next year’s number of new contributors. TOPIC: Trend component 8. The average SAT verbal score for students from one high school over the last ten exams is 508, 490, 502, 505, 493, 506, 492, 490, 503, 501 Do the scores support an increasing or a decreasing trend? TOPIC: Trend component 9. Use the following to forecast a value for period 14, a second quarter. T = 16.32 - .18(t) C2 = .91 S2 = .75 TOPIC: Time series 10. The number of properties newly listed with a real estate agency in each quarter over the last four years is

given. Calculate the seasonal index values.

Year Quarter 1 2 3 4

1 73 81 76 77 2 89 87 85 92 3 123 115 108 131 4 92 93 87 101

TOPIC: Seasonal component 11. Quarterly billing for water usage is shown below.

Year Quarter 1 2 3 4 Winter 64 66 68 73 Spring 103 103 104 120

Summer 152 160 162 176 Fall 73 72 78 88

a. Find the seasonal index for each quarter. b. Deseasonalize the data. c. Find the trend line. d. Assume there is no cyclical component and forecast the summer billing for year 5. TOPIC: Trend and seasonal components



12. A customer comment phone line is staffed from 8:00 a.m. to 4:30 p.m. five days a week. Records are available that show the number of calls received every day for the last five weeks.

Week Day Number Week Day Number

1 M 28 4 M 35 T 12 T 17 W 16 W 16 TH 15 TH 20 F 23 F 29

2 M 29 5 M 37 T 10 T 19 W 14 W 18 TH 14 TH 21 F 26 F 28

3 M 32 T 15 W 15 TH 18 F 27

a. Use this information to calculate a seasonal index. b. Deseasonalize the data. c. Find the trend line. d. Assume there is no cyclical component and forecast the calls for week 6. TOPIC: Trend and seasonal components 13. Monthly sales at a coffee shop have been analyzed. The seasonal index values are

Month Index Jan 1.38 Feb 1.42 Mar 1.35 Apr 1.03 May .99 June .62 July .51 Aug .58 Sept .82 Oct .82 Nov .92 Dec 1.56

and the trend line is 74123 + 26.9(t). Assume there is no cyclical component and forecast sales for year 8

(months 97 - 108). TOPIC: Seasonal component



14. A 24-hour coffee/donut shop makes donuts every eight hours. The manager must forecast donut demand so

that the bakers have the fresh ingredients they need. Listed below is the actual number of glazed donuts (in dozens) sold in each of the preceding 13 eight-hour shifts.

Date Shift Demand(dozens)

June 3 Day 59 Evening 47 Night 35




June 7 Day 69

Forecast the demand for glazed donuts for the three shifts of June 8 and the three shifts of June 9. TOPIC: Trend and seasonal components 15. In order to forecast the attendance at an annual tennis tournament, a model has been developed which uses

attendance from the previous year and the amount spent for advertising this year. From the years shown in the table, forecast the attendance for years 2-5 and calculate the forecast error.

Year

Attendance Advertising Expenditure

Forecast

Error

1 8363 750 2 9426 1250 3 9318 3200 4 10206 4500 5 11018 5600

The multiple regression model is Attendance = 6738 + .23($) + .25 (Attlag) TOPIC: Multiple regression 16. The number of plumbing repair jobs performed by Auger's Plumbing Service in each of the last nine

months are listed below.

Month Jobs Month Jobs Month Jobs March 353 June 374 September 399 April 387 July 396 October 412 May 342 August 409 November 408

a. Assuming a linear trend function, forecast the number of repair jobs Auger's will perform in

December using the least squares method. b. What is your forecast for December using a three-period weighted moving average with weights

of .6, .3, and .1? How does it compare with your forecast from part (a)? TOPIC: Trend component 17. Quarterly revenues (in $1,000,000's) for a national restaurant chain for a five-year period were as follows:



Quarter Year 1 Year 2 Year 3 Year 4 Year 5 1 33 42 54 70 85 2 36 40 53 67 82 3 35 42 54 70 87 4 38 47 62 77 99

a. Determine the appropriate seasonal index values for this time series. b. Determine the appropriate trend value for this time series. c. Forecast the revenues for the next four quarters.

TOPIC: Trend and seasonal components

18. Business at Terry's Tie Shop can be viewed as falling into three distinct seasons: (1) Christmas (November-December); (2) Father's Day (late May - mid-June); and (3) all other times. Average weekly sales (in $'s) during each of these three seasons during the past four years has been as follows:

Season Year 1 Year 2 Year 3 Year 4

1 1856 1995 2241 2280 2 2012 2168 2306 2408 3 985 1072 1105 1120

Determine a forecast for the average weekly sales in year 5 for each of the three seasons.

TOPIC: Trend and seasonal components 19. Connie Harris, in charge of office supplies at First Capital Mortgage Corp., would like to predict the

quantity of paper used in the office photocopying machines per month. She believes that the number of loans originated in a month influence the volume of photocopying performed. She has compiled the following recent monthly data:

Number of Loans Sheets of Photocopy

Originated in Month Paper Used (000's) 25 16 25 13 35 18 40 25 40 21 45 22 50 24 60 25

a. Develop the least-squares estimated regression equation that relates sheets of photocopy paper

used to loans originated. b. Use the regression equation developed in part (a) to forecast the amount of paper used in a month

when 65 loan originations are expected. TOPIC: Using regression analysis as a causal forecasting method 20. Sales (in thousands) of the new Thorton Model 506 convection oven over the eight-week period

since its introduction have been as follows:

Week Sales 1 18.6 2 21.4 3 25.2 4 22.4 5 24.6 6 19.2 7 21.7



8 23.8

a. Which exponential smoothing model provides better forecasts, one using α = .6 or α = .2? Compare them using mean squared error.

b. Using the two forecast models in part (a), what are the forecasts for week 9? TOPIC: Exponential smoothing and forecast accuracy


1. FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 3 TIME PERIODS

Time Period Actual Value Forecast Forecast Error 1 2 3 4 265 278.33 -13.33 5 314 254.00 60.00 6 323 285.67 37.33 7 299 300.67 -1.67 8 259 312.00 -53.00 9 287 293.67 -6.67

10 302 281.67 20.33 THE FORECAST FOR PERIOD 11 282.67 2. FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 4 TIME PERIODS

Time Period Actual Value Forecast Forecast Error 1 2 3 4 5 5,982 6,363.50 -381.50 6 6,519 6,522.50 -3.50 7 6,283 6,199.25 83.75 8 5,577 6,141.75 -564.75 9 6,712 6,090.25 621.75

10 7,345 6,272.75 1,072.25 THE FORECAST FOR PERIOD 11 6479.25



3. FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.4

Time Period Actual Value Forecast Forecast Error 1 2 27 15.00 12.00 3 26 19.80 6.20 4 24 22.28 1.72 5 18 22.97 -4.97 6 21 20.98 0.02 7 26 20.99 5.01 8 19 22.99 -3.99 9 15 21.40 -6.40

10 28 18.84 9.16 11 25 22.50 2.50 12 26 23.50 2.50 13 17 24.50 -7.50 14 23 21.50 1.50

THE FORECAST FOR PERIOD 15 22.10 4. FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8

Time Period Actual Value Forecast Forecast Error 1 2 68 47.00 21.00 3 65 63.80 1.20 4 92 64.76 27.24 5 98 86.55 11.45 6 121 95.71 25.29 7 146 115.94 30.06

THE FORECAST FOR PERIOD 8 139.99



5. FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.2

Time Period Actual Value Forecast Forecast Error 1 2 46 58.00 -12.00 3 55 55.60 -0.60 4 39 55.48 -16.48 5 42 52.18 -10.18 6 63 50.15 12.85 7 54 52.72 1.28 8 55 52.97 2.03 9 61 53.38 7.62

10 52 54.90 -2.90 THE MEAN SQUARE ERROR 84.12 THE FORECAST FOR PERIOD 11 54.32 FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8

Time Period Actual Value Forecast Forecast Error 1 2 46 58.00 -12.00 3 55 48.40 6.60 4 39 53.68 -14.68 5 42 41.94 0.06 6 63 41.99 21.01 7 54 58.80 -4.80 8 55 54.96 0.04 9 61 54.99 6.01

10 52 59.80 -7.80 THE MEAN SQUARE ERROR 107.17 THE FORECAST FOR PERIOD 11 53.56

Based on MSE, smoothing with α = .2 provides a better model. 6. 264 + .72(20) = 278.4 7. FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION: T = 24 + 15.345 t where T = trend value of the time series in period t



Time Period Actual Value Forecast Forecast Error 1 63 39.35 23.66 2 58 54.69 3.31 3 61 70.04 -9.03 4 72 85.38 -13.38 5 98 100.73 -2.73 6 103 116.07 -13.07 7 121 131.42 -10.42 8 147 146.76 0.24 9 163 162.11 0.89

10 198 177.45 20.55 THE MEAN SQUARE ERROR 154.11 THE FORECAST FOR PERIOD 11 192.80 8. FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION: T = 500.933 - 0.352 t where T = trend value of the time series in period t The negative slope for the trend indicates that scores are decreasing. (However, we have not tested for the

significance of this coefficient.) 9. Value = 9.4185



10.

Year

Quarter

Listings Centered

Moving Average

Seasonal-Irregular 1 1 73 2 89 3 123 95.250 1.291 4 92 96.000 0.958

2 1 81 94.750 0.855 2 87 93.875 0.927 3 115 93.375 1.232 4 93 92.500 1.005

3 1 76 91.375 0.832 2 85 89.750 0.947 3 108 89.125 1.212 4 87 90.125 0.965

4 1 77 93.875 0.820 2 92 98.500 0.934 3 131 4 101

Season Adjusted Seasonal Index

Quarter 1 0.8372 Quarter 2 0.9377 Quarter 3 1.2474 Quarter 4 0.9779

11. a.

Season Seasonal Index 1 0.666 2 1.035 3 1.568 4 0.730

b.

Year Quarter Original Deseasonalized 1 Winter 64 96.096 Spring 103 99.517 Summer 152 96.939 Fall 73 100.000

2 Winter 66 99.099 Spring 103 99.517 Summer 160 102.041 Fall 72 98.630



c. The trend line is 92.686 + 1.32325(19) d. The forecast for summer of year 5 is 184.74



12. a.

FORECASTING WITH TREND AND SEASONAL COMPONENTS ***********************************************************

Season Seasonal Index

1 1.549 2 .690 3 .742 4 .792 5 1.228

b. Deaseasonalized data

Time Actual Deseasonalized1 28 18.0762 12 17.3913 16 21.5634 15 18.9395 23 18.7306 29 18.7227 10 14.4938 14 18.8689 14 17.67710 26 21.17311 32 20.65812 15 21.73913 15 20.21614 18 22.72715 27 21.98716 35 22.59517 17 24.63818 16 21.56319 20 25.25320 29 23.61621 37 23.88622 19 27.53623 18 24.25924 21 26.51525 28 22.801

c. The trend line is 16.918 + .347(t) d. THE MEAN SQUARE ERROR 1.97 THE FORECAST FOR PERIOD 26 40.17 THE FORECAST FOR PERIOD 27 18.13 THE FORECAST FOR PERIOD 28 19.75 THE FORECAST FOR PERIOD 29 21.36 THE FORECAST FOR PERIOD 30 33.55



13.

SI Month Trend Forecast1.38 97 76994.2 106252.01.42 98 77023.8 109373.81.35 99 77053.4 104022.11.03 100 77083.0 79395.50.99 101 77112.6 76341.50.62 102 77142.2 47828.20.51 103 77171.8 39357.60.58 104 77201.4 44776.80.82 105 77231.0 63329.40.82 106 77260.6 63353.70.92 107 77290.2 71107.01.56 108 77319.8 120618.9

14. F6/8,N = 42.15; F6/8,D = 70.14; F6/8,E = 52.17; F6/9,N = 43.46; F6/9,D = 72.30; F6/9,E = 53.76 15.

Year

Attendance

Advertising Expenditure

Forecast

Error

1 8363 750 2 9426 1250 9116 310 3 9318 3200 9830 -512 4 10206 4500 10102 104 5 11018 5600 10577 441

16. a. T10 = 349.667 + (10)(7.4) = 423.667

b. F10 = .1(399) + .3(412) + .6(408) = 408.3 17. a. S1 = 1.069, S2 = 0.973, S3 = 0.954, S4 = 1.004

b. b1 = 3.361 c. F21 = 100.50, F22 = 94.72, F23 = 96.15, F24 = 104.53

18. S1 = 1.178, S2 = 1.236, S3 = 0.586

b1 = 33.96 F13 = 2382, F14 = 2541, F15 = 1225

19. a. Forecast = 7.5 + .325x

b. Forecast = 28,625 sheets 20. a. For α = .6, MSE = 9.12; for α = .2, MSE = 10.82

b. For α = .6, F9 = 22.86; for α = .2, F9 = 21.72


1

7

Introduction to Linear Programming

MULTIPLE CHOICE

1. The maximization or minimization of a quantity is the a. goal of management science. b. decision for decision analysis. c. constraint of operations research. d. objective of linear programming. ANSWER: d TOPIC: Introduction 2. Decision variables a. tell how much or how many of something to produce, invest, purchase, hire, etc. b. represent the values of the constraints. c. measure the objective function. d. must exist for each constraint. ANSWER: a TOPIC: Objective function 3. Which of the following is a valid objective function for a linear programming problem?

a. Max 5xy b. Min 4x + 3y + (2/3)z c. Max 5x2 + 6y2 d. Min (x1 + x2)/x3

ASNWER: b TOPIC: Objective function 4. Which of the following statements is NOT true?

a. A feasible solution satisfies all constraints. b. An optimal solution satisfies all constraints. c. An infeasible solution violates all constraints. d. A feasible solution point does not have to lie on the boundary of the feasible region.

ANSWER: c TOPIC: Graphical solution


2 Chapter 7 Introduction to Linear Programming

5. A solution that satisfies all the constraints of a linear programming problem except the nonnegativity constraints is called

a. optimal. b. feasible. c. infeasible. d. semi-feasible. ANSWE R: c TOPIC: Graphical solution 6. Slack a. is the difference between the left and right sides of a constraint. b. is the amount by which the left side of a < constraint is smaller than the right side. c. is the amount by which the left side of a > constraint is larger than the right side. d. exists for each variable in a linear programming problem. ANSWER: b TOPIC: Slack variables 7. To find the optimal solution to a linear programming problem using the graphical method

a. find the feasible point that is the farthest away from the origin. b. find the feasible point that is at the highest location. c. find the feasible point that is closest to the origin. d. None of the alternatives is correct.

ANSWER: d TOPIC: Extreme points 8. Which of the following special cases does not require reformulation of the problem in order to obtain a

solution? a. alternate optimality b. infeasibility c. unboundedness d. each case requires a reformulation.

ANSWER: a TOPIC: Special cases 9. The improvement in the value of the objective function per unit increase in a right-hand side is the

a. sensitivity value. b. dual price. c. constraint coefficient. d. slack value.

ANSWER: b TOPIC: Right-hand sides 10. As long as the slope of the objective function stays between the slopes of the binding constraints

a. the value of the objective function won’t change. b. there will be alternative optimal solutions. c. the values of the dual variables won’t change. d. there will be no slack in the solution.

ANSWER: c TOPIC: Objective function


Chapter 7 Introduction to Linear Programming 3

11. Infeasibility means that the number of solutions to the linear programming models that satisfies all constraints is a. at least 1. b. 0. c. an infinite number. d. at least 2.

ANSWER: b TOPIC: Alternate Optimal Solution 12. A constraint that does not affect the feasible region is a a. non-negativity constraint. b. redundant constraint. c. standard constraint. d. slack constraint. ANSWER: b TOPIC: Feasible regions 13. Whenever all the constraints in a linear program are expressed as equalities, the linear program is said to be

written in a. standard form. b. bounded form. c. feasible form. d. alternative form. ANSWER: a TOPIC: Slack variables 14. All of the following statements about a redundant constraint are correct EXCEPT a. A redundant constraint does not affect the optimal solution. b. A redundant constraint does not affect the feasible region. c. Recognizing a redundant constraint is easy with the graphical solution method. d. At the optimal solution, a redundant constraint will have zero slack. ANSWER: d TOPIC: Slack variables 15. All linear programming problems have all of the following properties EXCEPT a. a linear objective function that is to be maximized or minimized. b. a set of linear constraints. c. alternative optimal solutions. d. variables that are all restricted to nonnegative values. ANSWER: c TOPIC: Problem formulation

TRUE/FALSE

1. Increasing the right-hand side of a nonbinding constraint will not cause a change in the optimal solution. ANSWER: False TOPIC: Introduction 2. In a linear programming problem, the objective function and the constraints must be linear functions of the

decision variables. ANSWER: True TOPIC: Mathematical statement of the RMC Problem



3. In a feasible problem, an equal-to constraint cannot be nonbinding. ANSWER: True TOPIC: Graphical solution 4. Only binding constraints form the shape (boundaries) of the feasible region. ANSWER: False TOPIC: Graphical solution 5. The constraint 5x1 - 2x2 < 0 passes through the point (20, 50). ANSWER: True TOPIC: Graphing lines 6. A redundant constraint is a binding constraint. ANSWER: False TOPIC: Slack variables 7. Because surplus variables represent the amount by which the solution exceeds a minimum target, they are

given positive coefficients in the objective function. ANSWER: False TOPIC: Slack variables 8. Alternative optimal solutions occur when there is no feasible solution to the problem. ANSWER: False TOPIC: Alternative optimal solutions 9. A range of optimality is applicable only if the other coefficient remains at its original value. ANSWER: True TOPIC: Simultaneous changes 10. Because the dual price represents the improvement in the value of the optimal solution per unit increase in

right-hand-side, a dual price cannot be negative. ANSWER: False TOPIC: Right-hand sides 11. Decision variables limit the degree to which the objective in a linear programming problem is satisfied. ANSWER: False TOPIC: Introduction 12. No matter what value it has, each objective function line is parallel to every other objective function line in

a problem. ANSWER: True TOPIC: Graphical solution 13. The point (3, 2) is feasible for the constraint 2x1 + 6x2 ≤ 30. ANSWER: True TOPIC: Graphical solution 14. The constraint 2x1 - x2 = 0 passes through the point (200,100). ANSWER: False TOPIC: A note on graphing lines 15. The standard form of a linear programming problem will have the same solution as the original problem. ANSWER: True TOPIC: Surplus variables



16. An optimal solution to a linear programming problem can be found at an extreme point of the feasible region for the problem.

ANSWER: True TOPIC: Extreme Points 17. An unbounded feasible region might not result in an unbounded solution for a minimization or

maximization problem. ANSWER: True TOPIC: Special cases: unbounded 18. An infeasible problem is one in which the objective function can be increased to infinity. ANSWER: False TOPIC: Special cases: infeasibility 19. A linear programming problem can be both unbounded and infeasible. ANSWER: False TOPIC: Special cases: infeasibility and unbounded 20. It is possible to have exactly two optimal solutions to a linear programming problem. ANSWER: False TOPIC: Special cases: alternative optimal solutions

SHORT ANSWER

1. Explain the difference between profit and contribution in an objective function. Why is it important for the

decision maker to know which of these the objective function coefficients represent? TOPIC: Objective function 2. Explain how to graph the line x1 - 2x2 > 0. TOPIC: Graphing lines 3. Create a linear programming problem with two decision variables and three constraints that will include

both a slack and a surplus variable in standard form. Write your problem in standard form. TOPIC: Standard form 4. Explain what to look for in problems that are infeasible or unbounded. TOPIC: Special cases 5. Use a graph to illustrate why a change in an objective function coefficient does not necessarily lead to a

change in the optimal values of the decision variables, but a change in the right-hand sides of a binding constraint does lead to new values.

TOPIC: Graphical sensitivity analysis 6. Explain the concepts of proportionality, additivity, and divisibility. TOPIC: Notes and comments



PROBLEMS

1. Solve the following system of simultaneous equations. 6X + 2Y = 50 2X + 4Y = 20 TOPIC: Simultaneous equations 2. Solve the following system of simultaneous equations. 6X + 4Y = 40 2X + 3Y = 20 TOPIC: Simultaneous equations 3. Consider the following linear programming problem

Max 8X + 7Y s.t. 15X + 5Y < 75 10X + 6Y < 60 X + Y < 8 X, Y ≥ 0 a. Use a graph to show each constraint and the feasible region. b. Identify the optimal solution point on your graph. What are the values of X and Y at the optimal

solution? c. What is the optimal value of the objective function? TOPIC: Graphical solution 4. For the following linear programming problem, determine the optimal solution by the graphical solution

method Max −X + 2Y s.t. 6X – 2Y ≤ 3 −2X + 3Y ≤ 6 X + Y ≤ 3 X, Y ≥ 0 TOPIC: Graphical solution



5. Use this graph to answer the questions.

0

5

10

15

0 5 10 15

I

IIIII

IVV

A

B

C

D

E

Max 20X + 10Y s.t. 12X + 15Y < 180 15X + 10Y < 150 3X - 8Y < 0 X , Y > 0 a. Which area (I, II, III, IV, or V) forms the feasible region? b. Which point (A, B, C, D, or E) is optimal? c. Which constraints are binding? d. Which slack variables are zero? TOPIC: Graphical solution 6. Find the complete optimal solution to this linear programming problem. Min 5X + 6Y s.t. 3X + Y > 15 X + 2Y > 12 3X + 2Y > 24 X , Y > 0 TOPIC: Graphical solution 7. Find the complete optimal solution to this linear programming problem. Max 5X + 3Y s.t. 2X + 3Y < 30 2X + 5Y < 40 6X - 5Y < 0 X , Y > 0 TOPIC: Graphical solution



8. Find the complete optimal solution to this linear programming problem. Max 2X + 3Y s.t. 4X + 9Y < 72 10X + 11Y < 110 17X + 9Y < 153 X , Y > 0 TOPIC: Graphical solution 9. Find the complete optimal solution to this linear programming problem. Min 3X + 3Y s.t. 12X + 4Y > 48 10X + 5Y > 50 4X + 8Y > 32 X , Y > 0 TOPIC: Graphical solution 10. For the following linear programming problem, determine the optimal solution by the graphical solution

method. Are any of the constraints redundant? If yes, then identify the constraint that is redundant. Max X + 2Y s.t. X + Y < 3 X − 2Y > 0 Y < 1 X, Y ≥ 0 TOPIC: Graphical solution 11. Maxwell Manufacturing makes two models of felt tip marking pens. Requirements for each lot of pens are

given below.

Fliptop Model Tiptop Model Available Plastic 3 4 36 Ink Assembly 5 4 40 Molding Time 5 2 30

The profit for either model is $1000 per lot. a. What is the linear programming model for this problem? b. Find the optimal solution. c. Will there be excess capacity in any resource? TOPIC: Modeling and graphical solution 12. The Sanders Garden Shop mixes two types of grass seed into a blend. Each type of grass has been rated

(per pound) according to its shade tolerance, ability to stand up to traffic, and drought resistance, as shown in the table. Type A seed costs $1 and Type B seed costs $2. If the blend needs to score at least 300 points for shade tolerance, 400 points for traffic resistance, and 750 points for drought resistance, how many pounds of each seed should be in the blend? Which targets will be exceeded? How much will the blend cost?



Type A Type B Shade Tolerance 1 1 Traffic Resistance 2 1 Drought Resistance 2 5

TOPIC: Modeling and graphical solution 13. Muir Manufacturing produces two popular grades of commercial carpeting among its many other products.

In the coming production period, Muir needs to decide how many rolls of each grade should be produced in order to maximize profit. Each roll of Grade X carpet uses 50 units of synthetic fiber, requires 25 hours of production time, and needs 20 units of foam backing. Each roll of Grade Y carpet uses 40 units of synthetic fiber, requires 28 hours of production time, and needs 15 units of foam backing. The profit per roll of Grade X carpet is $200 and the profit per roll of Grade Y carpet is $160. In the coming production period, Muir has 3000 units of synthetic fiber available for use. Workers have been scheduled to provide at least 1800 hours of production time (overtime is a possibility). The company has 1500 units of foam backing available for use. Develop and solve a linear programming model for this problem.

TOPIC: Modeling and graphical solution 14. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternate optimal

solutions? Explain. Min 1X + 1Y s.t. 5X + 3Y < 30 3X + 4Y > 36 Y < 7 X , Y > 0 TOPIC: Special cases 15. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternate optimal

solutions? Explain. Min 3X + 3Y s.t. 1X + 2Y < 16 1X + 1Y < 10 5X + 3Y < 45 X , Y > 0 TOPIC: Special cases 16. A businessman is considering opening a small specialized trucking firm. To make the firm profitable, it is

estimated that it must have a daily trucking capacity of at least 84,000 cu. ft. Two types of trucks are appropriate for the specialized operation. Their characteristics and costs are summarized in the table below. Note that truck 2 requires 3 drivers for long haul trips. There are 41 potential drivers available and there are facilities for at most 40 trucks. The businessman's objective is to minimize the total cost outlay for trucks.

Capacity Drivers Truck Cost (Cu. Ft.) Needed Small $18,000 2,400 1 Large $45,000 6,000 3

Solve the problem graphically and note there are alternate optimal solutions. Which optimal solution:

a. uses only one type of truck?



b. utilizes the minimum total number of trucks? c. uses the same number of small and large trucks?

TOPIC: Alternative optimal solutions 17. Consider the following linear program: MAX 60X + 43Y s.t. X + 3Y > 9 6X − 2Y = 12 X + 2Y < 10 X, Y > 0

a. Write the problem in standard form. b. What is the feasible region for the problem? c. Show that regardless of the values of the actual objective function coefficients, the optimal

solution will occur at one of two points. Solve for these points and then determine which one maximizes the current objective function.

TOPIC: Standard form and extreme points 18. Solve the following linear program graphically. MAX 5X + 7Y s.t. X < 6 2X + 3Y < 19 X + Y < 8 X, Y > 0 TOPIC: Graphical solution procedure 19. Given the following linear program: MIN 150X + 210Y s.t. 3.8X + 1.2Y > 22.8 Y > 6 Y < 15 45X + 30Y = 630 X, Y > 0

Solve the problem graphically. How many extreme points exist for this problem? TOPIC: Graphical solution procedure 20. Solve the following linear program by the graphical method. MAX 4X + 5Y s.t. X + 3Y < 22 -X + Y < 4 Y < 6 2X - 5Y < 0 X, Y > 0 TOPIC: Graphical solution procedure



SOLUTIONS TO PROBLEMS 1. X = 8, Y = 1 2. X = 4, Y = 4 3. a.

0

5

10

15

0 5 10 15

Feas.Reg.

b. The optimal solution occurs at the intersection of constraints 2 and 3. The point is X = 3, Y = 5. c. The value of the objective function is 59.



4. X = 0.6 and Y = 2.4

0

1

2

3

0 1 2 3

FeasibleRegion

5. a. Area III is the feasible region b. Point D is optimal c. Constraints 2 and 3 are binding d. S2 and S3 are equal to 0 6.

0

5

10

15

0 5 10 15

FeasibleRegion

The complete optimal solution is X = 6, Y = 3, Z = 48, S1 = 6, S2 = 0, S3 = 0



7.

0

5

10

15

20

0 5 10 15 20

The complete optimal solution is X = 15, Y = 0, Z = 75, S1 = 0, S2 = 10, S3 = 90 8.

0

5

10

15

20

0 5 10 15 20

The complete optimal solution is X = 4.304, Y = 6.087, Z = 26.87, S1 = 0, S2 = 0, S3 = 25.043



9.

0

5

10

15

0 5 10 15

The complete optimal solution is X = 4, Y = 2, Z = 18, S1 = 8, S2 = 0, S3 = 0

10. X = 2, and Y = 1 Yes, there is a redundant constraint; Y ≤ 1

0

1

2

3

0 1 2 3

FeasibleRegion



11. a. Let F = the number of lots of Fliptip pens to produce Let T = the number of lots of Tiptop pens to produce Max 1000F + 1000T s.t. 3F + 4T < 36 5F + 4T < 40 5F + 2T < 30 F , T > 0 b.

0

5

10

15

0 5 10 15

The complete optimal solution is F = 2, T = 7.5, Z = 9500, S1 = 0, S2 = 0, S3 = 5

c. There is an excess of 5 units of molding time available. 12. Let A = the pounds of Type A seed in the blend Let B = the pounds of Type B seed in the blend Min 1A + 2B s.t. 1A + 1B ≥ 300 2A + 1B ≥ 400 2A + 5B ≥ 750 A, B ≥ 0



0

100

200

300

400

0 100 200 300 400

The optimal solution is at A = 250, B = 50. Constraint 2 has a surplus value of 150. The cost is 350. 13. Let X = the number of rolls of Grade X carpet to make Let Y = the number of rolls of Grade Y carpet to make Max 200X + 160Y s.t. 50X + 40Y < 3000 25X + 28Y > 1800 20X + 15Y < 1500 X , Y > 0

0

10

20

30

40

50

60

70

80

90

100

0 10 20 30 40 50 60 70 80 90 100

F. R.

The complete optimal solution is X = 30, Y = 37.5, Z = 12000, S1 = 0, S2 = 0, S3 = 337.5

14. The problem is infeasible.



0

5

10

15

0 5 10 15

15. The problem has alternate optimal solutions.

0

5

10

15

0 5 10 15

16. a. 35 small, 0 large b. 5 small, 12 large

c. 10 small, 10 large 17. a. MAX 60X + 43Y S.T. X + 3Y - S1 = 9 6X - - 2Y = 12 X + 2Y + S3 = 10 X, Y, S1, S3 > 0

b. Line segment of 6X – 2Y = 12 between (22/7,24/7) and (27/10,21/10). c. Extreme points: (22/7,24/7) and (27/10,21/10). First one is optimal, giving Z = 336.



18. From the graph below we see that the optimal solution occurs at X = 5, Y = 3, and Z = 46.

8

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

2X + 3Y < 19

Y

X

X + Y < 8

MAX 5X + 7Y

X < 6

Optimal X = 5, Y = 3 Z = 46

19. Two extreme points exist (Points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760

(Point B).

22

20

18

16

14

12

10

8

6

4

2

2 4 6 8 10 12 14 16 18 20 22

Y

X

3.8X + 1.2Y > 22.8

45X + 30Y = 630

Y < 15

Feasible region is line segment between points A and B MIN Z = 150X + 210Y

Y > 6

B

A

C



20. Two extreme points exist (Points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760

(Point B).

10

8

6

4

2

2 4 6 8 10 12 14 16 18 20 22

-X + Y < 4

X + 3Y < 22

2X – 5Y < 0 Optimal X = 10, Y = 4 Z = 22

MAX Z = 4X + 5Y

Y < 6

Y

X


1

8

Linear Programming: Sensitivity Analysis and Interpretation of Solution

MULTIPLE CHOICE

1. To solve a linear programming problem with thousands of variables and constraints

a. a personal computer can be used. b. a mainframe computer is required. c. the problem must be partitioned into subparts. d. unique software would need to be developed.

ANSWER: a TOPIC: Computer solution 2. A negative dual price for a constraint in a minimization problem means

a. as the right-hand side increases, the objective function value will increase. b. as the right-hand side decreases, the objective function value will increase. c. as the right-hand side increases, the objective function value will decrease. d. as the right-hand side decreases, the objective function value will decrease.

ANSWER: a TOPIC: Dual price 3. If a decision variable is not positive in the optimal solution, its reduced cost is

a. what its objective function value would need to be before it could become positive. b. the amount its objective function value would need to improve before it could become positive. c. zero. d. its dual price.

ANSWER: b TOPIC: Reduced cost 4. A constraint with a positive slack value a. will have a positive dual price. b. will have a negative dual price. c. will have a dual price of zero. d. has no restrictions for its dual price. ANSWER: c TOPIC: Slack and dual price


2 Chapter 8 LP Sensitivity Analysis and Interpretation of Solution

5. The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the a. optimal solution. b. dual solution. c. range of optimality. d. range of feasibility.

ANSWER: c TOPIC: Range of optimality 6. The range of feasibility measures

a. the right-hand-side values for which the objective function value will not change. b. the right-hand-side values for which the values of the decision variables will not change. c. the right-hand-side values for which the dual prices will not change. d. each of the above is true.

ANSWER: c TOPIC: Range of feasibility 7. The 100% Rule compares a. proposed changes to allowed changes. b. new values to original values. c. objective function changes to right-hand side changes. d. dual prices to reduced costs. ANSWER: a TOPIC: Simultaneous changes 8. An objective function reflects the relevant cost of labor hours used in production rather than treating them

as a sunk cost. The correct interpretation of the dual price associated with the labor hours constraint is a. the maximum premium (say for overtime) over the normal price that the company would be

willing to pay. b. the upper limit on the total hourly wage the company would pay. c. the reduction in hours that could be sustained before the solution would change. d. the number of hours by which the right-hand side can change before there is a change in the

solution point. ANSWER: a TOPIC: Dual price 9. A section of output from The Management Scientist is shown here.

Variable Lower Limit Current Value Upper Limit 1 60 100 120

What will happen to the solution if the objective function coefficient for variable 1 decreases by 20? a. Nothing. The values of the decision variables, the dual prices, and the objective function will all

remain the same. b. The value of the objective function will change, but the values of the decision variables and the

dual prices will remain the same. c. The same decision variables will be positive, but their values, the objective function value, and the

dual prices will change. d. The problem will need to be resolved to find the new optimal solution and dual price. ANSWER: b TOPIC: Range of optimality


Chapter 8 LP Sensitivity Analysis and Interpretation of Solution 3

10. A section of output from The Management Scientist is shown here.

Constraint Lower Limit Current Value Upper Limit 2 240 300 420

What will happen if the right-hand-side for constraint 2 increases by 200?

a. Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same.

b. The value of the objective function will change, but the values of the decision variables and the dual prices will remain the same.

c. The same decision variables will be positive, but their values, the objective function value, and the dual prices will change.

d. The problem will need to be resolved to find the new optimal solution and dual price. ANSWER: d TOPIC: Range of feasibility 11. The amount that the objective function coefficient of a decision variable would have to improve before that

variable would have a positive value in the solution is the a. dual price. b. surplus variable. c. reduced cost. d. upper limit. ANSWER: c TOPIC: Interpretation of computer output 12. The dual price measures, per unit increase in the right hand side, a. the increase in the value of the optimal solution. b. the decrease in the value of the optimal solution. c. the improvement in the value of the optimal solution. d. the change in the value of the optimal solution. ANSWER: c TOPIC: Interpretation of computer output 13. Sensitivity analysis information in computer output is based on the assumption of a. no coefficient change. b. one coefficient change. c. two coefficient change. d. all coefficients change. ANSWER: b TOPIC: Simultaneous changes 14. When the cost of a resource is sunk, then the dual price can be interpreted as the a. minimum amount the firm should be willing to pay for one additional unit of the resource. b. maximum amount the firm should be willing to pay for one additional unit of the resource. c. minimum amount the firm should be willing to pay for multiple additional units of the resource. d. maximum amount the firm should be willing to pay for multiple additional units of the resource. ANSWER: b TOPIC: Dual price



15. The amount by which an objective function coefficient would have to improve before it would be possible for the corresponding variable to assume a positive value in the optimal solution is called the

a. reduced cost. b. relevant cost. c. sunk cost. d. dual price. ANSWER: a TOPIC: Reduced cost 16. Which of the following is not a question answered by sensitivity analysis? a. If the right-hand side value of a constraint changes, will the objective function value change? b. Over what range can a constraint’s right-hand side value without the constraint’s dual price

possibly changing? c. By how much will the objective function value change if the right-hand side value of a constraint

changes beyond the range of feasibility? d. By how much will the objective function value change if a decision variable’s coefficient in the

objective function changes within the range of optimality? ANSWER: c TOPIC: Interpretation of computer output

TRUE/FALSE

1. Output from a computer package is precise and answers should never be rounded. ANSWER: False TOPIC: Computer solution 2. The reduced cost for a positive decision variable is 0. ANSWER: True TOPIC: Reduced cost 3. When the right-hand sides of two constraints are each increased by one unit, the objective function value

will be adjusted by the sum of the constraints’ dual prices. ANSWER: False TOPIC: Simultaneous changes 4. If the range of feasibility indicates that the original amount of a resource, which was 20, can increase by 5,

then the amount of the resource can increase to 25. ANSWER: True TOPIC: Range of feasibility 5. The 100% Rule does not imply that the optimal solution will necessarily change if the percentage exceeds

100%. ANSWER: True TOPIC: Simultaneous changes 6. For any constraint, either its slack/surplus value must be zero or its dual price must be zero. ANSWER: True TOPIC: Dual price



7. A negative dual price indicates that increasing the right-hand side of the associated constraint would be detrimental to the objective.

ANSWER: True TOPIC: Dual price 8. Decision variables must be clearly defined before constraints can be written. ANSWER: True TOPIC: Model formulation 9. Decreasing the objective function coefficient of a variable to its lower limit will create a revised problem

that is unbounded. ANSWER: False TOPIC: Range of optimality 10. The dual price for a percentage constraint provides a direct answer to questions about the effect of increases

or decreases in that percentage. ANSWER: False TOPIC: Dual price 11. The dual price associated with a constraint is the improvement in the value of the solution per unit decrease

in the right-hand side of the constraint. ANSWER: False TOPIC: Interpretation of computer output 12. For a minimization problem, a positive dual price indicates the value of the objective function will increase. ANSWER: False TOPIC: Interpretation of computer output--a second example 13. There is a dual price for every decision variable in a model. ANSWER: False TOPIC: Interpretation of computer output 14. The amount of a sunk cost will vary depending on the values of the decision variables. ANSWER: False TOPIC: Cautionary note on the interpretation of dual prices 15. If the optimal value of a decision variable is zero and its reduced cost is zero, this indicates that alternative

optimal solutions exist. ANSWER: True TOPIC: Interpretation of computer output 16. Any change to the objective function coefficient of a variable that is positive in the optimal solution will

change the optimal solution. ANSWER: False TOPIC: Range of optimality 17. Relevant costs should be reflected in the objective function, but sunk costs should not. ANSWER: True TOPIC: Cautionary note on the interpretation of dual prices 18. If the range of feasibility for b1 is between 16 and 37, then if b1 = 22 the optimal solution will not change

from the original optimal solution. ANSWER: False TOPIC: Right-hand sides



19. The 100 percent rule can be applied to changes in both objective function coefficients and right-hand sides at the same time.

ANSWER: False TOPIC: Simultaneous changes 20. If the dual price for the right-hand side of a < constraint is zero, there is no upper limit on its range of

feasibility. ANSWER: True TOPIC: Right-hand sides

SHORT ANSWER

1. Describe each of the sections of output that come from The Management Scientist and how you would use

each. TOPIC: Interpretation of computer output 2. Explain the connection between reduced costs and the range of optimality, and between dual prices and the

range of feasibility. TOPIC: Interpretation of computer output 3. Explain the two interpretations of dual prices based on the accounting assumptions made in calculating the

objective function coefficients. TOPIC: Dual price 4. How can the interpretation of dual prices help provide an economic justification for new technology? TOPIC: Dual price 5. How is sensitivity analysis used in linear programming? Given an example of what type of questions that

can be answered. TOPIC: Sensitivity analysis 6. How would sensitivity analysis of a linear program be undertaken if one wishes to consider simultaneous

changes for both the right-hand-side values and objective function. TOPIC: Simultaneous sensitivity analysis

PROBLEMS

1. In a linear programming problem, the binding constraints for the optimal solution are 5X + 3Y < 30 2X + 5Y < 20

a. Fill in the blanks in the following sentence: As long as the slope of the objective function stays between _______ and _______, the current optimal solution point will remain optimal.

b. Which of these objective functions will lead to the same optimal solution? 1) 2X + 1Y 2) 7X + 8Y 3) 80X + 60Y 4) 25X + 35Y



TOPIC: Graphical sensitivity analysis 2. The optimal solution of the linear programming problem is at the intersection of constraints 1 and 2. Max 2x1 + x2 s.t. 4x1 + 1x2 < 400 4x1 + 3x2 < 600 1x1 + 2x2 ≤ 300 x1 , x2 > 0 a. Over what range can the coefficient of x1 vary before the current solution is no longer optimal? b. Over what range can the coefficient of x2 vary before the current solution is no longer optimal? c. Compute the dual prices for the three constraints. TOPIC: Graphical sensitivity analysis 3. The binding constraints for this problem are the first and second. Min x1 + 2x2 s.t. x1 + x2 ≥ 300 2x1 + x2 ≥ 400 2x1 + 5x2 < 750 x1 , x2 > 0 a. Keeping c2 fixed at 2, over what range can c1 vary before there is a change in the optimal solution

point? b. Keeping c1 fixed at 1, over what range can c2 vary before there is a change in the optimal solution

point? c. If the objective function becomes Min 1.5x1 + 2x2, what will be the optimal values of x1, x2, and

the objective function? d. If the objective function becomes Min 7x1 + 6x2, what constraints will be binding? e. Find the dual price for each constraint in the original problem. TOPIC: Graphical sensitivity analysis



4. Excel’s Solver tool has been used in the spreadsheet below to solve a linear programming problem with a maximization objective function and all < constraints.

Input Section

Objective Function CoefficientsX Y4 6

Constraints Avail.#1 3 5 60#2 3 2 48#3 1 1 20

Output Section

Variables 13.333333 4Profit 53.333333 24 77.333333

Constraint Usage Slack#1 60 1.789E-11#2 48 -2.692E-11#3 17.333333 2.6666667

a. Give the original linear programming problem. b. Give the complete optimal solution. TOPIC: Spreadsheet solution of LPs



5. Excel’s Solver tool has been used in the spreadsheet below to solve a linear programming problem with a minimization objective function and all > constraints.

Input Section

Objective Function CoefficientsX Y5 4

Constraints Req'd.#1 4 3 60#2 2 5 50#3 9 8 144

Output Section

Variables 9.6 7.2Profit 48 28.8 76.8

Constraint Usage Slack#1 60 1.35E-11#2 55.2 -5.2#3 144 -2.62E-11

a. Give the original linear programming problem. b. Give the complete optimal solution.

TOPIC: Spreadsheet solution of LPs 6. Use the spreadsheet and Solver sensitivity report to answer these questions.

a. What is the cell formula for B12? b. What is the cell formula for C12? c. What is the cell formula for D12? d. What is the cell formula for B15? e. What is the cell formula for B16? f. What is the cell formula for B17? g. What is the optimal value for x1? h. What is the optimal value for x2? i. Would you pay $.50 each for up to 60 more units of resource 1? j. Is it possible to figure the new objective function value if the profit on product 1 increases by a

dollar, or do you have to rerun Solver?



Input InformationVar. 1 Var. 2 (type) Avail.

Constraint 1 2 5 < 40Constraint 2 3 1 < 30Constraint 3 1 1 > 12

Profit 5 4

Output Information Variables Profit = Total

Resources Used Slk/Surp Constraint 1 Constraint 2 Constraint 3

Microsoft Excel 7.0 Sensitivity ReportWorksheet: [x3s7base.xls]Sheet1

Changing CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$12 Variables Variable 1 8.461538462 0 5 7 3.4$C$12 Variables Variable 2 4.615384615 0 4 8.5 2.333333333

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$15 constraint 1 Used 40 0.538461538 40 110 7$B$16 constraint 2 Used 30 1.307692308 30 30 4.666666667$B$17 constraint 3 Used 13.07692308 0 12 1.076923077 1E+30

TOPIC: Spreadsheet solution of LPs 7. Use the following Management Scientist output to answer the questions.

LINEAR PROGRAMMING PROBLEM

MAX 31X1+35X2+32X3 S.T. 1) 3X1+5X2+2X3>90 2) 6X1+7X2+8X3<150

3) 5X1+3X2+3X3<120



OPTIMAL SOLUTION

Objective Function Value = 763.333

Variable Value Reduced Cost X1 13.333 0.000 X2 10.000 0.000 X3 0.000 10.889

Constraint Slack/Surplus Dual Price

1 0.000 -0.778 2 0.000 5.556 3 23.333 0.000

OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper Limit

X1 30.000 31.000 No Upper Limit X2 No Lower Limit 35.000 36.167 X3 No Lower Limit 32.000 42.889

RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit

1 77.647 90.000 107.143 2 126.000 150.000 163.125 3 96.667 120.000 No Upper Limit

a. Give the solution to the problem. b. Which constraints are binding? c. What would happen if the coefficient of x1 increased by 3? d. What would happen if the right-hand side of constraint 1 increased by 10? TOPIC: Interpretation of Management Scientist output 8. Use the following Management Scientist output to answer the questions. MIN 4X1+5X2+6X3 S.T. 1) X1+X2+X3<85 2) 3X1+4X2+2X3>280 3) 2X1+4X2+4X3>320 Objective Function Value = 400.000





1 5.000 0.000 2 40.000 0.000 3 0.000 -1.250


Variable Lower Limit Current Value Upper Limit X1 2.500 4.000 No Upper Limit X2 0.000 5.000 6.000 X3 5.000 6.000 No Upper Limit


Constraint Lower Limit Current Value Upper Limit 1 80.000 85.000 No Upper Limit 2 No Lower Limit 280.000 320.000 3 280.000 320.000 340.000

a. What is the optimal solution, and what is the value of the profit contribution? b. Which constraints are binding? c. What are the dual prices for each resource? Interpret. d. Compute and interpret the ranges of optimality. e. Compute and interpret the ranges of feasibility. TOPIC: Interpretation of Management Scientist output 9. The following linear programming problem has been solved by The Management Scientist. Use the output

to answer the questions. LINEAR PROGRAMMING PROBLEM MAX 25X1+30X2+15X3 S.T. 1) 4X1+5X2+8X3<1200 2) 9X1+15X2+3X3<1500 OPTIMAL SOLUTION Objective Function Value = 4700.000



1 0.000 1.000 2 0.000 2.333




Variable Lower Limit Current Value Upper Limit X1 19.286 25.000 45.000 X2 No Lower Limit 30.000 40.000 X3 8.333 15.000 50.000


Constraint Lower Limit Current Value Upper Limit 1 666.667 1200.000 4000.000 2 450.000 1500.000 2700.000

a. Give the complete optimal solution. b. Which constraints are binding? c. What is the dual price for the second constraint? What interpretation does this have? d. Over what range can the objective function coefficient of x2 vary before a new solution point

becomes optimal? e. By how much can the amount of resource 2 decrease before the dual price will change? f. What would happen if the first constraint's right-hand side increased by 700 and the second's

decreased by 350? TOPIC: Interpretation of Management Scientist output 10. LINDO output is given for the following linear programming problem. MIN 12 X1 + 10 X2 + 9 X3 SUBJECT TO 2) 5 X1 + 8 X2 + 5 X3 >= 60 3) 8 X1 + 10 X2 + 5 X3 >= 80 END LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 80.000000

VARIABLE VALUE REDUCED COST X1 .000000 4.000000 X2 8.000000 .000000 X3 .000000 4.000000

ROW SLACK OR SURPLUS DUAL PRICE

2) 4.000000 .000000 3) .000000 -1.000000



NO. ITERATIONS= 1 RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ. COEFFICIENT RANGES

VARIABLE CURRENT COEFFICIENT

ALLOWABLE INCREASE

ALLOWABLE DECREASE

X1 12.000000 INFINITY 4.000000 X2 10.000000 5.000000 10.000000 X3 9.000000 INFINITY 4.000000

RIGHTHAND SIDE RANGES

ROW

CURRENT RHS

ALLOWABLE INCREASE

ALLOWABLE DECREASE

2 60.000000 4.000000 INFINITY 3 80.000000 INFINITY 5.000000

a. What is the solution to the problem? b. Which constraints are binding? c. Interpret the reduced cost for x1. d. Interpret the dual price for constraint 2. e. What would happen if the cost of x1 dropped to 10 and the cost of x2 increased to 12? TOPIC: Interpretation of LINDO output 11. The LP problem whose output follows determines how many necklaces, bracelets, rings, and earrings a

jewelry store should stock. The objective function measures profit; it is assumed that every piece stocked will be sold. Constraint 1 measures display space in units, constraint 2 measures time to set up the display in minutes. Constraints 3 and 4 are marketing restrictions.

LINEAR PROGRAMMING PROBLEM MAX 100X1+120X2+150X3+125X4 S.T. 1) X1+2X2+2X3+2X4<108 2) 3X1+5X2+X4<120 3) X1+X3<25 4) X2+X3+X4>50 OPTIMAL SOLUTION Objective Function Value = 7475.000

Variable Value Reduced Cost X1 8.000 0.000 X2 0.000 5.000 X3 17.000 0.000 X4 33.000 0.000



Constraint Slack/Surplus Dual Price 1 0.000 75.000 2 63.000 0.000 3 0.000 25.000 4 0.000 -25.000


Variable Lower Limit Current Value Upper Limit X1 87.500 100.000 No Upper Limit X2 No Lower Limit 120.000 125.000 X3 125.000 150.000 162.500 X4 120.000 125.000 150.000



1 100.000 108.000 123.750 2 57.000 120.000 No Upper Limit 3 8.000 25.000 58.000 4 41.500 50.000 54.000

Use the output to answer the questions.

a. How many necklaces should be stocked? b. Now many bracelets should be stocked? c. How many rings should be stocked? d. How many earrings should be stocked? e. How much space will be left unused? f. How much time will be used? g. By how much will the second marketing restriction be exceeded? h. What is the profit? i. To what value can the profit on necklaces drop before the solution would change? j. By how much can the profit on rings increase before the solution would change? k. By how much can the amount of space decrease before there is a change in the profit? l. You are offered the chance to obtain more space. The offer is for 15 units and the total price is

1500. What should you do? TOPIC: Interpretation of Management Scientist output 12. The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The

objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3. Use the output to answer these questions.

a. How much of ingredient 1 will be put into the blend? b. How much of ingredient 2 will be put into the blend? c. How much of ingredient 3 will be put into the blend? d. How much resource A is used? e. How much resource B will be left unused? f. What will the profit be?



g. What will happen to the solution if the profit from ingredient 2 drops to 4? h. What will happen to the solution if the profit from ingredient 3 increases by 1? i. What will happen to the solution if the amount of resource C increases by 2? j. What will happen to the solution if the minimum requirement for ingredient 3 increases

to 15?


MAX 4X1+6X2+7X3 S.T. 1) 3X1+2X2+5X3<120 2) 1X1+3X2+3X3<80

3) 5X1+5X2+8X3<160 4) +1X3>10

OPTIMAL SOLUTION


Variable Value Reduced Cost

X1 0.000 2.000 X2 16.000 0.000 X3 10.000 0.000


1 38.000 0.000 2 2.000 0.000 3 0.000 1.200 4 0.000 -2.600



X1 No Lower Limit 4.000 6.000 X2 4.375 6.000 No Upper Limit X3 No Lower Limit 7.000 9.600



1 82.000 120.000 No Upper Limit 2 78.000 80.000 No Upper Limit 3 80.000 160.000 163.333 4 8.889 10.000 20.000

TOPIC: Interpretation of Management Scientist output 13. The LP model and LINDO output represent a problem whose solution will tell a specialty retailer how

many of four different styles of umbrellas to stock in order to maximize profit. It is assumed that every one stocked will be sold. The variables measure the number of women's, golf, men's, and folding umbrellas,



respectively. The constraints measure storage space in units, special display racks, demand, and a marketing restriction, respectively.

MAX 4 X1 + 6 X2 + 5 X3 + 3.5 X4 SUBJECT TO

2) 2 X1 + 3 X2 + 3 X3 + X4 <= 120 3) 1.5 X1 + 2 X2 <= 54 4) 2 X2 + X3 + X4 <= 72 5) X2 + X3 >= 12 END OBJECTIVE FUNCTION VALUE 1) 318.00000

VARIABLE VALUE REDUCED COST X1 12.000000 .000000 X2 .000000 .500000 X3 12.000000 .000000 X4 60.000000 .000000


2) .000000 2.000000 3) 36.000000 .000000 4) .000000 1.500000 5) .000000 -2.500000

RANGES IN WHICH THE BASIS IS UNCHANGED:



ALLOWABLE INCREASE

ALLOWABLE DECREASE

X1 4.000000 1.000000 2.500000 X2 6.000000 .500000 INFINITY X3 5.000000 2.500000 .500000 X4 3.500000 INFINITY .500000


ROW

CURRENT RHS

ALLOWABLE INCREASE

ALLOWABLE DECREASE

2 120.000000 48.000000 24.000000 3 54.000000 INFINITY 36.000000 4 72.000000 24.000000 48.000000 5 12.000000 12.000000 12.000000

Use the output to answer the questions.

a. How many women's umbrellas should be stocked? b. How many golf umbrellas should be stocked? c. How many men's umbrellas should be stocked? d. How many folding umbrellas should be stocked?



e. How much space is left unused? f. How many racks are used? g. By how much is the marketing restriction exceeded? h. What is the total profit? i. By how much can the profit on women's umbrellas increase before the solution would change? j. To what value can the profit on golf umbrellas increase before the solution would change? k. By how much can the amount of space increase before there is a change in the dual price? l. You are offered an advertisement that should increase the demand constraint from 72 to 86 for a

total cost of $20. Would you say yes or no? TOPIC: Interpretation of LINDO output 14. Eight of the entries have been deleted from the LINDO output that follows. Use what you know about

linear programming to find values for the blanks. MIN 6 X1 + 7.5 X2 + 10 X3 SUBJECT TO 2) 25 X1 + 35 X2 + 30 X3 >= 2400 3) 2 X1 + 4 X2 + 8 X3 >= 400 END LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 612.50000

VARIABLE VALUE REDUCED COST X1 ________ 1.312500 X2 ________ ________ X3 27.500000 ________


2) ________ -.125000 3) ________ -.781250




ALLOWABLE INCREASE

ALLOWABLE DECREASE

X1 6.000000 _________ _________ X2 7.500000 1.500000 2.500000 X3 10.000000 5.000000 3.571429


ROW

CURRENT RHS

ALLOWABLE INCREASE

ALLOWABLE DECREASE

2 2400.000000 1100.000000 900.000000 3 400.000000 240.000000 125.714300



TOPIC: Interpretation of LINDO output 15. Portions of a Management Scientist output are shown below. Use what you know about the solution of

linear programs to fill in the ten blanks.


MAX 12X1+9X2+7X3 S.T. 1) 3X1+5X2+4X3<150 2) 2X1+1X2+1X3<64 3) 1X1+2X2+1X3<80 4) 2X1+4X2+3X3>116

OPTIMAL SOLUTION


Variable Value Reduced Cost X1 ______ 0.000 X2 24.000 ______ X3 ______ 3.500


1 0.000 15.000 2 ______ 0.000 3 ______ 0.000 4 0.000 ______



X1 5.400 12.000 No Upper Limit X2 2.000 9.000 20.000 X3 No Lower Limit 7.000 10.500



1 145.000 150.000 156.667 2 ______ ______ 64.000 3 ______ ______ 80.000 4 110.286 116.000 120.000

TOPIC: Interpretation of Management Scientist output



Note to Instructor: The following problem is suitable for a take-home or lab exam. The student must formulate the model, solve the problem with a computer package, and then interpret the solution to answer the questions. 16. A large sporting goods store is placing an order for bicycles with its supplier. Four models can be ordered:

the adult Open Trail, the adult Cityscape, the girl's Sea Sprite, and the boy's Trail Blazer. It is assumed that every bike ordered will be sold, and their profits, respectively, are 30, 25, 22, and 20. The LP model should maximize profit. There are several conditions that the store needs to worry about. One of these is space to hold the inventory. An adult’s bike needs two feet, but a child's bike needs only one foot. The store has 500 feet of space. There are 1200 hours of assembly time available. The child's bike need 4 hours of assembly time; the Open Trail needs 5 hours and the Cityscape needs 6 hours. The store would like to place an order for at least 275 bikes.

a. Formulate a model for this problem. b. Solve your model with any computer package available to you. c. How many of each kind of bike should be ordered and what will the profit be? d. What would the profit be if the store had 100 more feet of storage space? e. If the profit on the Cityscape increases to $35, will any of the Cityscape bikes be ordered? f. Over what range of assembly hours is the dual price applicable? g. If we require 5 more bikes in inventory, what will happen to the value of the optimal solution? h. Which resource should the company work to increase, inventory space or assembly time? TOPIC: Formulation and computer solution 17. A company produces two products made from aluminum and copper. The table below gives the unit

requirements, the unit production man-hours required, the unit profit and the availability of the resources (in tons).

Aluminum Copper Man-hours Unit Profit Product 1 1 0 2 50 Product 2 1 1 3 60 Available 10 6 24

The Management Scientist provided the following solution output:

OBJECTIVE FUNCTION VALUE = 540.000

VARIABLE VALUE REDUCED COST

X1 6.000 0.000 X2 4.000 0.000

CONSTRAINT SLACK/SURPLUS DUAL PRICE

1 .000 30.000 2 2.000 0.000 3 0.000 10.000

RANGES IN WHICH THE BASIS IS UNCHANGED:



ALLOWABLE INCREASE

ALLOWABLE DECREASE

X1 50.000 10.000 10.000 X2 60.000 15.000 10.000




CONSTRAINT CURRENT

RHS ALLOWABLE

INCREASE ALLOWABLE

DECREASE 1 10.000 2.000 1.000 2 6.000 INFINITY 2.000 3 24.000 2.000 4.000

a. What is the optimal production schedule? b. Within what range for the profit on product 2 will the solution in (a) remain optimal? What is the

optimal profit when c2 = 70? c. Suppose that simultaneously the unit profits on x1 and x2 changed from 50 to 55 and 60 to 65

respectively. Would the optimal solution change? d. Explain the meaning of the "DUAL PRICES" column. Given the optimal solution, why should

the dual price for copper be 0? e. What is the increase in the value of the objective function for an extra unit of aluminum? f. Man-hours were not figured into the unit profit as it must pay three workers for eight hours of

work regardless of the number of man-hours used. What is the dual price for man-hours? Interpret.

g. On the other hand, aluminum and copper are resources that are ordered as needed. The unit profit coefficients were determined by: (selling price per unit) - (cost of the resources per unit). The 10 units of aluminum cost the company $100. What is the most the company should be willing to pay for extra aluminum?

TOPIC: Interpretation of solution 18. Given the following linear program: MAX 5x1 + 7x2 s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > 0

The graphical solution to the problem is shown below. From the graph we see that the optimal solution occurs at x1 = 5, x2 = 3, and z = 46.



8

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

2X1 + 3X2 < 19

X2

X1

X1 + X2 < 8

MAX 5X1 + 7X2

X1 < 6

Optimal X1 = 5, X2 = 3 Z = 46

a. Calculate the range of optimality for each objective function coefficient. b. Calculate the dual price for each resource.

TOPIC: Introduction to sensitivity analysis 19. Consider the following linear program: MAX 3x1 + 4x2 ($ Profit) s.t. x1 + 3x2 < 12 2x1 + x2 < 8 x1 < 3 x1, x2 > 0


OPTIMAL SOLUTION


Variable Value Reduced Cost X1 2.400 0.000 X2 3.200 0.000


1 0.000 1.000 2 0.000 1.000 3 0.600 0.000





X1 1.333 3.000 8.000 X2 1.500 4.000 9.000



1 9.000 12.000 24.000 2 4.000 8.000 9.000 3 2.400 3.000 No Upper Limit

a. What is the optimal solution including the optimal value of the objective function? b. Suppose the profit on x1 is increased to $7. Is the above solution still optimal? What is the value

of the objective function when this unit profit is increased to $7? c. If the unit profit on x2 was $10 instead of $4, would the optimal solution change? d. If simultaneously the profit on x1 was raised to $5.5 and the profit on x2 was reduced to $3, would

the current solution still remain optimal? TOPIC: Interpretation of solution 20. Consider the following linear program: MIN 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2 > 0


OPTIMAL SOLUTION


Variable Value Reduced Cost X1 1.500 0.000 X2 2.000 0.000


1 2.500 0.000 2 0.000 -0.600 3 0.000 -4.500





X1 0.000 6.000 12.000 X2 4.500 9.000 No Upper Limit



1 5.500 8.000 No Upper Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

a. What is the optimal solution including the optimal value of the objective function? b. Suppose the unit cost of x1 is decreased to $4. Is the above solution still optimal? What is the

value of the objective function when this unit cost is decreased to $4? c. How much can the unit cost of x2 be decreased without concern for the optimal solution changing? d. If simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was reduced to $6, would the

current solution still remain optimal? e. If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal

solution? TOPIC: Interpretation of solution


1. a. -5/3 and -2/5 b. Objective functions 2), 3), and 4) 2. a. 1.33 < c1 < 4 b. . 5 < c2 < 1.5 c. Dual prices are .25, .25, 0 3. a. .8 < c1 < 2 b. 1 < c2 < 2.5 c. x1 = 250, x2 = 50, z = 475 d. Constraints 1 and 2 will be binding. e. Dual prices are .33, 0, .33 (The first and third values are negative.) 4. a. Max 4X + 6Y s.t. 3X + 5Y < 60 3X + 2Y < 48 1X + 1Y < 20 X , Y > 0 b. The complete optimal solution is X = 13.333, Y = 4, Z = 73.333, S1 = 0, S2 = 0, S3 = 2.667 5. a. Min 5X + 4Y s.t. 4X + 3Y > 60 2X + 5Y > 50 9X + 8Y > 144 X , Y > 0 b. The complete optimal solution is X = 9.6, Y = 7.2, Z = 76.8, S1 = 0, S2 = 5.2, S3 = 0



6. a. =B8*B11 b. =C8*C11 c. =B12+C12 d. =B4*B11+C4*C11 e. =B5*B11+C5*C11 f. =B6*B11+C6*C11 g. 8.46 h. 4.61 i. yes j. no 7. a. x1 = 13.33, x2 = 10, x3 = 0, s1 = 0, s2 = 0, s3 = 23.33, z = 763.33 b. Constraints 1 and 2 are binding. c. The value of the objective function would increase by 40. d. The value of the objective function would decrease by 7.78. 8. a. x1 = 0, x2 = 80, x3 = 0, s1 = 5, s2 = 40, s3 = 0, Z = 400 b. Constraint 3 is binding. c. Dual prices are 0, 0, and -1.25. They measure the improvement in Z per unit increase in each right-hand side. d. 2.5 < c1 < ∞ 0 < c2 < 6 5 < c3 < ∞ As long as the objective function coefficient stays within its range, the current optimal solution

point will not change, although Z could. e. 80 < b1 < ∞ -∞ < b2 < 320 280 < b3 < 340 As long as the right-hand side value stays within its range, the currently binding constraints will

remain so, although the values of the decision variables could change. The dual variable values will remain the same.

9. a. x1 = 140, x2 = 0, x3 = 80, s1 = 0, s2 = 0, z = 4700 b. Constraints 1 and 2 are binding. c. Dual price 2 = 2.33. A unit increase in the right-hand side of constraint 2 will increase the value

of the objective function by 2.33. d. As long as c2 < 40, the solution will be unchanged. e. 1050 f. The sum of percentage changes is 700/2800 + (-350)/(-1050) < 1 so the solution will not change. 10. a. x1 = 0, x2 = 8, x3 = 0, s1 = 4, s2 = 0, z = 80 b. Constraint 2 is binding. c. c1 would have to decrease by 4 or more for x1 to become positive. d. Increasing the right-hand side by 1 will cause a negative improvement, or increase, of 1 in this

minimization objective function. e. The sum of the percentage changes is (-2)/(-4) + 2/5 < 1 so the solution would not change. 11. a. 8 b. 0 c. 17 d. 33 e. 0 f. 57 g. 0 h. 7475



i. 87.5 j 12.5 k. 0 l. Say no. Although 15 units can be evaluated, their value (1125) is less than the cost (1500). 12. a. 0 b. 16 c. 10 d. 44 e. 2 f. 166 g. rerun h. Z = 176 i. Z = 168.4 j. Z = 153 13. a. 12 b. 0 c. 12 d. 60 e. 0 f. 18 g. 0 h. 318 i. 1 j. 6.5 k. 48 l. Yes. The dual price is 1.5 for 24 additional units. The value of the ad (14)(1.5)=21 exceeds the

cost of 20. 14. It is easiest to calculate the values in this order. x1 = 0, x2 = 45, reduced cost 2 = 0, reduced cost 3 = 0, row 2 slack = 0, row 3 slack = 0, c1 allowable

decrease = 1.3125, allowable increase = infinity 15. x3 = 0 because the reduced cost is positive. x1 = 24 after plugging into the objective function The second reduced cost is 0. s2 = 20 and s3 = 22 from plugging into the constraints. The fourth dual price is -16.5 from plugging into the dual objective function, which your students might not

understand fully until Chapter 6. The lower limit for constraint 2 is 44 and for constraint 3 is 58, from the amount of slack in each constraint.

There are no upper limits for these constraints.



16. a. MAX 30 X1 + 25 X2 + 22 X3 + 20 X4 SUBJECT TO 2) 2 X1 + 2 X2 + X3 + X4 <= 500 3) 5 X1 + 6 X2 + 4 X3 + 4 X4 <= 1200 4) X1 + X2 + X3 + X4 >= 275 b. OBJECTIVE FUNCTION VALUE 1) 6850.0000

VARIABLE VALUE REDUCED COST X1 100.000000 .000000 X2 .000000 13.000000 X3 175.000000 .000000 X4 .000000 2.000000


2) 125.000000 .000000 3) .000000 8.000000 4) .000000 -10.000000




ALLOWABLE INCREASE

ALLOWABLE DECREASE

X1 30.000000 INFINITY 2.500000 X2 25.000000 13.000000 INFINITY X3 22.000000 2.000000 2.000000 X4 20.000000 2.000000 INFINITY


ROW

CURRENT RHS

ALLOWABLE INCREASE

ALLOWABLE DECREASE

2 500.000000 INFINITY 125.000000 3 1200.000000 125.000000 100.000000 4 275.000000 25.000000 35.000000

c. Order 100 Open Trails, 0 Cityscapes, 175 Sea Sprites, and 0 Trail Blazers. Profit will be 6850. d. 6850 e. No. The $10 increase is below the reduced cost. f. 1100 to 1325 g. It will decrease by 50.

h. Assembly time. 17. a. 6 product 1, 4 product 2, Profit = $540

b. Between $50 and $75; at $70 the profit is $580



c. No; total % change is 83 1/3% < 100% d. Dual prices are the shadow prices for the resources; since there was unused copper (because S2 =

2), extra copper is worth $0 e. $30!f. $10; this is the amount extra man-hours are worth

g. The shadow price is the "premium" for aluminum -- would be willing to pay up to $10 + $30 = $40 for extra aluminum

18. a. Ranges of optimality: 14/3 < c1 < 7 and 5 < c2 < 15/2

b. Summarizing, the dual price for the first resource is 0, for the second resource is 2, and for the third is 1

19. a. x1 = 2.4 and x2 = 3.2, and z = $20.00.

b. Optimal solution will not change. Optimal profit will equal $29.60. c. Because 10 is outside the range of 1.5 to 9.0, the optimal solution likely would change. d. Sum of the change percentages is 50% + 40% = 90%. Since this does not exceed 100% the

optimal solution would not change. 20. a. x1 = 1.5 and x2 = 2.0, and the objective function value = 27.00.

b. 4 is within this range of 0 to 12, so the optimal solution will not change. Optimal total cost will be $24.00.

c. x2 can fall to 4.5 without concern for the optimal solution changing. d. Sum of the change percentages is 91.7%. This does not exceed 100%, so the optimal solution

would not change. e. The right-hand side remains within the range of feasibility, so there is no change in the optimal

solution. However, the objective function value increases by $4.50.


1

9

Linear Programming Applications in Marketing, Finance, and Operations Management

MULTIPLE CHOICE

1. Media selection problems usually determine

a. how many times to use each media source. b. the coverage provided by each media source. c. the cost of each advertising exposure. d. the relative value of each medium.

ANSWER: a TOPIC: Media selection 2. To study consumer characteristics, attitudes, and preferences, a company would engage in a. client satisfaction processing. b. marketing research. c. capital budgeting. d. production planning. ANSWER: b TOPIC: Marketing research 3. A marketing research application uses the variable HD to represent the number of homeowners interviewed

during the day. The objective function minimizes the cost of interviewing this and other categories and there is a constraint that HD > 100. The solution indicates that interviewing another homeowner during the day will increase costs by 10.00. What do you know?

a. the objective function coefficient of HD is 10. b. the dual price for the HD constraint is 10. c. the objective function coefficient of HD is -10. d. the dual price for the HD constraint is -10. ANSWER: d TOPIC: Marketing research 4. The dual price for a constraint that compares funds used with funds available is .058. This means that

a. the cost of additional funds is 5.8%. b. if more funds can be obtained at a rate of 5.5%, some should be. c. no more funds are needed. d. the objective was to minimize.

ANSWER: b TOPIC: Portfolio selection


2 Chapter 9 Linear Programming Applications

5. Let M be the number of units to make and B be the number of units to buy. If it costs $2 to make a unit and $3 to buy a unit and 4000 units are needed, the objective function is a. Max 2M + 3B b. Min 4000 (M + B) c. Max 8000M + 12000B d. Min 2M + 3B

ANSWER: d TOPIC: Make or buy 6. If Pij = the production of product i in period j, then to indicate that the limit on production of the company’s

three products in period 2 is 400, a. P21 + P22 + P23 < 400 b. P12 + P22 + P32 < 400 c. P32 < 400 d. P23 < 400 ANSWER: b TOPIC: Production scheduling 7. Let Pij = the production of product i in period j. To specify that production of product 1 in period 3 and in

period 4 differs by no more than 100 units, a. P13 - P14 < 100; P14 - P13 < 100 b. P13 - P14 < 100; P13 - P14 > 100 c. P13 - P14 < 100; P14 - P13 > 100 d. P13 - P14 > 100; P14 - P13 > 100 ANSWER: a TOPIC: Production scheduling 8. Let A, B, and C be the amounts invested in companies A, B, and C. If no more than 50% of the total

investment can be in company B, then a. B < 5 b. A - .5B + C < 0 c. .5A - B - .5C < 0 d. -.5A + .5B - .5C < 0

ANSWER: d TOPIC: Portfolio selection 9. Department 3 has 2500 hours. Transfers are allowed to departments 2 and 4, and from departments 1 and

2. If Ai measures the labor hours allocated to department i and Tij the hours transferred from department i to department j, then

a. T13 + T23 - T32 - T34 - A3 = 2500 b. T31 + T32 - T23 - T43 + A3 = 2500 c. A3 + T13 + T23 - T32 - T34 = 2500 d. A3 - T13 - T23 + T32 + T34 = 2500 ANSWER: d TOPIC: Work force assignment 10. Modern revenue management systems maximize revenue potential for an organization by helping to

manage a. pricing strategies. b. reservation policies. c. short-term supply decisions. d. All of the alternatives are correct.

ANSWER: d TOPIC: Revenue management


Chapter 9 Linear Programming Applications 3

TRUE/FALSE

1. Media selection problems can maximize exposure quality and use number of customers reached as a

constraint, or maximize the number of customers reached and use exposure quality as a constraint. ANSWER: True TOPIC: Media selection 2. Revenue management methodology was originally developed for the banking industry. ANSWER: False TOPIC: Revenue management 3. Portfolio selection problems should acknowledge both risk and return. ANSWER: True TOPIC: Portfolio selection 4. If an LP problem is not correctly formulated, the computer software will indicate it is infeasible when

trying to solve it. ANSWER: False TOPIC: Computer solutions 5. It is improper to combine manufacturing costs and overtime costs in the same objective function. ANSWER: False TOPIC: Make-or-buy 6. Production constraints frequently take the form: beginning inventory + sales - production = ending inventory ANSWER: False TOPIC: Production scheduling 7. If a real-world problem is correctly formulated, it is not possible to have alternative optimal solutions. ANSWER: False TOPIC: Problem formulation 8. To properly interpret dual prices, one must know how costs were allocated in the objective function. ANSWER: True TOPIC: Make-or-buy 9. A company makes two products from steel; one requires 2 tons of steel and the other requires 3 tons.

There are 100 tons of steel available daily. A constraint on daily production could be written as: 2x1 + 3x2 < 100.

ANSWER: True TOPIC: Production scheduling 10. Double-subscript notation for decision variables should be avoided unless the number of decision variables

exceeds nine. ANSWER: False TOPIC: Formulation notation 11. Using minutes as the unit of measurement on the left-hand side of a constraint and using hours on the right-

hand side is acceptable since both are a measure of time. ANSWER: False TOPIC: General



12. Compared to the problems in the textbook, real-world problems generally require more variables and constraints.

ANSWER: True TOPIC: General 13. For the multiperiod production scheduling problem in the textbook, period n - 1's ending inventory variable

was also used as period n's beginning inventory variable. ANSWER: True TOPIC: Production scheduling 14. A company makes two products, A and B. A sells for $100 and B sells for $90. The variable production

costs are $30 per unit for A and $25 for B. The company's objective could be written as: MAX 190x1 - 55x2.

ANSWER: False TOPIC: Production scheduling 15. The primary limitation of linear programming's applicability is the requirement that all decision variables

be nonnegative. ANSWER: False TOPIC: General 16. A decision maker would be wise to not deviate from the optimal solution found by an LP model because it

is the best solution. ANSWER: False TOPIC: General

SHORT ANSWER

1. Discuss the need for the use of judgment or other subjective methods in mathematical modeling. TOPIC: Media selection 2. What benefits exist in using linear programming for production scheduling problems? TOPIC: Production scheduling 3. Describe some common feature of multiperiod financial planning models. TOPIC: Financial planning 4. Why should decision makers who are primarily concerned with marketing or finance or production know

about linear programming? TOPIC: Introduction 5. Discuss several resource allocation problems that can be modeled by varying the workforce assignment

model. TOPIC: Workforce assignment model

PROBLEMS



1. A&C Distributors is a company that represents many outdoor products companies and schedules deliveries to discount stores, garden centers, and hardware stores. Currently, scheduling needs to be done for two lawn sprinklers, the Water Wave and Spring Shower models. Requirements for shipment to a warehouse for a national chain of garden centers are shown below.

Month Shipping Capacity

Product

Minimum Requirement

Unit Cost to Ship

Per Unit Inventory Cost

March 8000 Water Wave 3000 .30 .06 Spring Shower 1800 .25 .05

April 7000 Water Wave 4000 .40 .09 Spring Shower 4000 .30 .06

May 6000 Water Wave 5000 .50 .12 Spring Shower 2000 .35 .07

Let Sij be the number of units of sprinkler i shipped in month j, where i = 1 or 2, and j = 1, 2, or 3. Let Wij be the number of sprinklers that are at the warehouse at the end of a month, in excess of the minimum requirement. a Write the portion of the objective function that minimizes shipping costs. b. An inventory cost is assessed against this ending inventory. Give the portion of the objective

function that represents inventory cost. c. There will be three constraints that guarantee, for each month, that the total number of sprinklers

shipped will not exceed the shipping capacity. Write these three constraints. d. There are six constraints that work with inventory and the number of units shipped, making sure

that enough sprinklers are shipped to meet the minimum requirements. Write these six constraints.

TOPIC: Production scheduling 2. An ad campaign for a new snack chip will be conducted in a limited geographical area and can use TV

time, radio time, and newspaper ads. Information about each medium is shown below.

Medium Cost Per Ad # Reached Exposure Quality TV 500 10000 30

Radio 200 3000 40 Newspaper 400 5000 25

If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is $10000, develop the model that will maximize the number reached and achieve an exposure quality if at least 1000.

TOPIC: Media selection 3. Information on a prospective investment for Wells Financial Services is given below.

Period 1 2 3 4 Loan Funds Available 3000 7000 4000 5000 Investment Income (% of previous period’s investment)

110%

112%

113%

Maximum Investment 4500 8000 6000 7500 Payroll Payment 100 120 150 100

In each period, funds available for investment come from two sources: loan funds and income from the

previous period's investment. Expenses, or cash outflows, in each period must include repayment of the previous period's loan plus 8.5% interest, and the current payroll payment. In addition, to end the planning



horizon, investment income from period 4 (at 110% of the investment) must be sufficient to cover the loan plus interest from period 4. The difference in these two quantities represents net income, and is to be maximized. How much should be borrowed and how much should be invested each period?

TOPIC: Financial planning 4. Tots Toys makes a plastic tricycle that is composed of three major components: a handlebar-front wheel-

pedal assembly, a seat and frame unit, and rear wheels. The company has orders for 12,000 of these tricycles. Current schedules yield the following information.

Requirements Cost to Cost to Component Plastic Time Space Manufacture Purchase Front 3 10 2 8 12 Seat/Frame 4 6 2 6 9 Rear wheel (each) .5 2 .1 1 3 Available 50000 160000 30000

The company obviously does not have the resources available to manufacture everything needed for the

completion of 12000 tricycles so has gathered purchase information for each component. Develop a linear programming model to tell the company how many of each component should be manufactured and how many should be purchased in order to provide 12000 fully completed tricycles at the minimum cost.

TOPIC: Make or buy 5. The Tots Toys Company is trying to schedule production of two very popular toys for the next three

months: a rocking horse and a scooter. Information about both toys is given below.

Begin. Invty. Required Required Production Production Toy June 1 Plastic Time Cost Cost Rocking Horse 25 5 2 12 1 Scooter 55 4 3 14 1.2

Plastic Time Monthly Demand Monthly Demand Summer Schedule Available Available Horse Scooter June 3500 2100 220 450 July 5000 3000 350 700 August 4800 2500 600 520

Develop a model that would tell the company how many of each toy to produce during each month. You are to minimize total cost. Inventory cost will be levied on any items in inventory on June 30, July 31, or August 31 after demand for the month has been satisfied. Your model should make use of the relationship

Beginning Inventory + Production - Demand = Ending Inventory for each month. The company wants to end the summer with 150 rocking horses and 60 scooters as

beginning inventory for Sept. 1. Don't forget to define your decision variables. TOPIC: Production scheduling 6. Larkin Industries manufactures several lines of decorative and functional metal items. The most recent

order has been for 1200 door lock units for an apartment complex developer. The sales and production departments must work together to determine delivery schedules. Each lock unit consists of three components: the knob and face plate, the actual lock itself, and a set of two keys. Although the processes used in the manufacture of the three components vary, there are three areas where the production manager is concerned about the availability of resources. These three areas, their usage by the three components, and their availability are detailed in the table.



Resource Knob and Plate Lock Key (each) Available Brass Alloy 12 5 1 15000 units Machining 18 20 10 36000 minutes Finishing 15 5 1 12000 minutes

A quick look at the amounts available confirms that Larkin does not have the resources to fill this contract.

A subcontractor, who can make an unlimited number of each of the three components, quotes the prices below.

Component Subcontractor Cost Larkin Cost Knob and Plate 10.00 6.00 Lock 9.00 4.00 Keys (set of 2) 1.00 .50

Develop a linear programming model that would tell Larkin how to fill the order for 1200 lock sets at the

minimum cost. TOPIC: Make-or-buy 7. G and P Manufacturing would like to minimize the labor cost of producing dishwasher motors for a major

appliance manufacturer. Although two models of motors exist, the finished models are indistinguishable from one another; their cost difference is due to a different production sequence. The time in hours required for each model in each production area is tabled here, along with the labor cost.

Model 1 Model 2

Area A 15 3 Area B 4 10 Area C 4 8

Cost 80 65 Currently labor assignments provide for 10,000 hours in each of Areas A and B and 18000 hours in Area C.

If 2000 hours are available to be transferred from area B to Area A, 3000 hours are available to be transferred from area C to either Areas A or B, develop the linear programming model whose solution would tell G&P how many of each model to produce and how to allocate the workforce.

TOPIC: Workforce assignment 8. FarmFresh Foods manufactures a snack mix called TrailTime by blending three ingredients: a dried fruit

mixture, a nut mixture, and a cereal mixture. Information about the three ingredients (per ounce) is shown below.

Ingredient Cost Volume Fat Grams Calories Dried Fruit .35 1/4 cup 0 150 Nut Mix .50 3/8 cup 10 400 Cereal Mix .20 1 cup 1 50

The company needs to develop a linear programming model whose solution would tell them how many

ounces of each mix to put into the TrailTime blend. TrailTime is packaged in boxes that will hold between three and four cups. The blend should contain no more than 1000 calories and no more than 25 grams of fat. Dried fruit must be at least 20% of the volume of the mixture, and nuts must be no more than 15% of the weight of the mixture. Develop a model that meets these restrictions and minimizes the cost of the blend.

TOPIC: Blending



9. The Meredith Ribbon Company produces paper and fabric decorative ribbon which it sells to paper

products companies and craft stores. The demand for ribbon is seasonal. Information about projected demand and production for a particular type of ribbon is given.

Demand (yards) Production Cost Per Yard Production Capacity (yards)

Quarter 1 10,000 .03 30,000 Quarter 2 18,000 .04 20,000 Quarter 3 16,000 .06 20,000 Quarter 4 30,000 .08 15,000

An inventory holding cost of $.005 is levied on every yard of ribbon carried over from one quarter to the next.

a. Define the decision variables needed to model this problem. b. The objective is to minimize total cost, the sum of production and inventory holding cost. Give

the objection function. c. Write the production capacity constraints. d. Write the constraints that balance inventory, production, and demand for each quarter. Assume

there is no beginning inventory in quarter 1. e. To attempt to balance the production and avoid large changes in the workforce, production in

period 1 must be within 5000 yards of production in period 2. Write this constraint.

TOPIC: Production scheduling 10. Island Water Sports is a business that provides rental equipment and instruction for a variety of water sports

in a resort town. On one particular morning, a decision must be made of how many Wildlife Raft Trips and how many Group Sailing Lessons should be scheduled. Each Wildlife Raft Trip requires one captain and one crew person, and can accommodate six passengers. The revenue per raft trip is $120. Ten rafts are available, and at least 30 people are on the list for reservations this morning. Each Group Sailing Lesson requires one captain and two crew people for instruction. Two boats are needed for each group. Four students form each group. There are 12 sailboats available, and at least 20 people are on the list for sailing instruction this morning. The revenue per group sailing lesson is $160. The company has 12 captains and 18 crew available this morning. The company would like to maximize the number of customers served while generating at least $1800 in revenue and honoring all reservations.

TOPIC: Scheduling 11. Evans Enterprises has bought a prime parcel of beachfront property and plans to build a luxury hotel. After

meeting with the architectural team, the Evans family has drawn up some information to make preliminary plans for construction. Excluding the suites, which are not part of this decision, the hotel will have four kinds of rooms: beachfront non-smoking, beachfront smoking, lagoon view non-smoking, and lagoon view smoking. In order to decide how many of each of the four kinds of rooms to plan for, the Evans family will consider the following information. a. After adjusting for expected occupancy, the average nightly revenue for a beachfront non-smoking

room is $175. The average nightly revenue for a lagoon view non-smoking room is $130. Smokers will be charged an extra $15.

b. Construction costs vary. The cost estimate for a lagoon view room is $12,000 and for a beachfront room is $15,000. Air purifying systems and additional smoke detectors and sprinklers ad $3000 to the cost of any smoking room. Evans Enterprises has raised $6.3 million in construction guarantees for this portion of the building.

c. There will be at least 100 but no more than 180 beachfront rooms. d. Design considerations require that the number of lagoon view rooms be at least 1.5 times the

number of beachfront rooms, and no more than 2.5 times that number. e. Industry trends recommend that the number of smoking rooms be no more than 50% of the

number of non-smoking rooms.



Develop the linear programming model to maximize revenue. TOPIC: Product mix 12. Super City Discount Department Store is open 24 hours a day. The number of cashiers need in each four

hour period of a day is listed below.

Period Cashiers Needed 10 p.m. to 2 a.m. 8 2 a.m. to 6 a.m. 4

6 a.m. to 10 a.m. 7 10 a.m. to 2 p.m. 12 2 p.m. to 6 p.m. 10

6 p.m. to 10 p.m. 15 If cashiers work for eight consecutive hours, how many should be scheduled to begin working in each

period in order to minimize the number of cashiers needed? TOPIC: Staff scheduling 13. Winslow Savings has $20 million available for investment. It wishes to invest over the next four months in

such a way that it will maximize the total interest earned over the four month period as well as have at least $10 million available at the start of the fifth month for a high rise building venture in which it will be participating.

For the time being, Winslow wishes to invest only in 2-month government bonds (earning 2% over the 2-month period) and 3-month construction loans (earning 6% over the 3-month period). Each of these is available each month for investment. Funds not invested in these two investments are liquid and earn 3/4 of 1% per month when invested locally.

Formulate and solve a linear program that will help Winslow Savings determine how to invest over the next four months if at no time does it wish to have more than $8 million in either government bonds or construction loans.

TOPIC: Portfolio selection 14. National Wing Company (NWC) is gearing up for the new B-48 contract. Currently NWC has 100 equally

qualified workers. Over the next three months NWC has made the following commitments for wing production:

Month Wing Production May 20 June 24 July 30

Each worker can either be placed in production or can train new recruits. A new recruit can be trained to be an apprentice in one month. The next month, he, himself, becomes a qualified worker (after two months from the start of training). Each trainer can train two recruits. The production rate and salary per employee is estimated below.

Employee Production Rate (Wings/Month) Salary Per Month Production .6 $3,000

Trainer .3 3,300 Apprentice .4 2,600

Recruit .05 2,200 At the end of July, NWC wishes to have no recruits or apprentices but have at least 140 full-time workers. Formulate and solve a linear program for NWC to accomplish this at minimum total cost.



TOPIC: Production scheduling 15. The SMM Company, which is manufacturing a new instant salad machine, has $280,000 to spend on

advertising. The product is only to be test marketed initially in the Dallas area. The money is to be spent on an advertising blitz during one weekend (Friday, Saturday, and Sunday) in January, and SMM is limited to television advertising.

The company has three options available: daytime advertising, evening news advertising and the Super Bowl. Even though the Super Bowl is a national telecast, the Dallas Cowboys will be playing in it, and hence, the viewing audience will be especially large in the Dallas area. A mixture of one-minute TV spots is desired. The table below gives pertinent data:

Estimated New Audience Cost Per Ad Reached With Each Ad Daytime $ 5,000 3,000 Evening News $ 7,000 4,000 Super Bowl $100,000 75,000

SMM has decided to take out at least one ad in each option. Further, there are only two Super Bowl ad spots available. There are 10 daytime spots and 6 evening news spots available daily. SMM wants to have at least 5 ads per day, but spend no more than $50,000 on Friday and no more than $75,000 on Saturday. Formulate and solve a linear program to help SMM decide how the company should advertise over the weekend.

TOPIC: Media selection 16. John Sweeney is an investment advisor who is attempting to construct an "optimal portfolio" for a client

who has $400,000 cash to invest. There are ten different investments, falling into four broad categories that John and his client have identified as potential candidates for this portfolio.

The following table lists the investments and their important characteristics. Note that Unidyde Equities (stocks) and Unidyde Debt (bonds) are two separate investments, whereas First General REIT is a single investment that is considered both an equities and a real estate investment.

Exp. Annual Liquidity Risk Category Investment After Tax Return Factor Factor Equities Unidyne Corp. 15.0% 100 60 Col. Mustard Restaurant 17.0% 100 70 First General REIT 17.5% 100 75 Debt Metropolitan Electric 11.8% 95 20 Unidyne Corp. 12.2% 92 30 Lemonville Transit 12.0% 79 22 Real Estate Fairview Apartment Partnership 22.0% 0 50 First General REIT (see above) (see above) (see above) Money T-Bill Account 9.6% 80 0 Money Market Fund 10.5% 100 10 All Saver’s Certificate 12.6% 0 0

Formulate and solve a linear program to accomplish John's objective as an investment advisor which is to construct a portfolio that maximizes his client's total expected after-tax return over the next year, subject to a number of constraints placed upon him by the client for the portfolio:

1. Its (weighted) average liquidity factor must be at least 65. 2. The (weighted) average risk factor must be no greater than 55. 3. At most, $60,000 is to be invested in Unidyde stocks or bonds. 4. No more than 40% of the investment can be in any one category except the money category. 5. No more than 20% of the investment can be in any one investment except the money market fund. 6. At least $1,000 must be invested in the money market fund. 7. The maximum investment in All Saver's Certificates is $15,000. 8. The minimum investment desired for debt is $90,000. 9. At least $10,000 must be placed in a T-Bill account.



TOPIC: Portfolio selection 17. BP Cola must decide how much money to allocate for new soda and traditional soda advertising over the

coming year. The advertising budget is $10,000,000. Because BP wants to push its new sodas, at least one-half of the advertising budget is to be devoted to new soda advertising. However, at least $2,000,000 is to be spent on its traditional sodas. BP estimates that each dollar spent on traditional sodas will translate into 100 cans sold, whereas, because of the harder sell needed for new products, each dollar spent on new sodas will translate into 50 cans sold.

To attract new customers BP has lowered its profit margin on new sodas to 2 cents per can as compared to 4 cents per can for traditional sodas. How should BP allocate its advertising budget if it wants to maximize its profits while selling at least 750 million cans?

TOPIC: Advertising budget allocation


1. a. Min .3S11 + .25S21 + .40S12 + .30S22 + .50S13 + .35S23

b. Min .06W11 + .05W21 + .09W12 + .06W22 + .12W13 + .07W23 c. S11 + S21 ≤ 8000

S12 + S22 ≤ 7000 S13 + S23 ≤ 6000

d. S11 - W11 = 3000 S21 - W21 = 1800 W11 + S12 - W12 = 4000 W21 + S22 - W22 = 4000 W12 + S13 - W13 = 5000 W22 + S23 - W23 = 2000

2. Let T = the number of TV ads

Let R = the number of radio ads Let N = the number of newspaper ads Max 10000T + 3000R + 5000N s.t. 500T + 200R + 400N ≤ 10000

30T + 40R + 25N ≥ 1000 T - R ≤ 4 T, R, N ≥ 0



3. Let Lt = loan in period t, t = 1,...,4 It = investment in period t, t = 1,...,4 Max 1.1I4 - 1.085L4 s.t. L1 < 3000 I1 < 4500 L1 - I1 = 100 L2 < 7000 I2 < 8000 L2 + 1.1I1 -1.085L1 - I2 = 120 L3 < 4000 I3 < 6000 L3 + 1.12I2 - 1.085L2 - I3 = 150 L4 < 5000 I4 < 7500 L4 + 1.13I3 -1.085L3 -I4 = 100 1.10I4 -1.085L4 > 0 Lt, It > 0 4. Let FM = number of fronts made SM = number of seats made WM = number of wheels made FP = number of fronts purchased SP = number of seats purchased WP = number of wheels purchased Min 8FM + 6SM + 1WM + 12FP + 9SP + 3WP s.t. 3FM + 4SM + .5WM < 50000 10FM + 6SM + 2WM < 160000 2FM + 2SM + .1WM < 30000 FM + FP > 12000 SM + SP > 12000 WM + WP > 24000 FM, SM, WM, FP, SP, WP > 0 5. Let Pij = number of toy i to produce in month j Sij = surplus (inventory) of toy i at end of month j Min 12P11 + 12P12 + 12P13 + 14P21 + 14P22 + 14P23 + 1S11 + 1S12 + 1S13 + 1.2S21 + 1.2S22 + 1.2S23 s.t. P11 - S11 = 195 S11 + P12 - S12 = 350 S12 + P13 - S13 = 600 S13 > 150 P21 - S21 = 395 S21 + P22 - S22 = 700 S22 + P23 - S23 = 520 S23 > 60 5P11 + 4P21 < 3500 5P12 + 4P22 < 5000 5P13 + 4P23 < 4800 2P11 + 3P21 < 2100 2P12 + 3P22 < 3000 2P13 + 3P23 < 2500 Pij, Sij > 0



6. Let PM = the number of knob and plate units to make PB = the number of knob and plate units to buy LM = the number of lock units to make LB = the number of lock units to buy KM = the number of key sets to make KB = the number of key sets to buy Min 6PM + 10PB + 4LM + 9LB + .5KM + 1KB s.t. 12PM + 5LM + 2KM < 15000 18PM + 20LM + 20KM < 36000 15PM + 5LM + 2KM < 12000 PM + PB > 1200 LM + LB > 1200 KM + KB > 1200 PM, PB, LM, LB, KM, KB > 0 7. Let P1 = the number of model 1 motors to produce P2 = the number of model 2 motors to produce AA = the number of hours allocated to area A AB = the number of hours allocated to area B AC = the number of hours allocated to area C TBA = the number of hours transferred from B to A TCA = the number of hours transferred from C to A TCB = the number of hours transferred from C to B Min 80P1 + 65P2 s.t. 15P1 + 3P2 - AA < 0 4P1 + 10P2 - AB < 0 4P1 + 8P2 - AC < 0 AA - TBA - TCA = 10000 AB - TCB + TBA = 10000 AC + TCA + TCB = 18000 TBA < 2000 TCA + TCB < 3000 all variables > 0 8. Let D = the number of ounces of dried fruit mix in the blend N = the number of ounces of nut mix in the blend C = the number of ounces of cereal mix in the blend Min .35D + .50N + .20C s.t. .25D + .375N + C > 3 . 25D + .375N + C < 4 150D + 400N + 50C < 1000 10N + C < 25 .2D - .075N - .2C > 0 -.15D + .85N - .15C < 0 D , N , C > 0 9. a. Let Pi = the production in yards in quarter i

Let Si = the ending surplus (inventory) in quarter i b. Min .03P1 + .04P2 + .06P3 + .08P4 + .005(S1 + S2 + S3 + S4 )



c. P1 ≤ 30000 P2 ≤ 20000 P3 ≤ 20000 P4 ≤ 15000 d. P1 - S1 = 10000 S1 + P2 - S2 = 18000 S2 + P3 - S3 = 16000 S3 + P4 - S4 = 30000 P1 ≤ 30000 P2 ≤ 20000 P3 ≤ 20000 P4 ≤ 15000

10. Let R = the number of Wildlife Raft Trips to schedule S = the number of Group Sailing Lessons to schedule Max 6R + 4S s.t. R + S < 12 R + 2S < 18 6R > 30 4S > 20 120R + 160S > 1800 R < 10 2S < 12 R , S > 0 11. Let BN = the number of beachfront non-smoking rooms BS = the number of beachfront smoking rooms LN = the number of lagoon view non-smoking rooms LS = the number of lagoon view smoking rooms Max 175BN + 190BS + 130LN + 145LS s.t. 15000BN + 18000BS + 12000LN + 15000LS < 6,300,000 BN + BS > 100 BN + BS < 180 -1.5BN - 1.5BS + LN + LS > 0 -2.5BN - 2.5BS + LN + LS < 0 - .5BN + BS - .5LN + LS < 0 BN, BS, LN, LS > 0 12. Let TNP = the number of cashiers who begin working at 10 p.m. TWA = the number of cashiers who begin working at 2 a.m. SXA = the number of cashiers who begin working at 6 a.m. TNA = the number of cashiers who begin working at 10 a.m. TWP = the number of cashiers who begin working at 2 p.m. SXP = the number of cashiers who begin working at 6 p.m. Min TNP + TWA + SXA + TNA + TWP + SXP s.t. TNP + TWA > 4 TWA + SXA > 7 SXA + TNA > 12 TNA + TWP > 10



TWP + SXP > 15 SXP + TNP > 8 all variables > 0 13. Define the decision variables

gj = amount of new investment in government bonds in month j cj = amount of new investment in construction loans in month j lj = amount invested locally in month j, where j = 1,2,3,4 Define the objective function MAX .02g1 + .02g2 + .02g3 + .02g4 + .06c1 + .06c2 + .06c3 + .06c4 + .0075l1 + .0075l2 + .0075l3 +

.0075l4 Define the constraints

(1) g1 + c1 + l1 = 20,000,000 (2) g2 + c2 + l2 = 1.0075l1 or g2 + c2 - 1.0075l1 + l2 = 0 (3) g3 + c3 + l3 = 1.02g1 + 1.0075l2 or - 1.02g1 + g3 + c3 - 1.0075l2 + l3 = 0 (4) g4 + c4 + l4 = 1.06c1 + 1.02g2 + 1.0075l3 or - 1.02g2 + g4 - 1.06c1 + c4 - 1.0075l3 + l4 = 0 (5) 1.06c2 + 1.02g3 + 1.0075l4 > 10,000,000 (6) g1 < 8,000,000 (7) g1 + g2 < 8,000,000 (8) g2 + g3 < 8,000,000 (9) g3 + g4 < 8,000,000 (10) c1 < 8,000,000 (11) c1 + c2 < 8,000,000 (12) c1 + c2 + c3 < 8,000,000 (13) c2 + c3 + c4 < 8,000,000 Nonnegativity: gj, cj, lj > 0 for j = 1,2,3,4

The Management Scientist provided the following solution:


VARIABLE VALUE REDUCED COST G1 8000000.000 0.000 G2 0.000 0.000 G3 5108613.923 0.000 G4 2891386.077 0.000 C1 8000000.000 0.000 C2 0.000 0.045 C3 0.000 0.008 C4 8000000.000 0.000 I1 4000000.000 0.000 I2 4030000.000 0.000 I3 7111611.077 0.000 I4 4753562.083 0.000

14. Pi = the number of producers in month i (where i = 1,2,3)

Ti = the number of trainers in month i (where i = 1,2) Ai = the number of apprentices in month i (where i = 2,3) Ri = the number of recruits in month i (where i = 1,2)



MIN 3000P1 + 3300T1 + 2200R1 + 3000P2 + 3300T2 + 2600A2+2200R2 + 3000P3 + 2600A3 s.t. .6P1 + .3T1 +.05R1 > 20 .6P1 + .3T1 + .05R1 + .6P2 + .3T2 + .4A2 + .05R2 > 44 .6P1+.3T1+.05R1+.6P2+.3T2+.4A2+.05R2+.6P3+.4A3 > 74 P1 - P2 + T1 - T2 = 0 P2 - P3 + T2 + A2 = 0 A2 - R1 = 0 A3 - R2 = 0 2T1 - R1 > 0 2T2 - R2 > 0 P1 + T1 = 100 P3 + A3 > 140 Pj, Tj, Aj, Rj > 0 for all j

Solution: P1 = 100, T1 = 0, R1 = 0, P2 = 80, T2 = 20, A2 = 0, R2 = 40, P3 = 100, A3 = 40

Total cost = $1,098,000. 15. Define the decision variables

x1 = the number of day ads on Friday x2 = the number of day ads on Saturday x3 = the number of day ads on Sunday x4 = the number of evening ads on Friday x5 = the number of evening ads on Saturday x6 = the number of evening ads on Sunday x7 = the number of Super Bowl ads

Define the objective function MAX 3000x1 +3000x2 +3000x3 +4000x4 +4000x5 +4000x6 +75000x7

Define the constraints (1) x1 + x2 + x3 > 1 (2) x4 + x5 + x6 > 1 (3) x7 > 1 (4) x1 < 10 (5) x2 < 10 (6) x3 < 10 (7) x4 < 6 (8) x5 < 6 (9) x6 < 6 (10) x7 < 2 (11) x1 + x4 > 5 (12) x2 + x5 > 5 (13) x3 + x6 + x7 > 5 (14) 5000x1 + 7000x4 < 50000 (15) 5000x2 + 7000x5 < 75000 (16) 5000x1 + 5000x2 + 5000x3 + 7000x4 + 7000x5 + 7000x6 + 100000x7 < 280000 xj > 0 j = 1,...,7






X1 7.600 0.000 X2 5.000 0.000 X3 2.000 0.000 X4 0.000 0.000 X5 0.000 0.000 X6 1.000 0.000 X7 2.000 0.000

16. Xj = $ invested in investment j; where j = 1(Uni Eq.), 2(Col. Must.), 3(1st Gen REIT), 4(Met. Elec.), 5(Uni Debt), 6(Lem. Trans.), 7(Fair. Apt.), 8(T-Bill), 9(Money Market), 10(All Saver's)

MAX .15X1 + .17X2 + .175X3 + .118X4 + .122X5 + .12X6 + .22X7 + .096X8 + .105X9 + .126X10 S.T. X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 = 400,000 100X1 + 100X2 + 100X3 + 95X4 + 92X5 + 79X6 + 80X8 + 100X9 > 65(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10)

60X1 + 70X2 + 75X3 + 20X4 + 30X5 + 22X6 + 50X7 + 10X9 < 55(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10)

X1 + X5 < 60,000 X1 + X2 + X3 < 160,000 X4 + X5 + X6 < 160,000 X3 + X7 < 160,000 X1 < 80,000 X2 < 80,000 X3 < 80,000 X4 < 80,000 X5 < 80,000 X6 < 80,000 X7 < 80,000 X8 < 80,000 X9 > 1,000 X10 < 15,000 X4 + X5 + X6 > 90,000 X8 > 10,000

Xj > 0 j = 1,...,10

Solution: X1 = 0; X2 = 80,000; X3 = 80,000; X4 = 0; X5 = 60,000; X6 = 74,000; X7 = 80,000; X8 = 10,000; X9 = 1,000; X10 = 15,000; Total return = $64,355. 17. X1 = amount invested in new soda advertising

X2 = amount invested in traditional soda advertising MAX X1 + 4X2 S.T. X1 + X2 < 10,000,000 X1 > 5,000,000



X2 > 2,000,000 50X1 + 100X2 > 750,000,000 X1, X2 > 0 Answer: spend $5,000,000 on new soda ad, spend $5,000,000 on traditional ad, profit = $25 million


1

10

Distribution and Network Problems

MULTIPLE CHOICE

1. The problem which deals with the distribution of goods from several sources to several destinations is the a. maximal flow problem b. transportation problem c. assignment problem d. shortest-route problem ANSWER: b TOPIC: Transportation problem 2. The parts of a network that represent the origins are a. the capacities b. the flows c. the nodes d. the arcs ANSWER: c TOPIC: Transportation problem 3. The objective of the transportation problem is to

a. identify one origin that can satisfy total demand at the destinations and at the same time minimize total shipping cost.

b. minimize the number of origins used to satisfy total demand at the destinations. c. minimize the number of shipments necessary to satisfy total demand at the destinations. d. minimize the cost of shipping products from several origins to several destinations. ANSWER: d TOPIC: Transportation problem 4. The number of units shipped from origin i to destination j is represented by a. xij. b. xji. c. cij. d. cji. ANSWER: a TOPIC: Transportation problem


2 Chapter 10 Distribution and Network Problems

5. Which of the following is not true regarding the linear programming formulation of a transportation problem?

a. Costs appear only in the objective function. b. The number of variables is (number of origins) x (number of destinations). c. The number of constraints is (number of origins) x (number of destinations). d. The constraints’ left-hand side coefficients are either 0 or 1. ANSWER: c TOPIC: Transportation problem 6. The difference between the transportation and assignment problems is that a. total supply must equal total demand in the transportation problem

b. the number of origins must equal the number of destinations in the transportation problem c. each supply and demand value is 1 in the assignment problem

d. there are many differences between the transportation and assignment problems ANSWER: c TOPIC: Assignment problem 7. In the general linear programming model of the assignment problem, a. one agent can do parts of several tasks. b. one task can be done by several agents. c. each agent is assigned to its own best task. d. one agent is assigned to one and only one task. ANSWER: d TOPIC: Assignment problem 8. The assignment problem is a special case of the a. transportation problem. b. transshipment problem. c. maximal flow problem. d. shortest-route problem. ANSWER: a TOPIC: Assignment problem 9. Which of the following is not true regarding an LP model of the assignment problem? a. Costs appear in the objective function only. b. All constraints are of the > form. c. All constraint left-hand side coefficient values are 1. d. All decision variable values are either 0 or 1. ANSWER: b TOPIC: Assignment problem 10. The assignment problem constraint x31 + x32 + x33 + x34 < 2 means a. agent 3 can be assigned to 2 tasks. b. agent 2 can be assigned to 3 tasks. c. a mixture of agents 1, 2, 3, and 4 will be assigned to tasks. d. there is no feasible solution. ANSWER: a TOPIC: Assignment problem 11. Arcs in a transshipment problem

a. must connect every node to a transshipment node. b. represent the cost of shipments. c. indicate the direction of the flow. d. All of the alternatives are correct.

ANSWER: c TOPIC: Transshipment problem


Chapter 10 Distribution and Network Problems 3

12. Constraints in a transshipment problem

a. correspond to arcs. b. include a variable for every arc. c. require the sum of the shipments out of an origin node to equal supply. d. All of the alternatives are correct.

ANSWER: b TOPIC: Transshipment problem 13. In a transshipment problem, shipments

a. cannot occur between two origin nodes. b. cannot occur between an origin node and a destination node. c. cannot occur between a transshipment node and a destination node. d. can occur between any two nodes.

ANSWER: d TOPIC: Transshipment problem 14. Consider a shortest route problem in which a bank courier must travel between branches and the main

operations center. When represented with a network, a. the branches are the arcs and the operations center is the node.

b. the branches are the nodes and the operations center is the source. c. the branches and the operations center are all nodes and the streets are the arcs. d. the branches are the network and the operations center is the node.

ANSWER: c TOPIC: Shortest-route problem 15. The shortest-route problem finds the shortest-route a. from the source to the sink. b. from the source to any other node. c. from any node to any other node. d. from any node to the sink. ANSWER: b TOPIC: Shortest-route problem 16. Consider a maximal flow problem in which vehicle traffic entering a city is routed among several routes

before eventually leaving the city. When represented with a network, a. the nodes represent stoplights. b. the arcs represent one way streets. c. the nodes represent locations where speed limits change. d. None of the alternatives is correct. ANSWER: b TOPIC: Maximal flow problem 17. We assume in the maximal flow problem that a. the flow out of a node is equal to the flow into the node. b. the source and sink nodes are at opposite ends of the network. c. the number of arcs entering a node is equal to the number of arcs exiting the node. d. None of the alternatives is correct. ANSWER: a TOPIC: Maximal flow problem 18. If a transportation problem has four origins and five destinations, the LP formulation of the problem will

have a. 5 constraints b. 9 constraints c. 18 constraints



d. 20 constraints ANSWER: b TOPIC: Transportation problem

TRUE/FALSE

1. Whenever total supply is less than total demand in a transportation problem, the LP model does not

determine how the unsatisfied demand is handled. ANSWER: True TOPIC: Transportation problem 2. Converting a transportation problem LP from cost minimization to profit maximization requires only

changing the objective function; the conversion does not affect the constraints. ANSWER: True TOPIC: Transportation problem 3. A transportation problem with 3 sources and 4 destinations will have 7 decision variables. ANSWER: False TOPIC: Transportation problem 4. If a transportation problem has four origins and five destinations, the LP formulation of the problem will

have nine constraints. ANSWER: True TOPIC: Transportation problem 5. The capacitated transportation problem includes constraints which reflect limited capacity on a route. ANSWER: True TOPIC: Transportation problem 6. When the number of agents exceeds the number of tasks in an assignment problem, one or more dummy

tasks must be introduced in the LP formulation or else the LP will not have a feasible solution. ANSWER: False TOPIC: Assignment problem 7. A transshipment constraint must contain a variable for every arc entering or leaving the node. ANSWER: True TOPIC: Transshipment problem 8. The shortest-route problem is a special case of the transshipment problem. ANSWER: True TOPIC: Shortest-route problem 9. Transshipment problem allows shipments both in and out of some nodes while transportation problems do

not. ANSWER: True TOPIC: Transportation and transshipment problems 10. A dummy origin in a transportation problem is used when supply exceeds demand. ANSWER: False TOPIC: Transportation problem 11. When a route in a transportation problem is unacceptable, the corresponding variable can be removed from

the LP formulation.



ANSWER: True TOPIC: Transportation problem 12. In the LP formulation of a maximal flow problem, a conservation-of-flow constraint ensures that an arc’s

flow capacity is not exceeded. ANSWER: False TOPIC: Maximal flow problem 13. The maximal flow problem can be formulated as a capacitated transshipment problem. ANSWER: True TOPIC: Maximal flow problem 14. The direction of flow in the shortest-route problem is always out of the origin node and into the destination

node. ANSWER: True TOPIC: Shortest-route problem 15. A transshipment problem is a generalization of the transportation problem in which certain nodes are

neither supply nodes nor destination nodes. ANSWER: True TOPIC: Transshipment problem 16. The assignment problem is a special case of the transportation problem in which all supply and demand

values equal one. ANSWER: True TOPIC: Assignment problem 17. A transportation problem with 3 sources and 4 destinations will have 7 variables in the objective function. ANSWER: False TOPIC: Assignment problem 18. Flow in a transportation network is limited to one direction. ANSWER: True TOPIC: Transportation problem 19. In a transportation problem with total supply equal to total demand, if there are four origins and seven

destinations, and there is a unique optimal solution, the optimal solution will utilize 11 shipping routes. ANSWER: False TOPIC: Transportation problem 20. In the general assignment problem, one agent can be assigned to several tasks. ANSWER: True TOPIC: Assignment problem

SHORT ANSWER

1. Explain how the general linear programming model of the assignment problem can be modified to handle

problems involving a maximization function, unacceptable assignments, and supply not equally demand. TOPIC: Assignment problem 2. Define the variables and constraints necessary in the LP formulation of the transshipment problem. TOPIC: Transshipment problem



3. Explain what adjustments are made to the transportation linear program when there are unacceptable

routes. TOPIC: Transportation problem 4. Is it a coincidence to obtain integer solutions to network problems? Explain. TOPIC: Network problems 5. How is the assignment linear program different from the transportation model? TOPIC: Transportation and assignment problems 6. Define the variables and constraints necessary in the LP formulation of the maximal flow problem. TOPIC: Maximal flow problem 7. How is the shortest-route problem like the transshipment problem? TOPIC: Shortest-route problem

PROBLEMS

1. Write the LP formulation for this transportation problem.

100

200

150

50

1

2

3

4

A 250

B 250

5

6 4

2

3

6

7

9

TOPIC: Transportation problem



2. Draw the network for this transportation problem. Min 2XAX + 3XAY + 5XAZ+ 9XBX + 12XBY + 10XBZ s.t. XAX + XAY + XAZ < 500 XBX + XBY + XBZ < 400 XAX + XBX = 300 XAY + XBY = 300 XAZ + XBZ = 300 Xij > 0 TOPIC: Transportation problem 3. Canning Transport is to move goods from three factories to three distribution centers. Information about

the move is given below. Give the network model and the linear programming model for this problem.

Source Supply Destination Demand A 200 X 50 B 100 Y 125 C 150 Z 125

Shipping costs are:

Destination Source X Y Z

A 3 2 5 B 9 10 -- C 5 6 4 (Source B cannot ship to destination Z)

TOPIC: Transportation problem 4. The following table shows the unit shipping cost between cities, the supply at each source city, and the

demand at each destination city. The Management Scientist solution is shown. Report the optimal solution.

Destination Source Terre Haute Indianapolis Ft. Wayne South Bend Supply St. Louis 8 6 12 9 100 Evansville 5 5 10 8 100 Bloomington 3 2 9 10 100 Demand 150 60 45 45

TRANSPORTATION PROBLEM *****************************

OBJECTIVE: MINIMIZATION



SUMMARY OF ORIGIN SUPPLIES ******************************** ORIGIN SUPPLY ---------- ----------- 1 100 2 100 3 100 SUMMARY OF DESTINATION DEMANDS *************************************** DESTINATION DEMAND ------------------- ------------- 1 150 2 60 3 45 4 45 SUMMARY OF UNIT COST OR REVENUE DATA ********************************************* FROM TO DESTINATION ORIGIN 1 2 3 4 ---------- ----- ----- ----- ----- 1 8 6 12 9 2 5 5 10 8 3 3 2 9 10 OPTIMAL TRANSPORTATION SCHEDULE **************************************** SHIP FROM TO DESTINATION ORIGIN 1 2 3 4 ---------- ----- ----- ----- ----- 1 0 10 45 45 2 100 0 0 0 3 50 50 0 0

TOTAL TRANSPORTATION COST OR REVENUE IS 1755 TOPIC: Transportation problem 5. After some special presentations, the employees of the AV Center have to move overhead projectors back

to classrooms. The table below indicates the buildings where the projectors are now (the sources), where they need to go (the destinations), and a measure of the distance between sites.

Destination Source Business Education Parsons Hall Holmstedt

Hall Supply

Baker Hall 10 9 5 2 35 Tirey Hall 12 11 1 6 10 Arena 15 14 7 6 20



Demand 12 20 10 10

a. If you were going to write this as a linear programming model, how many decision variables would there be, and how many constraints would there be?

The solution to this problem is shown below. Use it to answer the questions b - e.

TRANSPORTATION PROBLEM ***************************** OPTIMAL TRANSPORTATION SCHEDULE ****************************************

SHIP FROM TO DESTINATION ORIGIN 1 2 3 4 ----------- ------ ------ ------ ------ 1 12 20 0 3 2 0 0 10 0 3 0 0 0 7

TOTAL TRANSPORTATION COST OR REVENUE IS 358 NOTE: THE TOTAL SUPPLY EXCEEDS THE TOTAL DEMAND BY 13 ORIGIN EXCESS SUPPLY ---------- ----------------------- 3 13

b. How many projectors are moved from Baker to Business? c. How many projectors are moved from Tirey to Parsons? d. How many projectors are moved from the Arena to Education? e. Which site(s) has (have) projectors left?

TOPIC: Transportation problem 6. Show both the network and the linear programming formulation for this assignment problem.

Task Person A B C D

1 9 5 4 2 2 12 6 3 5 3 11 6 5 7

TOPIC: Assignment problem 7. Draw the network for this assignment problem.

Min 10x1A + 12x1B + 15x1C + 25x1D + 11x2A + 14x2B + 19x2C + 32x2D + 18x3A + 21x3B + 23x3C + 29x3D + 15x4A + 20x4B + 26x4C + 28x4D

s.t. x1A + x1B + x1C + x1D = 1 x2A + x2B + x2C + x2D = 1 x3A + x3B + x3C + x3D = 1

x4A + x4B + x4C + x4D = 1 x1A + x2A + x3A + x4A = 1 x1B + x 2B + x3B + x4B = 1 x1C + x2C + x3C + x4C = 1 x1D + x2D + x3D + x4D = 1



TOPIC: Assignment problem

8. A professor has been contacted by four not-for-profit agencies that are willing to work with student consulting teams. The agencies need help with such things as budgeting, information systems, coordinating volunteers, and forecasting. Although each of the four student teams could work with any of the agencies, the professor feels that there is a difference in the amount of time it would take each group to solve each problem. The professor’s estimate of the time, in days, is given in the table below. Use the computer solution to see which team works with which project.

Projects

Team Budgeting Information Volunteers Forecasting A 32 35 15 27 B 38 40 18 35 C 41 42 25 38 D 45 45 30 42

ASSIGNMENT PROBLEM ************************ OBJECTIVE: MINIMIZATION SUMMARY OF UNIT COST OR REVENUE DATA ********************************************* TASK AGENT 1 2 3 4

---------- ----- ----- ----- ----- 1 32 35 15 27 2 38 40 18 35 3 41 42 25 38 4 45 45 30 42

OPTIMAL ASSIGNMENTS COST/REVENUE ************************ *************** ASSIGN AGENT 3 TO TASK 1 41 ASSIGN AGENT 4 TO TASK 2 45

ASSIGN AGENT 2 TO TASK 3 18 ASSIGN AGENT 1 TO TASK 4 27

------------------------------------------- ----- TOTAL COST/REVENUE 131 TOPIC: Assignment problem



9. Write the linear program for this transshipment problem.

1

2

3

4

5

6

7

500

400

300

2

3

2 3

5

6

8

10

5 9

12

15

600

600

TOPIC: Transshipment problem 10. Peaches are to be transported from three orchard regions to two canneries. Intermediate stops at a

consolidation station are possible.

Orchard Supply Station Cannery Capacity Riverside 1200 Waterford Sanderson 2500 Sunny Slope 1500 Northside Millville 3000 Old Farm 2000

Shipment costs are shown in the table below. Where no cost is given, shipments are not possible. Where

costs are shown, shipments are possible in either direction. Draw the network model for this problem.

R SS OF W N S M Riverside 1 5 3 Sunny Side 4 5 Old Farm 6 3 Waterford 2 2 4 Northside 5 9 Sanderson 2 Millville

TOPIC: Transshipment problem 11. RVW (Restored Volkswagens) buys 15 used VW's at each of two car auctions each week held at different

locations. It then transports the cars to repair shops it contracts with. When they are restored to RVW's specifications, RVW sells 10 each to three different used car lots. There are various costs associated with the average purchase and transportation prices from each auction to each repair shop. Also there are transportation costs from the repair shops to the used car lots. RVW is concerned with minimizing its total cost given the costs in the table below.

a. Given the costs below, draw a network representation for this problem.

Repair Shops Used Car Lots S1 S2 L1 L2 L3 Auction 1 550 500 S1 250 300 500



Auction 2 600 450 S2 350 650 450

b. Formulate this problem as a transshipment linear programming model. TOPIC: Transshipment problem 12. Consider the network below. Formulate the LP for finding the shortest-route path from node 1 to node 7.

1

2

3

4

5

6

7

10

4

8

12

6 3

4

9

7 3

TOPIC: Shortest-route problem 13. Consider the following shortest-route problem involving six cities with the distances given. Draw the

network for this problem and formulate the LP for finding the shortest distance from City 1 to City 6.

Path Distance 1 to 2 3 1 to 3 2 2 to 4 4 2 to 5 5 3 to 4 3 3 to 5 7 4 to 6 6 5 to 6 2 TOPIC: Shortest-route problem



14. A beer distributor needs to plan how to make deliveries from its warehouse (Node 1) to a supermarket (Node 7), as shown in the network below. Develop the LP formulation for finding the shortest route from the warehouse to the supermarket.

1

3

2 3 4

5

6 7

5 6

5 4

3

12

8

3

TOPIC: Shortest-route problem 15. Consider the following shortest-route problem involving seven cities. The distances between the cities are

given below. Draw the network model for this problem and formulate the LP for finding the shortest route from City 1 to City 7.

Path Distance 1 to 2 6 1 to 3 10 1 to 4 7 2 to 3 4 2 to 5 5 3 to 4 5 3 to 5 2 3 to 6 4 4 to 6 8 5 to 7 7 6 to 7 5

TOPIC: Shortest-route problem 16. The network below shows the flows possible between pairs of six locations. Formulate an LP to find the

maximal flow possible from Node 1 to Node 6.



1

2

3

4

5

6

18

15

9 4

20

10

10 10

25

14

8 8

10

TOPIC: Maximal flow problem 17. A network of railway lines connects the main lines entering and leaving a city. Speed limits, track

reconstruction, and train length restrictions lead to the flow diagram below, where the numbers represent how many cars can pass per hour. Formulate an LP to find the maximal flow in cars per hour from Node 1 to Node F.

1

2

3

4

8 F

5

6

7 9

500

400

600 300

600

300

400 300 200 350

500

150 400 200

450

300 300

500

TOPIC: Maximal flow problem 18. Fodak must schedule its production of camera film for the first four months of the year. Film demand (in

1,000s of rolls) in January, February, March and April is expected to be 300, 500, 650 and 400, respectively. Fodak's production capacity is 500 thousand rolls of film per month. The film business is highly competitive, so Fodak cannot afford to lose sales or keep its customers waiting. Meeting month i 's demand with month i +1's production is unacceptable.

Film produced in month i can be used to meet demand in month i or can be held in inventory to meet demand in month i +1 or month i +2 (but not later due to the film's limited shelflife). There is no film in inventory at the start of January.

The film's production and delivery cost per thousand rolls will be $500 in January and February. This cost will increase to $600 in March and April due to a new labor contract. Any film put in inventory requires additional transport costing $100 per thousand rolls. It costs $50 per thousand rolls to hold film in inventory from one month to the next.



a. Modeling this problem as a transshipment problem, draw the network representation. b. Formulate and solve this problem as a linear program.

TOPIC: Production and inventory application 19. Find the maximal flow from node 1 to node 7 in the following network.

2 5

1 4 7

3 6

4

4

3

3

2

3

4

2

3 3

3

1

5 5

1

6

3

TOPIC: Maximal flow problem 20. A foreman is trying to assign crews to produce the maximum number of parts per hour of a certain product.

He has three crews and four possible work centers. The estimated number of parts per hour for each crew at each work center is summarized below. Solve for the optimal assignment of crews to work centers.

Work Center WC1 WC2 WC3 WC4 Crew A 15 20 18 30 Crew B 20 22 26 30 Crew C 25 26 27 30

TOPIC: Assignment problem


1. Min 5X1A + 6X1B + 4X2A + 2X2B + 3X3A + 6X3B + 9X4A + 7X4B

s.t. X1A + X1B < 100 X2A + X2B < 200 X3A + X3B < 150 X4A + X4B < 50 X1A + X2A + X3A + X4A = 250 X1B + X2B + X3B + X4B = 250 all Xij > 0 2.



500 A

400 B

X 300

Y 300

Z 300

2 3

5

9 12

10

3.

200 A

100 B

150 C

50 Dum

X 250

Y 125

Z 125

3

2

9

10

5

6 4

0 0

0

5

Min 3XAX + 2XAY + 5XAZ + 9XBX + 10XBY + 5XCX + 6XCY + 4XCZ s.t. XAX + XAY + XAZ < 200 XBX + XBY < 100 XCX + XCY + XCZ < 150 XDX + XDY + XDZ < 50 XAX + XBX + XCX + XDX = 250 XAY + XBY + XCY + XDY = 125 XAZ + XBZ + XCZ + XDZ = 125 Xij > 0 4. Ship 10 from St. Louis to Indianapolis, 45 from St. Louis to Ft. Wayne, 45 from St. Louis to South Bend,

100 from Evansville to Terre Haute, 50 from Bloomington to Terre Haute, and 50 from Bloomington to Indianapolis. The total cost is 1755.

5. a. 12 decision variables, 7 constraints b. 12 c. 10 d. 0 e. Arena



6.

1 1

1 2

1 3

4

A 1

B 1

C 1

D 1

9 5 4 2

12 6 3 5

11 6 5

7 0 0

0 0 1

Let Xij = 1 if person i is assigned to job j = 0 otherwise Min 9X1A + 5X1B + 4X1C + 2X1D + 12X2A + 6X2B + 3X2C + 5X2D + 11X3A + 6X3B + 5X3C + 7X3D s.t. X1A + X1B + X1C + X1D < 1 X2A + X2B + X2C + X2D < 1 X3A + X3B + X3C + X3D < 1 X4A + X4B + X4C + X4D < 1 X1A + X2A + X3A + X4A = 1 X1B + X2B + X3B + X4B = 1 X1C + X2C + X3C + X4C = 1 X1D + X2D + X3D + X4D = 1



7. 10

12 15 25

11 14

19 32 18 21

23 29

15 20

26

28

1

1

1

1

1

1

1

1

1

2

3

4 D

C

B

A

8. Team A works with the forecast, Team B works with volunteers, Team C works with budgeting, and Team

D works with information. The total time is 131. 9. Min 3x16 + 2x14 + 3x15 + 5x24 + 6x25 + 2x32 + 8x34 + 10x35 + 5x46 + 9x47 + 12x56 + 15x57 s.t. x16 + x14 + x35 ≤ 500 x24 + x25 - x23 ≤ 400 x32 + x34 + x35 ≤ 300 x46 + x47 - (x14 + x24 + x34) = 0 x56 + x57 - (x15 + x25 + x35) = 0 x16 + x46 + x56 = 600 x56 + x57 = 600 10.

1200 R

1500 SS

2000 OF

W S 2500

N M 3000

3

51

4

5

63

2

25

9

4

2

11. a.



A1 15

15

10

10

10

550

500

600

450

250

500

300

350

650

450

A2

S1

S2

L1

L2

L3

b. Denote A1 as node 1, A2 as node 2, S1 as node 3, S2 as node 4, L1 as node 5, L2 as node 6, and L3 as node 7 Min 550X13 + 500X14 + 600X23 + 450X24 + 250X35 + 300X36 + 500X37 + 350X45 + 650X46 + 450X47 s.t. X13 + X14 < 15 X23 + X24 < 15 X13 + X23 - X35 - X36 - X37 = 0 X14 + X24 - X45 - X46 - X47 = 0 X35 + X45 = 10 X36 + X46 = 10 X37 + X47 = 10 Xij > 0 for all i,j 12.

Min 10X12 + 12X13 + 4X24 + 8X25 + 7X35 + 9X36 + 4X42 + 3X45 + 6X47 + 8X52 + 7X53 + 3X54 + 4X57 + 9X63 + 3X67 s.t. X12 + X13 = 1 − X12 + X24 + X25 – X42 − X52 = 0 − X13 + X35 + X36 – X53 − X63 = 0 − X24 + X42 + X45 + X47 − X54 = 0 − X25 – X35 − X45 + X52 + X53 + X57 = 0 − X36 + X63 + X67 = 0 X47 + X57 + X67 = 1 Xij > 0 for all i,j

13.



6 1

2

5 4

5

3 4

2

6

7

2

3

3

Min 3X12 + 2X13 + 4X24 + 5X25 + 3X34 + 7X35

+ 4X42 + 3X43 + 6X46 + 5X52 + 7X53 + 2X56 s.t. X12 + X13 = 1 − X12 + X24 + X25 – X42 − X52 = 0 − X13 + X34 + X35 – X43 − X53 = 0 − X24 – X34 + X42 + X43 + X46 = 0 − X25 – X35 + X52 + X53 + X56 = 0 X46 + X56 = 1 Xij > 0 for all i,j 14.

Min 3X12 + 3X15 + 12X16 + 5X23 + 5X32 + 6X34 + 6X43 + 4X46 + 5X47 + 8X56 + 4X64 + 8X65 + 3X67 s.t. X12 + X15 + X16 = 1 − X12 + X23 = 0 − X23 + X32 + X34 = 0 − X34 + X43 + X46 + X47 − 4X64 = 0 − X15 + X56 = 0 − X16 − X46 − X56 + X64 + X65 + X67 = 0 X47 + X67 = 1 Xij > 0 for all i,j



15.

7 1

2

4 5

6

6 5

7

2 7

8

4 5

3

4

10

5

Min 6X12 + 10X13 + 7X14 + 4X23 + 5X25 + 4X32 + 5X34 + 2X35 + 4X36 + 5X43 + 8X46 + 5X52 + 2X53 + 7X57 + 4X63 + 8X64 + 5X67 s.t. X12 + X13 + X14 = 1 − X12 + X23 + X25 – X32 − X52 = 0 − X13 − X23 + X32 + X34 + X35 + X36 – X43 − X53 − X63 = 0 − X14 – X34 + X43 + X46 − X64 = 0 − X25 – X35 + X52 + X53 + X57 = 0 − X36 − X46 + X63 + X64 + X67 = 0 X57 + X67 = 1 Xij > 0 for all i,j 16.

Min X61

s.t. X12 + X13 + X15 − X61 = 0 X23 + X24 – X12 − X32 = 0 X32 + X34 + X35 – X13 − X23 − X53 = 0 X43 + X46 − X24 − X34 = 0 X53 + X56 − X15 − X35 = 0 X61 − X36 − X46 − X56 = 0 X12 < 18 X13 < 20 X15 < 10 X23 < 9 X24 < 15 X32 < 4 X34 < 10 X35 < 8 X43 < 10 X46 < 14 X53 < 8 X56 < 10 Xij > 0 for all i,j



17. Min XF1

s.t. X12 + X15 + X16 − XF1 = 0 X23 + X24 – X12 = 0 X34 − X23 = 0 X48 + X4F − X24 − X34 − X84 = 0 X57 − X15 = 0 X67 + X69 − X16 − X76 = 0 X76 + X79 − X57 − X67 = 0 X84 + X89 + X8F − X48 − X98 = 0 X98 + X9F − X69 − X79 − X89 = 0 XF1 − X4F − X8F − X9F = 0 X12 < 500 X15 < 300 X16 < 600 X23 < 300 X24 < 400 X34 < 150 X48 < 400 X4F < 600 X57 < 400 X67 < 300 X69 < 500 X76 < 200 X79 < 350 X84 < 200 X89 < 300 X8F < 450 X98 < 300 X9F < 500 Xij > 0 for all i,j 18. a.

2

3

4

7

6

5

11

10

9

81

FEBRUARYPRODUCTION

JANUARYPRODUCTION

MARCHPRODUCTION

JANUARYDEMAND

FEBRUARYDEMAND

MARCHDEMAND

APRILDEMAND

APRILPRODUCTION

MONTH 1ENDING INVENTORY



500

500

500

500

300

500

650

400

500

600

500

600

700

600

50

100

50

100

50

600

b. Define the decision variables: xij = amount of film “moving” between node i and node j



Define objective: MIN 600x15 + 500x18 + 600x26 + 500x29 + 700x37 + 600x310 + 600x411 + 50x59 + 100x510 + 50x610 + 100x611 + 50x711 Define the constraints: Amount (1000’s of rolls) of film produced in January: x15 + x18 < 500 Amount (1000’s of rolls) of film produced in February: x26 + x29 < 500 Amount (1000’s of rolls) of film produced in March: x37 + x310 < 500 Amount (1000’s of rolls) of film produced in April: x411 < 500 Amount (1000’s of rolls) of film in/out of January inventory: x15 – x59 – x510 = 0 Amount (1000’s of rolls) of film in/out of February inventory: x26 – x610 – x611 = 0 Amount (1000’s of rolls) of film in/out of March inventory: x37 – x711 = 0 Amount (1000’s of rolls) of film satisfying January demand: x18 = 300 Amount (1000’s of rolls) of film satisfying February demand x29 + x59 = 500 Amount (1000’s of rolls) of film satisfying March demand: x310 + x510 + x610 = 650 Amount (1000’s of rolls) of film satisfying April demand: x411 + x611 + x711 = 400 Non-negativity of variables: xij > 0, for all i and j.




X15 150.000 0.000 X18 300.000 0.000 X26 0.000 100.000 X29 500.000 0.000 X37 0.000 250.000 X310 500.000 0.000 X411 400.000 0.000 X59 0.000 0.000 X510 150.000 0.000 X610 0.000 0.000 X611 0.000 150.000 X711 0.000 0.000

19. Decision variables:

xij = amount of flow from node i to node j Objective function:

Maximize the flow through the network: Max x71 Constraints: Conservation of flow for each node: (1) x12 + x13 + x14 – x71 = 0 (node 1) (2) x24 + x25 – x12 – x42 – x52 = 0 (node 2) (3) x34 + x36 – x13 – x43 = 0 (node 3) (4) x42 + x43 + x45 + x46 + x47 – x14 – x24 – x34 – x54 – x64 = 0 (node 4) (5) x52 + x54 + x57 – x25 – x45 = 0 (node 5) (6) x64 + x67 – x36 – x46 = 0 (node 6) (7) x71 – x47 – x57 – x67 = 0 (node 7) Capacity for each arc: (8) x12 < 4 (14) x36 < 6 (20) x52 < 3 (9) x13 < 3 (15) x42 < 3 (21) x54 < 4 (10) x14 < 4 (16) x43 < 5 (22) x57 < 2



(11) x24 < 2 (17) x45 < 3 (23) x64 < 1 (12) x25 < 3 (18) x46 < 1 (24) x67 < 5 (13) x34 < 3 (19) x47 < 3

Nonnegativity: All xij > 0

The LP was solved using The Management Scientist . Two solutions are given below. Objective Function Value = 10.000

Variable Solution 1 Solution 2 x12 3.000 3.000 x13 3.000 3.000 x14 4.000 4.000 x24 1.000 1.000 x25 2.000 2.000 x36 5.000 4.000

x43 2.000 1.000 x46 0.000 1.000

x47 3.000 3.000 x57 2.000 2.000 x67 5.000 5.000 x71 10.000 10.000 20. OBJECTIVE FUNCTION VALUE = 82.000

VARIABLE VALUE REDUCED COST AA1 0.000 12.000 AA2 0.000 0.000 AA3 0.000 0.000 AA4 1.000 1.000 AB1 0.000 2.000 AB2 0.000 4.000 AB3 1.000 0.000 AB4 0.000 0.000 AC1 0.000 12.000 AC2 1.000 0.000 AC3 0.000 0.000 AC4 0.000 1.000 AD1 1.000 2.000 AD2 0.000 4.000 AD3 0.000 0.000 AD4 0.000 0.000

CONSTRAINT SLACK/SURPLUS DUAL PRICE

1 0.000 18.000 2 0.000 23.000 3 0.000 24.000 4 0.000 -1.000 5 0.000 1.000 6 0.000 2.000 7 0.000 3.000 8 0.000 12.000

An optimal solution is:

Crew Work Center Parts/Hour Crew A WC4 30



Crew B WC3 26 Crew C WC2 26 ---------- WC1 --- Total Parts Per Hour = 82


1

11

Integer Linear Programming

MULTIPLE CHOICE

1. Which of the following is the most useful contribution of integer programming?

a. finding whole number solutions where fractional solutions would not be appropriate b. using 0-1 variables for modeling flexibility c. increased ease of solution d. provision for solution procedures for transportation and assignment problems

ANSWER: b TOPIC: Introduction 2. In a model, x1 > 0 and integer, x2 > 0, and x3 = 0, 1. Which solution would not be feasible? a. x1 = 5, x2 = 3, x3 = 0 b. x1 = 4, x2 = .389, x3 = 1 c. x1 = 2, x2 = 3, x3 = .578 d. x1 = 0, x2 = 8, x3 = 0 ANSWER: c TOPIC: Introduction 3. Rounded solutions to linear programs must be evaluated for

a. feasibility and optimality. b. sensitivity and duality. c. relaxation and boundedness. d. each of the above is true.

ANSWER: a TOPIC: LP relaxation 4. Rounding the solution of an LP Relaxation to the nearest integer values provides a. a feasible but not necessarily optimal integer solution. b. an integer solution that is optimal. c. an integer solution that might be neither feasible nor optimal. d. an infeasible solution. ANSWER: c TOPIC: Graphical solution


2 Chapter 11 Integer Linear Programming

5. The solution to the LP Relaxation of a maximization integer linear program provides a. an upper bound for the value of the objective function. b. a lower bound for the value of the objective function. c. an upper bound for the value of the decision variables d. a lower bound for the value of the decision variables ANSWER: a TOPIC: Graphical solution 6. The graph of a problem that requires x1 and x2 to be integer has a feasible region a. the same as its LP relaxation. b. of dots. c. of horizontal stripes. d. of vertical stripes. ANSWER: b TOPIC: Graphical solution 7. The 0-1 variables in the fixed cost models correspond to a. a process for which a fixed cost occurs. b. the number of products produced. c. the number of units produced. d. the actual value of the fixed cost. ANSWER: a TOPIC: Fixed costs 8. Sensitivity analysis for integer linear programming a. can be provided only by computer. b. has precisely the same interpretation as that from linear programming. c. does not have the same interpretation and should be disregarded. d. is most useful for 0 - 1 models. ANSWER: c TOPIC: Sensitivity analysis 9. Let x1 and x2 be 0 - 1 variables whose values indicate whether projects 1 and 2 are not done or are done.

Which answer below indicates that project 2 can be done only if project 1 is done? a. x1 + x2 = 1 b. x1 + x2 = 2 c. x1 - x2 < 0 d. x1 - x2 > 0 ANSWER: d TOPIC: Conditional and corequisite constraints 10. Let x1 , x2 , and x3 be 0 - 1 variables whose values indicate whether the projects are not done (0) or are done

(1). Which answer below indicates that at least two of the projects must be done? a. x1 + x2 + x3 > 2 b. x1 + x2 + x3 < 2 c. x1 + x2 + x3 = 2 d. x1 - x2 = 0 ANSWER: a TOPIC: k out of n alternatives constraints


Chapter 11 Integer Linear Programming 3

11. If the acceptance of project A is conditional on the acceptance of project B, and vice versa, the appropriate constraint to use is a

a. multiple-choice constraint. b. k out of n alternatives constraint. c. mutually exclusive constraint. d. corequisite constraint. ANSWER: d TOPIC: Modeling flexibility provided by 0-1 integer variables 12. In an all-integer linear program, a. all objective function coefficients must be integer. b. all right-hand side values must be integer. c. all variables must be integer. d. all objective function coefficients and right-hand side values must be integer. ANSWER: c TOPIC: Types of integer linear programming models 13. To perform sensitivity analysis involving an integer linear program, it is recommended to a. use the dual prices very cautiously. b. make multiple computer runs. c. use the same approach as you would for a linear program. d. use LP relaxation. ANSWER: b TOPIC: A cautionary note about sensitivity analysis 14. Modeling a fixed cost problem as an integer linear program requires a. adding the fixed costs to the corresponding variable costs in the objective function. b. using 0-1 variables. c. using multiple-choice constraints. d. using LP relaxation. ANSWER: b TOPIC: Applications involving 0-1 variables 15. Most practical applications of integer linear programming involve a. only 0-1 integer variables and not ordinary integer variables. b. mostly ordinary integer variables and a small number of 0-1 integer variables. c. only ordinary integer variables. d. a near equal number of ordinary integer variables and 0-1 integer variables. ANSWER: a TOPIC: Applications involving 0-1 variables

TRUE/FALSE

1. The LP Relaxation contains the objective function and constraints of the IP problem, but drops all integer

restrictions. ANSWER: True TOPIC: LP relaxation 2. In general, rounding large values of decision variables to the nearest integer value causes fewer problems

than rounding small values. ANSWER: True TOPIC: LP relaxation



3. The solution to the LP Relaxation of a minimization problem will always be less than or equal to the value

of the integer program minimization problem. ANSWER: False TOPIC: Graphical solution 4. If the optimal solution to the LP relaxation problem is integer, it is the optimal solution to the integer linear

program. ANSWER: True TOPIC: LP relaxation 5. Slack and surplus variables are not useful in integer linear programs. ANSWER: False TOPIC: Capital budgeting 6. A multiple choice constraint involves selecting k out of n alternatives, where k > 2. ANSWER: False TOPIC: Multiple choice constraint 7. In a model involving fixed costs, the 0 - 1 variable guarantees that the capacity is not available unless the

cost has been incurred. ANSWER: True TOPIC: Fixed costs 8. If x1 + x2 < 500y1 and y1 is 0 - 1, then if y1 is 0, x1 and x2 will be 0. ANSWER: True TOPIC: Distribution system design 9. The constraint x1 + x2 + x3 + x4 < 2 means that two out of the first four projects must be selected. ANSWER: False TOPIC: k out of n alternatives constraint 10. The constraint x1 - x2 = 0 implies that if project 1 is selected, project 2 cannot be. ANSWER: False TOPIC: Conditional and corequisite constraints 11. The product design and market share optimization problem presented in the textbook is formulated as a 0-

1 integer linear programming model. ANSWER: True TOPIC: Product design and market share optimization problem 12. The objective of the product design and market share optimization problem presented in the textbook is to

choose the levels of each product attribute that will maximize the number of sampled customers preferring the brand in question.

ANSWER: True TOPIC: Product design and market share optimization problem 13. If a problem has only less-than-or-equal-to constraints with positive coefficients for the variables, rounding

down will always provide a feasible integer solution. ANSWER: True TOPIC: Rounding to obtain an integer solution 14. Dual prices cannot be used for integer programming sensitivity analysis because they are designed for

linear programs. ANSWER: True TOPIC: A cautionary note about sensitivity analysis



15. Some linear programming problems have a special structure that guarantees the variables will have integer

values. ANSWER: True TOPIC: Introduction to integer linear programming 16. Generally, the optimal solution to an integer linear program is less sensitive to the constraint coefficients

than is a linear program. ANSWER: False TOPIC: A cautionary note about sensitivity analysis 17. The classic assignment problem can be modeled as a 0-1 integer program. ANSWER: True TOPIC: Applications involving 0-1 variables 18. If Project 5 must be completed before Project 6, the constraint would be x5 – x6 < 0. ANSWER: False TOPIC: Conditional and co-requisite constraints 19. If the LP relaxation of an integer program has a feasible solution, then the integer program has a feasible

solution. ANSWER: False TOPIC: Rounding to obtain an integer solution 20. Multiple choice constraints involve binary variables. ANSWER: True TOPIC: Multiple choice and mutually exclusive constraints

SHORT ANSWER

1. The use of integer variables creates additional restrictions but provides additional flexibility. Explain. TOPIC: Introduction 2. Why are 0 - 1 variables sometimes called logical variables? TOPIC: Introduction 3. Give a verbal interpretation of each of these constraints in the context of a capital budgeting problem. a. x1 - x2 > 0 b. x1 - x2 = 0 c. x1 + x2 + x3 < 2 TOPIC: Capital budgeting 4. Explain how integer and 0-1 variables can be used in an objective function to minimize the sum of fixed

and variable costs for production on two machines. TOPIC: Distribution system design 5. Explain how integer and 0-1 variables can be used in a constraint to enable production. TOPIC: Distribution system design



PROBLEMS

1. Solve the following problem graphically. Max 5X + 6Y s.t. 17X + 8Y < 136 3X + 4Y < 36 X, Y > 0 and integer a. Graph the constraints for this problem. Indicate all feasible solutions.

b. Find the optimal solution to the LP Relaxation. Round down to find a feasible integer solution. Is this solution optimal?

c. Find the optimal solution. TOPIC: Graphical solution 2. Solve the following problem graphically. Max X + 2Y s.t. 6X + 8Y < 48 7X + 5Y > 35 X, Y > 0

Y integer a. Graph the constraints for this problem. Indicate all feasible solutions.

b. Find the optimal solution to the LP Relaxation. Round down to find a feasible integer solution. Is this solution optimal?

c. Find the optimal solution. TOPIC: Graphical solution 3. Solve the following problem graphically. Min 6X + 11Y s.t. 9X + 3Y > 27 7X + 6Y > 42 4X + 8Y > 32 X, Y > 0 and integer a. Graph the constraints for this problem. Indicate all feasible solutions.

b. Find the optimal solution to the LP Relaxation. Round up to find a feasible integer solution. Is this solution optimal?

c. Find the optimal solution. TOPIC: Graphical solution 4. Consider a capital budgeting example with five projects from which to select. Let xi = 1 if project i is

selected, 0 if not, for i = 1,...,5. Write the appropriate constraint(s) for each condition. Conditions are independent.

a. Choose no fewer than three projects. b. If project 3 is chosen, project 4 must be chosen. c. If project 1 is chosen, project 5 must not be chosen. d. Projects cost 100, 200, 150, 75, and 300 respectively. The budget is 450.

d. No more than two of projects 1, 2, and 3 can be chosen. TOPIC: Capital budgeting



5. Grush Consulting has five projects to consider. Each will require time in the next two quarters according to

the table below.

Project Time in first quarter Time in second quarter Revenue A 5 8 12000 B 3 12 10000 C 7 5 15000 D 2 3 5000 E 15 1 20000

Revenue from each project is also shown. Develop a model whose solution would maximize revenue, meet the time budget of 25 in the first quarter and 20 in the second quarter, and not do both projects C and D.

TOPIC: Capital budgeting 6. The Westfall Company has a contract to produce 10,000 garden hoses for a large discount chain. Westfall

has four different machines that can produce this kind of hose. Because these machines are from different manufacturers and use differing technologies, their specifications are not the same.

Machine Fixed Cost to Set

Up Production Run Variable Cost

Per Hose

Capacity 1 750 1.25 6000 2 500 1.50 7500 3 1000 1.00 4000 4 300 2.00 5000

a. This problem requires two different kinds of decision variables. Clearly define each kind. b. The company wants to minimize total cost. Give the objective function. c. Give the constraints for the problem. d. Write a constraint to ensure that if machine 4 is used, machine 1 cannot be. TOPIC: Fixed cost 7. Hansen Controls has been awarded a contract for a large number of control panels. To meet this demand, it

will use its existing plants in San Diego and Houston, and consider new plants in Tulsa, St. Louis, and Portland. Finished control panels are to be shipped to Seattle, Denver, and Kansas City. Pertinent information is given in the table.

Sources

Construction

Cost

Shipping Cost to Destination:

Capacity

Seattle

Denver Kansas

City San Diego ---- 5 7 8 2,500 Houston ---- 10 8 6 2,500 Tulsa 350,000 9 4 3 10,000 St. Louis 200,000 12 6 2 10,000 Portland 480,000 4 10 11 10,000 Demand 3,000 8,000 9,000

Develop a model whose solution would reveal which plants to build and the optimal shipping schedule. TOPIC: Distribution system design



8. Simplon Manufacturing must decide on the processes to use to produce 1650 units. If machine 1 is used, its production will be between 300 and 1500 units. Machine 2 and/or machine 3 can be used only if machine 1's production is at least 1000 units. Machine 4 can be used with no restrictions.

Machine Fixed cost

Variable cost

Minimum Production

Maximum Production

1 500 2.00 300 1500 2 800 0.50 500 1200 3 200 3.00 100 800 4 50 5.00 any any

(HINT: Use an additional 0 - 1 variable to indicate when machines 2 and 3 can be used.) TOPIC: Distribution system design 9. Your express package courier company is drawing up new zones for the location of drop boxes for

customers. The city has been divided into the seven zones shown below. You have targeted six possible locations for drop boxes. The list of which drop boxes could be reached easily from each zone is listed below.

Zone Can Be Served By Locations: Downtown Financial 1, 2, 5, 6 Downtown Legal 2, 4, 5 Retail South 1, 2, 4, 6 Retail East 3, 4, 5 Manufacturing North 1, 2, 5 Manufacturing East 3, 4 Corporate West 1, 2, 6

Let xi = 1 if drop box location i is used, 0 otherwise. Develop a model to provide the smallest number of

locations yet make sure that each zone is covered by at least two boxes. TOPIC: Applications of integer linear programming 10. Consider the problem faced by a summer camp recreation director who is trying to choose activities for a

rainy day. Information about possible choices is given in the table below.

Category

Activity

Time

(minutes)

Popularity with Campers

Popularity with Counselors

Art 1 - Painting 30 4 2 2 - Drawing 20 5 2 3 - Nature craft 30 3 1 Music 4 - Rhythm band 20 5 5 Sports 5 - Relay races 45 2 1 6 - Basketball 60 1 3 Computer 7 - Internet 45 1 1 8 - Creative writing 30 4 3 9 - Games 40 1 2

a. Give a general definition of the variables necessary in this problem so that each activity can be

considered for inclusion in the day’s schedule. b. The popularity ratings are defined so that 1 is the most popular. If the objective is to keep the

campers happy, what should the objective function be? Write constraints for these restrictions:



c. At most one art activity can be done. d. No more than two computer activities can be done. e. If basketball is chosen, then the music must be chosen. f. At least 120 minutes of activities must be selected. g. No more than 165 minutes of activities may be selected. h. To keep the staff happy, the counselor rating should be no higher than 10.

TOPIC: Applications of integer programming 11. Tower Engineering Corporation is considering undertaking several proposed projects for the next fiscal

year. The projects, the number of engineers and the number of support personnel required for each project, and the expected profits for each project are summarized in the following table:

Project

1 2 3 4 5 6 Engineers Required 20 55 47 38 90 63 Support Personnel Required 15 45 50 40 70 70 Profit ($1,000,000s) 1.0 1.8 2.0 1.5 3.6 2.2

Formulate an integer program that maximizes Tower's profit subject to the following management constraints:

1) Use no more than 175 engineers 2) Use no more than 150 support personnel 3) If either project 6 or project 4 is done, both must be done 4) Project 2 can be done only if project 1 is done

5) If project 5 is done, project 3 must not be done and vice versa 6) No more than three projects are to be done.

TOPIC: Applications of integer programming 12. Given the following all-integer linear program: MAX 3x1 + 2x2 s. t. 3x1 + x2 < 9 x1 + 3x2 < 7 -x1 + x2 < 1 x1, x2 > 0 and integer

a Solve the problem as a linear program ignoring the integer constraints. Show that the optimal solution to the linear program gives fractional values for both x1 and x2.

b. What is the solution obtained by rounding fractions greater than of equal to 1/2 to the next larger number? Show that this solution is not a feasible solution.

c. What is the solution obtained by rounding down all fractions? Is it feasible? d. Enumerate all points in the linear programming feasible region in which both x1 and x2 are

integers, and show that the feasible solution obtained in (c) is not optimal and that in fact the optimal integer is not obtained by any form of rounding.

TOPIC: Rounding to obtain an integer solution 13. Tom's Tailoring has five idle tailors and four custom garments to make. The estimated time (in hours) it

would take each tailor to make each garment is listed below. (An 'X' in the table indicates an unacceptable tailor-garment assignment.)



Tailor Garment 1 2 3 4 5 Wedding gown 19 23 20 21 18 Clown costume 11 14 X 12 10 Admiral’s uniform 12 8 11 X 9 Bullfighter’s outfit X 20 20 18 21

Formulate and solve an integer program for determining the tailor-garment assignments that minimize the total estimated time spent making the four garments. No tailor is to be assigned more than one garment and each garment is to be worked on by only one tailor.

TOPIC: Assignment problem 14. Market Pulse Research has conducted a study for Lucas Furniture on some designs for a new commercial

office desk. Three attributes were found to be most influential in determining which desk is most desirable: number of file drawers, the presence or absence of pullout writing boards, and simulated wood or solid color finish. Listed below are the part-worths for each level of each attribute provided by a sample of 7 potential Lucas customers.

File Drawers Pullout Writing Boards Finish

Consumer 0 1 2 Present Absent Simul. Wood Solid Color 1 5 26 20 18 11 17 10 2 18 11 5 12 16 15 26 3 4 16 22 7 13 11 19 4 12 8 4 18 9 22 14 5 19 9 3 4 14 30 19 6 6 15 21 8 17 20 11 7 9 6 3 13 5 16 28

Suppose the overall utility (sum of part-worths) of the current favorite commercial office desk is 50 for

each customer. What is the product design that will maximize the share of choices for the seven sample participants? Formulate and solve, using Lindo or Excel, this 0 – 1 integer programming problem.

TOPIC: Product design and market share optimization 15. Kloos Industries has projected the availability of capital over each of the next three years to be $850,000,

$1,000,000, and $1,200,000, respectively. It is considering four options for the disposition of the capital:

(1) Research and development of a promising new product (2) Plant expansion (3) Modernization of its current facilities (4) Investment in a valuable piece of nearby real estate

Monies not invested in these projects in a given year will NOT be available for following year's investment in the projects. The expected benefits three years hence from each of the four projects and the yearly capital outlays of the four options are summarized in the table below in $1,000,000's. In addition, Kloos has decided to undertake exactly two of the projects, and if plant expansion is selected, it will also modernize its current facilities.

Capital Outlays Projected Options Year 1 Year 2 Year 3 Benefits New Product R&D .35 .55 .75 5.2 Plant Expansion .50 .50 0 3.6



Modernization .35 .40 .45 3.2 Real Estate .50 0 0 2.8

Formulate and solve this problem as a binary programming problem.

TOPIC: Capital budgeting


1. a. The feasible region is those integer values in the space labeled feasible region.

0

5

10

15

20

0 5 10 15

FeasibleRegion

b. Optimal LP relaxed occurs at X = 5.818, Y = 4.636, with Z = 56.909. Rounded down solution

occurs at X = 5, Y = 4, Z= 49. c. Optimal solution is at X = 4, Y = 6, and Z = 56. 2. a. The feasible region consists of the portions of the horizontal lines that lie within the area labeled F. R.



0

5

10

0 5 10

F. R.

b. The optimal relaxed solution is at X = 1.538, Y = 4.846 where Z = 11.231. The rounded solution is

X = 1.538, Y = 4. c. The optimal solution is at X = 2.667, Y = 4, Z = 10.667. 3. a. The feasible region is the set of integer points in the area labeled feasible region.

0

5

10

0 5 10

Feasible Region

b. The optimal relaxed solution is at X = 4.5, Y = 1.75, and Z = 46.25.

The rounded solution is X = 5, Y = 2. c. The optimal solution is at X = 6, Y = 1, and Z = 47. 4. a. x1 + x2 + x3 + x4 + x5 > 3 b. x3 - x4 < 0 c. x1 + x5 < 1 d. 100x1 + 200x2 + 150x3 + 75x4 + 300x5 < 450 e. x1 + x2 + x3 < 2



5. Let A = 1 if project A is selected, 0 otherwise; same for B, C, D, and E Max 12000A + 10000B + 15000C + 5000D + 20000E s.t. 5A + 3B + 7C + 2D + 15E ≤ 25 8A + 12B + 5C + 3D + 1E ≤ 20 C + D ≤ 1 6. a. Let Pi = the number of hoses produced on machine i Ui = 1 if machine i is used, = 0 otherwise b. Min 750U1 + 500U2 + 1000U3 + 300U4 + 1.25P1 + 1.5P2 + P3 + 2P4 c. P1 < 6000U1 P2 < 7500U2 P3 < 4000U3 P4 < 5000U4 P1 + P2 + P3 + P4 > 10000

c. U1 + U4 < 1 7. Let Pij = the number of panels shipped from source i to destination j Bi = 1 if plant i is built, = 0 otherwise (i = 3, 4, 5) Min 350000B3 + 200000B4 + 480000B5 + 5P11 + 7P12 + 8P13 + 10P21 + 8P22 + 6P23 + 9P31 + 4P32 + 3P33 + 12P41 + 6P42 + 2P43 + 4P51 + 10P52 + 11P53 s.t. P11 + P12 + P13 < 2500 P21 + P22 + P23 < 2500 P31 + P32 + P33 < 10000B3 P41 + P42 + P43 < 10000B4 P51 + P52 + P53 < 10000B5 P11 + P21 + P31 + P41 + P51 = 3000 P12 + P22 + P32 + P42 + P52 = 8000 P13 + P23 + P33 + P43 + P53 = 9000 8. Let Ui = the number of units made by machine i Si = 1 if machine i is used (requiring a set-up), = 0 otherwise K = 1 if machine 1 produces at least 1000 units, = 0 otherwise Min 500S1 + 2U1 + 800S2 + 5U2 + 200S3 + 3U3 + 50S4 + 5U4 s.t. U1 > 300S1 U1 < 1500S1 U1 > 1000K S2 < K S3 < K U2 > 500S2 U2 < 1200S2 U3 > 100S3 U3 < 800S3 U4 < 1650S4 U1 + U2 + U3 + U4 = 1650 9. Min Σxi s.t. x1 + x2 + x5 + x6 > 2 x2 + x4 + x5 > 2 x1 + x2 + x4 + x6 > 2 x3 + x4 + x5 > 2 x1 + x2 + x5 > 2 x3 + x4 > 2



x1 + x2 + x6 > 2 10. a. Let xi = 1 if activity i is chosen, 0 if not, for i = 1, … , 9 b. Max 4x1 + 5x2 + 3x3 + 5x4 + 2x5 + 1x6 + 1x7 + 4x8 + 1x9 c. x1 + x2 + x3 ≤ 1 d. x7 + x8 + x9 ≤ 2 e. x6 ≤ x4 f. 30x1 + 20x2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≥ 120 g. 30x1 + 20x2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≤ 165

h. 2x1 + 2x2 + 1x3 + 5x4 + 1x5 + 3x6 + 1x7 + 3x8 + 2x9 ≤ 10 11. Max P

1 + 1.8P

2 + 2P

3 + 1.5P

4 + 3.6P

5 + 2.2P

6

s.t. 20P1 + 55P

2 + 47P

3 + 38P

4 + 90P

5 + 63P

6 < 175

15P1 + 45P

2 + 50P

3 + 40P

4 + 70P

5 + 70P

6 < 150

P4 - P

6 = 0

P1 - P

2 > 0

P3 + P

5 < 1

P1 + P

2 + P

3 + P

4 + P

5 + P

6 < 3

Pi = 0 or 1

12. a. From the graph on the next page, the optimal solution to the linear program is x1 = 2.5, x2 = 1.5, z

= 10.5. b. By rounding the optimal solution of x1 = 2.5, x2 = 1.5 to x1 = 3, x2 = 2, this point lies outside the

feasible region. c. By rounding the optimal solution down to x1 = 2, x2 = 1, we see that this solution indeed is an

integer solution within the feasible region, and substituting in the objective function, it gives z = 8. d. There are eight feasible integer solutions in the linear programming feasible region with z values

as follows:

x1 x2 z

1. 0 0 0 2. 1 0 3 3. 2 0 6 4. 3 0 9 optimal 5. 0 1 2 6. 1 1 5 7. 2 1 8 part (c) solution 8. 1 2 7

x1 = 3, x2 = 0 is the optimal solution. Rounding the LP solution (x1 = 2.5, x2 = 1.5) would not have been optimal.



5

4

3

2

1

1 2 3 4 5 6 7

LP Optimal (2.5, 1.5)

X1 + 3X2 < 7

MAX 3X1 + 2X2

(3, 2)

-X1 + X2 < 1

3X1 + X2 < 9

(2, 1)

X2

X1

13. Define the decision variables: xij = 1 if garment i is assigned to tailor j; = 0 otherwise. Number of decision variables = [(number of garments)(number of tailors)] - (number of unacceptable assignments) = [4(5)] - 3 = 17.

Define the objective function: Minimize total time spent making garments: MIN 19x11 + 23x12 + 20x13 + 21x14 + 18x15 + 11x21 + 14x22 + 12x24 + 10x25 + 12x31 + 8x32 + 11x33 + 9x35 + 20x42 + 20x43 + 18x44 + 21x45

Define the constraints: Exactly one tailor per garment: No more than one garment per tailor: 1) x11 + x12 + x13 + x14 + x15 = 1 5) x11 + x21 + x31 < 1 2) x21 + x22 + x24 + x25 = 1 6) x12 + x22 + x32 + x42 < 1 3) x31 + x32 + x33 + x35 = 1 7) x13 + x33 + x43 < 1 4) x42 + x43 + x44 + x45 = 1 8) x14 + x24 + x44 < 1 9) x15 + x25 + x35 + x45 < 1 Nonnegativity: xij > 0 for i = 1,..,4 and j = 1,..,5

Optimal Solution: Assign wedding gown to tailor 5 Assign clown costume to tailor 1 Assign admiral uniform to tailor 2 Assign bullfighter outfit to tailor 4 Total time spent = 55 hours

14. Define the decision variables: There are 7 lij decision variables, one for each level of attribute. lij = 1 if Lucas chooses level i for attribute j; 0 otherwise. There are 7 Yk decision variables, one for each consumer in the sample. Yk = 1 if consumer k chooses the Lucas brand, 0 otherwise.

Define the objective function: Maximize the number of consumers preferring the Lucas brand desk.



MAX Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 Define the constraints: There is one constraint for each consumer in the sample. 5l11 + 26l21 + 20l31 + 18l12 + 11l22 + 17l13 + 10l23 – 50Y1 > 1 18l11 + 11l21 + 5l31 + 12l12 + 16l22 + 15l13 + 26l23 – 50Y2 > 1 4l11 + 16l21 + 22l31 + 7l12 + 13l22 + 11l13 + 19l23 – 50Y3 > 1 12l11 + 8l21 + 4l31 + 18l12 + 9l22 + 22l13 + 14l23 – 50Y4 > 1 19l11 + 9l21 + 3l31 + 4l12 + 14l22 + 30l13 + 19l23 – 50Y5 > 1 6l11 + 15l21 + 21l31 + 8l12 + 17l22 + 20l13 + 11l23 – 50Y6 > 1 9l11 + 6l21 + 3l31 + 13l12 + 5l22 + 16l13 + 28l23 – 50Y7 > 1

There is one constraint for each attribute. l11 + l21 + l31 = 1 l12 + l22 = 1 l13 + l23 = 1

Optimal Solution: Lucas should choose these product features: 1 file drawer (I21 = 1)

No pullout writing boards (I22 = 1) Simulated wood finish (I13 = 1) Three sample participants would choose the Lucas design: Participant 1 (Y1 = 1) Participant 5 (Y5 = 1) Participant 6 (Y6 = 1)

15. MAX 5.2X1 + 3.6X2 + 3.2X3 + 2.8X4

S.T. .35X1 + .50X2 + .35X3 + .50X4 < 0.85 (First Year) .55X1 + .50X2 + .40X3 < 1.00 (Second Year) .75X1 + .45X3 < 1.20 (Third Year) X1 + X2 + X3 + X4 = 2 - X2 + X3 > 0 Xi = 0 or 1

Solution: X1 = 1, X2 = 0, X3 = 1, X4 = 0, Total projected benefits = $8.4 million.


1

12

Project Scheduling: PERT/CPM

MULTIPLE CHOICE

1. PERT and CPM

a. are most valuable when a small number of activities must be scheduled. b. have different features and are not applied to the same situation. c. do not require a chronological relationship among activities. d. have been combined to develop a procedure that uses the best of each.

ANSWER: d TOPIC: Introduction 2. Which is not a significant challenge of project scheduling? a. deadlines exist. b. activities are independent. c. many employees could be required. d. delays are costly. ANSWER: b TOPIC: Introduction 3. Arcs in a project network indicate a. completion times. b. precedence relationships. c. activities. d. the critical path. ANSWER: b TOPIC: Critical path 4. The critical path a. is any path that goes from the starting node to the completion node. b. is a combination of all paths. c. is the shortest path. d. is the longest path. ANSWER: d TOPIC: Critical path


2 Chapter 12 Project Scheduling: PERT/CPM

5. The earliest start time rule a. compares the starting times of all activities for successors of an activity. b. compares the finish times for all immediate predecessors of an activity. c. determines when the project can begin. d. determines when the project must begin. ANSWER: b TOPIC: Critical path 6. Activities following a node a. can begin as soon as any activity preceding the node has been completed. b. have an earliest start time equal to the largest of the earliest finish times for all activities entering

the node. c. have a latest start time equal to the largest of the earliest finish times for all activities entering the

node. d. None of the alternatives is correct. ANSWER: b TOPIC: Critical path 7. Activities G, P, and R are the immediate predecessors for activity W. If the earliest finish times for the

three are 12, 15, and 10, then the earliest start time for W a. is 10. b. is 12. c. is 15. d. cannot be determined. ANSWER: c TOPIC: Critical path 8. Activities K, M and S immediately follow activity H, and their latest start times are 14, 18, and 11. The

latest finish time for activity H a. is 11. b. is 14. c. is 18. d. cannot be determined. ANSWER: a TOPIC: Critical path 9. When activity times are uncertain, a. assume they are normally distributed. b. calculate the expected time, using (a + 4m + b)/6. c. use the most likely time. d. calculate the expected time, using (a + m + b)/3. ANSWER: b TOPIC: Uncertain activity times 10. To determine how to crash activity times a. normal activity costs and costs under maximum crashing must be known. b. shortest times with crashing must be known. c. realize that new paths may become critical. d. All of the alternatives are true. ANSWER: d TOPIC: Crashing activity times


Chapter 12 Project Scheduling: PERT/CPM 3

11. Slack equals a. LF – EF. b. EF – LF. c. EF – LS. d. LF – ES. ANSWER: b TOPIC: Determining the critical path 12. Activities with zero slack a. can be delayed. b. must be completed first. c. lie on a critical path. d. have no predecessors. ANSWER: c TOPIC: Determining the critical path 13. In deciding which activities to crash, one must a. crash all critical activities. b. crash largest-duration activities. c. crash lowest-cost activities. d. crash activities on the critical path(s) only. ANSWER: d TOPIC: Crashing activity times 14. For an activity with more than one immediate predecessor activity, which of the following is used to

compute its earliest finish (EF) time? a. the largest EF among the immediate predecessors. b. the average EF among the immediate predecessors.

c. the largest LF among the immediate predecessors. d. the difference in EF among the immediate predecessors. ANSWER: a TOPIC: Determining the critical path 15. Which of the following is always true about a critical activity?

a. LS = EF. b. LF = LS.

c. ES = LS. d. EF = ES. ANSWER: c TOPIC: Determining the critical path 16. For an activity with more than one immediate successor activity, its latest-finish time is equal to the

a. largest latest-finish time among its immediate successors. b. smallest latest-finish time among its immediate successors.

c. largest latest-start time among its immediate successors. d. smallest latest-start time among its immediate successors. ANSWER: d TOPIC: Determining the critical path



17. Which of the following is a general rule for crashing activities? a. Crash only non-critical activities. b. Crash activities with zero slack. c. Crash activities with the greatest number of predecessors. d. Crash the path with the fewest activities. ANSWER: b TOPIC: Crashing activity times

TRUE/FALSE

1. Critical activities are those that can be delayed without delaying the entire project. ANSWER: False TOPIC: Introduction 2. PERT and CPM are applicable only when there is no dependence among activities. ANSWER: False TOPIC: Introduction 3. A path through a project network must reach every node. ANSWER: False TOPIC: Critical path 4. A critical activity can be part of a noncritical path. ANSWER: True TOPIC: Critical path 5. When activity times are uncertain, an activity's most likely time is the same as its expected time. ANSWER: False TOPIC: Variability in project completion time 6. The earliest finish time for the final activity is the project duration. ANSWER: True TOPIC: Critical path 7. The length of time an activity can be delayed without affecting the project completion time is the slack. ANSWER: True TOPIC: Critical path 8. When activity times are uncertain, total project time is normally distributed with mean equal to the sum of

the means of all of the critical activities. ANSWER: True TOPIC: Variability in project completion time 9. Crashing refers to an unanticipated delay in a critical path activity that causes the total time to exceed its

limit. ANSWER: False TOPIC: Crashing activity times 10. Constraints in the LP models for crashing decisions are required to compare the activity’s earliest finish

time with the earliest finish time of each predecessor. ANSWER: True TOPIC: Crashing activity times



11. The project manager should monitor the progress of any activity with a large time variance even if the

expected time does not identify the activity as a critical activity. ANSWER: True TOPIC: Variability in project completion time 12. The variance in the project completion time is the sum of the variances of all activities in the project. ANSWER: False TOPIC: Variability in project completion time 13. The latest finish time for an activity is the largest of the latest start times for all activities that immediately

follow the activity. ANSWER: False TOPIC: Earliest and latest times 14. The earliest start time for an activity is equal to the smallest of the earliest finish times for all its immediate

predecessors. ANSWER: False TOPIC: Earliest and latest times 15. The linear programming model for crashing presented in the textbook assumes that any portion of the

activity crash time can be achieved for a corresponding portion of the activity crashing cost. ANSWER: True TOPIC: Linear programming model for crashing 16. All activities on a critical path have zero slack time. ANSWER: True TOPIC: Critical path 17. The difference between an activity’s earliest finish time and latest finish time equals the difference between

its earliest start time and latest start time. ANSWER: True TOPIC: Earliest and latest times 18. It is possible to have more than one critical path at a time. ANSWER: True TOPIC: Critical path

SHORT ANSWER

1. Name at least three managerial situations where answers are provided by project management solutions. TOPIC: Introduction 2. Explain how and why all predecessor activities must be considered when finding the earliest start time. TOPIC: Earliest and latest times 3. Explain how and why all successor activities must be considered when finding the latest finish time. TOPIC: Earliest and latest times 4. Once the earliest and latest times are calculated, how is the critical path determined?



TOPIC: Critical path 5. Why should projects be monitored after the critical path is found? TOPIC: Critical path

PROBLEMS

1. From this schedule of activities, draw the PERT/CPM network.

Activity

Immediate Predecessor

A --- B A C B D B E A F C, D G E, F

TOPIC: PERT/CPM networks 2. From this PERT/CPM network, determine the list of activities and their predecessors.

C A E G Start D H B F TOPIC: PERT/CPM networks 3. A cookie recipe gives the following numbered steps. 1. Preheat oven. 2. Grease cookie sheets. 3. Cream shortening and sugar. 4. Add eggs and flavoring. 5. Measure and sift dry ingredients. 6. Add dry ingredients to mixture. 7. Drop by spoonfuls onto sheets and bake for 10 minutes. Although the steps are numbered, they do not always reflect immediate precedence relationships. Develop

a table that lists the immediate predecessors for each activity. TOPIC: Precedence relationships

Finish



4. A senior MIS design class project team has developed the following schedule of activities for their project, using their best estimate of completion times. Both written and oral reports are required. Draw the project network. Can they complete the project in the 38 class days remaining until the end of the semester?

Activity Time Immediate Predecessor

A. Find client 4 --- B. Write prospectus 2 A C. Obtain approval from client and professor 3 B D. Complete programming 12 C E. Do industry background research 10 --- F. Write final paper 6 D, E G. Write oral report 5 D, E

TOPIC: Critical path

5. A project network is shown below. Use a forward and a backward pass to determine the critical path, and

then fill out the table below.

Start

A 5

B 4

C 8

D 4

E 12

F 10

G 3

H 2

I 5

Finish

Activity

Precedence Activities

Activity Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A B C D E F G H I

Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in fewer than 25 weeks more or less than 50%?



TOPIC: Variability in project completion time 6. A project network is shown below. Use a forward and a backward pass to determine the critical path, and

then fill out the table below.

Start

A 5

B 4

C 8

D 4

E 12

F 10

G 3

H 2

I 5

Finish

Activity


Activity Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A B C D E F G H I

Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in more than 28 weeks more or less than 50%?

TOPIC: Variability in project completion time



7. Use the following network of related activities with their duration times to complete a row for each activity under the column headings below.

Start

Finish

A8

B2

C5

D3

E7

G12

F6

H10

Activity Immediate

Predecessors Activity

Time

ES

LS

EF

LF

Slack Critical Path?

A B C D E F G H

TOPIC: Critical path 8. Use the following network of related activities with their duration times to complete a row for each activity

under the column headings below.

Start A2

B4

C3

D7

E6

F5

G8

H5

I4

J6

Finish

Activity

Immediate Predecessors

Activity Time

ES

LS

EF

LF

Slack

Critical Path?

A



B C D E F G H I J

TOPIC: Critical path 9. Given the following network with activities and times estimated in days,

Activity

Optimistic

Most Probable

Pessimistic

D

E

F

G

H J

KI Finish

A

B

C

Start



A 2 5 6 B 1 3 7 C 6 7 10 D 5 12 14 E 3 4 5 F 8 9 12 G 4 6 8 H 3 6 8 I 5 7 12 J 12 13 14 K 1 3 4

a. What are the critical path activities? b. What is the expected time to complete the project? c. What is the probability the project will take more than 28 days to complete? TOPIC: Critical path with uncertain times 10. The critical path for this network is A - E - F and the project completion time is 22 weeks.

Start

A12

D5

E4

B14

C8

F6

G10

Finish

Activity

Normal Time

Crash Time

Normal Cost

Crash Cost

A 12 8 8,000 12,000 B 14 10 5,000 7,500 C 8 8 10,000 10,000 D 5 3 6,000 8,000 E 4 3 5,000 7,000 F 6 5 9,000 12,000 G 10 8 5,000 8,000

If a deadline of 17 weeks is imposed, give the linear programming model for the crashing decision.

TOPIC: Crashing activity times 11. For the project represented below, determine the earliest and latest start and finish times for each activity as

well as the expected overall completion time.



C

Start

E

D

G

A

Finish

H

F

B

Activity Duration ES EF LS LF Slack A 4 B 3 C 4 D 2 E 5 F 2 G 5 H 6

TOPIC: Determining the critical path 12. Consider the following PERT/CPM network with estimated times in weeks. The project is scheduled to

begin on May 1.

C4

Start

E1

D2

H7

A1 Finish

G3

F2

B3

The three-time estimate approach was used to calculate the expected times and the following table gives the variance for each activity:

Activity Variance Activity Variance A 1.1 E 0.3 B 0.5 F 0.6 C 1.2 G 0.6 D 0.8 H 1.0

a. Give the expected project completion date and the critical path. b. By what date are you 99% sure the project will be completed?

TOPIC: Variability in project completion time 13. A project consists of five activities. Naturally the paint mixing precedes the painting activities. Also, both

ceiling painting and floor sanding must be done prior to floor buffing.



Optimistic Most Probable Pessimistic Activity Time (hr) Time (hr) Time (hr) Floor sanding 3 4 5 Floor buffing 1 2 3 Paint mixing 0.5 1 1.5 Wall painting 1 2 9 Ceiling painting 1 5.5 7

a. Construct the PERT/CPM network for this problem. b. What is the expected completion time of this project? c. What is the probability that the project can be completed within 9 hr.?

TOPIC: Variability in project completion time 14. Consider a project that has been modeled as follows:

Activity Immediate Predecessors Duration (hr) A --- 7 B --- 10 C A 4 D A 30 E A 7 F B,C 12 G B,C 15 H E,F 11 I E,F 25 J E,F 6 K D,H 21 L G,J 25

a. Draw the PERT/CPM network for this project and determine project's expected completion time

and its critical path. b. Can activities E and G be performed simultaneously without delaying the minimum project

completion time? c. Can one person perform A, G, and I without delaying the project? d. By how much can activities G and L be delayed without delaying the entire project? e. How much would the project be delayed if activity G was delayed by 7 hours and activity L was

delayed by 4 hours? Explain. TOPIC: Determining the critical path 15. Joseph King has ambitions to be mayor of Williston, North Dakota. Joe has determined the breakdown of

the steps to the nomination and has estimated normal and crash costs and times for the campaign as follows (times are in weeks).

Normal Crash Immediate Activity Time Cost Time Cost Predecessors A. Solicit Volunteers 6 $5,000 4 $ 9,000 --- B. Initial “Free” Exposure 3 $4,000 3 $ 4,000 --- C. Raise Money 9 $4,000 6 $10,000 A D. Organize Schedule 4 $1,000 2 $ 2,000 A E. Hire Advertising Firm 2 $1,500 1 $ 2,000 B F. Arrange TV Interview 3 $4,000 1 $ 8,000 B G. Advertising Campaign 5 $7,000 4 $12,000 C,E H. Personal Campaigning 7 $8,000 5 $20,000 D,F



Joe King is not a wealthy man and would like to organize a 16-week campaign at minimum cost. Write and solve a linear program to accomplish this task.

TOPIC: Crashing activity times


1.

A B

C

D

E

F G FinishStart

2.

Activity

Immediate Predecessor

A --- B --- C A D A, B E C, D F D G E H F, G

3.

Activity Immediate

Predecessor 1 --- 2 --- 3 --- 4 3 5 --- 6 4, 5 7 1, 2, 6

4.



Start

A 0 4 4 0 4

B 4 6 2 4 6

C 6 9 3 6 9

D 9 12 12 9 21

E 0 10 10 11 21

F 21 27 6 21 27

G 21 26 5 22 27

Finish

and the critical path is A-B-C-D-F 5.

Activity


Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A -- 5 0 0 5 5 0 Yes B -- 4 0 1 4 5 1 C -- 8 0 0 8 8 0 Yes D A, B 4 5 5 9 9 0 Yes E B, C 12 8 8 20 20 0 Yes F D 10 9 9 19 19 0 Yes G A, F 3 19 19 22 22 0 Yes H E, F 2 20 20 22 22 0 Yes I G, H 5 22 22 27 27 0 Yes

The probability is less than 50% because 25 weeks is less than the mean time of 27 weeks.

6.

Activity


Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A -- 5 0 13 5 18 13 B -- 4 0 14 4 18 14 C -- 8 0 0 8 8 0 Yes D A, B 4 5 18 9 22 13 E C 12 8 8 20 20 0 Yes F C 10 8 10 18 20 2 G D, E 3 20 22 23 25 2 H E 2 20 23 22 25 3 I E, F 5 20 20 25 25 0 Yes

The probability is less than 50% because 28 weeks is more than the mean time of 25 weeks.

7.

Activity


Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A -- 8 0 0 8 8 0 Yes B -- 2 0 1 2 3 1 C B 5 2 3 7 8 1 D A, C 3 8 8 11 11 0 Yes E B 7 2 4 9 11 2 F B 5 2 7 8 13 5



G D, E 12 11 11 23 23 0 Yes H F 10 8 13 18 23 5

CRITICAL PATH: A-D-G PROJECT COMPLETION TIME = 23 8.

Activity


Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A -- 2 0 0 2 2 0 Yes B A 4 2 5 6 9 3 C A 3 2 3 5 6 1 D A 7 2 2 9 9 0 Yes E B 6 6 10 12 16 4 F B 5 6 9 11 14 3 G C 8 5 6 13 14 1 H D 5 9 9 14 14 0 Yes I E 4 12 16 16 20 4 J F, G, H 6 14 14 20 20 0 Yes

CRITICAL PATH: A-D-H-J PROJECT COMPLETION TIME = 20 9. a. and b.

Activity Expected Time Variance

A 4.67 0.44 B 3.33 1.00 C 7.33 0.44 D 11.17 2.25 E 4.00 0.11 F 9.33 0.44 G 6.00 0.44 H 5.83 0.69 I 7.50 1.36 J 13.00 0.11 K 2.83 0.25

Activity


Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

A --- 4.67 0.00 4.67 4.67 9.33 4.67 B --- 3.33 0.00 6.00 3.33 9.33 6.00 C --- 7.33 0.00 0.00 7.33 7.33 0.00 Yes D A 11.17 4.67 14.67 8.67 18.67 10.33 E A 4.00 4.67 14.67 8.67 18.67 10.00 F A, B 9.33 4.67 9.33 14.00 18.67 4.67 G C 6.00 7.33 12.67 13.33 18.67 5.33 H C 5.83 7.33 7.33 13.17 13.17 0.00 Yes I E, F, G 7.50 14.00 18.67 21.50 26.17 4.67 J H 13.00 13.17 13.17 26.17 26.17 0.00 Yes K D, J 2.83 26.17 26.17 29.00 29.00 0.00 Yes



CRITICAL PATH: C-H-J-K EXPECTED PROJECT COMPLETION TIME = 29 VARIANCE OF PROJECT COMPLETION TIME = 1.5 c. The probability that the project will take more than 28 days is P(z > (28 - 29) / 1.22) = P(z > -.82) = .7939 10. Let Ei = the earliest finish time for activity i Let Ci = the amount to crash activity i Min 1000CA + 625CB + 1000CD + 2000CE + 3000CF + 1500CG s.t. EA ≥ 0 + 12 - CA EB ≥ 0 + 14 - CB EC ≥ 0 + 8 - CC ED ≥ EA + 5 - CD EE ≥ EA + 4 - CE EF ≥ EE + 6 - CF EF ≥ EC + 6 - CF EG ≥ EC + 10 - CG EA ≤ 4 EB ≤ 4 ED ≤ 2 EE ≤ 1 EF ≤ 1 EG ≤ 2 11. [Activity, ES, EF, LS, LF, Slack]

[A, 0, 4, 0, 4, 0]; [B, 4, 7, 4, 7, 0]; [C, 4, 8, 6, 10, 2]; [D, 7, 9, 8, 10, 1]; [E, 7, 12, 7, 12, 0] [F, 9, 11, 10, 12, 1]; [G, 12, 17, 12, 17, 0]; [H, 9, 15, 11, 17, 2]

12. a. 16 weeks = Aug. 21 b. About Sept. 22 (20.66 weeks from May 1)

13. a.

START FINISHPAINTCEILING

PAINTWALLS

FLOORBUFFING

FLOORSANDING

PAINTMIXING

b. 8 hrs. c. .8264

14. a.



7 37 7 37

D30

Start

0 7 0 7

A7

23 34 26 37

H11

23 4833 58

I25

37 58 37 58

K21

11 23 14 26

F12

0 10 4 14

B10

Finish

11 26 18 33

G15

29 54 33 58

L25

23 29 27 33

J6

7 1110 14

C4

7 14 19 26

E7

Expected Completion Time = 58; Critical Path = A - D - K

b. Yes c. Yes d. Slack on G = 7 hours; Slack on L = 4 hours e. 4 hours; the slack times are not independent as G and L are on the same path.

15.

B

Start

D

E

A

Finish

H

C

F

G

Define Xi = earliest finish time for activity i Yi = the amount of time activity i is crashed MIN 2000YA+2000YC+500YD+500YE+2000YF+5000YG+6000YH S.T. XA > 0 + (6 - YA)

XB > 0 + 3 XC > XA + (9 - YC) XD > XA + (4 - YD) XE > XB + (2 - YE) XF > XB + (3 - YF) XG < 16 XG > XC + (5 - YG) XG > XE + (5 - YG)

XH > XD + (7 - YH) XH > XF + (7 - YH) XH < 16 YA < 2 YC < 3 YD < 2 YE < 1 YF < 2 YG < 1



YH < 2 Xi, Yj > 0 for all i

Solution: XA = 4, XB = 6, XC = 11, XD = 9, XE = 11, XF = 9, XG = 16, XH = 16, YA = 2, YC = 2, YD = 0, YE = 0, YF = 0, YG = 0, YH = 0, Total crash cost = $8,000.


1

15

Simulation

MULTIPLE CHOICE

1. A simulation model uses the mathematical expressions and logical relationships of the

a. real system. b. computer model. c. performance measures. d. estimated inferences.

ANSWER: a TOPIC: Introduction 2. Values for the probabilistic inputs to a simulation

a. are selected by the decision maker. b. are controlled by the decision maker. c. are randomly generated based on historical information. d. are calculated by fixed mathematical formulas.

ANSWER: c TOPIC: Introduction 3. A quantity that is difficult to measure with certainty is called a

a. risk analysis. b. project determinant. c. probabilistic input. d. profit/loss process.

ANSWER: c TOPIC: Risk analysis 4. A value for probabilistic input from a discrete probability distribution

a. is the value given by the RAND() function. b. is given by matching the probabilistic input with an interval of random numbers. c. is between 0 and 1. d. must be non-negative.

ANSWER: b TOPIC: Simulation approach


2 Chapter 15 Simulation

5. The number of units expected to be sold is uniformly distributed between 300 and 500. If r is a random number between 0 and 1, then the proper expression for sales is a. 200(r) b. r + 300 c. 300 + 500(r) d. 300 + r(200)

ANSWER: d TOPIC: Simulation approach 6. When events occur at discrete points in time

a. a simulation clock is required. b. the simulation advances to the next event. c. the model is a discrete-event simulation. d. All of the alternatives are correct.

ANSWER: d TOPIC: Discrete-event simulation 7. If customer 2 has a service time of 1.6, and if customer 3 has an interarrival time of 1.1 and a service time

of 2.3, when will customer 3’s service be completed? a. 5.0 b. 3.9 c. 3.4 d. There is not enough information to answer.

ANSWER: d TOPIC: Waiting line simulation 8. Common features of simulations--generating values from probability distributions, maintaining records,

recording data and summarizing results--led to the development of a. Excel and Lotus. b. BASIC, FORTRAN, PASCAL, and C. c. GPSS, SIMSCRIPT, SLAM, and Arena d. LINDO and The Management Scientist

ANSWER: c TOPIC: Computer implementation 9. In order to verify a simulation model

a. compare results from several simulation languages. b. be sure that the procedures for calculations are logically correct. c. confirm that the model accurately represents the real system. d. run the model long enough to overcome initial start-up results.

ANSWER: b TOPIC: Verification and validation 10. Simulation

a. does not guarantee optimality. b. is flexible and does not require the assumptions of theoretical models. c. allows testing of the system without affecting the real system. d. All of the alternatives are correct.

ANSWER: d TOPIC: Advantages and disadvantages 11. A simulation model used in situations where the state of the system at one point in time does not affect the

state of the system at future points in time is called a a. dynamic simulation model. b. static simulation model. c. steady-state simulation model.


Chapter 15 Simulation 3

d. discrete-event simulation model. ANSWER: b TOPIC: Static simulation models 12. The process of determining that the computer procedure that performs the simulation calculations is

logically correct is called a. implementation. b. validation. c. verification. d. repetition.

ANSWER: c TOPIC: Other simulation issues 13. Numerical values that appear in the mathematical relationships of a model and are considered known and

remain constant over all trials of a simulation are a. parameters. b. probabilistic input. c. controllable input. d. events.

ANSWER: a TOPIC: Other simulation issues 14. Which of the following statements is INCORRECT regarding the disadvantages of simulation?

a. Each simulation run only provides a sample of how the real system will operate. b. The summary of the simulation data only provides estimates about the real system. c. The process of developing a simulation model of a complex system can be time-consuming. d. The larger the number of probabilistic inputs a system has, the less likely a simulation will provide

the best approach for studying the system. ANSWER: d TOPIC: Advantages and disadvantages of using simulation 15. Which of the following statements is INCORRECT regarding the advantages of simulation?

a. Simulation is relatively easy to explain and understand. b. Simulation guarantees an optimal solution. c. Simulation models are flexible. d. A simulation model provides a convenient experimental laboratory for the real system.

ANSWER: b TOPIC: Advantages and disadvantages of using simulation 16. The word “uniform” in the term “uniform random numbers” means

a. all the numbers have the same number of digits. b. if one number is, say, 10 units above the mean, the next number will be 10 units below the mean. c. all the numbers are odd or all are even. d. each number has an equal probability of being drawn.

ANSWER: d TOPIC: Random numbers and generating probabilistic input values 17. A table of uniformly distributed random numbers should be read

a. from left to right. b. from top to bottom. c. diagonally, starting from the top left corner and moving to the bottom right. d. in any consistent sequence.

ANSWER: d TOPIC: Random numbers and generating probabilistic input values 18. The process of generating probabilistic inputs and computing the value of the output is called



a. simulation. b. verification. c. validation. d. implementation.

ANSWER: a TOPIC: Simulation

TRUE/FALSE

1. Simulation is an excellent technique to use when a situation is too complicated to use standard analytical

procedures. ANSWER: True TOPIC: Introduction 2. Simulation is a trial-and-error approach to problem solving. ANSWER: True TOPIC: Introduction 3. The degree of risk is associated with the probability or magnitude of loss. ANSWER: True TOPIC: Simulation approach 4. To use Excel to generate a normally distributed random variable, you must know the mean and standard

deviation of the distribution and have a random number between 0 and 1. ANSWER: True TOPIC: Simulation approach 5. Trials of a simulation show what would happen when values of the probabilistic input change. ANSWER: True TOPIC: Simulation approach 6. In a Monte Carlo simulation, each simulation trial is dependent upon the result of a previous trial. ANSWER: False TOPIC: Monte Carlo simulation 7. Verification is the process of ensuring that the simulation model provides an accurate representation of the

real system. ANSWER: False TOPIC: Verification and validation 8. In comparing different policies using simulation, one should use the same set of random numbers whenever

possible. ANSWER: True TOPIC: Waiting line simulation 9. Validation determines that the computer procedure is operating as it is intended to operate. ANSWER: False TOPIC: Verification and validation 10. A discrete-event simulation reviews the status of the system periodically, whether or not an event occurs. ANSWER: True



TOPIC: Waiting line simulation 11. A static simulation model is used in situations where the state of the system affects how the system changes

or evolves over time. ANSWER: False TOPIC: Waiting line simulation 12. For any waiting line system, (Average number of units in waiting line) = (Total waiting time) divided by

(Total time of simulation). ANSWER: True TOPIC: Waiting line simulation 13. The parameters of a simulation model are the controllable inputs. ANSWER: False TOPIC: Controllable inputs 14. Using simulation to perform risk analysis is like playing out many what-if scenarios by randomly

generating values for the probabilistic inputs. ANSWER: True TOPIC: Risk analysis 15. Computer-generated random numbers are normally distributed over the interval from 0 to 1. ANSWER: False TOPIC: Random numbers and generating probabilistic input values 16. Computer-generated random numbers are not technically random. ANSWER: True TOPIC: Random numbers and generating probabilistic input values 17. Simulation is an optimization technique. ANSWER: False TOPIC: Introduction 18. Simulation models that must take into account how the system changes or evolves over time are referred to

as dynamic simulation models. ANSWER: True TOPIC: Waiting line simulation 19. Computer-generated random numbers are normally distributed. ANSWER: False TOPIC: Random numbers and generating probabilistic input values 20. Each simulation run provides only a sample of how the real system will operate. ANSWER: True TOPIC: Advantages and disadvantages of using simulation

SHORT ANSWER

1. Simulation is to be used to study customer waiting patterns at several branches of an organization.

Acknowledging that arrivals and service times follow different distributions over the branches, of what use is the development of a general simulation model?



TOPIC: Advantages and disadvantages 2. Why would one want to use a general purpose programming language rather than a spreadsheet to develop

a simulation? TOPIC: Computer implementation 3. How are both analysts and managers involved in the validation process? TOPIC: Verification and validation 4. How can historical information be used to create discrete probability distributions? TOPIC: Simulation approach 5. Why is a flowchart useful in simulation? TOPIC: Simulation approach PROBLEMS 1. For the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the

nearest 10).

Units Sold Number of Times 30 8 40 12 50 15 60 10 70 5

a. Determine the relative frequency for each number of units sold. b. Suppose that the following random numbers were obtained using Excel:

.12 .96 .53 .80 .95 .10 .40 .45 .77 .29 Use these random numbers to simulate 10 days of sales.

TOPIC: Simulation approach: discrete-event 2. The drying rate in an industrial process is dependent on many factors and varies according to the following

distribution.

Minutes Relative Frequency 3 .14 4 .30 5 .27 6 .18 7 .11

a. Compute the mean drying time. b. Using these random numbers, simulate the drying time for 12 processes.

.33 .09 .19 .81 .12 .88 .53 .95 .77 .61 .91 .47



c. What is the average drying time for the 10 processes you simulated? TOPIC: Simulation approach: discrete-event 3. Greenfields is a mail order seed and plant business. The size of orders is uniformly distributed over the

interval from $25 to $80. Use the following random numbers to generate the size of 10 orders.

.41 .99 .07 .05 .38 .77 .19 .12 .58 .60 TOPIC: Simulation approach: uniform distribution 4. The time required to set up lighting for a portrait studio is uniformly distributed between 12 and 20

minutes. Use the following random numbers to generate the setup time for 10 customers.

.27 .53 .06 .92 .16 .74 .06 .29 .82 .23 TOPIC: Simulation approach: uniform distribution 5. Estimates of the financial information for a new product show the following information:

Units Sold Probability Fixed cost $8,000 600 .35 Variable cost $6 / unit 800 .45 Revenue $22 / unit 1000 .20

Use the random numbers .51, .97, .58, .22, and .16 to simulate five trials. What is the net profit for each

trial? TOPIC: Simulation approach: profit 6. Seventy-five percent of calls arriving at a help line can be handled by the person who answers the phone,

but the remaining 25% of them will need to be referred to someone else. Assume that every call requires one minute of attention by the person who answers the phone (either to answer the question or to figure out how the referral should be handled). Calls that are referred need an additional amount of time, as given in the table below.

Time Required Probability

3 minutes .25 5 minutes .35 10 minutes .40

Callers are served on a first come, first served basis, and are put on hold until the line is free. Use the

random numbers to simulate what happens to 10 callers. (Use the random numbers in order - from left to right, first row first – as you need them.) What percentage of your callers needs to be referred? Of those who had to be referred, what is the average referral time?

.82 .39 .16 .79 .56 .62 .13 .04 .42 .81 .85 .32 .64 .90 .73 .02 .76 .03 .86 .67

TOPIC: Simulation approach: two events



7. An airline reservation system first asks customers whether they want to schedule a domestic or an international flight. Sixty-five percent of the reservations are for domestic flights. The time distribution of advance sales is also important, and it is given below.

Domestic Flights International Flights

Make Reservations Rel. Freq. RN Range Make Reservations Rel. Freq. RN Range Less than 1 week in advance

.25 Less than 1 week in advance

.12

1 week to 2 months in advance

.35 1 week to 2 months in advance

.35

Over 2 months in advance

.40 2 months to 6 months in advance

.40

Over 6 months In advance

.13

Flight Type Rel. Freq. RN Range Domestic .65 International .35

a. Place the appropriate random number ranges in the tables above. b. Set up and perform a simulation for three customers. Determine whether they want a domestic or

international flight, and how far in advance the reservation is being made. Use random numbers from this list: .632 .715 .998 .671 .744 .021

TOPIC: Simulation approach: two events 8. On a visit to an amusement park you pass someone who has just ridden a roller coaster and asks you for

directions to the First Aid Station. Realizing that traffic at the First Aid Station would be something to study with simulation, you gather some information. Two EMTs staff the station, and patients wait and go to the first one available. People coming there can be divided into two groups: those who need something minor (e.g. Tylenol, a band-aid) or those who need more help. Assume those in the first group constitute 25% of the patients and take 5 minutes to have their problem solved. Those in the second group need an uncertain amount of time, as given by a probability distribution. Develop a flowchart for this simulation problem.

TOPIC: Simulation flowchart 9. Using the spreadsheet below, give the cell address which would have the formula shown.

Cell Formula Belongs in Cell

=VLOOKUP(B18,$B$10:$C$12,2)

=VLOOKUP(D23,$F$11:$G$14,2)

=K19*($I$16-I19)

=VLOOKUP(H27,$B$10:$C$12,2)

=AVERAGE(L18:L27)



TOPIC: Simulation with Excel NOTE TO INSTRUCTOR: The following problem requires the use of a computer and is suitable for a lab or take-

home exam. 10. Arrivals to a truck repair facility have an interarrival time that is uniformly distributed between 20 and 50

minutes. Service times are normally distributed with mean 30 minutes and standard deviation 10 minutes. Develop a spreadsheet model to simulate the arrival of 100 trucks. Collect information on the time the repair facility is idle and on the average waiting time for trucks.

TOPIC: Spreadsheet simulation, uniform and normal distributions 11. As the owner of a rent-a-car agency you have determined the following statistics:

Potential Rentals Daily

Probability

Rental Duration

Probability

0 .10 1 day .50 1 .15 2 day .30 2 .20 3 days .15 3 .30 4 days .05 4 .25

The gross profit is $40 per car per day rented. When there is demand for a car when none is available there is a goodwill loss of $80 and the rental is lost. Each day a car is unused costs you $5 per car. Your firm initially has 4 cars.



a. Conduct a 10-day simulation of this business using Row #1 below for demand and Row #2 below for rental length. Row #1: 63 88 55 46 55 69 13 17 36 81

Row #2: 59 09 57 87 07 92 29 28 64 36 b. If your firm can obtain another car for $200 for 10 days, should you take the extra car?

TOPIC: Simulation approach: two events 12. Susan Winslow has two alternative routes to travel from her home in Olport to her office in Lewisburg.

She can travel on Freeway 5 to Freeway 57 or on Freeway 55 to Freeway 91. The time distributions are as follows:

Freeway 5 Freeway 57 Freeway 55 Freeway 91

Relative Relative Relative Relative Time Frequency Time Frequency Time Frequency Time Frequency

5 .30 4 .10 6 .20 3 .30 6 .20 5 .20 7 .20 4 .35 7 .40 6 .35 8 .40 5 .20 8 .10 7 .20 9 .20 6 .15 8 .15

Do a five-day simulation of each of the two combinations of routes using the random numbers below. Based on this simulation, which routes should Susan take if her objective is to minimize her total travel time?

Freeway 5 63 88 55 46 55 Freeway 57 59 09 57 87 07 Freeway 55 71 95 83 44 34 Freeway 91 51 79 09 67 15

TOPIC: Simulation approach: multiple events 13. Three airlines compete on the route between New York and Los Angeles. Stanton Marketing has

performed an analysis of first class business travelers to determine their airline choice. Stanton has modeled this choice as a Markov process and has determined the following transition probabilities.

Next Airline

Last Airline A B C A .50 .30 .20 B .30 .45 .25 C .10 .35 .55

a. Show the random number assignments that can be used to simulate the first class business

traveler's next airline when her last airline is A, B, and C. b. Assume the traveler used airline C last. Simulate which airline the traveler will be using over her

next 25 flights. What percent of her flights are on each of the three airlines? Use the following random numbers, going from left to right, top to bottom.

71 95 83 44 34 49 88 56 05 39 75 12 03 59 29



77 76 57 15 53 37 46 85 24 53


1. a.

Units Sold

Relative Frequency

Cumulative Frequency

Interval of Random Numbers

30 .12 .12 0.00 but less than 0.12 40 .24 .36 0.12 but less than 0.36 50 .32 .68 0.36 but less than 0.68 60 .26 .94 0.68 but less than 0.94 70 .06 1.00 0.94 but less than 1.00

b.

Random Number Units Sold .12 40 .96 70 .53 50 .80 60 .95 70 .10 30 .40 50 .45 50 .77 60 .29 40

2. a. 4.82 b.

Random Number Drying Time .33 4 .09 3 .19 4 .81 6 .12 3 .88 6 .53 5 .95 7 .77 6 .61 5 .91 7 .47 5

c. The average is 5.083



3.

Random Number Order size0.41 47.550.99 79.450.07 28.850.05 27.750.38 45.900.77 67.350.19 35.450.12 31.600.58 56.900.60 58.00

4.

Random Number Time0.27 14.160.53 16.240.06 12.480.92 19.360.16 13.280.74 17.920.06 12.480.29 14.320.82 18.560.23 13.84

5.

Trial 1 2 3 4 5Random Number 0.51 0.97 0.58 0.22 0.16Units sold 800 1000 800 600 600Revenue 17600 22000 17600 13200 13200Variable cost 4800 6000 4800 3600 3600Fixed cost 8000 8000 8000 8000 8000Net Profit 4800 8000 4800 1600 1600

6.

Call RN Referred? RN Time1 0.82 yes 0.39 52 0.16 no3 0.79 yes 0.56 54 0.62 no5 0.13 no6 0.04 no7 0.42 no8 0.81 yes 0.85 109 0.85 yes 0.32 510 0.64 no



40% of the callers were referred. The average referral time was 6.25 minutes. 7. a.

Domestic Flights International Flights Make Reservations Rel. Freq. RN Range Make Reservations Rel. Freq. RN Range Less than 1 week in advance

.25 .00 to < .25 Less than 1 week in advance

.12 .00 to < .12

1 week to 2 months in advance

.35 .25 to < .60 1 week to 2 months in advance

.35 .12 to < .47


.40 .6 to < 1.00 2 months to 6 months in advance

.40 .47 to < .87


.13 .87 to < 1.00

Flight Type Rel. Freq. RN Range Domestic .65 .00 to < .65 International .35 .65 to < 1.00

b.

Observation RN Flight Type RN Booking 1 .632 Domestic .715 Over 2 months 2 .998 International .671 2 to 6 months 3 .774 International .021 Less than 1 week

8. Let i = patient counter AT(i) = arrival time IAT(i) = interarrival time Start(i) = time patient i begins service ST(i) = service time Finish(i) = time patient leaves CT1 = time that EMT 1 completes current service CT2 = time that EMT 2 completes current service



Initialize Model i = 0, all above = 0

New arrival: i = i + 1

Simulate IAT(i)

AT(i) = AT(i-1) + IAT(i)

Simulate type of service

Is need minor? yes no ST(i) = 5 Simulate ST(i) yes no

Is AT(i) > CT1? Is AT(i) > CT2? yes no Enter EMT1 Enter EMT2 Start(i) = AT(i) Start(i) = AT(i) Is CT1 < CT2? Finish(i) = Start(i) + ST(i) Finish(i) = Start(i) + ST(i) CT1 = Finish(i) CT2 = Finish(i) yes no Start(i) = CT1 Start(i) = CT2 Finish(i)=Start(i)+ST(i) Finish(i)=Start(i)+ST(i) CT1 = Finish(i) CT2 = Finish(i)

Next arrival



9. Cell Formula Belongs in Cell

=VLOOKUP(B18,$B$10:$C$12,2) C18 =VLOOKUP(D23,$F$11:$G$14,2) E23 =K19*($I$16-I19) L19 =VLOOKUP(H27,$B$10:$C$12,2) I27 =AVERAGE(L18:L27) L28

10. Representative formulas for cell references are shown in comments.

A B C D E F G H I J K12 Problem 10 Solution34 Interarrival time is uniform (20, 50)56 Service time is normal (30, 10)78 Start the clock at 0 minutes9

1011 Arrival RN IAT AT RN ST Wait for Start at Depart at Idle12 1 0.01 20.24 20.24 0.96 47.38 0.00 20.24 67.62 20.2413 2 0.00 20.11 40.35 0.66 34.14 27.27 67.62 74.50 0.0014 3 0.36 30.68 71.03 0.76 37.02 3.46 74.50 108.05 0.0015 4 0.06 21.87 92.90 0.96 47.73 15.15 108.05 140.63 0.0016 5 0.55 36.52 129.42 0.97 49.24 11.22 140.63 178.66 0.0017 6 0.04 21.27 150.69 0.66 34.21 27.97 178.66 184.90 0.0018 7 0.55 36.44 187.12 0.73 36.18 0.00 187.12 223.30 2.2319 8 0.78 43.29 230.41 0.12 18.31 0.00 230.41 248.71 7.1120 9 0.47 34.10 264.51 0.70 35.20 0.00 264.51 299.71 15.7921 10 0.77 43.06 307.56 0.33 25.50 0.00 307.56 333.06 7.8522232425

C12:=20 + b12*30

D14:=d13 + c14

F12:=norminv(e12,30,10)

J13:=h13 + i12

G14:=h14 + d14

H16:=max(d16,i15)

I19:=d19 + f19

11. a. 10-day profit is $885 b. 10-day profit is $970. Take the extra car. 12. Freeways 5-57 have 62 minutes; Freeways 55-91 have 61 minutes; Select 55-91. 13. a.

A Last B Last C Last RNs RNs RNs

A Next 00 - 49 A Next 00 - 29 A Next 00 - 09 B Next 50 - 79 B Next 30 – 74 B Next 10 - 44 C Next 80 - 99 C Next 75 - 99 C Next 45 - 99

b. 20% of flights on Airline A 48% of flights on Airline B 32% of flights on Airline C


601

CHAPTER 16 SIMPLE LINEAR REGRESSION AND CORRELATION

SECTIONS 1 - 2

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 1. The regression line y = 3 + 2x has been fitted to the data points (4, 8), (2, 5), and (1, 2).

The sum of the squared residuals will be: a. 7 b. 15 c. 8 d. 22 ANSWER: d

2. If an estimated regression line has a y-intercept of 10 and a slope of 4, then when x = 2

the actual value of y is: a. 18 b. 15 c. 14 d. unknown ANSWER: d


602 Chapter Sixteen

3. Given the least squares regression line y = 5 –2x: a. the relationship between x and y is positive b. the relationship between x and y is negative c. as x increases, so does y d. as x decreases, so does y ANSWER: b

4. A regression analysis between weight (y in pounds) and height (x in inches) resulted in

the following least squares line: y = 120 + 5x. This implies that if the height is increased by 1 inch, the weight, on average, is expected to: a. increase by 1 pound b. decrease by 1 pound c. increase by 5 pounds d. increase by 24 pounds ANSWER: c

5. A regression analysis between sales (in $1000) and advertising (in $100) resulted in the

following least squares line: y = 75 +6x. This implies that if advertising is $800, then the predicted amount of sales (in dollars) is: a. $4875 b. $123,000 c. $487,500 d. $12,300 ANSWER: b

6. A regression analysis between sales (in $1000) and advertising (in $) resulted in the

following least squares line: y = 80,000 + 5x. This implies that an: a. increase of $1 in advertising is expected, on average, to result in an increase of $5 in

sales b. increase of $5 in advertising is expected, on average, to result in an increase of $5,000

in sales c. increase of $1 in advertising is expected, on average, to result in an increase of

$80,005 in sales d. increase of $1 in advertising is expected, on average, to result in an increase of $5,000

in sales ANSWER: d

7. Which of the following techniques is used to predict the value of one variable on the

basis of other variables? a. Correlation analysis b. Coefficient of correlation c. Covariance d. Regression analysis ANSWER: d


Simple Linear Regression and Correlation 603

8. The residual is defined as the difference between: a. the actual value of y and the estimated value of y b. the actual value of x and the estimated value of x c. the actual value of y and the estimated value of x d. the actual value of x and the estimated value of y ANSWER: a

9. In the simple linear regression model, the y-intercept represents the:

a. change in y per unit change in x b. change in x per unit change in y c. value of y when x = 0 d. value of x when y = 0 ANSWER: c

10. In the first order linear regression model, the population parameters of the y-intercept and

the slope are estimated respectively, by: a. 0b and 1b b. 0b and 1β c. 0β and 1b d. 0β and 1β ANSWER: a

11. In the simple linear regression model, the slope represents the:

a. value of y when x = 0 b. average change in y per unit change in x c. value of x when y = 0 d. average change in x per unit change in y ANSWER: b

12. In regression analysis, the residuals represent the:

a. difference between the actual y values and their predicted values b. difference between the actual x values and their predicted values c. square root of the slope of the regression line d. change in y per unit change in x ANSWER: a

13. In the first-order linear regression model, the population parameters of the y-intercept and

the slope are, respectively, a. 0b and 1b b. 0b and 1β c. 0β and 1b d. 0β and 1β ANSWER: d


604 Chapter Sixteen

14. In a simple linear regression problem, the following statistics are calculated from a sample of 10 observations: ∑ −− ))(( yyxx = 2250, xs = 10, ∑x = 50, ∑ y = 75. The least squares estimates of the slope and y-intercept are respectively: a. 1.5 and 0.5 b. 2.5 and 1.5 c. 1.5 and 2.5 d. 2.5 and –5.0 ANSWER: d

15. If a simple linear regression model has no y-intercept, then:

a. all values of x are zero b. all values of y are zero c. when y = 0 so does x d. when x = 0 so does y ANSWER: d

16. In the least squares regression line y = 3 - 2x, the predicted value of y equals:

a. 1.0 when x = -1.0 b. 2.0 when x = 1.0 c. 2.0 when x = -1.0 d. 1.0 when x = 1.0 ANSWER: d

17. The least squares method for determining the best fit minimizes:

a. total variation in the dependent variable b. sum of squares for error c. sum of squares for regression d. All of the above ANSWER: b

18. What do we mean when we say that a simple linear regression model is “statistically”

useful? a. All the statistics computed from the sample make sense b. The model is an excellent predictor of y c. The model is “practically” useful for predicting y d. The model is a better predictor of y than the sample y ANSWER: d



TRUE / FALSE QUESTIONS 19. An inverse relationship between an independent variable x and a dependent variably y

means that as x increases, y decreases, and vice versa. ANSWER: T 20. A direct relationship between an independent variable x and a dependent variably y

means that the variables x and y increase or decrease together. ANSWER: T 21. Another name for the residual term in a regression equation is random error. ANSWER: T 22. A simple linear regression equation is given by 5.25 3.8y x= + . The point estimate of y

when x = 4 is 20.45. ANSWER: T 23. The vertical spread of the data points about the regression line is measured by the y-

intercept. ANSWER: F 24. The method of least squares requires that the sum of the squared deviations between

actual y values in the scatter diagram and y values predicted by the regression line be minimized.

ANSWER: T 25. A regression analysis between sales (in $1000) and advertising (in $) resulted in the

following least squares line: y = 60 + 5x. This implies that an increase of $1 in advertising is expected to result in an increase of $65 in sales.

ANSWER: F 26. A regression analysis between weight ( y in pounds) and height ( x in inches) resulted in

the following least squares line: y = 135 + 6 x . This implies that if the height is increased by 1 inch, the weight is expected to increase by an average of 6 pounds.

ANSWER: T 27. The residual ir is defined as the difference between the actual value iy and the estimated

value ˆiy . ANSWER: T 28. The regression line y = 2 + 3x has been fitted to the data points (4,11), (2,7), and (1,5).

The sum of squares for error will be 10.0. ANSWER: T


606 Chapter Sixteen

29. A regression analysis between sales (in $1000) and advertising (in $100) resulted in the following least squares line: y = 77 +8x. This implies that if advertising is $600, then the predicted amount of sales (in dollars) is $125,000.

ANSWER: T 30. The residuals are observations of the error variable ε . Consequently, the minimized sum

of squared deviations is called the sum of squares for error, denoted SSE. ANSWER: T 31. Statisticians have shown that sample y -intercept 0b and sample slope coefficient 1b are

unbiased estimators of the population regression parameters 0β and 1β , respectively. ANSWER: T 32. If cov(x, y) = 7.5075 and 2

xs = 3.5, then the sample slope coefficient is 2.145. ANSWER: T 33. The first – order linear model is sometimes called the simple linear regression model. ANSWER: T 34. To create a deterministic model, we start with a probabilistic model that approximates the

relationship we want to model. ANSWER: F 35. The residual represents the discrepancy between the observed dependent variable and its Predicted or estimated average value. ANSWER: T



STATISTICAL CONCEPTS & APPLIED QUESTIONS FOR QUESTIONS 36 AND 37, USE THE FOLLOWING NARRATIVE: Narrative: Car Speed and Gas Mileage An economist wanted to analyze the relationship between the speed of a car (x) and its gas mileage (y). As an experiment a car is operated at several different speeds and for each speed the gas mileage is measured. These data are shown below. Speed 25 35 45 50 60 65 70 Gas Mileage 40 39 37 33 30 27 25

36. {Car Speed and Gas Mileage Narrative} Determine the least squares regression line.

ANSWER: =y 50.6563 – 0.3531x

37. {Car Speed and Gas Mileage Narrative} Estimate the gas mileage of a car traveling 70

mph. ANSWER: When x = 70, y = 25.9393 mpg 38. The following 10 observations of variables x and y were collected.

x 1 2 3 4 5 6 7 8 9 10 y 25 22 21 19 14 15 12 10 6 2

Find the least squares regression line, and the estimated value of y when x = 3 ANSWER: =y 27.733-2.389x. When x = 3, y = 20.566

39. A scatter diagram includes the following data points:

x 3 2 5 4 5 y 8 6 12 10 14

Two regression models are proposed: (1) =y 1.2 + 2.5x, and (2) =y 5.5 + 4.0x. Using

the least squares method, which of these regression models provide the better fit to the data? Why?

ANSWER: SSE = 4.95 and 593.25 for models 1 and 2, respectively. Therefore, model (1) fits the

data better than model (2).


608 Chapter Sixteen

40. Consider the following data values of variables x and y.

a. Determine the least squares regression line. b. Find the predicted value of y for x = 9. c. What does the value of the slope of the regression line tell you?

ANSWER: a. =y 0.934 + 2.637x b. When x = 9, y = 24.667 c. If x increases by one unit, y on average will increase by 2.637.

FOR QUESTIONS 41 THROUGH 45, USE THE FOLLOWING NARRATIVE: Narrative: Sunshine and Skin Cancer A medical statistician wanted to examine the relationship between the amount of sunshine (x) in hours, and incidence of skin cancer (y). As an experiment he found the number of skin cancers detected per 100,000 of population and the average daily sunshine in eight counties around the country. These data are shown below. Average Daily Sunshine 5 7 6 7 8 6 4 3 Skin Cancer per 100,000 7 11 9 12 15 10 7 5

41. {Sunshine and Skin Cancer Narrative} Determine the least squares regression line. ANSWER:

=y -1.115 + 1.846x 42. {Sunshine and Skin Cancer Narrative} Draw a scatter diagram of the data and plot the

least squares regression line on it. ANSWER:

x 2 4 6 8 10 13 y 7 11 17 21 27 36

Average Daily Sunshine Line Fit Plot

0

4

8

12

16

0 2 4 6 8 10

Average Daily Sunshine

Ski

n C

ance

r

Skin Cancer

Predicted Skin Cancer

Linear (Predicted SkinCancer)



43. {Sunshine and Skin Cancer Narrative} Estimate the number of skin cancer per 100,000 of population for 6 hours of sunshine.

ANSWER: When x = 6, y = 9.961 44. {Sunshine and Skin Cancer Narrative} What does the value of the slope of the regression

line tell you? ANSWER: If the amount of sunshine x increases by one hour, the amount of skin cancer y increases

by an average of 1.846 per 100,000 of population. 45. {Sunshine and Skin Cancer Narrative} Calculate the residual corresponding to the pair (x,

y) = (8, 15). ANSWER: e = y - y = 15 – 13.653 = 1.347 FOR QUESTIONS 46 THROUGH 49, USE THE FOLLOWING NARRATIVE: NARRATIVE: Sales and Experience The general manager of a chain of furniture stores believes that experience is the most important factor in determining the level of success of a salesperson. To examine this belief she records last month’s sales (in $1,000s) and the years of experience of 10 randomly selected salespeople. These data are listed below. Salesperson Years of Experience Sales

1 0 7 2 2 9 3 10 20 4 3 15 5 8 18 6 5 14 7 12 20 8 7 17 9 20 30 10 15 25


610 Chapter Sixteen

46. {Sales and Experience Narrative} Draw a scatter diagram of the data to determine whether a linear model appears to be appropriate.

ANSWER:

It appears that a linear model is appropriate. 47. {Sales and Experience Narrative} Determine the least squares regression line. ANSWER:

=y 8.63 + 1.0817x 48. {Sales and Experience Narrative} Interpret the value of the slope of the regression line. ANSWER:

For each additional year of experience, monthly sales of a salesperson increase by an average of $1,081.7.

49. {Sales and Experience Narrative} Estimate the monthly sales for a salesperson with 16

years of experience. ANSWER: When x =16, y = 25.94 FOR QUESTIONS 50 THROUGH 53, USE THE FOLLOWING NARRATIVE: Narrative: Income and Education A professor of economics wants to study the relationship between income (y in $1000s) and education (x in years). A random sample eight individuals is taken and the results are shown below. Education 16 11 15 8 12 10 13 14 Income 58 40 55 35 43 41 52 49

Scatter Diagram

05

101520253035

0 5 10 15 20 25

Years of Experience

Sal

es



50. {Income and Education Narrative} Draw a scatter diagram of the data to determine whether a linear model appears to be appropriate.

ANSWER:

It appears that a linear model is appropriate. 51. {Income and Education Narrative} Determine the least squares regression line.

ANSWER: =y 10.6165 + 2.9098x

52. {Income and Education Narrative} Interpret the value of the slope of the regression line.

ANSWER: For each additional year of education, the income increases by an average of $2,909.80.

53. {Income and Education Narrative} Estimate the income of an individual with 15 years of

education. ANSWER: When x = 15, y = 54.264 (in $1000s) or $54,264.0

FOR QUESTIONS 54 THROUGH 57, USE THE FOLLOWING NARRATIVE: Narrative: Game Winnings and Education An ardent fan of television game shows has observed that, in general, the more educated the contestant, the less money he or she wins. To test her belief she gathers data about the last eight winners of her favorite game show. She records their winnings in dollars and the number of years of education. The results are as follows.

Scatter Diagram

30

40

50

60

6 8 10 12 14 16 18

Years of Education

Inco

me


612 Chapter Sixteen

Contestant Years of Education Winnings 1 11 750 2 15 400 3 12 600 4 16 350 5 11 800 6 16 300 7 13 650 8 14 400

54. {Game Winnings and Education Narrative} Draw a scatter diagram of the data to

determine whether a linear model appears to be appropriate.

ANSWER:

It appears that a linear model is appropriate. 55. {Game Winnings and Education Narrative} Determine the least squares regression line.

ANSWER: =y 1735 – 89.1667x 56. {Game Winnings and Education Narrative} Interpret the value of the slope of the

regression line.

ANSWER: For each additional year of education a contestant has, his or her winnings on TV game

shows decreases by an average of approximately $89.20.

Scatter Diagram

200

400

600

800

1000

8 10 12 14 16 18

Years of Education

Win

ning

s



57. {Game Winnings and Education Narrative} Estimate the game winnings for a contestant with 15 years of education.

ANSWER: When x = 15, y = $397.50 FOR QUESTIONS 58 THROUGH 61, USE THE FOLLOWING NARRATIVE: Narrative: Movie Revenues A financier whose specialty is investing in movie productions has observed that, in general, movies with “big-name” stars seem to generate more revenue than those movies whose stars are less well known. To examine his belief he records the gross revenue and the payment (in $ millions) given to the two highest-paid performers in the movie for ten recently released movies.

Movie Cost of Two Highest Paid Performers

Gross Revenue

1 5.3 48 2 7.2 65 3 1.3 18 4 1.8 20 5 3.5 31 6 2.6 26 7 8.0 73 8 2.4 23 9 4.5 39 10 6.7 58

58. {Movie Revenues Narrative} Draw a scatter diagram of the data to determine whether a

linear model appears to be appropriate. ANSWER: It appears that a linear model is appropriate.

Scatter Diagram

01020304050607080

0 2 4 6 8 10

Payment to Top Tw o Stars

Gro

ss R

even

ue


614 Chapter Sixteen

59. {Movie Revenues Narrative} Determine the least squares regression line. ANSWER:

=y 4.225 + 8.285x 60. {Movie Revenues Narrative} Interpret the value of the slope of the regression line. ANSWER:

For each million dollar paid to the two highest paid performers, the gross revenue of the movie increases by an average of $8.285 million.

61. {Movie Revenues Narrative} Estimate the gross revenue of a movie if the two highest

paid performers received 6 million dollars. ANSWER: When x = 6, y = $53.935 million FOR QUESTIONS 62 THROUGH 65, USE THE FOLLOWING NARRATIVE: NARRATIVE: Cost of Books The editor of a major academic book publisher claims that a large part of the cost of books is the cost of paper. This implies that larger books will cost more money. As an experiment to analyze the claim, a university student visits the bookstore and records the number of pages and the selling price of twelve randomly selected books. These data are listed below.

Book Number of Pages Selling Price ($) 1 844 55 2 727 50 3 360 35 4 915 60 5 295 30 6 706 50 7 410 40 8 905 53 9 1058 65 10 865 54 11 677 42 12 912 58

62. {Cost of Books Narrative} Determine the least squares regression line.

ANSWER: =y 19.387 + .0414x



63. {Cost of Books Narrative} Draw a scatter diagram of the data and plot the least squares regression line on it.

ANSWER:

64. {Cost of Books Narrative} Interpret the value of the slope of the regression line.

ANSWER: For every additional page, the price of a book increases by an average of about 4 cents.

65. {Cost of Books Narrative} Estimate the selling price for a 650 pages book. ANSWER: When x = 650, y = $46.037 FOR QUESTIONS 66 THROUGH 68, USE THE FOLLOWING NARRATIVE: Narrative: Accidents and Precipitation A statistician investigating the relationship between the amount of precipitation (in inches) and the number of automobile accidents gathered data for 10 randomly selected days. The results

Day Precipitation Number of Accidents 1 0.05 5 2 0.12 6 3 0.05 2 4 0.08 4 5 0.10 8 6 0.35 14 7 0.15 7 8 0.30 13 9 0.10 7 10 0.20 10

Number of Pages Line Fit Plot

010203040506070

0 200 400 600 800 1000 1200

Number of Pages

Sel

ling

Pric

e

Selling Price

Predicted Selling Price

Linear (Predicted SellingPrice)


616 Chapter Sixteen

SUMMARY OUTPUT DESCRIPTIVE STATISTICS

Regression Statistics Age ConcertsMultiple R 0.80203 Mean 53 Mean 3.65R Square 0.64326 Standard Error 2.1849 Standard Error 0.3424Adjusted R Square 0.62344 Standard Deviation 9.7711 Standard Deviation 1.5313Standard Error 0.93965 Sample Variance 95.4737 Sample Variance 2.3447Observations 20 Count 20 Count 20

SPEARMAN RANK CORRELATION COEFFICIENT=0.8306

ANOVAdf SS MS F Significance F

Regression 1 28.65711 28.65711 32.45653 2.1082E-05Residual 18 15.89289 0.88294Total 19 44.55

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept -3.01152 1.18802 -2.53491 0.02074 -5.50746 -0.5156Age 0.12569 0.02206 5.69706 0.00002 0.07934 0.1720

66. {Accidents and Precipitation Narrative} Find the least squares regression line.

ANSWER: =y 2.3704 + 34.864x

67. {Accidents and Precipitation Narrative} Estimate the number of accidents in a day with

0.25 inches of precipitation ANSWER: When x = 0.25, y = 11.08 ≈ 11 accidents 68. {Accidents and Precipitation Narrative} What does the slope of the least squares

regression line tell you? ANSWER: For each additional inch of precipitation, the number of accidents on average increases by

34.864 (about 35 accidents). FOR QUESTIONS 69 THROUGH 73, USETHE FOLLOWING NARRATIVE: Narrative: Willie Nelson Concert At a recent Willie Nelson concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected: Age 62 57 40 49 67 54 43 65 54 41 Number of Concerts 6 5 4 3 5 5 2 6 3 1

Age 44 48 55 60 59 63 69 40 38 52 Number of Concerts 3 2 4 5 4 5 4 2 1 3

An Excel output follows :



69. {Willie Nelson Concert Narrative} Draw a scatter diagram of the data to determine whether a linear model appears to be appropriate to describe the relationship between the age and number of concerts attended by the respondents.

ANSWER:

A linear model appears to be appropriate to describe the relationship between the age and

number of concerts attended by the respondents. 70. {Willie Nelson Concert Narrative} Determine the least squares regression line.

ANSWER: =y -3.0115 + 0.1257x 71. {Willie Nelson Concert Narrative} Plot the least squares regression line on the scatter

diagram.

ANSWER:

Scatter Diagram

0

1

2

3

4

5

6

7

30 35 40 45 50 55 60 65 70 75

Age

Num

ber o

f Con

certs

Scatter Diagram with Trendline

0

1

2

3

4

5

6

7

30 35 40 45 50 55 60 65 70 75

Age

Num

ber o

f Con

certs


618 Chapter Sixteen

72. {Willie Nelson Concert Narrative} Interpret the value of the slope of the regression line.

ANSWER: For every additional year of age, the number of concerts attended increases on average by

0.1257. Equivalently we may say, for every additional 20 years of age, the number of concerts attended increases on average by about 2.50.

73. {Willie Nelson Concert Narrative} Estimate the number of Willie Nelson concerts

attended by a 64 year old person. ANSWER: When x = 64, y = 5.03 (about 5 concerts) FOR QUESTIONS 74 THROUGH 77, USE THE FOLLOWING NARRATIVE: Narrative: Oil Quality and Price Quality of oil is measured in API gravity degrees – the higher the degrees API, the higher the quality. The table shown below is produced by an expert in the field who believes that there is a relationship between quality and price per barrel.

A partial Minitab output follows: Descriptive Statistics Variable N Mean StDev SE Mean Degrees 13 34.60 4.613 1.280 Price 13 12.730 0.457 0.127 Covariances

Degrees Price Degrees 21.281667 Price 2.026750 0.208833

Oil degrees API Price per barrel (in $) 27.0 12.02 28.5 12.04 30.8 12.32 31.3 12.27 31.9 12.49 34.5 12.70 34.0 12.80 34.7 13.00 37.0 13.00 41.0 13.17 41.0 13.19 38.8 13.22 39.3 13.27



Regression Analysis Predictor Coef StDev T P Constant 9.4349 0.2867 32.91 0.000 Degrees 0.095235 0.008220 11.59 0.000 S = 0.1314 R-Sq = 92.46% R-Sq(adj) = 91.7% Analysis of Variance Source DF SS MS F P Regression 1 2.3162 2.3162 134.24 0.000 Residual Error 11 0.1898 0.0173 Total 12 2.5060 74. {Oil Quality and Price Narrative} Draw a scatter diagram of the data to determine

whether a linear model appears to be appropriate to describe the relationship between the quality of oil and price per barrel.

ANSWER:

A linear model appears to be appropriate to describe the relationship between the quality

of oil and price per barrel. 75. {Oil Quality and Price Narrative} Determine the least squares regression line.

ANSWER: =y 9.4349 + 0.095235x

Scatter Diagram

11.812

12.212.412.612.8

1313.213.4

20 25 30 35 40 45

Degrees

Pric

e


620 Chapter Sixteen

76. {Oil Quality and Price Narrative} Plot the least squares regression line on the scatter diagram.

ANSWER:

77. {Oil Quality and Price Narrative} Interpret the value of the slope of the regression line.

ANSWER: For every additional API gravity degree, the price of oil per barrel increases by an

average of 9.52 cents.

Scatter Diagram

11.812

12.212.412.612.8

1313.213.413.6

20 25 30 35 40 45

Degrees

Pric

e



SECTIONS 3 - 4

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 78. In a simple linear regression problem, the following sum of squares are produced:

∑ =− 200)( 2yyi , ∑ =− 50)ˆ( 2ii yy , and ∑ =− 150)ˆ( 2yyi . The percentage of the

variation in y that is explained by the variation in x is: a. 25% b. 75% c. 33% d. 50% ANSWER: b

79. In simple linear regression, most often we perform a two-tail test of the population slope 1β to determine whether there is sufficient evidence to infer that a linear relationship

exists. The null hypothesis is stated as: a. 0: 10 =βH b. 110 : bH =β c. rH =10 : β d. =10 : βH sρ ANSWER: a

80. Testing whether the slope of the population regression line could be zero is equivalent to

testing whether the: a. sample coefficient of correlation could be zero b. standard error of estimate could be zero c. population coefficient of correlation could be zero d. sum of squares for error could be zero ANSWER: c

81. Given that =2xs 500, 7502 =ys , cov (x, y) = 100, and n = 6, the standard error of estimate is: a. 12.247 b. 24.933 c. 30.2076 d. 11.180 ANSWER: c


622 Chapter Sixteen

82. The symbol for the population coefficient of correlation is: a. r b. ρ c. r 2 d. 2ρ ANSWER: b

83. Given that the sum of squares for error is 60 and the sum of squares for regression is 140,

then the coefficient of determination is: a. 0.429 b. 0.300 c. 0.700 d. 0.837 ANSWER: c

84. A regression line using 25 observations produced SSR = 118.68 and SSE = 56.32. The

standard error of estimate was: a. 2.1788 b. 1.5648 c. 1.5009 d. 2.2716 ANSWER: b

85. The symbol for the sample coefficient of correlation is:

a. r a. ρ b. 2r c. 2ρ ANSWER: a

86. Given the least squares regression line y = -2.48 + 1.63x, and a coefficient of

determination of 0.81, the coefficient of correlation is: a. -0.85 b. 0.85 c. -0.90 d. 0.90 ANSWER: d

87. Which value of the coefficient of correlation r indicates a stronger correlation than 0.65?

a. 0.55 b. -0.75 c. 0.60 d. -0.45 ANSWER: b



88. If the coefficient of determination is 0.975, then the slope of the regression line: a. must be positive b. must be negative c. could be either positive or negative d. None of the above. ANSWER: c

89. In regression analysis, if the coefficient of determination is 1.0, then:

a. the sum of squares for error must be 1.0 b. the sum of squares for regression must be 1.0 c. the sum of squares for error must be 0.0 d. the sum of squares for regression must be 0.0 ANSWER: c

90 The sample correlation coefficient between x and y is 0.375. It has been found out that

the p– value is 0.744 when testing : 0oH ρ = against the one-sided alternative 1 : 0H ρ < . To test the : 0oH ρ = against the two-sided alternative 1 : 0H ρ ≠ at a significance level of 0.193, the p – value is a. 0.372 b. 1.488 c. 0.256 d. 0.512 ANSWER: d

91. Correlation analysis is used to determine:

a. the strength of the relationship between x and y b. the least squares estimates of the regression parameters c. the predicted value of y for a given value of x d. the coefficient of determination ANSWER: a

92. If the coefficient of correlation is –0.80 then, the percentage of the variation in y that is

explained by the variation in x is: a. 80% b. 64% c. –80% d. –64% ANSWER: b

93. If all the points in a scatter diagram lie on the least squares regression line, then the

coefficient of correlation must be: a. 1.0 b. –1.0 c. either 1.0 or –1.0 d. 0.0 ANSWER: c


624 Chapter Sixteen

94. If the coefficient of correlation is –0.60, then the coefficient of determination is: a. -0.60 b. -0.36 c. 0.36 d. 0.40 ANSWER: c

95. In regression analysis, if the coefficient of correlation is –1.0, then:

a. the sum of squares for error is –1.0 b. the sum of squares for regression is 1.0 c. the sum of squares for error and sum of squares for regression are equal d. the sum of squares for regression and total variation in y are equal ANSWER: d

96. If the coefficient of correlation between x and y is close to 1.0, this indicates that:

a. y causes x to happen b. x causes y to happen c. both (a) and (b) d. there may or may not be any causal relationship between x and y ANSWER: d

97. For the values of the coefficient of determination listed below, which one implies the

greatest value of the sum of squares for regression given that the total variation in y is 1800? a. 0.69 b. 0.96 c. 0.58 d. 0.85 ANSWER: b

98. When all the actual and predicted values of y are equal, the standard error of estimate will

be: a. 1.0 b. –1.0 c. 0.0 d. 2.0 ANSWER: c

99. Which of the following statistics and procedures can be used to determine whether a linear model should be employed? a. The standard error of estimate b. The coefficient of determination c. The t-test of the slope d. All of the above ANSWER: d



100. In testing the hypotheses: 0: 10 =βH vs. 0: 11 ≠βH , the following statistics are available: n = 10, 8.10 −=b , 1 2.45b = ,

1bs = 1.20, and y = 6. The value of the test

statistic is: a. 2.042 b. 0.306 c. –1.50 d. -0.300 ANSWER: a

101. The standard error of estimate εs is given by:

a. SSE/(n – 2) b. )2/( −nSSE c. )2/( −nSSE

d. SSE/ 2−n ANSWER: c

102. If the standard error of estimate εs = 20 and n = 10, then the sum of squares for error,

SSE, is: a. 400 b. 3200 c. 4000 d. 40000 ANSWER: b

103. The smallest value that the standard error of estimate εs can assume is:

a. –1 b. 0 c. 1 d. –2 ANSWER: b

104. If cov(x, y) = 1260, 16002 =xs and ,12252 =ys then the coefficient of determination is:

a. 0.7875 b. 1.0286 c. 0.8100 d. 0.7656 ANSWER: c


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105. The standard error of estimate εs is a measure of the: a. variation of y around the regression line b. variation of x around the regression line c. variation of y around the mean y d. variation of x around the mean x ANSWER: a

106. The Pearson coefficient of correlation r equals 1 when there is no:

a. explained variation b. unexplained variation c. y-intercept in the model d. outliers ANSWER: b

107. In regression analysis, the coefficient of determination 2R measures the amount of

variation in y that is: a. caused by the variation in x b. explained by the variation in x c. unexplained by the variation in x d. None of the above ANSWER: b

108. If we are interested in determining whether two variables are linearly related, it is necessary to: a. perform the t-test of the slope 1β b. perform the t-test of the coefficient of correlation ρ c. either (a) or (b) since they are identical d. calculate the standard error of estimate εs ANSWER: c

109. In a regression problem the following pairs of (x,y) are given: (3,1), (3,-1), (3,0), (3,-2) and (3,2). That indicates that the: a. correlation coefficient is –1 b. correlation coefficient is 0 c. correlation coefficient is 1 d. coefficient of determination is between –1 and 1 ANSWER: b

110. In a regression problem, if the coefficient of determination is 0.95, this means that:

a. 95% of the y values are positive b. 95% of the variation in y can be explained by the variation in x c. 95% of the x values are equal d. 95% of the variation in x can be explained by the variation in y ANSWER: b



111. The sample correlation coefficient between x and y is 0.375. It has been found out that the p – value is 0.256 when testing : 0oH ρ = against the two-sided alternative

1 : 0H ρ ≠ . To test : 0oH ρ = against the one-sided alternative 1 : 0H ρ > at a significant level of 0.193, the p – value will be equal to a. 0.128 b. 0.512 c. 0.744 d. 0.872 ANSWER: a

112. In simple linear regression, which of the following statements indicate no linear

relationship between the variables x and y? a. Coefficient of determination is 1.0 b. Coefficient of correlation is 0.0 c. Sum of squares for error is 0.0 d. Sum of squares for regression is relatively large ANSWER: b

113. If the sum of squared residuals is zero, then the:

a. coefficient of determination must be 1.0 b. coefficient of correlation must be 1.0 c. coefficient of determination must be 0. 0 d. coefficient of correlation must be 0.0 ANSWER: a

114. In a regression problem, if all the values of the independent variable are equal, then the

coefficient of determination must be: a. 1.0 b. 0.5 c. 0.0 d. –1.0 ANSWER: c

115. The standard error of the estimate is a measure of

a. total variation of the y variable b. the variation around the sample regression line c. explained variation d. the variation of the x variable ANSWER: b


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116. In simple linear regression, the coefficient of correlation r and the least squares estimate 1b of the population slope 1β :

a. must be equal b. must have opposite signs c. must have the same sign d. may have opposite signs or the same sign ANSWER: c

117. The coefficient of determination ( 2R ) tells us

a. that the coefficient of correlation is larger than 1 b. whether r has any significance c. that we should not partition the total variation d. the proportion of total variation in y that is explained by x ANSWER: d

118. In performing a regression analysis involving two numerical variables, we are assuming:

a. the variances of x and yare equal b. the variation around the line of regression is the same for each x value c. that x and y are independent d. All of the above ANSWER: b

119. Which of the following assumptions concerning the probability distribution of the

random error term is stated incorrectly? a. The distribution is normal b. The mean of the distribution is 0 c. The variance of the distribution increases as x increases d. The errors are independent ANSWER: c

120. If the correlation coefficient (r) = 1.00, then

a. The y – intercept ( ob ) must equal 0 b. The explained variation equals the unexplained variation c. There is no unexplained variation d. There is no explained variation ANSWER: c

121. In a simple linear regression problem, r and 1b

a. may have opposite signs b. must have the same sign c. must have opposite signs d. must be equal ANSWER: b



122. The sample correlation coefficient between x and y is 0.375. It has been found out that the p – value is 0.256 when testing : 0oH ρ = against a two-sided alternative 1 : 0H ρ ≠ . To test : 0oH ρ = against the one-sided alternative 1 : 0H ρ < at a significance level of 0.193, the p - value will be equal to a. 0.128 b. 0.512 c. 0.744 d. 0.872 ANSWER: d

123. Which of the following in not a required condition for the error variable ε in the simple

linear regression model? a. The probability distribution of ε is normal. b. The mean of the probability distribution of ε is zero. c. The standard deviation εσ of ε is a constant no matter what the value of x. d. The values of ε are auto correlated. ANSWER: d

124. Testing for existence of correlation is equivalent to

a. testing for the existence of the slope ( 1β ) b. testing for the existence of the Y – intercept ( oβ ) c. the confidence interval estimate for predicting Y d. None of the above ANSWER: a

125. The coefficient of determination 2R measures the amount of:

a. variation in y that is explained by variation in x b. variation in x that is explained by variation in y c. variation in y that is unexplained by variation in x d. variation in x that is unexplained by variation in y ANSWER: a

126. If the coefficient of correlation is 0.90, then the percentage of the variation in the

dependent variable y that is explained by the variation in the independent variable x is: a. 90% b. 81% c. 0.90% d. 0.81% ANSWER: b


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127. If a researcher wanted to find out if alcohol consumptions and grade point average on a 4 – point scale are linearly related, he would perform a a. 2χ test for the difference in two proportions b. 2χ test for independence c. a z test for the difference in two proportions d. a t test for no linear relationship between the two variables ANSWER: d



TRUE / FALSE QUESTIONS 128. If the value of the sum of squares for error SSE equals zero, then the coefficient of

determination must equal zero. ANSWER: F 129. When the actual values y of a dependent variable and the corresponding predicted values

y are the same, the standard error of the estimate will be 1.0. ANSWER: F 130. The value of the sum of squares for regression SSR can never be smaller than 0.0. ANSWER: T 131. The value of the sum of squares for regression SSR can never be smaller than 1. ANSWER: F 132. If all the values of an independent variable x are equal, then regressing a dependent

variable y on x will result in a coefficient of determination of zero. ANSWER: T 133. In a simple linear regression model, testing whether the slope 1β of the population

regression line could be zero is the same as testing whether or not the population coefficient of correlation ρ equals zero.

ANSWER: T 134. When the actual values y of a dependent variable and the corresponding predicted values

y are the same, the standard error of estimate sε will be 0.0. ANSWER: T 135. If there is no linear relationship between two variables x and y , the coefficient of

determination must be ± 1.0. ANSWER: F 136. The value of the sum of squares for regression SSR can never be larger than the value of

sum of squares for error SSE. ANSWER: F 137. When the actual values y of a dependent variable and the corresponding predicted values

y are the same, the standard error of estimate sε will be -1.0. ANSWER: F 138. In a simple linear regression problem, the least squares line is y = -3.75 + 1.25 x , and

the coefficient of determination is 0.81. The coefficient of correlation must be –0.90. ANSWER: F


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139. In simple linear regression, the divisor of the standard error of estimate sε is n – 2. ANSWER: T 140. In a regression problem the following pairs of (x, y) are given: (4,-2), (4,-1), (4,0), (4,1)

and (4,2). That indicates that the coefficient of correlation is –1. ANSWER: F 141. The value of the sum of squares for regression SSR can never be larger than the value of

total sum of squares SST. ANSWER: T 142. In regression analysis, if the coefficient of determination is 1.0, then the coefficient of

correlation must be 1.0. ANSWER: F 143. Correlation analysis is used to determine the strength of the relationship between an

independent variable x and dependent variable y. ANSWER: T 144. If the coefficient of correlation is –0.81, then the percentage of the variation in y that is

explained by the regression line is 81%. ANSWER: F 145. If all the points in a scatter diagram lie on the least squares regression line, then the

coefficient of correlation must be 1.0. ANSWER: F 146. If the standard error of estimate sε = 20 and n = 8, then the sum of squares for error SSE

is 2,400. ANSWER: T 147. The probability distribution of the error variable ε is normal, with mean E(ε ) = 0, and

standard deviation εσ =1. ANSWER: F 148. In a simple linear regression problem, if the coefficient of determination is 0.95, this

means that 95% of the variation in the independent variable x can be explained by regression line.

ANSWER: F 149. Given that cov(x, y) = 10, 2

ys = 15, 2xs = 8, and n = 12, the value of the standard error of

estimate sε is 2.75. ANSWER: F



150. If the error variable ε is normally distributed, the test statistic for testing 0 1: 0H β = is Student t distributed with n – 2 degrees of freedom.

ANSWER: T 151. Given that cov(x, y) = 8.5, 2

ys = 8, and 2xs = 10, then the value of the coefficient of

determination is 0.95. ANSWER: F 152. The coefficient of determination is the coefficient of correlation squared. That is, 2 2R r= ANSWER: T 153. Given that SSE = 60 and SSR = 540, the proportion of the variation in y that is explained

by the variation in x is 0.90. ANSWER: T 154. Given that SSE = 84 and SSR = 358.12, the coefficient of correlation (also called the

Pearson coefficient of correlation) must be 0.90. ANSWER: F 155. Except for the values r = -1, 0, and 1, we cannot be specific in our interpretation of the

coefficient of correlation r. However, when we square it we produce a more meaningful statistic.

ANSWER: T 156. A zero population correlation coefficient between a pair of random variables means that

there is no linear relationship between the random variables. ANSWER: T 157. Given that cov(x, y) = 8, 2

ys = 14, 2xs = 10, and n = 6, the value of the sum of squares for

error SSE is 38. ANSWER: T 158. A store manager gives a pre-employment examination to new employees. The test is

scored from 1 to 100. He has data on their sales at the end of one year measured in dollars. He wants to know if there is any linear relationship between pre-employment examination score and sales. An appropriate test to use is the t test on the population correlation coefficient.

ANSWER: T


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STATISTICAL CONCEPTS & APPLIED QUESTIONS

FOR QUESTIONS 159 THROUGH 164, USE THE FOLLOWING NARRATIVE: Narrative: Car Speed and Gas Mileage An economist wanted to analyze the relationship between the speed of a car (x) and its gas mileage (y). As an experiment a car is operated at several different speeds and for each speed the gas mileage is measured. These data are shown below. Speed 25 35 45 50 60 65 70 Gas Mileage 40 39 37 33 30 27 25

159. {Car Speed and Gas Mileage Narrative} Calculate the standard error of estimate, and

describe what this statistic tells you about the regression line. ANSWER: =εs 1.448; the model’s fit to these data is good. 160. {Car Speed and Gas Mileage Narrative} Do these data provide sufficient evidence at the

5% significance level to infer that a linear relationship exists between higher speeds and lower gas mileage?

ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,10t = 2.228 Test statistic: t = -9.754 Conclusion: Reject the null hypothesis. Yes, these data provide sufficient evidence at the 5% significance level to infer that a linear relationship exists between higher speeds and lower gas mileage.

161. {Car Speed and Gas Mileage Narrative} Predict with 99% confidence the gas mileage of a car traveling 55 mph. ANSWER:

31.236 ± 6.284. Thus, LCL = 24.952, and UCL = 37.52 162. {Car Speed and Gas Mileage Narrative} Calculate the Pearson coefficient of correlation. ANSWER: r = -0.975 163. {Car Speed and Gas Mileage Narrative} What does the coefficient of correlation tell you

about the direction and strength of the relationship between the two variables? ANSWER: There is a very strong negative linear relationship between car speed and gas mileage.



164. {Car Speed and Gas Mileage} Calculate the coefficient of determination and interpret its value.

ANSWER: 2R = 0.95. This means that 95% of the total variation in gas mileage can be explained by

the speed of the car. 165. The following 10 observations of variables x and y were collected.

x 1 2 3 4 5 6 7 8 9 10 y 25 22 21 19 14 15 12 10 6 2

a. Calculate the standard error of estimate. b. Test to determine if there is enough evidence at the 5% significance level to indicate

that x and y are negatively linearly related. c. Calculate the coefficient of correlation, and describe what this statistic tells you about

the regression line. ANSWER: a. =εs 1.322 b. 0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.05,8t =1.86 Test statistic: t = -16.402

Conclusion: Reject the null hypothesis. Yes, there is enough evidence at the 5% significance level to indicate that x and y are negatively linearly related.

c. r = -0.9854. This indicates a very strong negative linear relationship between the two variables.

166. Consider the following data values of variables x and y.

x 2 4 6 8 10 13 y 7 11 17 21 27 36

a. Calculate the coefficient of determination, and describe what this statistic tells you

about the relationship between the two variables. b. Calculate the Pearson coefficient of correlation. What sign does it have? Why? c. What does the coefficient of correlation calculated Tell you about the direction and

strength of the relationship between the two variables?

ANSWER: a. =2R 0.995. This means that 99.5% of the variation in the dependent variable y is

explained by the variation in the independent variable x. b. r = 0.9975. It is positive since the slope of the regression line is positive. c. There is a very strong (almost perfect) positive linear relationship between the two

variables.


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FOR QUESTIONS 167 THROUGH 171, USE THE FOLLOWING NARRATIVE: Narrative: Sunshine and Skin Cancer A medical statistician wanted to examine the relationship between the amount of sunshine (x) and incidence of skin cancer (y). As an experiment he found the number of skin cancers detected per 100,000 of population and the average daily sunshine in eight counties around the country. These data are shown below. Average Daily Sunshine 5 7 6 7 8 6 4 3 Skin Cancer per 100,000 7 11 9 12 15 10 7 5

167. {Sunshine and Skin Cancer Narrative} Calculate the standard error of estimate, and

describe what this statistic tells you about the regression line. ANSWER: =εs 0.9608; the model’s fit to these data is good. 168. {Sunshine and Skin Cancer Narrative} Can we conclude at the 1% significance level that

there is a linear relationship between sunshine and skin cancer? ANSWER: 0:0 =ρH vs. 0:1 ≠ρH

Rejection region: | t | > 0.005,6t =3.707 Test statistic: t = 8.485

Conclusion: Reject the null hypothesis. Yes, we conclude at the 1% significance level that there is a linear relationship between sunshine and skin cancer.

169. {Sunshine and Skin Cancer Narrative} Calculate the coefficient of determination and

interpret it. ANSWER:

=2R 0.9231. This means that 92.31% of the variation in the incidence of skin cancer is explained by the variation in the amount of sunshine.

170. {Sunshine and Skin Cancer Narrative} Calculate the Pearson coefficient. What sign does

it have? Why? ANSWER: R = 0.9608. It is positive since the slope of the regression line ( 1b = 1.846) is positive. 171. {Sunshine and Skin Cancer Narrative} What does the coefficient of correlation calculated

Tell you about the direction and strength of the relationship between the two variables? ANSWER:

There is a very strong (almost perfect) positive linear relationship between the two variables.



FOR QUESTIONS 172 THROUGH 177, USE THE FOLLOWING NARRATIVE: Narrative: Sales and Experience The general manager of a chain of furniture stores believes that experience is the most important factor in determining the level of success of a salesperson. To examine this belief she records last month’s sales (in $1,000s) and the years of experience of 10 randomly selected salespeople. These data are listed below. Salesperson Years of Experience Sales

1 0 7 2 2 9 3 10 20 4 3 15 5 8 18 6 5 14 7 12 20 8 7 17 9 20 30 10 15 25

172. {Sales and Experience Narrative} Determine the standard error of estimate and describe

what this statistic tells you about the regression line. ANSWER: =εs 1.5724; the model’s fit is good.

173. (Sales and Experience Narrative} Determine the coefficient of determination and discuss

what its value tells you about the two variables.

ANSWER: 2R = 0.9536, which means that 95.36% of the variation in sales is explained by the

variation in years of experience of the salesperson. 174. {Sales and Experience Narrative} Calculate the Pearson correlation coefficient. What

sign does it have? Why?

ANSWER: =r 0.9765. It has a positive sign since the slope of the regression line ( 1b = 1.0817) is

positive.


638 Chapter Sixteen

175. {Sales and Experience Narrative} Conduct a test of the population coefficient of correlation to determine at the 5% significance level whether a linear relationship exists between years of experience and sales. ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,8t =2.306 Test statistic: t = 12.8258 Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of experience and sales.

176. {Sales and Experience Narrative} Conduct a test of the population slope to determine at the 5% significance level whether a linear relationship exists between years of experience and sales. ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.025,8t =2.306 Test statistic: t = 12.8258

Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of experience and sales.

177. {Sales and Experience Narrative} Do the tests of ρ and 1β in the previous two questions provide the same results? Explain. ANSWER:

Yes; both tests have the same value of the test statistic, the same rejection region, and of course the same conclusion. This is not a coincidence; the two tests are identical.

FOR QUESTIONS 178 THROUGH 183, USE THE FOLLOWING NARRATIVE: Narrative: Income and Education A professor of economics wants to study the relationship between income (y in $1000s) and education (x in years). A random sample eight individuals is taken and the results are shown below. Education 16 11 15 8 12 10 13 14 Income 58 40 55 35 43 41 52 49

178. {Income and Education Narrative} Determine the standard error of estimate and describe what this statistic tells you about the regression line. ANSWER:

=εs 2.436; the model’s fit to these data is good.



179. {Income and Education Narrative} Determine the coefficient of determination and discuss what its value tells you about the two variables.

ANSWER:

2R = 0.9223, which means that 92.03% of the variation in income is explained by the variation in years of education.

180. {Income and Education Narrative} Calculate the Pearson correlation coefficient. What



positive. 181. {Income and Education Narrative} Conduct a test of the population coefficient of

correlation to determine at the 5% significance level whether a linear relationship exists between years of education and income. ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,6t =2.447 Test statistic: t = 8.439

Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of education and income.

182. {Income and Education Narrative} Conduct a test of the population slope to determine at the 5% significance level whether a linear relationship exists between years of education and income.

ANSWER:

0: 10 =βH , 0: 11 ≠βH Rejection region: | t | > 0.025,6t =2.447 Test statistic: t = 8.439 Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of education and income.

183. {Income and Education Narrative} Do the tests of ρ and 1β in the previous two provide the same results? Explain.

ANSWER: Yes; both tests have the same value of the test statistic, the same rejection region, and of

course the same conclusion. This is not a coincidence; the two tests are identical.


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FOR QUESTIONS 184 THROUGH 189, USE THE FOLLOWING NARRATIVE: Narrative: Game Winnings and Education An ardent fan of television game shows has observed that, in general, the more educated the contestant, the less money he or she wins. To test her belief she gathers data about the last eight winners of her favorite game show. She records their winnings in dollars and the number of years of education. The results are as follows. Contestant Years of Education Winnings

1 11 750 2 15 400 3 12 600 4 16 350 5 11 800 6 16 300 7 13 650 8 14 400

184. {Game Winnings and Education Narrative} Determine the standard error of estimate and

describe what this statistic tells you about the regression line.

ANSWER: =εs 59.395; the model’s fit to these data is good.

185. {Game Winnings and Education Narrative} Determine the coefficient of determination

and discuss what its value tells you about the two variables.

ANSWER: 2R = 0.9185, which means that 91.85% of the variation in TV game shows’ winnings is

explained by the variation in years of education. 186. {Game Winnings and Education Narrative} Calculate the Pearson correlation coefficient.

What sign does it have? Why?

ANSWER: =r -0.9584. It has a negative sign since the slope of the regression line ( 1b = -89.1667) is

negative. 187. {Game Winnings and Education Narrative} Conduct a test of the population coefficient

of correlation to determine at the 5% significance level whether a linear relationship exists between years of education and TV game shows’ winnings. ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,6t =2.447 Test statistic: t = -8.2227



Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of education and TV game shows’ winnings.

188. {Game Winnings and Education Narrative} Conduct a test of the population slope to

determine at the 5% significance level whether a linear relationship exists between years of education and TV game shows’ winnings.

ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.025,6t =2.447 Test statistic: t = -8.2227

Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between years of education and TV game shows’ winnings.

189. {Game Winnings and Education Narrative} Do the tests ρ and 1β in the previous two

questions provide the same results? Explain. ANSWER: Yes. This is not a coincidence; the two tests are identical.

FOR QUESTIONS 190 THROUGH 195, USE THE FOLLOWING NARRATIVE: Narrative: Movie Revenues A financier whose specialty is investing in movie productions has observed that, in general, movies with “big-name” stars seem to generate more revenue than those movies whose stars are less well known. To examine his belief he records the gross revenue and the payment (in $ millions) given to the two highest-paid performers in the movie for ten recently released movies.


Gross Revenue

1 5.3 48 2 7.2 65 3 1.3 18 4 1.8 20 5 3.5 31 6 2.6 26 7 8.0 73 8 2.4 23 9 4.5 39 10 6.7 58

190. {Movie Revenues Narrative} Determine the standard error of estimate and describe what

this statistic tells you about the regression line.


642 Chapter Sixteen

ANSWER:

=εs 2.0247; the model’s fit to these is good. 191. {Movie Revenues Narrative} Determine the coefficient of determination and discuss


ANSWER: 2R = 0.9908, which means that 99.08% of the variation in gross revenue is explained by

the variation in payment to the highest performers. 192. {Movie Revenues Narrative} Calculate the Pearson correlation coefficient. What sign

does it have? Why?


positive. 193. {Movie Revenues Narrative} Conduct a test of the population coefficient of correlation

to determine at the 5% significance level whether a linear relationship exists between payment to the two highest-paid performers and gross revenue. ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,8t =2.306 Test statistic: t = 29.304

Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between payment to the two highest-paid performers and gross revenue.

194. {Movie Revenues Narrative} Conduct a test of the population slope to determine at the 5% significance level whether a linear relationship exists between payment to the two highest-paid performers and gross revenue.

ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.025,8t =2.306 Test statistic: t = 29.304 Conclusion: Reject the null hypothesis. Yes, a linear relationship exists between payment to the two highest-paid performers and gross revenue.

195. {Movie Revenues Narrative} Do the ρ and 1β tests in the previous questions provide the same results? Explain.



ANSWER: Yes; both tests have the same value of the test statistic, the same rejection region, and of course the same conclusion. This is not a coincidence; the two tests are identical.

FOR QUESTIONS 196 AND 197, USE THE FOLLOWING NARRATIVE: Narrative: Cost of Books The editor of a major academic book publisher claims that a large part of the cost of books is the cost of paper. This implies that larger books will cost more money. As an experiment to analyze the claim, a university student visits the bookstore and records the number of pages and the selling price of twelve randomly selected books. These data are listed below.


196. {Cost of Books Narrative} Determine the coefficient of determination and discuss what

its value tells you. ANSWER: 2R =0.9378, which means that 93.78% of the variation in the price of books is explained

by the variation in the number of pages. 197. {Cost of Books Narrative} Can we infer at the 5% significance level that the editor is

correct?

ANSWER: 0: 10 =βH vs. 0: 11 ≠βH

Rejection region: | t | > 0.025,10t = 2.228 Test statistic: t = 12.2814 Conclusion: Reject the null hypothesis. Yes, we can infer at the 5% significance level that the editor is correct

FOR QUESTIONS 198 THROUGH 202, USE THE FOLLOWING NARRATIVE: Narrative: Automobile Accidents and Precipitation A statistician investigating the relationship between the amount of precipitation (in inches) and the number of automobile accidents gathered data for 10 randomly selected days. The results


644 Chapter Sixteen

Day Precipitation Number of Accidents

1 0.05 5 2 0.12 6 3 0.05 2 4 0.08 4 5 0.10 8 6 0.35 14 7 0.15 7 8 0.30 13 9 0.10 7 10 0.20 10

198. {Automobile Accidents and Precipitation Narrative} Calculate the standard error of

estimate, and describe what this statistic tells you about the regression line.

ANSWER: =εs 1.3207; the model’s fit to these is good.

199. {Automobile Accidents and Precipitation Narrative} Determine the coefficient of

determination and discuss what its value tells you about the two variables.

ANSWER: 2R =0.893, which means that 89.3% of the variation in the number of accidents is

explained by the variation in the amount of precipitation. 200. {Automobile Accidents and Precipitation Narrative} Conduct a test of the population

slope to determine whether these data allow us to conclude at the 10% significance level that the amount of precipitation and the number of accidents are linearly related? ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.05,8t =1.86 Test statistic: t = 8.1709 Conclusion: Reject the null hypothesis. Yes, these data allow us to conclude at the 10% significance level that the amount of precipitation and the number of accidents are linearly related

201. {Automobile Accidents and Precipitation Narrative} Conduct a test of the population

coefficient of correlation to determine whether these data allow us to conclude at the 10% significance level that the amount of precipitation and the number of accidents are linearly related.









ANSWER: 1: 0 vs. : 0oH Hρ ρ= ≠

Rejection region: | t | > 0.05,8t =1.86 Test statistic: t = 8.1709

Conclusion: Reject the null hypothesis. Yes, these data allow us to conclude at the 10% significance level that the amount of precipitation and the number of accidents are linearly related.

202. {Automobile Accidents and Precipitation Narrative} Do the 1β and ρ tests in the

previous two questions provide the same results? Explain

ANSWER: Yes, the two tests are identical to each other. FOR QUESTIONS 203 THROUGH 208, USE THE FOLLOWING NARRATIVE: Narrative: Willie Nelson Concert At a recent Willie Nelson concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected: Age 62 57 40 49 67 54 43 65 54 41 Number of Concerts 6 5 4 3 5 5 2 6 3 1



203. {Willie Nelson Concert Narrative} Determine the standard error of estimate and describe what this statistic tells you about the model’s fit.


646 Chapter Sixteen

ANSWER:

=εs 0.9396, and since the sample mean y = 3.65, we would have to admit that the standard error of estimate is not very small. On the other hand, it is not a large number either. Because there is no predefined upper limit on sε , it is difficult in this problem to assess the model in this way. However, using other criteria, it seems that the model’s fit to these data is reasonable.

204. {Willie Nelson Concert Narrative} Determine the coefficient of determination and

discuss what its value tells you about the two variables.

ANSWER: 2R =0.64326, which means that 64.326% of the variation in number of concerts attended

is explained by the variation in age of the attendees. 205. {Willie Nelson Concert Narrative} Calculate the Pearson correlation coefficient. What


ANSWER: =r 0.80204. It has a positive sign since the slope of the regression line, 1b , is positive.

206. {Willie Nelson Concert Narrative} Conduct a test of the population coefficient of

correlation to determine at the 5% significance level whether a linear relationship exists between age and number of concerts attended. ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,18t =2.101

Test statistic: 2( 2) /(1 )t r n r= − − = 5.6971 Conclusion: Reject the null hypothesis. Yes

207. {Willie Nelson Concert Narrative} Conduct a test of the population slope to determine at

the 5% significance level whether a linear relationship exists between age and number of concerts attended.

ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.025,18t =2.101 Test statistic: t = 5.6971

Conclusion: Reject the null hypothesis. Yes, we can infer that at the 5% significance level that a linear relationship exists between age and number of concerts attended.



208. {Willie Nelson Concert Narrative} Do the ρ and 1β tests in the previous two questions provide the same results? Explain. ANSWER: Yes; both tests have the same value of the test statistic, the same rejection region, and of course the same conclusion. This is not a coincidence; the two tests are identical.

FOR QUESTIONS 209 THROUGH 214, USE THE FOLLOWING NARRATIVE: Narrative: Oil Quality and Price Quality of oil is measured in API gravity degrees – the higher the degrees API, the higher the quality. The table shown below is produced by an expert in the field who believes that there is a relationship between quality and price per barrel.

A partial statistical software output follows: Descriptive Statistics Variable N Mean StDev SE Mean Degrees 13 34.60 4.613 1.280 Price 13 12.730 0.457 0.127 Covariances

Degrees Price Degrees 21.281667 Price 2.026750 0.208833 Regression Analysis Predictor Coef StDev T P Constant 9.4349 0.2867 32.91 0.000 Degrees 0.095235 0.008220 11.59 0.000 S = 0.1314 R-Sq = 92.46% R-Sq(adj) = 91.7%



648 Chapter Sixteen

Analysis of Variance Source DF SS MS F P Regression 1 2.3162 2.3162 134.24 0.000 Residual Error 11 0.1898 0.0173 Total 12 2.5060 209. {Oil Quality and Price Narrative} Determine the standard error of estimate and describe

what this statistic tells you. ANSWER:

=εs 0.1314. Since the sample mean y = 12.73, the standard error of estimate is judged to be small, and we may say that the model fits the data well.

210. {Oil Quality and Price Narrative} Determine the coefficient of determination and discuss


ANSWER: 2R =0.9246, which means that 92.46% of the variation in the oil price per barrel is

explained by the variation in the API degrees. 211. {Oil Quality and Price Narrative} Calculate the Pearson correlation coefficient. What


ANSWER: =r 0.9616. It has a positive sign since the slope of the regression line, 1b , is positive.

212. {Oil Quality and Price Narrative} Conduct a test of the population coefficient of

correlation to determine at the 5% significance level whether a linear relationship exists between the quality of oil and price per barrel.

ANSWER:

0:0 =ρH vs. 0:1 ≠ρH Rejection region: | t | > 0.025,11t =2.201

Test statistic: 2( 2) /(1 )t r n r= − − = 11.61 Conclusion: Reject the null hypothesis. Yes, we can infer that at the 5% significance level

that a linear relationship exists between the quality of oil and price per barrel.



213. {Oil Quality and Price Narrative} Conduct a test of the population slope to determine at the 5% significance level whether a linear relationship exists between the quality of oil and price per barrel.

ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: | t | > 0.025,11t =2.201 Test statistic: t = 11.59 (from Minitab output)

Conclusion: Reject the null hypothesis. Yes, we can infer at the 5% significance level that a linear relationship exists between the quality of oil and price per barrel.

214. {Oil Quality and Price Narrative} Do the 1 and ρ β tests in the previous two questions provide the same results? Explain. ANSWER: Yes; both tests have the same value of the test statistic (the small difference between 11.61 and 11.59 is due to rounding in Minitab output), the same rejection region, and of course the same conclusion. This is not a coincidence; the two tests are identical.


650 Chapter Sixteen

SECTION 6

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 215. In order to estimate with 95% confidence the expected value of y for a given value of x

in a simple linear regression problem, a random sample of 10 observations is taken. Which of the following t-table values listed below would be used? a. 2.228 b. 2.306 c. 1.860 d. 1.812 ANSWER: b

216. Given a specific value of x and confidence level, which of the following statements is

correct? a. The confidence interval estimate of the expected value of y can be calculated but the

prediction interval of y for the given value of x cannot be calculated. b. The confidence interval estimate of the expected value of y will be wider than the

prediction interval. c. The prediction interval of y for the given value of x can be calculated but the

confidence interval estimate of the expected value of y cannot be calculated. d. The confidence interval estimate of the expected value of y will be narrower than the

prediction interval. ANSWER: d

217. In order to predict with 90% confidence the expected value of y for a given value of x in a

simple linear regression problem, a random sample of 10 observations is taken. Which of the following t-table values listed below would be used? a. 2.228 b. 2.306 c. 1.860 d. 1.812 ANSWER: c

218. The confidence interval estimate of the expected value of y for a given value y x,

compared to the prediction interval of y for the same given value of x and confidence level, will be a. wider b. narrower c. the same d. impossible to know

ANSWER: b



219. In order to predict with 99% confidence the expected value of y for a given value of x in a simple linear regression problem, a random sample of 10 observations is taken. Which of the following t-table values listed below would be used? a. 1.860 b. 2.306 c. 2.896 d. 3.355 ANSWER: d

220. The width of the confidence interval estimate for the predicted value of y depends on

a. the standard error of the estimate b. the value of x for which the prediction is being made c. the sample size d. All of the above ANSWER: d


simple linear regression problem, a random sample of 15 observations is taken. Which of the following t-table values listed below would be used? a. 1.350 b. 1.771 c. 2.160 d. 2.650 ANSWER: a


simple linear regression problem, a random sample of 15 observations is taken. Which of the following t-table values listed below would be used? a. 1.350 b. 1.771 c. 2.160 d. 2.650 ANSWER: d


652 Chapter Sixteen

TRUE / FALSE QUESTIONS 223. In developing a 95% confidence interval for the expected value of y from a simple linear

regression problem involving a sample of size 10, the appropriate table value would be 1.86.

ANSWER: F 224. In developing a 80% prediction interval for the particular value of y from a simple linear

regression problem involving a sample of size 12, the appropriate table value would be 1.372

ANSWER: T 225. In developing 90% prediction interval for the particular value of y from a simple linear


ANSWER: F 226. In order to predict with 95% confidence a particular value of y for a given value of x in

a simple linear regression problem, a random sample of 20 observations is taken. The appropriate table value that would be used is 2.101.

ANSWER: T 227. The confidence interval estimate of the expected value of y will be narrower than the

prediction interval for the same given value of x and confidence level. This is because there is less error in estimating a mean value as opposed to predicting an individual value.

ANSWER: T 228. The confidence interval estimate of the expected value of y will be wider than the

prediction interval for the same given value of x and confidence level. This is because there is more error in estimating a mean value as opposed to predicting an individual value.

ANSWER: F 229. In developing a 90% confidence interval for the expected value of y from a simple linear

regression problem involving a sample of size 15, the appropriate table value would be 1.761.

ANSWER: F 230. In developing a 99% confidence interval for the expected value of y from a simple linear


ANSWER: T 231. The prediction interval for a particular value of y is always wider than the confidence

interval for mean value of y, given the same data set, x value, and confidence level. ANSWER: T



BASIC TECHNIQUES & APPLIED QUESTIONS 232. A medical statistician wanted to examine the relationship between the amount of

sunshine (x) and incidence of skin cancer (y). As an experiment he found the number of skin cancers detected per 100,000 of population and the average daily sunshine in eight counties around the country. These data are shown below.

Average Daily Sunshine 5 7 6 7 8 6 4 3 Skin Cancer per 100,000 7 11 9 12 15 10 7 5

Predict with 95% confidence the skin cancers per 100,000 in a county with a daily

average of 6.5 hours of sunshine. ANSWER:

10.884 ± 2.525. Thus, LCL= 8.359, and UCL = 13.409 FOR QUESTIONS 233 THROUGH 235, USE THE FOLLOWING NARRATIVE: Narrative: Sales and Experience The general manager of a chain of furniture stores believes that experience is the most important factor in determining the level of success of a salesperson. To examine this belief she records last month’s sales (in $1,000s) and the years of experience of 10 randomly selected salespeople. These data are listed below. Salesperson Years of Experience Sales

1 0 7 2 2 9 3 10 20 4 3 15 5 8 18 6 5 14 7 12 20 8 7 17 9 20 30 10 15 25

233. {Sales and Experience Narrative} Predict with 95% confidence the monthly sales of a

salesperson with 10 years of experience. ANSWER:

19.447 ± 3.819. Thus LCL = 15.628 (in $1000s), and UCL = 23.266 (in $1000s) 234. {Sales and Experience Narrative} Estimate with 95% confidence the average monthly

sales of all salespersons with 10 years of experience. ANSWER:

19.447 ± 1.199. Thus LCL = 18.248 (in $1000s), and UCL = 20.646 (in $1000s)


654 Chapter Sixteen

235. {Sales and Experience Narrative} Which interval in the previous two questions is narrower: the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?

ANSWER: The confidence interval estimate of the expected value of y is narrower than the

prediction interval for the same given value of x (10 years) and some confidence level. This is because there is less error in estimating a mean value as opposed to predicting an individual value.

FOR QUESTIONS 236 THROUGH 238, USE THE FOLLOWING NARRATIVE: Narrative: Income and Education A professor of economics wants to study the relationship between income (y in $1000s) and education (x in years). A random sample eight individuals is taken and the results are shown below. Education 16 11 15 8 12 10 13 14 Income 58 40 55 35 43 41 52 49

236. {Income and Education Narrative} Predict with 95% confidence the income of an

individual with 10 years of education. ANSWER:

39.715 ± 2.710. Thus, LCL = 37.005 (in $1000s), and UCL = 42.425 (in $1000s) 237. {Income and Education Narrative} Estimate with 95% confidence the average income of

all individuals with 10 years of education. ANSWER:

39.715 ± 1.188. Thus, LCL = 38.527 (in $1000s), and UCL = 40.903 (in $1000s) 238. {Income and Education Narrative} Which interval in the previous two questions is

narrower: the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?





FOR QUESTIONS 239 THROUGH 242, USE THE FOLLOWING NARRATIVE: Narrative: Movie Revenues An ardent fan of television game shows has observed that, in general, the more educated the contestant, the less money he or she wins. To test her belief she gathers data about the last eight winners of her favorite game show. She records their winnings in dollars and the number of years of education. The results are as follows. Contestant Years of Education Winnings

1 11 750 2 15 400 3 12 600 4 16 350 5 11 800 6 16 300 7 13 650 8 14 400

239. {Movie Revenues Narrative} Predict with 95% the winnings of a contestant who has 15

years of education.

ANSWER: 397.500 ± 159.213. Thus, LCL = $238.287, and UCL = $556.713

240. {Movie Revenues Narrative} Predict with 95% the winnings of a contestant who has 10

years of education.


241. {Movie Revenues Narrative} Estimate with 95% confidence the average winnings of all

contestants who have 15 years of education.


242. {Movie Revenues Narrative} Estimate with 95% confidence the average winnings of all

contestants who have 10 years of education. ANSWER: 397.500 ± 106.141. Thus, LCL = $291.359, and UCL = $503.641


656 Chapter Sixteen



Gross Revenue

1 5.3 48 2 7.2 65 3 1.3 18 4 1.8 20 5 3.5 31 6 2.6 26 7 8.0 73 8 2.4 23 9 4.5 39 10 6.7 58

243. {Movie Revenues Narrative} Predict with 95% confidence the gross revenue of a movie

whose top two stars earn $5.0 million. ANSWER:

45.65 ± 4.916. Thus, LCL = 40.734 (in $1,000,000s), and UCL = 50.566 (in $1,000,000s)

244. {Movie Revenues Narrative} Estimate with 95% confidence the average gross revenue of

a movie whose top two stars earn $5.0 million. ANSWER:

45.65 ± 1.54. Thus, LCL= 44.11 (in $1,000,000s), and UCL = 47.19 (in $1,000,000s) 245. {Movie Revenues Narrative} Which interval in the previous two questions is narrower:

the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?





FOR QUESTIONS 246 THROUGH 248, USE THE FOLLOWING NARRATIVE: Narrative: Cost of Books The editor of a major academic book publisher claims that a large part of the cost of books is the cost of paper. This implies that larger books will cost more money. As an experiment to analyze the claim, a university student visits the bookstore and records the number of pages and the selling price of twelve randomly selected books. These data are listed below.


246. {Cost of Books Narrative} Predict with 90% confidence the selling price of a book with

900 pages.


247. {Cost of Books Narrative} Estimate with 90% confidence the average selling price of all

books with 900 pages.


248. {Cost of Books Narrative} Which interval in the previous two questions is narrower: the

confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?




658 Chapter Sixteen

FOR QUESTIONS 249 THROUGH 251, USE THE FOLLOWING NARRATIVE: Narrative: Automobile Accidents and Precipitation A statistician investigating the relationship between the amount of precipitation (in inches) and the number of automobile accidents gathered data for 10 randomly selected days. The results Day Precipitation Number of Accidents

1 0.05 5 2 0.12 6 3 0.05 2 4 0.08 4 5 0.10 8 6 0.35 14 7 0.15 7 8 0.30 13 9 0.10 7 10 0.20 10

249. {Automobile Accidents and Precipitation Narrative} Predict with 95% confidence the

number of accidents that occur when there is 0.40 inches of rain.

ANSWER: 16.316 ± 4.032. Thus, LCL = 12.284, and UCL = 20.348

250. {Automobile Accidents and Precipitation Narrative} Estimate with 95% confidence the

average daily number of accidents when the daily precipitation is 0.25 inches. ANSWER: 11.086 ± 1.377. Thus, LCL = 9.709, and UCL = 12.463

251. {Automobile Accidents and Precipitation Narrative} Which interval in the previous two

questions is narrower: the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?











FOR QUESTIONS 252 THROUGH 254, USE THE FOLLOWING NARRATIVE: Narrative: Willie Nelson Concert At a recent Willie Nelson concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected: Age 62 57 40 49 67 54 43 65 54 41 Number of Concerts 6 5 4 3 5 5 2 6 3 1



252. {Willie Nelson Concert Narrative} Predict with 95% confidence the number of concerts

attended by a 45 years-old individual. ANSWER:

2.645 ± 2.057. Thus, LCL = 0.588, and UCL = 4.702 253. {Willie Nelson Concert Narrative} Estimate with 95% confidence the average number of

concerts attended by all 45 year-old individuals. ANSWER:

2.645 ± 0.577. Thus, LCL = 2.068, and UCL = 3.222


660 Chapter Sixteen

254. {Willie Nelson Concert Narrative} Which interval in the previous two questions is narrower: the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?



FOR QUESTIONS 255 THROUGH 257, USE THE FOLLOWING NARRATIVE: Narrative: Oil Quality and Price Quality of oil is measured in API gravity degrees – the higher the degrees API, the higher the quality. The table shown below is produced by an expert in the field who believes that there is a relationship between quality and price per barrel.

A partial Minitab output follows: Descriptive Statistics Variable N Mean StDev SE Mean Degrees 13 34.60 4.613 1.280 Price 13 12.730 0.457 0.127 Covariances

Degrees Price Degrees 21.281667 Price 2.026750 0.208833




Regression Analysis Predictor Coef StDev T P Constant 9.4349 0.2867 32.91 0.000 Degrees 0.095235 0.008220 11.59 0.000 S = 0.1314 R-Sq = 92.46% R-Sq(adj) = 91.7% Analysis of Variance Source DF SS MS F P Regression 1 2.3162 2.3162 134.24 0.000 Residual Error 11 0.1898 0.0173 Total 12 2.5060 255. {Oil Quality and Price Narrative} Predict with 95% confidence the oil price per barrel for

an API degree of 35. ANSWER:

12.768 ± (2.201)(0.1314)(1.038) = 12.768 ± 0.300 . Thus, LCL = 12.468, and UCL = 13.068

256. {Oil Quality and Price Narrative} Estimate with 95% confidence the average oil price per

barrel for an API degree of 35. ANSWER:

12.768 ± (2.201)(0.1314)(0.2785) = 12.768 ± 0.081. Thus, LCL = 12.687, and UCL = 12.849

257. {Oil Quality and Price Narrative} Which interval in the previous two questions is

narrower: the confidence interval estimate of the expected value of y or the prediction interval for the same given value of x (10 years) and same confidence level? Why?




662 Chapter Sixteen

SECTION 7

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 258. The standardized residual is defined as:

a. residual divided by the standard error of estimate b. residual multiplied by the square root of the standard error of estimate c. residual divided by the square of the standard error of estimate d. residual multiplied by the standard error of estimate ANSWER: a

259. The least squares method requires that the variance 2

εσ of the error variable ε is a constant no matter what the value of x is. When this requirement is violated, the condition is called: a. non-independence of ε b. homoscedasticity c. heteroscedasticity d. influential observation ANSWER: c

260. When the variance 2

εσ of the error variable ε is a constant no matter what the value of x is, this condition is called: a. homocausality b. heteroscedasticity c. homoscedasticity d. heterocausality ANSWER: c

261. If the plot of the residuals is fan shaped, which assumption of regression analysis if

violated? a. Normality b. Homoscedasticity c. Independence of errors d. No assumptions are violated, the graph should resemble a fan ANSWER: b



262. In regression analysis we use the Spearman rank correlation coefficient to measure and test to determine whether a relationship exists between the two variables if a. one or both variables may be ordinal b. both variables are interval but the normality requirement is not met c. both (a) and (b) d. neither (a) nor (b) ANSWER: c

263. The sample Spearman rank correlation coefficient, where a and b are the ranks of x and y,

respectively, is given by a. ( ) ( )cov , / /s a br a b s s= b. ( ) ( )cov , /s a br a b s s= − c. ( ) ( )cov , /s a br a b s s= + d. ( ) ( )cov , /s a br a b s s= ⋅ ANSWER: d


664 Chapter Sixteen

TRUE / FALSE QUESTIONS 264. The variance of the error variable 2

εσ is required to be constant. When this requirement is satisfied, the condition is called homoscedasticity.

ANSWER: T 265. The variance of the error variable 2

εσ is required to be constant. When this requirement is violated, the condition is called heteroscedasticity.

ANSWER: T 266. We standardize residuals in the same way we standardize all variables, by subtracting the

mean and dividing by the variance. ANSWER: F 267. An outlier is an observation that is unusually small or unusually large. ANSWER: T 268. One method of diagnosing heteroscedasticity is to plot the residuals against the predicted

values of y, then look for a change in the spread of the plotted values. ANSWER: T 269. Regardless of the value of x, the standard deviation of the distribution of y values about

the regression line is the same. This assumption of equal standard deviations about the regression line is called residual analysis.

ANSWER: F 270. Data that exhibit an autocorrelation effect violate the regression assumption of

independence. ANSWER: T 271. When n is greater than 30, the sample Spearman rank correlation coefficient sr is

approximately normally distributed with mean of 0 and standard deviation of 1. ANSWER: F 272. Given that n = 37, and the value of sample Spearman rank correlation coefficient sr =

0.35, the value of the test statistic for testing : 0o sH ρ = is z = 2.10 ANSWER: T 273. Another name for Pearson coefficient of correlation is the Spearman rank correlation

coefficient. ANSWER: F




FOR QUESTIONS 274 THROUGH 278, USE THE FOLLOWING NARRATIVE: Narrative: Sales and Experience The general manager of a chain of furniture stores believes that experience is the most important factor in determining the level of success of a salesperson. To examine this belief she records last month’s sales (in $1,000s) and the years of experience of 10 randomly selected salespeople. These data are listed below. Salesperson Years of Experience Sales

1 0 7 2 2 9 3 10 20 4 3 15 5 8 18 6 5 14 7 12 20 8 7 17 9 20 30 10 15 25

274. {Sales and Experience Narrative} Use the regression equation ˆ 8.63 1.0817y x= + to

determine the predicted values of y. ANSWER:

y : 8.630, 10.793, 19.447, 11.875, 17.284, 14.039, 21.610, 16.202, 30.264, and 24.856 275. {Sales and Experience Narrative} Use the predicted and actual values of y to calculate

the residuals. ANSWER:

ir : –1.630, -1.793, 0.553, 3.125, 0.716, -0.039, -1.610, 0.798. –0.264, and 0.144


666 Chapter Sixteen

276. {Sales and Experience Narrative} Plot the residuals against the predicted values of y. Does the variance appear to be constant?

ANSWER:

It appears that heteroscedasticity is not a problem. 277. {Sales and Experience Narrative} Compute the standardized residuals. ANSWER:

–1.100, -1.210, 0.373, 2.108, 0.483, -0.026, -1.086, 0.538, -0.178, and 0.097 278. {Sales and Experience Narrative} Identify possible outliers. ANSWER: The point (3, 15) is a possible outlier since its standardized residual 2.108 exceeds 2.0. FOR QUESTIONS 279 THROUGH 283, USE THE FOLLOWING NARRATIVE: Narrative: Income and Education A professor of economics wants to study the relationship between income (y in $1000s) and education (x in years). A random sample eight individuals is taken and the results are shown below. Education 16 11 15 8 12 10 13 14 Income 58 40 55 35 43 41 52 49

279. {Income and Education Narrative} Use the regression equation ˆ 10.6165 2.9098y x= + to

determine the predicted values of y. ANSWER: y : 57.173, 42.624, 54.263, 33.895, 45.534, 39.714, 48.444, and 51.353

Residuals versus Predicted

-3

-2

-1

0

1

2

3

4

0 5 10 15 20 25 30 35

Predicted Values

Res

idua

ls



280. {Income and Education Narrative} Use the predicted and actual values of y to calculate the residuals. ANSWER: ir : 0.877, -2.624, 0.737, 1.105, -2.534, 1.286, 3.556, and –2.353.

281. {Income and Education Narrative} Plot the residuals against the predicted values of y.

Does the variance appear to be constant? ANSWER:

It appears that heteroscedasticity is not a problem. 282. {Income and Education Narrative} Compute the standardized residuals.

ANSWER:

0.367, -1.164, 0.327, 0.490, -1.124, 0.570, 1.577, and –1.044 283. {Income and Education Narrative} Identify possible outliers.

ANSWER:

No outliers exist, since no observation has standard residual whose absolute value exceeds 2.0.

FOR QUESTIONS 284 THROUGH 288, USE THE FOLLOWING NARRATIVE: Narrative: Game Winnings and Education An ardent fan of television game shows has observed that, in general, the more educated the contestant, the less money he or she wins. To test her belief she gathers data about the last eight winners of her favorite game show. She records their winnings in dollars and the number of years of education. The results are as follows.


-4

-2

0

2

4

0 10 20 30 40 50 60 70

Predicted Values

Res

idul

as


668 Chapter Sixteen

Contestant Years of Education Winnings 1 11 750 2 15 400 3 12 600 4 16 350 5 11 800 6 16 300 7 13 650 8 14 400

284. {Game Winnings and Education Narrative} Use the regression equation

ˆ 1735 89.1667y x= − to determine the predicted values of y. ANSWER:

y : 754.167, 397.500, 665.000, 308.333, 754.167, 308.333, 575.833, and 486.667 285. {Game Winnings and Education Narrative} Use the predicted and actual values of y to

calculate the residuals. ANSWER:

ir : –4.167, 2.500, -65.000, 41.667, 45.833, -8.333, 74.167, and –86.667 286. {Game Winnings and Education Narrative} Plot the residuals against the predicted

values y . Does the variance appear to be constant. ANSWER:

The variance appears to be constant.


-100-75-50-25

0255075

100

0 100 200 300 400 500 600 700 800

Predicted Values

Res

idua

ls



287. {Game Winnings and Education Narrative} Compute the standardized residuals. ANSWER:

The standardized residuals are: –0.076, 0.045, -1.182, 0.758, 0.833, -0.152, 1.349, and –1.576.

288. {Game Winnings and Education Narrative} Identify possible outliers. ANSWER:

No outliers exist, since no observation has standard residual whose absolute value exceeds 2.0.



Gross Revenue

1 5.3 48 2 7.2 65 3 1.3 18 4 1.8 20 5 3.5 31 6 2.6 26 7 8.0 73 8 2.4 23 9 4.5 39 10 6.7 58

289. {Movie Revenues Narrative} Use the regression equation ˆ 4.225 8.285y x= + to determine

the predicted values of y.

ANSWER: y : 48.137, 63.878, 14.996, 19.139, 33.223, 25.767, 70.506, 24.110, 41.508, and 59.736.

290. {Movie Revenues Narrative} Use the predicted and actual values of y to calculate the

residuals.

ANSWER: ir : -0.137, 1.122, 3.004, 0.861, -2.223, 0.233, 2.494, -1.110, –2.508, and –1.736


670 Chapter Sixteen

291. {Movie Revenues Narrative} Plot the residuals against the predicted values of y. Does the variance appear to be constant.

ANSWER:

It appears that heteroscedasticity is not a problem. 292. {Movie Revenues Narrative} Compute the standardized residuals.

ANSWER: The standardized residuals are: –0.072, 0.588, 1.574, 0.451, -1.165, 0.122, 1.306, -0.581, -1.314, and –0.909.

293. {Movie Revenues Narrative} Identify possible outliers.

ANSWER: No outliers exist, since no observation has standardized residual whose absolute value exceeds 2.0.

FOR QUESTIONS 294 THROUGH 301, USE THE FOLLOWING NARRATIVE: Narrative: Willie Nelson Concert At a recent Willie Nelson concert, a survey was conducted that asked a random sample of 20 people their age and how many concerts they have attended since the first of the year. The following data were collected: Age 62 57 40 49 67 54 43 65 54 41 Number of Concerts 6 5 4 3 5 5 2 6 3 1



-4

-2

0

2

4

0 10 20 30 40 50 60 70 80

Predicted Values

Res

idua

ls










294. {Willie Nelson Concert Narrative} Use the regression equation ˆ 3.0115 0.1257y x= − + to

determine the predicted values of y.

ANSWER: The predicted values y are: 4.781 4.153 2.016 3.147 5.410 3.776 2.393 5.158 3.776 2.142 2.519 3.022 3.901 4.530 4.404 4.907 5.661 2.016 1.765 3.524

295. {Willie Nelson Concert Narrative} Use the predicted values and the actual values of y to

calculate the residuals. ANSWER: The residuals ˆr y y= − are: 1.219 0.847 1.984 -0.147 -0.410 1.224 -0.393 0.842 -0.776 -1.142 0.481 -1.022 0.099 0.470 -0.404 0.093 -1.661 -0.016 -0.765 -0.524


672 Chapter Sixteen

296. {Willie Nelson Concert Narrative} Plot the residuals in against the predicted values y . ANSWER: 297. {Willie Nelson Concert Narrative} Does it appear that heteroscedasticity is a problem?

Explain. ANSWER: The variance of the error variable appears to be constant; therefore heteroscedasticity is

not a problem. 298. {Willie Nelson Concert Narrative} Draw a histogram of the residuals. ANSWER:


-2.000

-1.000

0.000

1.000

2.000

3.000

0 1 2 3 4 5 6

Predicted

Res

idua

ls

Histogram

0

24

68

10

-1 0 1 2

Residuals

Frequency



299. {Willie Nelson Concert Narrative} Does it appear that the errors are normally distributed? Explain.

ANSWER: The histogram is positively skewed. The errors may not be normally distributed. 300. {Willie Nelson Concert Narrative} Use the residuals to compute the standardized

residuals. ANSWER: The standardized residuals /r sε are: 1.297 0.902 2.111 -0.157 -0.436 1.303 -0.418 0.896 -0.826 -1.215 0.512 -1.087 0.105 0.500 -0.430 0.099 -1.768 -0.017 -0.814 -0.558 301. {Willie Nelson Concert Narrative} Identify possible outliers. ANSWER: There are no outliers since none of the 20 observations has a standardized residual whose

absolute value exceeds 2.0. FOR QUESTIONS 302 THROUGH 309, USE THE FOLLOWING NARRATIVE: Narrative: Oil Quality and Price Quality of oil is measured in API gravity degrees – the higher the degrees API, the higher the quality. The table shown below is produced by an expert in the field who believes that there is a relationship between quality and price per barrel.

A partial Minitab output follows:



674 Chapter Sixteen

Descriptive Statistics Variable N Mean StDev SE Mean Degrees 13 34.60 4.613 1.280 Price 13 12.730 0.457 0.127 Covariances Degrees Price Degrees 21.281667 Price 2.026750 0.208833 Regression Analysis Predictor Coef StDev T P Constant 9.4349 0.2867 32.91 0.000 Degrees 0.095235 0.008220 11.59 0.000 S = 0.1314 R-Sq = 92.46% R-Sq(adj) = 91.7% Analysis of Variance Source DF SS MS F P Regression 1 2.3162 2.3162 134.24 0.000 Residual Error 11 0.1898 0.0173 Total 12 2.5060 302. {Oil Quality and Price Narrative} Use the regression equation ˆ 9.4349 0.095235y x= + to

determine the predicted values of y.

ANSWER: The predicted values y are: 12.006, 12.149, 12.368, 12.416, 12.473, 12.721, 12.673,

12.740, 12.959, 13.340, 13.340, 13.130, and 13.178. 303. {Oil Quality and Price Narrative} Use the predicted values and the actual values of y to

calculate the residuals. ANSWER: The residuals ˆr y y= − are: 0.014, -0.109, -0.048, -0.146, 0.017, -0.021, 0.127, 0.260,

0.041, -0.170, -0.150, 0.090, and 0.092.



304. {Oil Quality and Price Narrative} Plot the residuals against the predicted values y . ANSWER: 305. {Oil Quality and Price Narrative} Does it appear that heteroscedasticity is a problem?

Explain. ANSWER: The variance of the error variable appears to be constant; therefore heteroscedasticity is

not a problem. 306. {Oil Quality and Price Narrative} Draw a histogram of the residuals. ANSWER:

13.413.213.012.812.612.412.212.0

0.3

0.2

0.1

0.0

-0.1

-0.2

Fitted Value

Res

idua

l

Residuals Versus the Fitted Values(response is Price)

0.30.20.10.0-0.1-0.2

5

4

3

2

1

0

Residual

Freq

uenc

y

Histogram of the Residuals(response is Price)


676 Chapter Sixteen

307. {Oil Quality and Price Narrative} Does it appear that the errors are normally distributed? Explain.

ANSWER: The histogram is fairly symmetric; therefore we may conclude that the errors are

normally distributed. 308. {Oil Quality and Price Narrative} Use the residuals to compute the standardized

residuals. ANSWER: The standardized residuals /r sε are: 0.105, -0.830, -0.366, -1.109, 0.130, -0.156, 0.967,

1.982, 0.315, -1.290, -1.138, 0.685, and 0.703. 309. Identify possible outliers. ANSWER: There are no outliers since none of the 13 observations has a standardized residual whose

absolute value exceeds 2.0. However, observation 9 with standardized residual of 1.982 may be an outlier.


677

CHAPTER 17

MULTIPLE REGRESSION

SECTIONS 1 - 3

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 1. In a multiple regression analysis, if the model provides a poor fit, this indicates that:

a. the sum of squares for error will be large b. the standard error of estimate will be large c. the multiple coefficient of determination will be close to zero d. All of the above

ANSWER: d 2. In a multiple regression analysis, when there is no linear relationship between each of the

independent variables and the dependent variable, then a. multiple t-tests of the individual coefficients will likely show some are significant b. we will conclude erroneously that the model has some validity c. the chance of erroneously concluding that the model is useful is substantially less

with the F-test than with multiple t-tests d. All of the above statements are correct

ANSWER: d


678 Chapter Seventeen

3. In testing the validity of a multiple regression model, a large value of the F-test statistic indicates that: a. most of the variation in the independent variables is explained by the variation in y b. most of the variation in y is explained by the regression equation c. most of the variation in y is unexplained by the regression equation d. the model provides a poor fit

ANSWER: b 4. Which of the following statements regarding multicollinearity is not true?

a. It exists in virtually all multiple regression models. b. It is also called collinearity and intercorrelation. c. It is a condition that exists when the independent variables are highly correlated with

the dependent variable. d. It does not affect the F-test of the analysis of variance.

ANSWER: c 5. In a multiple regression analysis involving 25 data points, the standard error of estimate

squared is calculated as 8.12 =εs and the sum of squares for error as SSE = 36. Then, the number of the independent variables must be: a. 6 b. 5 c. 4 d. 3

ANSWER: c 6. When the independent variables are correlated with one another in a multiple regression

analysis, this condition is called: a. heteroscedasticity b. homoscedasticity c. multicollinearity d. elasticity

ANSWER: c 7. In a multiple regression model, the mean of the probability distribution of the error

variable ε is assumed to be: a. 1.0 b. 0.0 c. Any value greater than 1 d. k, where k is the number of independent variables included in the model

ANSWER: b


Multiple Regression 679

8. The adjusted multiple coefficient of determination is adjusted for the: a. number of regression parameters including the y-intercept b. number of dependent variables and the sample size c. number of independent variables and the sample size d. coefficient of correlation and the significance level

ANSWER: c 9. In a multiple regression model, the standard deviation of the error variable ε is assumed

to be: a. constant for all values of the independent variables b. constant for all values of the dependent variable c. 1.0 d. not enough information is given to answer this question

ANSWER: a 10. In multiple regression analysis, the ratio MSR/MSE yields the:

a. t-test statistic for testing each individual regression coefficient b. F-test statistic for testing the validity of the regression equation c. multiple coefficient of determination d. adjusted multiple coefficient of determination

ANSWER: b 11. In a multiple regression analysis involving 6 independent variables, the sum of squares

are calculated as: Total variation in Y = SSY = 900, SSR = 600 and SSE = 300. Then, the value of the F-test statistic for this model is: a. 150 b. 100 c. 50 d. None of the above

ANSWER: d 12. In order to test the validity of a multiple regression model involving 5 independent

variables and 30 observations, the numerator and denominator degrees of freedom for the critical value of F are, respectively, a. 5 and 30 b. 6 and 29 c. 5 and 24 d. 6 and 25

ANSWER: c 13. In multiple regression models, the values of the error variable ε are assumed to be:

a. autocorrelated b. dependent of each other c. independent of each other d. always positive

ANSWER: c



14. A multiple regression model involves 5 independent variables and a sample of 10 data

points. If we want to test the validity of the model at the 5% significance level, the critical value is: a. 6.26 b. 3.33 c. 9.36 d. 4.24

ANSWER: a 15. A multiple regression model involves 10 independent variables and 30 observations. If

we want to test at the 5% significance level the parameter 4β , the critical value will be: a. 2.093 b. 1.697 c. 2.228 d. 1.729

ANSWER: a 16. In a multiple regression analysis involving k independent variables and n data points, the

number of degrees of freedom associated with the sum of squares for error is: a. k-1 b. n-k c. n-1 d. n-k-1

ANSWER: d 17. A multiple regression model has the form 321 4538ˆ xxxy −++= . As 3x increases by

one unit, with 1x and 2x held constant, the y on average is expected to: a. increase by 1 unit b. increase by 12 units c. decrease by 4 units d. decrease by 16 units

ANSWER: c 18. The problem of multicollinearity arises when the:

a. dependent variables are highly correlated with one another b. independent variables are highly correlated with one another c. independent variables are highly correlated with the dependent variable d. None of the above

ANSWER: b



19. To test the validity of a multiple regression model, we test the null hypothesis that the regression coefficients are all zero by applying the: a. t-test b. z-test c. F-test d. All of the above

ANSWER: c 20. To test the validity of a multiple regression model involving two independent variables,

the null hypothesis is that: a. 210 βββ == b. 021 == ββ c. 21 ββ = d. 21 ββ ≠

ANSWER: b 21. If multicollinearity exists among the independent variables included in a multiple

regression model, then: a. regression coefficients will be difficult to interpret b. standard errors of the regression coefficients for the correlated independent variables

will increase c. multiple coefficient of determination will assume a value close to zero d. both (a) and (b) are correct statements

ANSWER: d 22. Which of the following is not true when we add an independent variable to a multiple

regression model? a. Adjusted coefficient of determination can assume a negative value b. Unadjusted coefficient of determination always increases c. Unadjusted coefficient of determination may increase or decrease d. Adjusted coefficient of determination may increase

ANSWER: c 23. A multiple regression model has the form 22110ˆ xbxbby ++= . The coefficient 1b is

interpreted as the: a. change in y per unit change in 1x b. change in y per unit change in 1x , holding 2x constant c. change in y per unit change in 1x , when 1x and 2x values are correlated d. change in the average value of y per unit change in 1x , holding 2x constant

ANSWER: d



24. A multiple regression analysis involving three independent variables and 25 data points results in a value of 0.769 for the unadjusted multiple coefficient of determination. Then, the adjusted multiple coefficient of determination is: a. 0.385 b. 0.877 c. 0.591 d. 0.736

ANSWER: d 25. The coefficient of multiple determination ranges from:

a. 1.0 to ∞ b. 0.0 to 1.0 c. 1.0 to k, where k is the number of independent variables in the model d. 1.0 to n, where n is the number of observations in the dependent variable

ANSWER: b 26. For a multiple regression model, the following statistics are given: Total variation in Y =

SSY = 500, SSE = 80, and n = 25. Then, the coefficient of determination is: a. 0.84 b. 0.16 c. 0.3125 d. 0.05

ANSWER: a 27. For a multiple regression model the following statistics are given: Total variation in Y =

SSY = 250, SSE = 50, k = 4, and n = 20. Then, the coefficient of determination adjusted for the degrees of freedom is: a. 0.800 b. 0.747 c. 0.840 d. 0.775

ANSWER: b 28. A multiple regression model has the form: 21 6225.5ˆ xxy ++= . As 2x increases by one

unit, holding 1x constant, then the value of y will increase by: a. 2 units b. 7.25 units c. 6 units on average d. None of the above

ANSWER: c



29. The graphical depiction of the equation of a multiple regression model with k independent variables (k > 1) is referred to as: a. a straight line b. response variable c. response surface d. a plane only when k = 3

ANSWER: c 30. A multiple regression model has:

a. only one independent variable b. only two independent variables c. more than one independent variable d. more than one dependent variable

ANSWER: c 31. If all the points for a multiple regression model with two independent variables were on

the regression plane, then the multiple coefficient of determination would equal: a. 0 b. 1 c. 2, since there are two independent variables d. any number between 0 and 2

ANSWER: b 32. If none of the data points for a multiple regression model with two independent variables

were on the regression plane, then the multiple coefficient of determination would be: a. –1.0 b. 1.0 c. any number between –1 and 1, inclusive d. any number greater than or equal to zero but smaller than 1

ANSWER: d 33. The multiple coefficient of determination is defined as:

a. SSE/SSY b. MSE/MSR c. 1- (SSE/SSY) d. 1- (MSE/MSR)

ANSWER: c 34. In a multiple regression model, the following statistics are given: SSE = 100, 2 0.995R = ,

k = 5, and n = 15. Then, the multiple coefficient of determination adjusted for degrees of freedom is: a. 0.955 b. 0.930 c. 0.900 d. 0.855

ANSWER: b



35. In a multiple regression model, the error variable ε is assumed to have a mean of:

a. –1.0 b. 0.0 c. 1.0 d. Any value smaller than –1.0

ANSWER: b 36. For the following multiple regression model: 321 5432ˆ xxxy ++−= , a unit increase in

1x , holding 2x and 3x constant, results in: a. an increase of 3 units in the value of y b. a decrease of 3 units in the value of y c. a decrease of 3 units on average in the value of y d. an increase of 8 units in the value of y

ANSWER: c 37. In a multiple regression model, the probability distribution of the error variable ε is

assumed to be: a. normal b. nonnormal c. positively skewed d. negatively skewed

ANSWER: a 38. Which of the following measures can be used to assess the multiple regression model’s

fit? a. sum of squares for error b. sum of squares for regression c. standard error of estimate d. single t-test

ANSWER: c 39. In a multiple regression analysis involving 40 observations and 5 independent variables,

the following statistics are given: Total variation in Y = SSY = 350 and SSE = 50. Then, the multiple coefficient of determination is: a. 0.8408 b. 0.8571 c. 0.8469 d. 0.8529

ANSWER: b



40. In a multiple regression analysis involving 20 observations and 5 independent variables, the following statistics are given: Total variation in Y = SSY = 250 and SSE = 35. The multiple coefficient of determination adjusted for degrees of freedom is: a. 0.810 b. 0.860 c. 0.835 d. 0.831

ANSWER: a 41. In testing the validity of a multiple regression model involving 10 independent variables

and 100 observations, the numerator and denominator degrees of freedom for the critical value of F will be, respectively, a. 9 and 90 b. 10 and 100 c. 9 and 10 d. 10 and 89

ANSWER: d 42. In multiple regression analysis involving 10 independent variables and 100 observations,

the critical value of t for testing individual coefficients in the model will have: a. 100 degrees of freedom b. 10 degrees of freedom c. 89 degrees of freedom d. 9 degrees of freedom

ANSWER: c 43. For a multiple regression model,

a. SSY = SSR – SSE b. SSE = SSR – SSY c. SSR = SSE – SSY d. SSY = SSE + SSR

ANSWER: d 44. In a regression model involving 50 observations, the following estimated regression

model was obtained: 1 2 3ˆ 10.5 3.2 5.8 6.5y x x x= + + + . For this model, the following statistics are given: SSR = 450 and SSE = 175. Then, the value of MSR is: a. 12.50 b. 275 c. 150 d. 3.804

ANSWER: c



45. In a regression model involving 30 observations, the following estimated regression

model was obtained: 321 2.18.260ˆ xxxy −++= . For this model, the following statistics were given: Total variation in Y = SSY = 800 and SSE = 200. Then, the value of the F statistic for testing the validity of this model is: a. 26.00 b. 7.69 c. 3.38 d. 0.039

ANSWER: a 46. Most statistical software provide p-value for testing each coefficient in the multiple

regression model. In the case of 2b , this represents the probability that: a. 02 =b b. 02 =β c. || 2b could be this large if 02 =β d. || 2b could be this large if 02 ≠β

ANSWER: c 47. In a regression model involving 60 observations, the following estimated regression

model was obtained: 321 378.0679.070.04.51ˆ xxxy −++= , and the following statistics were given: SSY = 119,724 and SSR = 29,029.72. Then, the value of MSE is: a. 1619.541 b. 9676.572 c. 1995.400 d. 5020.235

ANSWER: a 48. In testing the validity of a multiple regression model in which there are four independent

variables, the null hypothesis is: a. 1: 43210 ==== ββββH b. 432100 : βββββ ====H c. 0: 43210 ==== ββββH d. 0: 432100 ≠==== βββββH

ANSWER: c



49. For a set of 20 data points, a statistical software listed the estimated multiple regression

equation as 321 2872261.8ˆ xxxy +++−= , and also has listed the t statistic for testing the significance of each regression coefficient. Using the 5% significance level for testing whether 72 =b differs significantly from zero, the critical region will be that the absolute value of t is greater than or equal to: a. 1.746 b. 2.120 c. 1.337 d. 1.333

ANSWER: b 50. For the multiple regression model: 321 10152575ˆ xxxy +−+= , if 2x were to increase by

5, holding 1x and 3x constant, the value of y will: a. increase by 5 b. increase by 75 c. decrease on average by 5 d. decrease on average by 75

ANSWER: d 51. In a multiple regression analysis, there are 20 data points and 4 independent variables,

and the sum of the squared differences between observed and predicted values of y is 180. The multiple standard error of estimate will be: a. 6.708 b. 3.464 c. 9.000 d. 3.000

ANSWER: b 52. A multiple regression analysis includes 4 independent variables results in sum of squares

for regression of 1200 and sum of squares for error of 800. Then, the multiple coefficient of determination will be: a. 0.667 b. 0.600 c. 0.400 d. 0.200

ANSWER: b



53. A multiple regression analysis includes 20 data points and 4 independent variables produced the following statistics: Total variation in Y = SSY = 200 and SSR = 160. Then, the multiple standard error of estimate will be: a. 0.80 b. 3.266 c. 3.651 d. 1.633

ANSWER: d 54. In a multiple regression analysis involving 25 data points and 5 independent variables,

the sum of squares terms are calculated as Total variation in Y = SSY = 500, SSR = 300, and SSE = 200. In testing the validity of the regression model, the F value of the test statistic will be: a. 5.70 b. 2.50 c. 1.50 d. 0.176

ANSWER: a 55. A multiple regression equation includes 5 independent variables, and the coefficient of

determination is 0.81. The percentage of the variation in y that is explained by the regression equation is: a. 81% b. 90% c. 86% d. about 16%

ANSWER: a 56. In a simple linear regression problem, the following pairs of ( ˆ,i iy y ) are given: (6.75,

7.42), (8.96, 8.06), (10.30, 11.65), and (13.24, 12.15). Then, the sum of squares for error is a. 39.2500 b. -0.0300 c. 4.2695 d. 39.2800 ANSWER: c



57. In a multiple regression problem involving two independent variables, if 1b is computed to be + 2.0, it meant that the a. relationship between 1x and y is significant b. estimated average of y increases by two units for each increase of one unit of 1x

holding 2x constant c. estimated average of y increases by two units for each increase of one unit of 1x ,

without regard to 2x d. estimated average of y is two when 1x equals 0 ANSWER: b

58. In a multiple regression model, the value of the coefficient of multiple determination has

to fall between a. – 1 and + 1 b. 0 and + 1 c. – 1 and 0 d. Any pair of real numbers ANSWER: b

59. In a multiple regression model, which of the following is correct regarding the value of

the value of 2R adjusted for the degrees of freedom? a. It can be negative b. It has to be positive c. It has to be larger than the coefficient of multiple determination d. It can be larger than 1 ANSWER: a

60. An interaction term in a multiple regression model with two independent variables may

be used when a. the coefficient of determination is small b. there is a curvilinear relationship between the dependent and independent variables c. neither one of the two independent variables contribute significantly to the regression

model d. the relationship between 1x and y changes for differing values of 2x ANSWER: d

61. In a multiple regression model, the adjusted 2R

a. cannot be negative b. can sometimes be negative c. can sometimes be greater than + 1 d. has to fall between 0 and + 1 ANSWER: b



62. The coefficient of multiple determination 2R a. measures the variation around the predicted regression equation b. measures the proportion of variation in y that is explained by 1x and 2x c. measures the proportion of variation in y that is explained by 1x holding 2x constant d. will have the same sign as 1b ANSWER: b

63. If a group of independent variables are not significant individually but are significant as a

group at a specified level of significance, this is most likely due to a. autocorrelation b. the presence of dummy variables c. the absence of dummy variables d. multicollinearity ANSWER: d



TRUE / FALSE QUESTIONS

64. Multiple regression is the process of using several independent variables to predict a

number of dependent variables. ANSWER: F 65. In multiple regression, the descriptor “multiple” refers to more than one dependent

variable. ANSWER: F 66. For each x term in the multiple regression equation, the corresponding β is referred to as

a partial regression coefficient. ANSWER: T 67. In a multiple regression problem, the regression equation is 1 260.6 5.2 0.75y x x− += . The

estimated value for y when 1 3x = and 2 4x = is 48. ANSWER: T 68. In reference to the equation 1 2ˆ 0.80 0.12 0.08y x x= − + + , the value –0.80 is the y intercept. ANSWER: T 69. In testing the significance of a multiple regression model in which there are three

independent variables, the null hypothesis is 0 1 2 3:H β β β= = . ANSWER: F 70. In a multiple regression problem involving 24 observations and three independent

variables, the estimated regression equation is 1 2 372 3.2 1.5y x x x+= + − . For this model, SST = 800 and SSE = 245. Then, the value of the F statistic for testing the significance of the model is 15.102.

ANSWER: T 71. A multiple regression equation includes 5 independent variables, and the coefficient of

determination is 0.81. Then, the percentage of the variation in y that is explained by the regression equation is 90%.

ANSWER: F 72. In a multiple regression analysis involving 4 independent variables and 30 data points,

the number of degrees of freedom associated with the sum of squares for error, SSE, is 25.

ANSWER: T



73. In order to test the significance of a multiple regression model involving 4 independent

variables and 25 observations, the numerator and denominator degrees of freedom for the critical value of F are 3 and 21, respectively.

ANSWER: F 74. In multiple regression analysis, the adjusted multiple coefficient of determination is

adjusted for the number of independent variables and the sample size. ANSWER: T 75. A multiple regression analysis includes 25 data points and 4 independent variables

produces SST = 400 and SSR = 300. Then, the multiple standard error of estimate is 5. ANSWER: F 76. Multicollinearity is present if the dependent variable is linearly related to one of the

explanatory variables. ANSWER: F 77. In a multiple regression analysis involving 50 observations and 5 independent variables,

SST = 475 and SSE = 71.25. Then, the multiple coefficient of determination is 0.85. ANSWER: T 78. A multiple regression model has the form 1 26.75 2.25 3.5y x x+= + . As 1x increases by

one unit, holding 2x constant, the value of y will increase by 9 units. ANSWER: F 79. In reference to the multiple regression model 1 2 340 15 10 5y x x x+ − += , if 2x were to

increase by five units, holding 1x and 3x constant, then, the value of y would decrease on average by 50 units.

ANSWER: T 80. A multiple regression model involves 40 observations and 4 independent variables

produces SST = 100,000 and SSR = 80,400. Then, the value of MSE is 560. ANSWER: T 81. In order to test the significance of a multiple regression model involving 5 independent

variables and 30 observations, the numerator and denominator degrees of freedom for the critical value of F are 5 and 24, respectively.

ANSWER: T 82. In reference to the equation 1 2ˆ 0.80 0.12 0.08y x x= − + + , the value 0.12 is the average

change in y per unit change in 1x , when 2x is held constant. ANSWER: T



83. In multiple regression, if the error sum of squares SSE equals the total variation in y, then the value of the test statistic F is zero.

ANSWER: T 84. In reference to the equation 1 2ˆ 1.86 0.51 0.60y x x= − + , the value 0.60 is the average change

in y per unit change in 2x , regardless of the value of 1x . ANSWER: F 85. Most statistical software print a second 2R statistic, called the coefficient of

determination adjusted for degrees of freedom, which has been adjusted to take into account the sample size and the number of independent variables.

ANSWER: T 86. In multiple regression, the standard error of estimate is defined by /( )SSE n ksε −= ,

where n is the sample size and k is the number of independent variables. ANSWER: F 87. In regression analysis, the total variation in the dependent variable y, measured by

2( )iy y−∑ , can be decomposed into two parts: the explained variation, measured by SSR, and the unexplained variation, measured by SSE.

ANSWER: T 88. In multiple regression, a large value of the test statistic F indicates that most of the

variation in y is unexplained by the regression equation and that the model is useless. A small value of F indicates that most of the variation in y is explained by the regression equation and that the model is useful.

ANSWER: F 89. When an additional explanatory variable is introduced into a multiple regression model,

coefficient of multiple determination adjusted for degrees of freedom can never decrease. ANSWER: F 90. In multiple regression analysis, when the response surface (the graphical depiction of the

regression equation) hits every single point, the sum of squares for error SSE = 0, the standard error of estimate sε = 0, and the coefficient of determination 2R = 1.

ANSWER: T 91. In a multiple regression analysis involving k independent variables, the t-tests of the

individual coefficients allows us to determine whether 0iβ ≠ (for i = 1, 2, …., k), which tells us whether a linear relationship exists between ix and y.

ANSWER: T



92. In multiple regression analysis, the problem of multicollinearity affects the t-tests of the individual coefficients as well as the F-test in the analysis of variance for regression, since the F-test combines these t-tests into a single test.

ANSWER: F 93. A multiple regression model is assessed to be good if the error sum of squares SSE and

the standard error of estimate sε are both small, the coefficient of multiple determination 2R is close to 1, and the value of the test statistic F is large.

ANSWER: T 94. The most commonly method to remedy non-normality or heteroscedasticity in regression

analysis is to transform the dependent variable, y. The most commonly used transformations are log provided ( 0)y y y! = ≥ , 2y y! = , provided( 0)y y y! = ≥ , and

1y y! = . ANSWER: T 95. In multiple regression analysis, and because of a commonly occurring problem called

multicollinearity, the t-tests of the individual coefficients may indicate that some independent variables are not linearly related to the dependent variable, when in fact they are.

ANSWER: T 96. Multicollinearity is present when there is a high degree of correlation between the

dependent variable and any of the independent variables. ANSWER: F 97. The coefficient of multiple determination 2R measures the proportion of variation in y

that is explained by the explanatory variables included in the model. ANSWER: T 98. When an additional explanatory variable is introduced into a multiple regression model,

the coefficient of multiple determination will never decrease. ANSWER: T 99. In regression analysis, we judge the magnitude of the standard error of estimate relative

to the values of the dependent variable, and particularly to the mean of y. ANSWER: T 100. In calculating the standard error of the estimate, s MSEε = , there are(n – k – 1) degrees

of freedom, where n is the sample size and k is the number of independent variables in the model.

ANSWER: T 101. A multiple regression is called “multiple” because it has several explanatory variables. ANSWER: T



102. The coefficient of multiple determination measures the proportion or percentage of the total variation in the dependent variable y that is explained by the regression plane.

ANSWER: T 103. When an explanatory variable is dropped from a multiple regression model, the adjusted

coefficient of determination can increase. ANSWER: T 104. The coefficient of multiple determination is calculated by dividing the regression sum of

squares by the total sum of squares (SSR/SST) and subtracting that value from 1 ANSWER: F 105. In a multiple regression model involving 5 independent variables, if the sum of the

squared residuals is 847 and the data set contains 40 points, then, the value of the standard error of the estimate is 24.911.

ANSWER: F 106. One of the consequences of multicollinearity in multiple regression is biased estimates on

the slope coefficients. ANSWER: F 107. When an explanatory variable is dropped from a multiple regression model, the

coefficient of multiple determination can increase. ANSWER: F 108. Multicollinearity is a situation in which two or more of the independent variables are

highly correlated with each other. ANSWER: T 109. You have just run a regression in which the coefficient of multiple determination is 0.78.

To determine if this indicates that the independent variables explain a significant portion of the variation in the dependent variable, you would perform an F – test.

ANSWER: T 110. From the coefficient of multiple determination, we cannot detect the strength of the

relationship between the dependent variable y and any individual independent variable. ANSWER: T 111. The total sum of squares (SST) in a regression model will never exceed the regression

sum of squares (SSR). ANSWER: F 112. A regression had the following results: SST = 92.25, SSE = 34.55. It can be said that

37.45% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: F



113. An interaction term in a multiple regression model involving two independent variables

may be used when the relationship between 1x and y changes for differing values of 2x . ANSWER: T 114. Multicollinearity is present when there is a high degree of correlation between the

independent variables included in the regression model. ANSWER: T 115. The interpretation of the slope is different in a multiple linear regression model as

compared to a simple linear regression model. ANSWER: T 116. A multiple regression is called “multiple” because it has several data points, and multiple

dependent variables. ANSWER: F 117. A high value of the coefficient of multiple determination significantly above 0 in multiple

regression, accompanied by insignificant t – values on all parameter estimates, very often indicates a high correlation between independent variables in the model.

ANSWER: T 118. One of the consequences of multicollinearity in multiple regression is inflated standard

errors in some or all of the estimated slope coefficients. ANSWER: T 119. A regression analysis showed that SST = 112.18 and SSE = 33.65. It can be said that

70% of the variation in the dependent variable is explained by the independent variables in the regression.

ANSWER: T 120. A multiple regression model has the form 22110ˆ xbxbby ++= . The coefficient 1b is

interpreted as the average change in y per unit change in 1x . ANSWER: F 121. When an explanatory variable is dropped from a multiple regression model, the adjusted

coefficient of multiple of multiple determination can increase. ANSWER: T 122. The parameter estimates are biased when multicollinearity is present in a multiple

regression equation. ANSWER: F



123. In trying to obtain a model to estimate grades on a statistics test, a professor wanted to include, among other factors, whether the person had taken the course previously. To do this, the professor included a dummy variable in her regression that was equal to 1 if the person had previously taken the course, and 0 otherwise. The interpretation of the coefficient associated with this dummy variable would be the average amount the repeat students tended to be above or below non-repeaters, with all other factors the same.

ANSWER: T 124. When an additional explanatory variable is introduced into a multiple regression model,

the adjusted coefficient of multiple determination can never decrease. ANSWER: F 125. If we have taken into account all relevant explanatory variables, the residuals from a

multiple regression should be random. ANSWER: T 126. When an additional explanatory variable is introduced into a multiple regression model,

the coefficient of multiple determination will increase. ANSWER: T 127. Multicollinearity will result in excessively low standard errors of the parameter estimates

reported in the regression output. ANSWER: F 128. A multiple regression model is assessed to be perfect if the error sum of squares SSE = 0,

the standard error of estimate sε = 0, the coefficient of multiple determination 2R =1, and the value of the test statistic F = ∞ .

ANSWER: T 129. A multiple regression model is assessed to be poor if the error sum of squares SSE , and

the standard error of estimate sε are both large, the coefficient of multiple determination 2R is close to 0, and the value of the test statistic F is small.

ANSWER: T




130. Consider the following statistics of a multiple regression model: Total variation in y =

SSY = 1000, SSE = 300, n = 50, and k = 4 . a. Determine the standard error of estimate b. Determine the multiple coefficient of determination c. Determine the F-statistics

ANSWER:

a. =εs 2.582 b. =2R 70% c. F = MSR/MSE = 26.25

131. Consider the following statistics of a multiple regression model: n = 25, k = 5, =1b -6.31,

and =1bs 2.98. Can we conclude at the 1% significance level that 1x and y are linearly

related? ANSWER:

0: 10 =βH vs. :1H ≠1β 0 Rejection region: | t | > 0.005,19t = 2.861, Test statistic: t = -2.117 Conclusion: Don’t reject the null hypothesis. No 132. The computer output for the multiple regression model εβββ +++= 22110 xxy is shown

below. However, because of a printer malfunction some of the results are not shown. These are indicated by the boldface letters a to i. Fill in the missing results (up to three decimal places).

Predictor Coef StDev T Constant a 6.15 4.11

1x 3.51 b 1.25

2x -0.71 0.30 c S = d R-Sq = e

ANALYSIS OF VARIANCE

Source of Variation df SS MS F Regression 2 412 g i Error 37 f h Total 39 974



ANSWER: a = 25.277 b = 2.808 c = -2.367 d = 3.897 e = .423 f = 562 g = 206 h = 15.189 i = 13.5623 FOR QUESTIONS 133 THROUGH 140, USE THE FOLLOWING NARRATIVE: Narrative: Life Expectancy An actuary wanted to develop a model to predict how long individuals will live. After consulting a number of physicians, she collected the age at death (y), the average number of hours of exercise per week ( 1x ), the cholesterol level ( 2x ), and the number of points that the individual’s blood pressure exceeded the recommended value ( 3x ). A random sample of 40 individuals was selected. The computer output of the multiple regression model is shown below. THE REGRESSION EQUATION IS =y 321 016.0021.079.18.55 xxx −−+

Predictor Coef StDev T Constant 55.8 11.8 4.729

1x 1.79 0.44 4.068

2x -0.021 0.011 -1.909

3x -0.016 0.014 -1.143 S = 9.47 R-Sq = 22.5% ANALYSIS OF VARIANCE

133. {Life Expectancy Narrative} Is there enough evidence at the 10% significance level to

infer that the model is useful in predicting length of life? ANSWER:

0: 3210 === βββH :1H At least one iβ is not equal to zero.

Rejection region: F > 0.05,3,36F ≈= 2.84 Test statistic: F = 3.477

Conclusion: Reject the null hypothesis. Yes, there enough evidence at the 10% significance level to infer that the model is useful in predicting length of life.

Source of Variation df SS MS F Regression 3 936 312 3.477 Error 36 3230 89.722 Total 39 4166



134. {Life Expectancy Narrative} Is there enough evidence at the 1% significance level to infer that the average number of hours of exercise per week and the age at death are linearly related?

ANSWER: 0: 10 =βH vs. :1H ≠1β 0

Rejection region: | t | > 0.005,36t ≈2.724 Test statistic: t = 4.068 Conclusion: Reject the null hypothesis. Yes, there enough evidence at the 1%

significance level to infer that the average number of hours of exercise per week and the age at death are linearly related.

135. {Life Expectancy Narrative} Is there enough evidence at the 5% significance level to

infer that the cholesterol level and the age at death are negatively linearly related? ANSWER:

0: 20 =βH vs. :1H <2β 0 Rejection region: t < - 0.05,36t ≈ -1.69 Test statistic: t = -1.909 Conclusion: Reject the null hypothesis. Yes, there enough evidence at the 5%

significance level to infer that the cholesterol level and the age at death are negatively linearly related.

136. {Life Expectancy Narrative} Is there sufficient evidence at the 5% significance level to

infer that the number of points that the individual’s blood pressure exceeded the recommended value and the age at death are negatively linearly related? ANSWER:

0: 30 =βH vs. :1H <3β 0 Rejection region: t < - 0.05,36t ≈ -1.69 Test statistic: t = -1.143 Conclusion: Don’t reject the null hypothesis. No, sufficient evidence at the 5%

significance level to infer that the number of points that the individual’s blood pressure exceeded the recommended value and the age at death are negatively linearly related.

137. {Life Expectancy Narrative} What is the coefficient of determination? What does this

statistic tell you? ANSWER: =2R 0.225. This means that 22.5% of the variation in the age at death is explained by the

three variables: the average number of hours of exercise per week, the cholesterol level, and the number of points that the individual’s blood pressure exceeded the recommended value, while 77.5% of the variation remains unexplained.



138. {Life Expectancy Narrative} Interpret the coefficient 1b . ANSWER: 1b = 1.79. This tells us for each additional hour increase of exercise per week, the age at

death on average is extended by 1.79 years (assuming that the other independent variables in the model are held constant).

139. {Life Expectancy Narrative} Interpret the coefficient 2b .

ANSWER: 2b = -0.021. This tells us that for each additional unit increase in the cholesterol level, the

age at death on average is shortened by .021 years or equivalently about a week (assuming that the other independent variables in the model are held constant).

140. {Life Expectancy Narrative} Interpret the coefficient 3b . ANSWER: 3b = 0.016. This tells us for each additional point increase of the individual’s blood

pressure that exceeded the recommended value, the age at death on average is shortened by 0.016 years or equivalent, about six days (assuming that the other independent variables in the model are held constant).

FOR QUESTIONS 141 THROUGH 147, USE THE FOLLOWING NARRATIVE: Narrative: Demographic Variables and TV A statistician wanted to determine if the demographic variables of age, education, and income influence the number of hours of television watched per week. A random sample of 25 adults was selected to estimate the multiple regression model: εββββ ++++= 3322110 xxxy , where y is the number of hours of television watched last week, 1x is the age (in years), 2x is the number of years of education, and 3x is income (in $1,000). The computer output is shown below. THE REGRESSION EQUATION IS =y 321 12.029.041.03.22 xxx −−+


1x 0.41 0.19 2.158

2x -0.29 0.13 -2.231

3x -0.12 0.03 -4.00 S = 4.51 R-Sq = 34.8%




Source of Variation df SS MS F Regression 3 227 75.667 3.730 Error 21 426 20.286 Total 24 653

141. {Demographic Variables and TV Narrative} Test the overall validity of the model at the

5% significance level. ANSWER:


Rejection region: F > 0.05,3,21F = 3.07 Test statistic: F = 3.73 Conclusion: Reject the null hypothesis. The model is valid at α = .05.

142. {Demographic Variables and TV Narrative} Is there sufficient evidence at the 1%

significance level to indicate that hours of television watched and age are linearly related? ANSWER:

0: 10 =βH vs. :1H ≠1β 0 Rejection region: | t | > 0.005,21t = 2.831 Test statistic: t = 2.158 Conclusion: Don’t reject the null hypothesis. No, sufficient evidence at the 1%

significance level to indicate that hours of television watched and age are linearly related. 143. {Demographic Variables and TV Narrative} Is there sufficient evidence at the 1%

significance level to indicate that hours of television watched and education are negatively linearly related? ANSWER:

0: 20 =βH vs. :1H <2β 0 Rejection region: t < - 0.01,21t = -2.518 Test statistic: t = -2.231 Conclusion: Don’t reject the null hypothesis. No, sufficient evidence at the 1%

significance level to indicate that hours of television watched and education are negatively linearly related.



144. {Demographic Variables and TV Narrative} What is the coefficient of determination?

What does this statistic tell you? ANSWER: =2R 0.348. This means that 34.8% of the variation in the number of hours of television

watched per week is explained by the three variables: age, number of years of education, and income, while 65.2% remains unexplained.

145. {Demographic Variables and TV Narrative} Interpret the coefficient 1b . ANSWER:

1b = 0.41. This tells us that for each additional year of age, the number of hours of television watched per week on average increases by 0.41 (assuming that the other independent variables in the model are held constant).

146. {Demographic Variables and TV Narrative} Interpret the coefficient 2b .

ANSWER: 2b = -0.29. This tells us that for each additional year of education, the number of hours of

television watched per week on average decreases by 0.29 (assuming that the other independent variables in the model are held constant).

147. {Demographic Variables and TV Narrative} Interpret the coefficient 3b . ANSWER:

3b = -0.12. This tells us that for each additional year of $1000 in income, the number of hours of television watched per week on average decreases by 0.12 (assuming that the other independent variables in the model are held constant).

FOR QUESTIONS 148 THROUGH 155, USE THE FOLLOWING NARRATIVE: Narrative: Family Expenditure on Clothes An economist wanted to develop a multiple regression model to enable him to predict the annual family expenditure on clothes. After some consideration, he developed the multiple regression model εββββ ++++= 3322110 xxxy , where y is the annual family clothes expenditure (in $1,000), 1x is the annual household income (in $1,000), 2x is the number of family members, and 3x is the number of children under 10 years of age. The computer output is shown below.



THE REGRESSION EQUATION IS =y 321 26.093.0091.074.1 xxx +++


1x 0.091 0.025 3.640

2x 0.93 0.290 3.207

3x 0.26 0.180 1.444 S = 2.06 R-Sq = 59.6%



148. {Family Expenditure on Clothes Narrative} Test the overall model’s validity at the 5%

significance level ANSWER:


Rejection region: F > 0.05,3,46F ≈ 2.84 Test statistic: F = 22.647 Conclusion: Reject the null hypothesis. Yes, the model is valid at α = .05.

149. {Family Expenditure on Clothes Narrative}Test at the 5% significance level to determine

whether annual household income and annual family clothes expenditure are positively linearly related.

ANSWER:

0: 10 =βH vs. :1H 1β > 0 Rejection region: t > 0.05,46t ≈ 1.68 Test statistic: t = 3.64 Conclusion: Reject the null hypothesis. Yes, annual household income and annual family

clothes expenditure are positively linearly related.



150. {Family Expenditure on Clothes Narrative} Test at the 1% significance level to

determine whether the number of family members and annual family clothes expenditure are linearly related. ANSWER:

0: 20 =βH vs. 1 2: 0H β ≠ Rejection region: | t | > 0.005,36t ≈2.69 Test statistic: t = 3.207 Conclusion: Reject the null hypothesis. Yes, the number of family members and annual

family clothes expenditure are linearly related. 151. {Family Expenditure on Clothes Narrative} Test at the 1% significance level to

determine whether the number of children under 10 years of age and annual family clothes expenditure are linearly related. ANSWER:

0 3: 0H β = vs. 1 3: 0H β ≠ Rejection region: | t | > 0.005,46 2.69t ≈ Test statistic: t = 1.444 Conclusion: Don’t reject the null hypothesis. No sufficient evidence to conclude that the

number of children under 10 years of age and annual family clothes expenditure are linearly related.

152. {Family Expenditure on Clothes Narrative}What is the coefficient of determination?

What does this statistic tell you? ANSWER:

=2R 0.596. This means that 59.6% of the variation in the annual family clothes expenditure is explained by the three variables: annual household income, number of family members, and number of children under 10 years of age, while 40.4% of the variation remains unexplained.

153. {Family Expenditure on Clothes Narrative} Interpret the coefficient 1.b ANSWER:

1b = 0.091. This tells us that for each additional $1000 in annual household income, the annual family clothes expenditure increases on average by $91, assuming that the number of family members, and the number of children under 10 years of age in the model are held constant.



154. {Family Expenditure on Clothes Narrative} Interpret the coefficient 2b .

ANSWER: 2b = 0.93. This tells us that for each additional family member, the annual family clothes

expenditure increases on average by $930, assuming that the annual household income, and the number of children under 10 years of age in the model are held constant.

155. {Family Expenditure on Clothes Narrative} Interpret the coefficient 3b .

ANSWER: 3b = 0.26. This tells us that for each additional child under the age of 10, the annual

family clothes expenditure increases on average by $260, assuming that the number of family members and the annual household income in the model are held constant.

FOR QUESTIONS 156 THROUGH 163, USE THE FOLLOWING NARRATIVE: Narrative: Student’s Final Grade A statistics professor investigated some of the factors that affect an individual student’s final grade in his course. He proposed the multiple regression model εββββ ++++= 3322110 xxxy , where y is the final mark (out of 100), 1x is the number of lectures skipped, 2x is the number of late assignments, and 3x is the mid-term test mark (out of 100). The professor recorded the data for 50 randomly selected students. The computer output is shown below. THE REGRESSION EQUATION IS y = 321 63.17.118.36.41 xxx +−− Predictor Coef StDev T Constant 41.6 17.8 2.337

1x -3.18 1.66 -1.916

2x -1.17 1.13 -1.035

3x 0.63 0.13 4.846

S = 13.74 R-Sq = 30.0%





156. {Student’s Final Grade Narrative} What is the coefficient of determination? What does this statistic tell you?

ANSWER: =2R 0.30. This means that 30% of the variation in the student’s final grade in statistics

is explained by the three variables: number of lectures skipped, number of late assignments, and mid-term test grade, while 70% remains unexplained.

157. {Student’s Final Grade Narrative} Do these data provide enough evidence to conclude at

the 5% significance level that the model is useful in predicting the final mark? ANSWER:


Rejection region: F > 0.05,3,46F ≈ 2.84 Test statistic: F = 6.558 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the final mark.

158. {Student’s Final Grade Narrative} Do these data provide enough evidence to conclude at

the 5% significance level that the final mark and the number of skipped lectures are linearly related?

ANSWER:

0: 10 =βH vs. :1H ≠1β 0 Rejection region: | t | > 0.025,46t ≈2.014 Test statistic: t = -1.916 Conclusion: Don’t reject the null hypothesis. No, enough evidence to conclude at the 5%

significance level that the final mark and the number of skipped lectures are linearly related.

159. {Student’s Final Grade Narrative} Do these data provide enough evidence at the 5%

significance level to conclude that the final mark and the number of late assignments are negatively linearly related? ANSWER:

0: 20 =βH vs. :1H <2β 0 Rejection region: t < - 0.05,46t ≈ -1.679 Test statistic: t = -1.035 Conclusion: Don’t reject the null hypothesis. No, enough evidence at the 5% significance

level to conclude that the final mark and the number of late assignments are negatively linearly related.



160. {Student’s Final Grade Narrative} Do these data provide enough evidence at the 1% significance level to conclude that the final mark and the mid-term mark are positively linearly related? ANSWER:

0: 30 =βH vs. :1H >3β 0 Rejection region: t > 0.01,46t ≈2.412 Test statistic: t = 4.846 Conclusion: Reject the null hypothesis. Yes, these data provide enough evidence at the

1% significance level to conclude that the final mark and the mid-term mark are positively linearly related.

161. {Student’s Final Grade Narrative} Interpret the coefficient 1b .

ANSWER: 1b = -3.18. This tells us that for each additional lecture skipped, the student’s final score

on average decreases by 3.18 points, assuming that the number of late assignments, and the mid-term test mark (out of 100) in the model are held constant.

162. {Student’s Final Grade Narrative} Interpret the coefficient 2b . ANSWER:

2b = -1.17. This tells us that for each additional late assignment, the student’s final score on average decreases by 1.17 points, assuming that the number of lectures skipped, and the mid-term test mark (out of 100) in the model are held constant.

163. {Student’s Final Grade Narrative} Interpret the coefficient 3b . ANSWER: 3b = 0.63. This tells us that for each additional mid-term test score (out of 100), the

student’s final score on average increases by 0.63 points assuming that the number of lectures skipped, and the number of late assignments in the model are held constant.



FOR QUESTIONS 164 THROUGH 182, USE THE FOLLOWING NARRATIVE: Narrative: Real Estate A real estate builder wishes to determine how house size is influenced by family income, family size, and education of the head of household. House size is measured in hundreds of square feet, income is measured in thousands of dollars, and education is measured in years. A partial computer output is shown below. SUMMARY OUTPUT Regression Statistics Multiple R 0.865 R Square 0.748 Adjusted R Square 0.726 Standard Error 5.195 Observations 50 ANOVA df SS MS F Signif F Regression 3605.7736 901.4434 0.0001 Residual 1214.2264 26.9828 Total 49 4820.0000

Coeff. St. Error t Stat P-value

Intercept – 1.6335 5.8078 – 0.281 0.7798 Family Income 0.4485 0.1137 3.9545 0.0003 Family Size 4.2615 0.8062 5.286 0.0001 Education – 0.6517 0.4319 – 1.509 0.1383 164. {Real Estate Narrative} What percentage of the variability in house size is explained by

income? ANSWER: 74.8% of the variability in house size is explained by income 165. {Real Estate Narrative} Which of the independent variables in the model are significant

at the 2% level? ANSWER: Family income and family size 166. {Real Estate Narrative} Which of the following values for the level of significance is the

smallest for which all explanatory variables are significant individually: α = .01, .05, .10, and .15?

ANSWER: .15α =



167. {Real Estate Narrative} When the builder used a simple linear regression model with

house size as the dependent variable and education as the independent variable, he obtained an 2r value of 23.0%. What additional percentage of the total variation in house size has been explained by including family size and income in the multiple regression?

ANSWER: 74.8% - 23.0% = 51.8%. This means that additional 51.8% of the total variation in house

size has been explained by including family size and income in the multiple regression. 168. {Real Estate Narrative} Which of the following values for the level of significance is the

smallest for which at least two explanatory variables are significant individually: α = .01, .05, .10, and .15?

ANSWER: .01α = 169. {Real Estate Narrative} Which of the following values for the level of significance is the

smallest for which the regression model as a whole is significant: α = .00005, .001, .01, and .05?

ANSWER: .001α = 170. {Real Estate Narrative} What is the predicted house size for an individual earning an

annual income of $40,000, having a family size of 4, and having 13 years of education? ANSWER: 2488 square feet 171. {Real Estate Narrative} What minimum annual income would an individual with a

family size of 4 and 16 years of education need to attain a predicted 10,000 square foot home?

ANSWER: $211,850 172. {Real Estate Narrative} What minimum annual income would an individual with a

family size of 9 and 10 years of education need to attain a predicted 5,000 square foot home?

ANSWER: $44,140



173. {Real Estate Narrative} One individual in the sample had an annual income of $100,000, a family size of 10, and an education of 16 years. This individual owned a home with an area of 7,000 square feet. What is the residual (in hundreds of square feet) for this data point?

ANSWER:

-5.40 174. {Real Estate Narrative} One individual in the sample had an annual income of $10,000, a

family size of 1, and an education of 8 years. This individual owned a home with an area of 1,000 square fee (House = 10.00). What is the residual (in hundreds of square feet) for this data point?

ANSWER: y - y = 70 – 75.404 = - 5.404 or – 540.4 square feet 175. {Real Estate Narrative} Suppose the builder wants to test whether the coefficient on

income is significantly different from 0. What is the value of the relevant t – statistic? ANSWER: t = 3.9549 176. {Real Estate Narrative} At the 0.01 level of significance, what conclusion should the

builder draw regarding the inclusion of income in the regression model? ANSWER: Income is significant in explaining house size and should be included in the model

because its p value of .0003 is less than 0.01. 177. {Real Estate Narrative} Suppose the builder wants to test whether the coefficient on

education is significantly different from 0. What is the value of the relevant t – statistic? ANSWER: t = - 1.509 178. {Real Estate Narrative} What is the value of the calculated F test statistic that is missing

from the output for testing whether the whole regression model is significant? ANSWER: F = 901.4434/26.9828 = 33.408



179. {Real Estate Narrative} At the 0.01 level of significance, what conclusion should the builder draw regarding the inclusion of education in the regression model?

ANSWER: Education is not significant in explaining house size and should not be included in the

model because its p value of 0.1383 is larger than 0.01 180. {Real Estate Narrative} What are the regression degrees of freedom that are missing from

the output? ANSWER: df = 3605.7736/901.4434 = 4 181. {Real Estate Narrative} What are the residual degrees of freedom that are missing from

the output? ANSWER: df = 1214.2264/26.9828 = 45 182. {Real Estate Narrative} The observed value of the F – statistic is missing from the

printout. What are the numerator and denominator degrees of freedom for this F – statistic?

ANSWER: df = 4 for the numerator, and 45 for the denominator 183. Three predictor variables are being considered for use in a linear regression model. Given

the correlation matrix below, does it appear that multicollinearity could be a problem?

1x 2x 3x

1x 1.000

2x 0.025 1.000

3x 0.968 0.897 1.000

ANSWER: It appears that multicollinearity could be a problem because 3x is highly correlated with

both 1x and 2x .



184. Discuss some of the signals for the presence of multicollinearity. ANSWER: There are several clues to the presence of multicollinearity:

a. An independent variable known to be an important predictor ends up having a partial regression coefficient that is not significant.

b. A partial regression coefficient exhibits the wrong sign. c. When an independent variable is added or deleted, the partial regression coefficients

for the other variables change dramatically. A more practical way to identify multicollinearity is through the examination of a correlation matrix, which is a matrix that shows the correlation of each variable with each of the other variables. A high correlation between two independent variables is an indication of multicollinearity.

185. A statistician estimated the multiple regression model: εβββ +++= 22110 xxy , with 45

observations. The computer output is shown below. However, because of a printer malfunction, some of the results are not shown. These are indicated by the boldface letters a to l. Fill in the missing results (up to three decimal places).

Predictor Coef StDev T Constant a 3.51 2.03

1x 21.6 b 4.73

2x -12.5 7.61 c S = d R-Sq = e


Source of Variation df SS MS F Regression f i j l Error g 388 k Total h 519

ANSWER: a = 7.125 b = 4.567 c = -1.643 d = 3.039 e = .252 f = 2 g = 42 h = 44 i = 131 j = 65.5 k = 9.238 l = 7.090



186. What is meant by multicollinearity? ANSWER: Multicollinearity is a condition which indicates that two or more of the independent

variables are highly correlated with each other. 187. A multiple regression equation has been developed for y = daily attendance at a

community swimming pool, 1x = temperature (degrees Fahrenheit), and 2x = weekend versus weekday, ( 2x =1 for Saturday and Sunday, and 0 for other days of the week.) For the regression equation shown below, interpret each partial regression coefficient:

1 2100 10 175ˆ .y x x+ += ANSWER: The partial regression coefficient for 1x implies that, holding the day of the week

constant, a one degree Fahrenheit increase in the temperature will result in an increase of 10 in attendance. The partial regression coefficient for 2x implies that the attendance increases by 75 people on Saturdays and Sundays (assuming a constant temperature).



SECTION 4 MULTIPLE CHOICE QUESTIONS

In the following multiple-choice questions, please circle the correct answer. 188. If the Durbin-Watson statistic has a value close to 0, which assumptions is violated?

a. Normality of the errors b. Independence of errors c. Homoscedasticity d. None of the above. ANSWER: b

189. If the Durbin-Watson statistic d has values smaller than 2, this indicates

a. a positive first – order autocorrelation b. a negative first – order autocorrelation c. no first – order autocorrelation at all d. None of the above. ANSWER: a

190. If the Durbin-Watson statistic d has values greater than 2, this indicates

a. a positive first – order autocorrelation b. a negative first – order autocorrelation c. no first – order autocorrelation at all d. None of the above. ANSWER: b

191. If the Durbin-Watson statistic has a value close to 4, which assumption is violated?

a. Normality of the errors b. Independence of errors c. Homoscedasticity d. None of the above

ANSWER: b 192. The range of the values of the Durbin-Watson statistic d is

a. – 4 ≤ d ≤ 4 b. – 2 ≤ d ≤ 2 c. 0 ≤ d ≤ 4 d. 0 ≤ d ≤ 2 ANSWER: c



193. Which of the following statements is false? a. Time series data refer to data that are gathered at a specific period of time b. First – order autocorrelation is a condition in which a relationship exists between

consecutive residuals ie and 1,ie − where i is the time period c. Time series data refer to data that are gathered sequentially over a series of time

periods d. None of the above ANSWER: a

194. The Durbin – Watson test is used to test for positive first – order autocorrelation by

comparing its statistic value d to the critical values Ld and Ud available in most statistics books. Which of the following statements is true? a. If d < Ld , we conclude that there is enough evidence to show that positive first –

order autocorrelation exists. b. If d > Ld , we conclude that there is not enough evidence to show that positive first –

order autocorrelation exists c. If L Ud d d≤ ≤ , we conclude that the test is inconclusive. d. All of the above

ANSWER: d 195. In reference to the Durbin – Watson statistic d and the critical values Ld and Ud , which

of the following statements is false? a. If d > 4 - Ld , we conclude that the negative first – order autocorrelation exists b. If d < 4 - Ud , we conclude that there is not enough evidence to show that negative

first – order autocorrelation exists c. If Ud ≤ d ≤ 4 - Ud , we conclude that there is no evidence of first – order

autocorrelation d. None of the above ANSWER: d

196. In reference to the Durbin – Watson statistic d and the critical values Ld and Ud , which

of the following statements is false? a. If d < Ld , we conclude that positive first – order autocorrelation exists b. If d > Ud , we conclude that there is not enough evidence to show that positive first –

order autocorrelation exists c. If d < Ld or d > 4 - Ld , we conclude that there is no evidence of first – order

autocorrelation d. None of the above ANSWER: c



TRUE / FALSE QUESTIONS

197. The Durbin-Watson d statistic is used to check the assumption of normality. ANSWER: F 198. The Durbin-Watson test allows the statistics practitioner to determine whether there is

evidence of first – order autocorrelation. ANSWER: T

199. The Durbin-Watson statistic d is defined as 21

2 1( ) /

n n

i i ii i

d e e e−= =

= −∑ ∑ , where ie is the

residual at time period i. ANSWER: F 200. The range of the values of the Durbin-Watson statistic d is 0 4.d≤ ≤ ANSWER: T 201. Time series data refer to data that are gathered sequentially over a series of time periods. ANSWER: T 202. Small values of the Durbin-Watson statistic d (d < 2) indicate a negative first – order

autocorrelation. ANSWER: F 203. Large values of the Durbin-Watson statistic d (d > 2) indicate a positive first – order

autocorrelation. ANSWER: F 204. If the value of the Durbin-Watson statistic d satisfies the inequality L Ud d d≤ ≤ , where Ld

and Ud are the critical values for d, then the test for positive first – order autocorrelation is inconclusive.

ANSWER: T 205. If the value of the Durbin-Watson test statistic d satisfies the inequality d > 4 - Ld is a

critical value of d, we conclude that positive first – order autocorrelation exists. ANSWER: F 206. If the value of the Durbin-Watson test statistic d satisfies the inequalities d < Ld or d > 4

- Ld , where Ld and Ud are the critical values of d, we conclude that positive first – order autocorrelation exists.

ANSWER: T




207. Test the hypotheses: :0H There is no first-order autocorrelation vs. :1H There is negative

first-order autocorrelation given that: Durbin–Watson Statistic d = 1.75, n = 20, k = 2, and =α 0.01.

ANSWER: =Ld 0.86 and =Ud 1.27 The decision is made as follows: If d > 4 - =Ld 3.14, reject the null hypothesis and conclude that negative autocorrelation

is present. If 2.73 = 4 - Ud ≤ ≤d 4 - Ld = 3.14, we say that the test is inconclusive. If d ≤ 4 - =Ud 2.73, we conclude that there is no evidence of negative autocorrelation. Since d = 1.75, we conclude that there is no evidence of negative autocorrelation. 208. Test the hypotheses :0H There is no first-order autocorrelation vs. :1H There is

positive first-order autocorrelation, given that: Durbin–Watson Statistic d = 1.12, n = 45, k = 5, and =α 0.05.

ANSWER: =Ld 1.29 and =Ud 1.78 The decision is made as follows: If d < =Ld 1.29, reject the null hypothesis and conclude that positive autocorrelation is

present. If 1.29 = Ld ≤≤ d =Ud 1.78, we say that the test is inconclusive. If ≥d =Ud 1.78, we conclude that there is no evidence of positive autocorrelation. Since d = 1.12, we reject the null hypothesis and conclude that positive autocorrelation is

present. 209. If the residuals in a regression analysis of time ordered data are not correlated, the value

of the Durbin-Watson d statistic should be near __________. ANSWER: 2



210. If the value of the Durbin-Watson statistic d is small (d < 2), this indicates a __________(positive/negative) first – order autocorrelation exists.

ANSWER: positive 211. Test the hypotheses :0H There is no first-order autocorrelation vs. :1H There is first-

order autocorrelation, given that: Durbin–Watson Statistic d = 1.89, n = 28, k = 3, and =α 0.05.

ANSWER: =Ld 0.97, and =Ud 1.41 The decision is made as follows: If d < =Ld 0.97 or d > 4 - =Ld 3.03, reject the null hypothesis and conclude that the

autocorrelation is present.. If 0.97 = Ld ≤≤ d =Ud 1.41, or 2.59 = 4 - Ud ≤≤ d 4 - =Ld 3.03, we say that the test

is inconclusive. If 1.41 = Ud ≤≤ d 4 - =Ud 2.59, we conclude that there is no evidence of autocorrelation Since d = 1.70, we conclude that there is no evidence of autocorrelation. 212. If the value of the Durbin-Watson statistic d is large (d > 2), this indicates a __________

(positive/negative) first – order autocorrelation exists. ANSWER: negative 213. To use the Durbin-Watson test to test for positive first – order autocorrelation, the null

hypothesis will be :oH __________ (there is, there is no) first – order autocorrelation. ANSWER: there is no 214. To use the Durbin-Watson test to test for negative first – order autocorrelation, the null

hypothesis will be :oH __________ (there is, there is no) first – order autocorrelation. ANSWER: there is no 215. The range of the values of the Durbin-Watson statistic d is __________. ANSWER: 0 4d≤ ≤



216. Given that the Durbin-Watson test is conducted to test for positive first – order autocorrelation with .05α = , n = 20, and there are two independent variables in the model, the critical values for the test are Ld = __________ and Ud = __________, respectively.

ANSWER: 1.10 and 1.54


721

CHAPTER 18

MODEL BUILDING

SECTIONS 1

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 1. The model 21322110 xxxxy ββββ +++= +ε is referred to as:

a. first order model with two predictor variables with no interaction b. first order model with two predictor variables with interaction c. second order model with three predictor variables with no interaction d. second order model with three predictor variables with interaction ANSWER: b

2. The model εββββ +++++= p

p xxxy !!!2210 is referred to as a polynomial

model with: a. one predictor variable b. p predictor variables c. (p + 1) predictor variables d. x predictor variables ANSWER: a


722 Chapter Eighteen

3. For the following regression equation 2121 35820ˆ xxxxy +++= , which combination of

1x and 2x , respectively, results in the largest average value of y? a. 3 and 5 b. 5 and 3 c. 6 and 3 d. 3 and 6 ANSWER: c

4. Which of the following statements is false regarding the graph of the third – order

polynomial model 2 30 1 2 3y x x xβ β β β ε= + + + + ?

a. When 3β is negative, y decreases over the range of x b. When 3β is positive, y increases over the range of x c. The number of real – life applications of this model is quite large. Statistics

practitioners often encounter problems involving more than one curvature reversal d. The coefficients, 1β and 2β determine the position of the curvature changes, while

the coefficient 0β determines the point at which the curve interacts the y – axis ANSWER: c

5. For the following regression equation 2121 641050ˆ xxxxy −−+= , a unit increase in 2x ,

while holding 1x constant at a value of 3, decreases the value of y on average by: a. 56 b. 22 c. 50 d. An amount that depends on the value of 2x ANSWER: b

6. Suppose that the sample regression line of the first order model is 21 328ˆ xxy ++= . If we

examine the relationship between y and 1x for four different values of 2x , we observe that the: a. effect of x 1 on y remains the same no matter what the value of x 2 b. effect of x 1 on y remains the same no matter what the value of x 1 c. only difference in the four equations produced is the coefficient of x 2 d. not enough information is given to answer this question ANSWER: a

7. Suppose that the sample regression equation of a second-order model is given by

245.015.050.2ˆ xxy ++= . Then, the value 4.60 is the: a. predicted value of y for any positive value of x b. predicted value of y when x = 2 c. estimated change in y when x increases by 1 unit d. intercept where the response surface strikes the x-axis ANSWER: b


Model Building 723

8. For the following regression equation 2121 5152075ˆ xxxxy +−+= , a unit increase in 2x , while holding 1x constant at 1, changes the value of y on average by: a. - 5 b. +5 c. 10 d. –10 ANSWER: d

9. For the following regression equation 2121 4512100ˆ xxxxy −+−= , a unit increase in 1x ,

while holding 2x constant at a value of 2, decreases the value of y on average by: a. 92 b. 85 c. 20 d. an amount that depends on the value of 1x ANSWER: c

10. For the following regression equation 21 4310ˆ xxy ++= , a unit increase in 2x increases

the value of y on average by: a. 4 b. 7 c. 17 d. an amount that depends on the value of 1x ANSWER: a

11. In first order model with two predictors 1x and 2x , an interaction term may be used when

the: a. relationship between the dependent variable and the independent variables is linear b. effect of 1x on the dependent variable is influenced by 2x c. effect of 2x on the dependent variable is influenced by 1x d. both (b) and (c) ANSWER: d

12. The model 2

210 xxy βββ ++= +ε is referred to as: a. simple linear regression model b. first order model with one predictor variable c. second order model with one predictor variable d. third order model with two predictor variables ANSWER: c



13. The model 22110 xxy βββ ++= +ε is used whenever the statistician believes that, on average, y is linearly related to: a. 1x and the predictor variables do not interact b. 2x and the predictor variables do not interact c. either (a) or (b) d. both (a) and (b) ANSWER: d

14. For the following regression equation 2121 45615ˆ xxxxy +++= , a unit increase in 1x

increases the value of y on average by: a. 5 b. 30 c. 26 d. an amount that depends on the value of 2x ANSWER: d

15. When we plot x versus y, the graph of the model 2

210 xxy βββ ++= +ε is shaped like a: a. straight line going upwards b. straight line going downwards c. circle d. parabola ANSWER: d

16. Suppose that the sample regression equation of a model is 2121 3410ˆ xxxxy −++= . If

we examine the relationship between 1x and y for three different values of 2x , we observe that the: a. three equations produced differ only in the intercept b. coefficient of 2x remains unchanged c. coefficient of 1x varies d. three equations produced differ not only in the intercept term but the coefficient of

1x , also varies ANSWER: d

17. The model 22110 xxy βββ ++= +ε is referred to as:

a. first order model with one predictor variable b. first order model with two predictor variables c. second order model with one predictor variable d. second order model with two predictor variables ANSWER: b


Model Building 725

18. Which of the following is not an advantage of multiple regression as compared with analysis of variance? a. Multiple regression can be used to estimate the relationship between the dependent

variable and independent variables. b. Multiple regression handles qualitative variables better than analysis of variance. c. Multiple regression handles problems with more than two independent variables

easier than analysis of variance. d. All of the above are advantages of multiple regression as compared with analysis of

variance. ANSWER: b

19. Suppose that the sample regression equation of a second-order model is given by

245.015.050.2ˆ xxy ++= . Then, the value 2.50 is the: a. intercept where the response surface strikes the y-axis b. intercept where the response surface strikes the x-axis c. predicted value of y d. predicted value of y when x = 1 ANSWER: a

20. Which of the following statements is false regarding the graph of the second – order

polynomial model 20 1 2y x xβ β β ε= + + + ?

a. When we plot x versus y, the graph is shaped like a parabola b. If 2β is negative, the graph is concave while if 2β is positive, the graph is convex c. The greater the absolute value of 2β , the greater the rate of curvature d. The greater the absolute value of 2β , the smaller the rate of curvature ANSWER: d



TRUE / FALSE QUESTIONS 21. Suppose that the sample regression equation of a model is 1 2 1 24 1.5 2y x x x x= + + − . If we

examine the relationship between 1x and y for four different values of 2x , we observe that the four equations produced differ only in the intercept term.

ANSWER: F 22. The model 0 1 1 2 2y x xβ β β= + + is used whenever the statistician believes that, on

average, y is linearly related to 1x and 2x and the predictor variables do not interact. ANSWER: T 23 The graph of the model 2

0 1 2ˆi i iy x xβ β β= + + is shaped like a straight line going upwards. ANSWER: F 24. The model 2

0 1 2p

i i i p i iy x x xβ β β β ε= + + + + +L L L is referred to as a polynomial model with one predictor variable.

ANSWER: T 25. Suppose that the sample regression line of the first order model is 1 24 3 2y x x= + + . If we

examine the relationship between y and 1x for three different values of 2x , we observe that the effect of 1x on y remains the same no matter what the value of 2x .

ANSWER: T 26. In the first-order model 1 2 1 275 12 5 3y x x x x− + −= , a unit increase in 1x , while holding 2x

constant at a value of 2, decreases the value of y on average by 8 units. ANSWER: F 27. The model 2

0 1 2y x xβ β β ε= + + + is referred to as simple linear regression model. ANSWER: F 28. In first - order model with two predictors 1x and 2x , an interaction term may be used

when the relationship between the dependent variable y and the predictor variables is linear.

ANSWER: F 29. In the first-order model 1 2 1 260 40 10 5y x x x x+ − += , a unit increase in 2x , while holding

1x constant at 1, changes the value of y on average by –5 units. ANSWER: T


Model Building 727

30. In the first-order regression model 1 2 1 212 6 8 4y x x x x+ + += , a unit increase in 1x increases the value of y on average by 6 units.

ANSWER: F 31. The model 0 1 1 2 2 3 1 2y x x x xβ β β β ε= + + + + is referred to as second-order model with two

predictor variables with interaction. ANSWER: F 32. In the first-order model 1 28 3 5y x x+ += , a unit increase in 2x , while holding 1x constant,

increases the value of y on average by 5 units. ANSWER: T 33. The model 0 1 1 2 2y x xβ β β ε= + + + is referred to as first-order model with two predictor

variables with no interaction. ANSWER: T 34. Regression analysis allows the statistics practitioner to use mathematical models to

realistically describe relationships between the dependent variable and independent variables.

ANSWER: T 35. A first – order polynomial model with one predictor variable is the familiar simple linear

regression model. ANSWER: T 36. Suppose that the sample regression equation of a model is 1 2 1 24.7 2.2 2.6y x x x x= + + − . If

we examine the relationship between y and 2x for 1x = 1, 2, and 3, we observe that the three equations produced not only differ in the intercept term, but the coefficient of 2x also varies.

ANSWER: T 37. We interpret the coefficients in a multiple regression model by holding all variables in the

model constant. ANSWER: F 38. In the first-order model 1 2 1 250 25 10 6y x x x x+ − −= , a unit increase in 2x , while holding

1x constant at a value of 3, decreases the value of y on average by 3 units. ANSWER: F




39. A regression analysis was performed to study the relationship between a dependent

variable and five independent variables. The following information was obtained from the regression analysis: 2 0.80R = , SSR = 9600, and n = 40.

a. Develop the ANOVA table. b. Test the overall validity of the model at the 5% significant level.

ANSWER:

a. Source of Variation df SS MS F Regression 5 9600 1920 27.2 Error 34 2400 70.588 Total 39 12000

b. 0: 543210 ===== βββββH :1H At least one β i does not equal 0 Rejection region: 0.05,5,34 2.53F F> ≈ Test statistic: F = 27.2 Conclusion: Reject the null hypothesis. The model is valid at .05.α =

40. A regression analysis involving 25 observations and four independent variables revealed

that the total variation in the dependent variable y is 1600 and that the mean squares for error is 20.

a. Develop the ANOVA table. b. Test the validity of the model at the 1% significance level. ANSWER: a.

Source of Variation df SS MS F Regression 4 1200 300 15 Error 20 400 20 Total 24 1600

b. 0: 43210 ==== ββββH :1H At least one β i does not equal 0 Rejection region: 0.01,4,20 4.43F F> ≈ Test statistic: F = 15 Conclusion: Reject the null hypothesis. The model is valid at .01.α =


Model Building 729

FOR QUESTIONS 41 THROUGH 49, USE THE FOLLOWING NARRATIVE: Narrative: Sales Training Consider the following data for two variables, x and y. The independent variable x represents the amount of training time (in hours) for a salesperson starting a new car dealership to adjust fully, and the dependent variable y represents the weekly sales (in $1000s).

x 10 14 16 20 25 30 35 40 50 y 12 20 23 27 36 45 40 28 30

Use statistical software to answer the following questions. 41. {Sales Training Narrative} Develop an estimated regression equation of the form

.ˆ 10 xbby += ANSWER:

=y 17.245 + 0.441x 42. {Sales Training Narrative} Estimate the value of y when x = 45 using the estimated linear

regression equation in the previous question. ANSWER: y = 37.09 43. {Sales Training Narrative} Determine if there is sufficient evidence at the 5%

significance level to infer that the relationship between x and y is significant. ANSWER:

0: 10 =βH vs. 0: 11 ≠βH Rejection region: 0.025,7| |t t> =2.365 Test statistic: t = 1.836

Conclusion: Don’t reject the null hypothesis. No, sufficient evidence at the 5% significance level to infer that the relationship in between x and y is significant.

44. {Sales Training Narrative} Find the coefficient of determination of this simple linear model. What this statistic tell you about the model?

ANSWER:

Since the coefficient of determination =2R 0.325, this means that only 32.5% of the variation in weekly sales is explained by this simple linear model, while 67.5% remains unexplained. We may conclude that the simple linear model does not provide a good fit to this data set.



45. {Sales Training Narrative} Develop a scatter diagram for the data. Does the scatter

diagram suggest an estimated regression equation of the form 2210ˆ xbxbby ++= ?

Explain. ANSWER:

Yes, the scatter diagram suggests that a curvilinear relationship may be appropriate. 46. {Sales Training Narrative} Develop an estimated regression equation of the form

2210ˆ xbxbby ++= .

ANSWER:

215.177 3.166 0.0464y x x− + −=

47. {Sales Training Narrative} Determine if there is sufficient evidence at the 5% significance level to infer that the quadratic relationship between y, x, and 2x in the previous question is significant.

ANSWER:

0: 210 == ββH vs. :1H At least one iβ is not equal to zero. Rejection region: 0.05,2,6 5.14F F> = Test statistic: F = 13.616 Conclusion: Reject the null hypothesis. Yes, the quadratic relationship between y, x, and

2x in the previous question is significant.

Scatter Diagram

05

101520253035404550

0 10 20 30 40 50 60

x

y


Model Building 731

48. {Sales Training Narrative} Determine the coefficient of determination quadratic model. What does this statistic tell you about this model?

ANSWER:

Since =2R 0.819, this means that 81.9% of the variation in y is explained by this quadratic model. Clearly the quadratic model provides better fit to the data than the simple linear model.

49. {Sales Training Narrative} Use the quadratic model to predict the value of y when x = 45. ANSWER:

When x = 45.0, y = 33.333. FOR QUESTIONS 50 THROUGH 54, USE THE FOLLOWING NARRATIVE: Narrative: Football Teams An avid football fan was in the process of examining the factors that determine the success or failure of football teams. He noticed that teams with many rookies and teams with many veterans seem to do quite poorly. To further analyze his beliefs he took a random sample of 20 teams and proposed a second-order model with one independent variable. The selected model is

εβββ +++= 2210 xxy , where y = winning team’s percentage, and x = average years of

professional experience. The computer output is shown below. THE REGRESSION EQUATION IS =y 248.096.56.32 xx −+

S = 16.1 R-Sq = 43.9%




x 5.96 2.41 2.473 2x -0.48 0.22 -2.182



50. {Football Teams Narrative} Do these results allow us to conclude at the 5% significance level that the model is useful in predicting the team’s winning percentage?

ANSWER: 0: 210 == ββH vs. :1H At least one iβ is not equal to zero Rejection region: F > 0.05,2,17F =3.59 Test statistics: F = 6.663 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the team’s

winning percentage. 51. {Football Teams Narrative} Test to determine at the 10% significance level if the linear

term should be retained. ANSWER: 0: 10 =βH vs. :1H 01 ≠β Rejection region: | t | > 0.05,17t = 1.74 Test statistics: t = 2.473 Conclusion: Reject the null hypothesis. Yes, the linear term should be retained in the

model. 52. {Football Teams Narrative}Test to determine at the 10% significance level if the 2x term

should be retained. ANSWER: 0: 20 =βH vs. :1H 02 ≠β Rejection region: | t | > 0.05,17t = 1.74 Test statistics: t = -2.182 Conclusion: Reject the null hypothesis. Yes, the 2x term should be retained. 53. {Football Teams Narrative} Predict the winning percentage for a football team with an

average of 6 years of professional experience. ANSWER: ˆ 51.08y = 54. {Football Teams Narrative} What is the coefficient of determination? Explain what this

statistic tells you about the model. ANSWER: 2R = 0.439, which means that 43.9% of the variation in winning team’s percentage is

explained by the model, while 56.1% remains unexplained.


Model Building 733

FOR QUESTIONS 55 THROUGH 63, USE THE FOLLOWING NARRATIVE: Narrative: Traffic Fatalities A traffic consultant has analyzed the factors that affect the number of traffic fatalities. She has come to the conclusion that two important variables are the number of cars and the number of tractor-trailer trucks. She proposed the model =y εββββββ ++++++ 215

224

21322110 xxxxxx

(the second-order model with interaction), where y = number of annual fatalities per county, 1x = number of cars registered in the county (in 10,000), and 2x = number of trucks registered in the county (in 1000). The computer output (based on a random sample of 35 counties) is shown below: THE REGRESSION EQUATION IS =y 21

22

2121 13.051.015.161.73.117.69 xxxxxx −−−++

S = 15.2 R-Sq = 47.2% ANALYSIS OF VARIANCE


55. {Traffic Fatalities Narrative} Is there enough evidence at the 5% significance level to

conclude that the model is useful in predicting the number of fatalities? ANSWER: 0: 543210 ===== βββββH :1H At least one iβ is not equal to zero Rejection region: F > 0.05,5,29F =2.55 Test statistics: F = 5.181 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the number

of fatalities.


1x 11.3 5.1 2.216

2x 7.61 2.55 2.984 21x -1.15 0.64 -1.797 22x -0.51 0.20 -2.55

21xx -0.13 0.10 -1.30



56. {Traffic Fatalities Narrative} Test at the 1% significance level to determine if the 1x term should be retained in the model. ANSWER:

0: 10 =βH vs. :1H 01 ≠β Rejection region: | t | > 0.005,29t = 2.756 Test statistics: t = 2.216 Conclusion: Don’t reject the null hypothesis. The 1x term should not be retained in the

model. 57. {Traffic Fatalities Narrative} Test at the 1% significance level to determine if the 2x

term should be retained in the model. ANSWER:

0: 20 =βH vs. :1H 02 ≠β Rejection region: | t | > 0.005,29t = 2.756 Test statistics: t = 2.984 Conclusion: Reject the null hypothesis. Yes, the 2x term should be retained in the model. 58. {Traffic Fatalities Narrative} Test at the 1% significance level to determine if the 2

1x term should be retained in the model. ANSWER:

0: 30 =βH vs. :1H 03 ≠β Rejection region: | t | > 0.005,29t = 2.756 Test statistics: t = -1.797 Conclusion: Don’t reject the null hypothesis. The 2

1x term should not be retained in the model.

59. {Traffic Fatalities Narrative} Test at the 1% significance level to determine if the 2

2x term should be retained in the model. ANSWER:

0: 40 =βH vs. :1H 04 ≠β Rejection region: | t | > 0.005,29t = 2.756 Test statistics: t = -2.55

Conclusion: Don’t reject the null hypothesis. The 22x term should not be retained in the

model.


Model Building 735

60. {Traffic Fatalities Narrative} Test at the 1% significance level to determine if the interaction term should be retained in the model. ANSWER:

0: 50 =βH vs. :1H 05 ≠β Rejection region: | t | > 0.005,29t = 2.756 Test statistics: t = -1.3 Conclusion: Don’t reject the null hypothesis. The interaction tern should not be retained. 61. {Traffic Fatalities Narrative} What does the coefficient of 2

1x tell you about the model?

ANSWER: =3b -1.15. This can be interpreted as for each additional car 2 registered in the county

(holding other variables constant), the average number of traffic fatalities falls by 1.15. 62. {Traffic Fatalities Narrative} What does the coefficient of 2

2x tell you about the model?

ANSWER: =4b -0.51. This can be interpreted as for each additional tractor-trailer truck 2 registered

in the county (holding other variables constant), the average number of traffic fatalities falls by 0.51.

63. {Traffic Fatalities Narrative} What is the multiple coefficient of determination? What

does this statistic tell you about the model? ANSWER:

=2R 0.472. It means that 47.2% of the number of traffic fatalities is explained by the model, while 52.8% remains unexplained.

FOR QUESTIONS 64 THROUGH 69, USE THE FOLLOWING NARRATIVE: Narrative: Price of Gold An economist is in the process of developing a model to predict the price of gold. She believes that the two most important variables are the price of a barrel of oil )( 1x and the interest rate ).( 2x She proposes the first-order model with interaction; εββββ ++++= 31322110 xxxxy . A

random sample of 20 daily observations was taken. The computer output is shown below.



THE REGRESSION EQUATION IS =y 2121 36.17.143.226.115 xxxx −++



64. {Price of Gold Narrative} Do these results allow us at the 5% significance level to

conclude that the model is useful in predicting the price of gold?

ANSWER: 0: 3210 === βββH , :1H At least one iβ is not equal to zero Rejection region: F > 0.05,3,16F =3.24 Test statistics: F = 6.626 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the price of

gold. 65. {Price of Gold Narrative} Is there sufficient evidence at the 1% significance level to

conclude that the price of a barrel of oil and the price of gold are linearly related? ANSWER: 0: 10 =βH vs. :1H 01 ≠β Rejection region: | t | > 0.005,16t = 2.921 Test statistics: t = 3.141 Conclusion: Reject the null hypothesis. Yes, we conclude that the price of a barrel of oil

and the price of gold are linearly related?


1x 22.3 7.1 3.141

2x 14.7 6.3 2.333

21xx -1.36 0.52 -2.615


Model Building 737

66. {Price of Gold Narrative} Is there sufficient evidence at the 1% significance level to conclude that the interest rate and the price of gold are linearly related?

ANSWER: 0: 20 =βH vs. :1H 02 ≠β Rejection region: | t | > 0.005,16t = 2.921 Test statistics: t = 2.333 Conclusion: Don’t reject the null hypothesis. No, there sufficient evidence at the 1%

significance level to conclude that the interest rate and the price of gold are linearly related.

67. {Price of Gold Narrative} Is there sufficient evidence at the 1% significance level to

conclude that the interaction term should be retained? ANSWER: 0: 30 =βH vs. :1H 03 ≠β Rejection region: | t | > 0.005,16t = 2.921 Test statistics: t = -2.615 Conclusion: Don’t reject the null hypothesis. No, sufficient evidence at the 1%

significance level to conclude that the interaction term should be retained. 68. {Price of Gold Narrative} Interpret the coefficient 1b .

ANSWER:

1b = 22.3; as the price of a barrel of oil increases by one unit (holding the interest rate constant), the price of gold on average increases 22.3 units.

69. {Price of Gold Narrative} Interpret the coefficient 2b . ANSWER: 2b = 14.7; as the interest rate increases by one unit (holding the price of a barrel of oil

constant), the price of gold on average increases by 14.7 units. 70. A first - order model was used in regression analysis involving 25 observations to study

the relationship between a dependent variable y and three independent variables 1x , 2x , and 3x . The analysis showed that the mean squares for regression is 160 and the sum of squares for error is 1050. In addition, the following is a partial computer printout.

Predictor Coef StDev Constant 25 4

1x 18 6

2x -12 4.8

3x 6 5



a. Develop the ANOVA table. b. Is there enough evidence at the 5% significance level to conclude that the model is

useful in predicting the value of y? c. Test at the 5% significance level to determine whether 1x is linearly related to y. d. Is there sufficient evidence at the 5% significance level to indicate that 2x is

negatively linearly related to y? e. Is there sufficient evidence at the 5% significance level to indicate that 3x is

positively linearly related to y? ANSWER: a.

Source of Variation df SS MS F Regression 3 480 160 3.2 Error 21 1050 50 Total 24 1530

b. 0: 3210 === βββH , :1H At least one 0≠iβ Rejection region: 0.05,3,21 3.07F F> = Test statistic: F = 3.20

Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the value of y.

c. 0: 10 =βH vs. 0: 11 ≠βH Rejection region: 0.025,21| | 2.08t t> = Test statistic: t = 3 Conclusion: Reject the null hypothesis. Yes, 1x is linearly related to y. d. 0: 20 =βH vs. 0: 21 <βH Rejection region: 0.05,21 1.721t t< − = − Test statistic: t = -2.5 Conclusion: Reject the null hypothesis. Yes, 2x is negatively linearly related to y. e. 0: 30 =βH vs. 0: 31 >βH Rejection region: 0.05,21 1.721t t> = Test statistic: t = 1.20 Conclusion: Don’t reject 0H . The variable 3x is not positively linearly related to y.


Model Building 739

SECTIONS 2 - 3

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 71. In explaining the amount of money spent on children’s clothes each month, which of the

following independent variables is best represented with an indicator variable? a. Age b. Height c. Gender d. Weight ANSWER: c

72. In explaining students’ test scores, which of the following independent variables would

not best be represented with indicator variables? a. Gender b. Race c. Number of hours studying for the test d. Marital status ANSWER: c

73. In explaining starting salaries for graduates of accountancy programs, which of the

following independent variables would not best be represented with dummy variables? a. Grade point average b. Gender c. Race d. Marital status ANSWER: a

74. For a sample of 500 college professors, the estimated regression equation is given by

Ixy 23275ˆ −−= , where y is retirement age, x is pre-retirement annual income (in $1000s), and I is an indicator variable that takes the value of 0 for male professors and 1 for female professors. Assume that there is a relationship between y, x and I. For male professors with pre-retirement income of $70,000, the average age of retirement is: a. 75 b. 70 c. 65 d. 60 ANSWER: c



75. In explaining the income earned by college graduates, which of the following independent variables is best represented by a dummy variable? a. Grade point average b. Age c. Number of years since graduating from high school d. College major ANSWER: d

76. Suppose that the estimated regression equation of 200 College of Business graduates is

given by Ixy 15002000000,20ˆ ++= , where y is the starting salary, x is the grade point average and I is an indicator variable which takes the value of 1 if the student is a finance major and 0 if not. A business administration major graduate with 3.5 grade point average would have an average starting salary of a. $22,000 b. $27,000 c. $28,500 d. $20,000 ANSWER: b


Ixy 23275ˆ −−= , where y is retirement age, x is pre-retirement annual income (in $1000s), and I is an indicator variable that takes the value of 0 for male professors and 1 for female professors. Assume that there is a relationship between y, x and I. For each additional thousand dollars of pre-retirement income, the average age at retirement for male professors: a. increases by 3 b. increases by 2 c. decreases by 3 d. decreases by 2 ANSWER: c


Ixy 23275ˆ −−= , where y is retirement age, x is pre-retirement annual income (in $1000s), and I is an indicator variable that takes the value of 0 for male professors and 1 for female professors. Assume that there is a relationship between y, x and I. For female professors with pre-retirement income of $70,000, the average age of retirement is: a. 70 b. 63 c. 60 d. 58 ANSWER: b


Model Building 741

79. If a qualitative independent variable has 4 possible categories, the number of dummy variables needed to uniquely represent these categories is a. 5 b. 4 c. 3 d. 2 ANSWER: c

80. An indicator variable is a variable that can assume:

a. one of two values (usually 0 and 1) b. one of three values (usually 0, 1 and 2) c. any number of values d. None of the above ANSWER: a

81. An indicator variable is also called:

a. a response variable b. a dummy variable c. a predictor variable d. a dependent variable ANSWER: b

82. In general, to represent a qualitative independent variable that has m possible categories,

we must create: a. (m + 1) indicator variables b. m indicator variables c. (m – 1) indicator variables d. (m – 2) indicator variables ANSWER: c

83. In regression analysis, indicator variables allow us to use:

a. quantitative variables b. qualitative variables c. only quantitative variables that interact d. only qualitative variables that interact ANSWER: b

84. If a categorical independent variable contains 5 categories, the number of dummy

variables needed to uniquely represent these categories is a. 1 b. 2 c. 3 d. 4 ANSWER: d



85. Suppose that we want to model the randomized block design of the analysis of variance with, say, three treatments and four blocks. We would create: a. 7 indicator variables b. 6 indicator variables c. 5 indicator variables d. 4 indicator variables ANSWER: c


Ixy 23275ˆ −−= , where y is retirement age, x is pre-retirement annual income (in $1000s), and I is an indicator variable that takes the value of 0 for male professors and 1 for female professors. Assume that there is a relationship between y, x and I. If Michael Harris has a pre-retirement income of $70,000, he will retire at age: a. 70 b. 63 c. 60 d. Cannot be determined ANSWER: d

87. A dummy variable is used as an independent variable in a regression model when

a. the variable involved is numerical b. the variable involved is categorical c. a curvilinear relationship is suspected d. two independent variables interact ANSWER: b

88. We can incorporate any qualitative variable into regression analysis by creating one or

more dummy variables, also known as a. dependent variables b. response variables c. indicator variables d. None of the above ANSWER: c

89. Which of the following statements about dummy variables is false?

a. We can incorporate any qualitative variable into regression analysis by creating one or more dummy variables

b. Dummy variables are also known as binary variables, categorical variables, or indicator variables.

c. These variables take on only two values, namely 0 or 1, and those values then indicate the absence or presence of a particular qualitative characteristic.

d. None of the above ANSWER: d


Model Building 743

TRUE / FALSE QUESTIONS 90. In regression analysis, a nominal independent variable such as color, with three different

categories such as red, white, and blue, is best represented by three indicator variables to represent the three colors.

ANSWER: F 91. In regression analysis, indicator variables are also called dependent variables. ANSWER: F 92. In explaining the amount of money spent on children’s toys during Christmas each year,

the independent variable “gender” is best represented by a dummy variable. ANSWER: T 93. In general, to represent a nominal independent variable that has c possible categories, we

would create (c –1) dummy variables. ANSWER: T 94. An indicator variable (also called a dummy variable) is a variable that can assume either

one of two values (usually 0 and 1), where one value represents the existence of a certain condition, and the other value indicates that the condition does not hold.

ANSWER: T 95. When a dummy variable is included in a multiple regression model, the interpretation of

the estimated slope coefficient does not make any sense anymore. ANSWER: F 96. Dummy variables are variables that can take on only two values (namely, 0 or 1) and that

are used to indicate the absence or presence of a particular qualitative characteristic. ANSWER: T 97. In order to represent a nominal variable with m categories, we must create m – 1 indicator

variables. The last category represented by 1 2 1..... 0mI I I −= = = is called the empty category.

ANSWER: F 98. In regression analysis, all the variables must e interval. But in many real – life cases, one

or more independent variables may be nominal. ANSWER: T 99. Although we cannot use nominal data in regression analysis, it is possible to include

nominal variables in the regression model. ANSWER: T



100. If the nominal variable “religious affiliation” contains 4 categories: Catholic, Protestant, Muslim, and others, then the number of dummy variables needed to uniquely represent these categories is 4; one variable for each category.

ANSWER: F


Model Building 745

STATISTICAL CONCEPTS & APPLIED QUESTIONS FOR QUESTIONS 101 THROUGH 107, USE THE FOLLOWING NARRATIVE: Narrative: Incomes of Professionals An economist is analyzing the incomes of professionals (physicians, dentists, and lawyers). He realizes that an important factor is the number of years of experience. However, he wants to know if there are differences among the three professional groups. He takes a random sample of 125 professionals and estimates the multiple regression model εββββ ++++= 3322110 xxxy , where y = annual income (in $1,000), 1x = years of experience, 2x = 1 if physician and 0 if not, and 3x = 1 if dentist and 0 if not. The computer output is shown below. THE REGRESSION EQUATION IS =y 321 44.716.1007.265.71 xxx −++

S = 42.6 R-Sq = 30.9%



101. {Incomes of Professionals Narrative} Estimate the annual income for a physician with 15

years of experience. ANSWER: ˆ 112.86y = (in $1,000) or equivalently $112,860. 102. {Incomes of Professionals Narrative} Estimate the annual income for a dentist with 15

years of experience. ANSWER: ˆ 95.26y = (in $1,000) or equivalently $95,260.


1x 2.07 0.81 2.556

2x 10.16 3.16 3.215

3x -7.44 2.85 -2.611



103. {Incomes of Professionals Narrative} Estimate the annual income for a lawyer with 15 years of experience.

ANSWER: ˆ 73.72y = (in $1,000) or equivalently $73,720. 104. {Incomes of Professionals Narrative} Do these results allow us to conclude at the 1%

significance level that the model is useful in predicting the income of professionals? ANSWER: 0: 3210 === βββH , :1H At least one iβ is not equal to zero Rejection region: F > 0.01,3,121F ≈ 3.95 Test statistics: F = 18.008 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the income

of professionals. 105. {Incomes of Professionals Narrative}Is there enough evidence at the 5% significance

level to conclude that income and experience are linearly related? ANSWER: 0: 10 =βH vs. :1H 01 ≠β Rejection region: | t | > 0.025,121t ≈ 1.98 Test statistics: t = 2.556 Conclusion: Reject the null hypothesis. Yes, we can conclude that income and experience

are linearly related. 106. {Incomes of Professionals Narrative}Is there enough evidence at the1% significant level

to conclude that physicians earn more on average than lawyers? ANSWER: 0: 20 =βH vs. :1H 2β > 0 Rejection region: t > 0.01,121t ≈ 2.358 Test statistics: t = 3.215 Conclusion: Reject the null hypothesis. Yes, we can conclude that physicians earn more

on average than lawyers.


Model Building 747

107. {Incomes of Professionals Narrative}Is there enough evidence at the 10% significance level to conclude that dentists earn less on average than lawyers?

ANSWER: 0: 30 =βH vs. :1H 3β < 0 Rejection region: t < - 0.10,121t ≈ -1.289 Test statistics: t = -2.611 Conclusion: Reject the null hypothesis. Yes, we can conclude that dentists earn less on

average than lawyers? FOR QUESTIONS 108 THROUGH 116 USE THE FOLLOWING NARRATIVE: Narrative: Senior Accounting Students’ Grades A professor of accounting wanted to develop a multiple regression model to predict the students’ grades in her fourth-year accounting course. She decides that the two most important factors are the student’s grade point average in the first three years and the student’s major. She proposes the model εββββ ++++= 3322110 xxxy , where y = Fourth-year accounting course mark (out of 100), 1x = G.P.A. in first three years (range from 0 to 12), 2x = 1 if student’s major is accounting and 0 if not, and 3x = 1 if student’s major is finance and 0 if not. The computer output is shown below. THE REGRESSION EQUATION IS =y 321 16.542.1073.614.9 xxx +++




1x 6.73 1.91 3.524

2x 10.42 4.16 2.505

3x 5.16 3.93 1.313



108. {Senior Accounting Students’ Grades Narrative} Predict the score (out of 100) in the fourth year accounting course for an accounting major student who has 10.95 G.P.A. in first three years (range from 0 to 12).

ANSWER: ˆ 93.25y = 109. {Senior Accounting Students’ Grades Narrative} Predict the score (out of 100) in the

fourth year accounting course for a finance major student who has 10.95 G.P.A. in first three years (range from 0 to 12).

ANSWER: ˆ 87.9y = 110. {Senior Accounting Students’ Grades Narrative} Predict the score (out of 100) in the

fourth year accounting course for an international business major student who has 10.95 G.P.A. in first three years (range from 0 to 12).

ANSWER: ˆ 82.83y = 111. {Senior Accounting Students’ Grades Narrative} Do these results allow us to conclude at

the 1% significance level that the model is useful in predicting the fourth-year accounting course mark?

ANSWER: 0: 3210 === βββH , :1H At least one iβ is not equal to zero Rejection region: F > 0.01,3,96F ≈3.95 Test statistics: F = 25.386 Conclusion: Reject the null hypothesis. Yes, the model is useful in predicting the fourth-

year accounting course mark. 112. {Senior Accounting Students’ Grades Narrative} Do these results allow us to conclude at

the 1% significance level that on average accounting majors outperform those whose majors are not accounting or finance?

ANSWER: 0: 20 =βH vs. :1H 2β > 0 Rejection region: t > 0.01,96t ≈2.364 Test statistics: t = 2.505 Conclusion: Reject the null hypothesis. Yes, on average accounting majors outperform

those whose majors are not accounting or finance.


Model Building 749

113. {Senior Accounting Students’ Grades Narrative} Do these results allow us to conclude at the 1% significance level that on average finance majors outperform those whose majors are not accounting or finance?

ANSWER: 0: 30 =βH vs. :1H 3β > 0 Rejection region: t > 0.01,96t ≈2.364 Test statistics: t = 1.313 Conclusion: Don’t reject the null hypothesis. No, these results do not allow us to

conclude at the 1% significance level that on average finance majors outperform those whose majors are not accounting or finance.

114. {Senior Accounting Students’ Grades Narrative} Do these results allow us to conclude at

the 1% significance level that grade point average in first three years is linearly related to fourth-year accounting course mark?

ANSWER: 0: 10 =βH vs. :1H 01 ≠β Rejection region: | t | > 0.005,96t ≈2.626 Test statistics: t = 3.524 Conclusion: Reject the null hypothesis. Yes, grade point average in first three years is

linearly related to fourth – year accounting course mark. 115. {Senior Accounting Students’ Grades Narrative} Interpret the coefficient 2b .

ANSWER: 2 6.73.b = This can be interpreted as accounting major’s fourth year accounting course

mark (out of 100), on average, is 6.75 higher than those who are not in accounting or finance (assuming that the other independent variables in the model are held constant)..

116 {Senior Accounting Students’ Grades Narrative} Interpret the coefficient 3b .

ANSWER:

3b = 5.16. This can be interpreted as finance major’s fourth year accounting course mark (out of 100), on average, is 5.16 higher than those who are not in accounting or finance (assuming that the other independent variables in the model are held constant)...

117. In general, to represent a nominal variable with m categories, we must create __________

indicator variables. The last category represented by 1 2 1....... 0mI I I −= = = is called the __________.

ANSWER: m – 1, omitted category



118. What is a dummy variable, and how is it useful to multiple regression? Give an example of three dummy variables that could be used in describing your home town.

ANSWER: A dummy variable is a variable that takes on a value of one or zero to indicate the

presence or absence of an attribute. Dummy variables can help explain some of the variation in y due to the presence or absence of a characteristic. Three dummy variables that can be used to describe one town versus another are URBAN (1 if urban, 0 otherwise), MANUF (1 if durable goods manufacturing is the major industry, 0 otherwise), and POPMIL (1 if the population is 1 million or more, 0 otherwise). Other dummy variables could include the presence of a major university, a major medical center, a major research institution, and many more.


Model Building 751

SECTIONS 4 - 6

MULTIPLE CHOICE QUESTIONS In the following multiple-choice questions, please circle the correct answer. 119. Stepwise regression is an iterative procedure that:

a. adds one independent variable at a time b. deletes one independent variable at a time c. Either (a) or (b) d. Both (a) and (b) ANSWER: d

120. In stepwise regression procedure, if two independent variables are highly correlated,

then: a. both variables will enter the equation b. only one variable will enter the equation c. neither variable will enter the equation d. not enough information is given to answer this question ANSWER: b

121. In stepwise regression procedure, the independent variable with the largest F-statistic, or

equally with the smallest p- value, is chosen as the first entering variable. The standard, also called the F-to-enter, is usually set at F equals: a. 4 b. 2 c. 1 d. 0 ANSWER: a

122. In multiple regression analysis, which procedure permits variables to enter and leave the

model at different stages of its development? a. Forward selection b. Residual analysis c. Backward elimination d. Stepwise regression ANSWER: d



123. One of the requirements of regression analysis is that the dependent variable must be a. discrete b. continuous c. interval d. nominal

ANSWER: c 124. If the probability of an event is .20, then the odds ratio in favor of the event occurring is

expressed as a. 1 to 1 b. 1 to 2 c. 1 to 3 d. 1 to 4

ANSWER: d 125. Financial analysts would like to predict whether a company will become bankrupt in the

next year, using independent variables such as company sales, product costs, market share, and profits. In this situation, which of the following methods is appropriate? a. The least squares technique b. Logistic regression c. Stepwise regression d. All of the above

ANSWER: b 126. If the odds ratio that Jessica achieving 25 or higher in the Admission College Test (ACT)

is 4 (also expressed as 4 to 1), then the probability that Jessica will achieve her goal is a. 0.80 b. 0.85 c. 0.90 d. 0.95

ANSWER: a 127. The coefficients of the logistic regression model ln(y) = 0 1 1 2 2 k kx x xβ β β β ε+ + + + +L L

are estimated using a statistical technique called a. least squares method b. maximum likelihood estimation c. logistic likelihood estimation d. All of the above

ANSWER: b


Model Building 753

128. Which of the following statements is false? a. Minitab statistical software package performs the required calculations for logistic

regression. b. Excel does not conduct logistic regression analysis c. Logistic regression does not allow us to use nominal independent variables d. Multiple regression allows us to use nominal independent variables

ANSWER: c 129. Which of the following statements is false?

a. In linear regression, the coefficients describe the relationship between each of the independent variables and the dependent variable.

b. In linear regression, a negative coefficient of an independent variable means that when that variable increases by one unit (holding all other independent variables constant), on average the dependent variable will decrease by the amount of the coefficient.

c. In linear regression, a positive coefficient of an independent variable means that when the variable increases by one unit (holding all other independent variables constant), on average the dependent variable will increase by one unit

d. Interpreting the coefficients of the logistic regression is somewhat more complex. ANSWER: c 130. In stepwise regression procedure, if two independent variables are highly correlated,

then: a. both variables will enter the equation b. only one variable will enter the equation c. neither variable will enter the equation d. None of the above.

ANSWER: b 131. Stepwise regression is an iterative procedure that:

a. adds one independent variable at a time b. deletes one independent variable at a time c. neither (a) nor (b) is correct d. both (a) and (b) are correct

ANSWER: d 132. Stepwise regression is especially useful when there are:

a. a great many independent variables b. few independent variables c. a great many dependent variables d. few dependent variables

ANSWER: a



TRUE / FALSE QUESTIONS 133. If the odds ratio that an overweight person who smokes 15 or more cigarettes per day

suffers a heart attack is 9, then the probability that the person will suffer a heart attack is 0.81.

ANSWER: F 134. Logistic regression allows us to use nominal independent variables. ANSWER: T 135. In stepwise regression procedure, if two independent variables are highly correlated, then

neither variable will enter the equation. ANSWER: F 136. Stepwise regression is especially useful when there are many independent variables. ANSWER: T 137. Stepwise regression is an iterative procedure that adds and deletes one independent

variable at a time. ANSWER: T 138. Stepwise multiple regression is a procedure that develops a multiple-regression equation

in carefully delineated steps, either by means of the forward-selection method or the backward-elimination method.

ANSWER: T 139. The stepwise regression procedure begins by computing the simple regression model for

each independent variable. ANSWER: T 140. The stepwise regression procedure begins by computing the multiple regression model

for all independent variables of interest. ANSWER: F 141. In stepwise regression, the independent variable with the largest F- statistic, or equally

the smallest p- value is chosen as the first entering variable. ANSWER: T 142. In stepwise regression, if two independent variables are highly correlated, both variables

must enter the model simultaneously. ANSWER: F 143. At each step of the stepwise regression procedure, the p- values of all variables are

computed and composed to the F-to-remove. If a variable’s F- statistic falls below this standard, it is removed from the equation.

ANSWER: T


Model Building 755

STATISTICAL CONCEPTS & APPLIED QUESTIONS 144. In multiple regression, which procedure permits variables to enter and leave the model at

different stages of its development? ANSWER: Stepwise regression 145. A logistic regression equation is ln 1 2 3 4ˆ( ) .15 .03 .02 .01 .01y x x x x= + + + + .

a. What is the estimated odds ratio for the event of interest occurring when 1 30x = ,

2 3 460, 8, and 4x x x= = = . b. What is the estimated probability of the event?

ANSWER:

a. ln( y ) = 2.37, hence the estimated odds ratio y = 10.697. b. Estimated probability = y / (1+ y ) = 0.9145

146. Discuss briefly the procedure that is employed in the building of a model. ANSWER:

a. Clearly define the dependent variable that you wish to analyze or predict b. Using your knowledge of the dependent variable, produce a list of predictors that may

be related to the dependent variable. c. Gather the required observations (at least 6 for each independent variable used in the

equation) for the potential models. d. Use your knowledge of the dependent variable and predictor variables to identify and

formulate several possible models. e. Use statistical software to estimate the possible models. f. Determine whether the required conditions are satisfied. If not, attempt to correct the

problem. At this point, you may have several “equal” models from which to choose. g. Use your judgment and the statistical output to select the best model.

147. What is stepwise regression, and when is it desirable to make use of this multiple

regression technique? ANSWER: Stepwise regression is a multiple regression estimation technique whereby independent

variables are added to the regression equation one at a time. The first x variable to enter the regression is the one that explains the greatest amount of variation in y. The second variable to enter is the one that explains the greatest amount of the remaining variation. The use of stepwise regression can reduce the possibility for multicollinearity since it is unlikely that two highly correlated x variables will be included in a multiple regression that is estimated using the stepwise technique. This technique is useful when there are a great many independent variables.



148. The two largest values in a correlation matrix are the 0.89 correlation between y and 3x , and the 0.83 correlation between y and 7x . During a stepwise regression analysis 3x is the first independent variable brought into the equation. Will 7x necessarily be next? If not, why not?

ANSWER: Predictor variable 7x will not necessarily be the next variable brought into the equation.

We do not know about the correlation between 3x and 7x , so we cannot determine whether 7x will explain the greatest amount of the remaining variation in y.

149. In general, on what basis are independent variables selected for entry into the equation

during stepwise regression? ANSWER: Independent variables are selected for entry into the equation during stepwise regression

based upon the amount of the remaining variation in y (the variation that has not already been explained by included variables) that a candidate variable can explain.


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