An Introduction to Management Science

(Ninth edition)

Chapter 9: Network Models

Problem 1:

Find the shortest route from node 1 to each of the other nodes in the transportation network shown.

5

7 3 6

9 2

3

18

3

Node Shortest route from Node 1

Distance

2 1-2 73 1-3 94 1-2-5-6-4 175 1-2-5 12

22

3

46

7

1

5

6 1-2-5-6 147 1-2-5-6-7 17

Problem 2:

For the Gorman Construction Company problem (see Figure 9.1), assume that Node 7 is the company’s warehouse and supply center. Several daily trips are commonly made from node 7 to the other nodes or construction sites. Using node 7 as the starting node, find the shortest route from this node to each of the other nodes in the network.

Node Shortest route from Node 7

Distance

1 7-6-5-3-1 222 7-4-2 113 7-6-5-3 124 7-4 55 7-6-5 86 7-6 6

Problem 3:

In the original Gorman Construction Company problem, we found the shortest distance from the office (node 1) to each of the other nodes or construction sites. Because some of the roads are highways and others are city streets, the shortest-distance routes between the office and the construction sites may not necessarily provide the quickest or shortest-time routes. Shown here is the Gorman road network with travel time values rather than distance values. Find the shortest route from Gorman’s office to each of the construction sites if the objective is to minimize travel time rather than distance.

25

12 11

20 18

6

16 5

5

4

6

7

3

2

1

15

Problem 4:

Find the shortest route between nodes

1 and 8 in the following network:

The shortest route between node 1 and 8:

1-4-5-6-8 3+3+2+2= distance: 10

Problem 5:

Find the shortest route between nodes 1 and 10 in the following network:

1-3-5-8-10 5+6+3+5= distance: 19

Chapter 10: Project scheduling: PERT/CPM

Problem 1:

The Mohawk Discount Store is designing a management training program for individuals at its corporate headquarter. The company wants to design the program so that trainees can complete it as quickly as

Node Shortest route from Node 7

Distance

2 1-2 203 1-3 164 1-2-4 325 1-3-5 316 1-3-5-6 367 1-2-4-7 43

possible. Important precedence relationship must be maintained between assignments or activities in the program. For example, a trainee cannot serve as an assistant to the store manager until the trainee has obtained experience in the credit department and at least one sales department. The following activities are the assignments that must be completed by each trainee in the program. Construct a project network for this problem. Do not perform any further analysis.

Activity A B C D E F G HImmediate Predecessor

- - A A,B A,B C D,F E,G

C F

A

D G

Start Finish

E H

B

Problem 2:

Bridge City Developer is coordinating the construction of an office complex. As part of the planning process, the company generated the following activity list. Draw a project network that can be used to assist in the scheduling of the project activities.

Activity A B C D E F G H I JImmediate Predecessor

- - - A,B A,B D E C C F,G,H,I

A D F

Start B E G J Finish

H

C I

Problem 3:

Construct a project network for the following project. The project is completed when activities F and G are both completed.

Activity A B C D E F GImmediate Predecessor

- - A A C,B C,B D,E

A C E

D

Start G Finish

F

B

Problem 4:

Assume that the project in Problem 3 has the following activity times (in months).

Activity A B C D E F GImmediate Predecessor

4 6 2 6 3 3 5

A 0 44 0 4

Start

C 4 62 5 7

E 6 93 7 10D 4 10

6 4 10

B 0 66 1 7

G 10 155 10 15

Finish

a) Find the critical path.b) The project must be completed in and year and a half. Do you anticipate difficulty in meeting the deadline? Explain.

a) The critical path A-D-G

b) No, I don’t find a difficulty, because it will be 15 months.

Problem 5:

MDS is a consulting company that specializes in the development of decision support systems. MDS has just obtained a contract to develop a computer system to assist the management of a large company in formulating its capital expenditure plan. The project leader has developed the following list of activities and immediate predecessors. Construct a project network for this problem.

Activity A B C D E F G H I JImmediate Predecessor

- - - B A B C,D B,E F,G H

A E H

Start B D J Finish

F I

C G

Chapter 14: Decision Analysis

F 6 93 12 15

Problem 1:

The following payoff table shows profit for a decision analysis problem with two decisions and three states of natures.

a. Construct a decision tree for this problemb. If the decision maker knows nothing about the probabilities of the three states of nature, what is

the recommended decision, using the optimistic, conservative, and minimax regret approaches?

Decision Alternative State of natureS1 S2 S3

D1 250 100 25D2 100 100 75

a. s1 250

D1 s2 100

s3 25

s1 100

s2 100

D2 s3 75

b.

Decision Maximum Profit Minimum profitS1 S3

D1 250 25D2 100 75

Optimistic approach (250 is the largest payoff) – Select D1

Conservative approach (75 is the best of the worst possible payoffs) – Select D2

Minimax regret approach D1 – 50; D2- 150; Select D1

Decision Maximum Profit Minimum profitS1 S3

1

2

3

D1 0 50D2 150 0

Problem 2:

Suppose that a decision maker faced with four decision alternatives and four states of nature develops the following profit payoff table.

Decision Alternative State of natureS1 S2 S3 S4

D1 14 9 10 5D2 11 10 8 7D3 9 10 10 11D4 8 10 11 13

a. If the decision maker knows nothing about the probabilities of the four states of nature, what is the recommended decision using the optimistic, conservative, and minimax regret approaches?

b. Which approach do you prefer? Explain. Is establishing the most appropriate approach before analyzing the problem important for the decision maker? Explain.

c. Assume that the payoff table provides cost rather than profit payoffs. What is the recommended decision using the optimistic, conservative, and minimax regret approach?

a.

Optimistic approach select D1, 14

Conservative approach select D3, 9

Minimax regret approach select D3, 5

Decision Alternative State of natureS1 S2 S3 S4

D1 0 1 1 8D2 3 0 2 6D3 5 0 1 2D4 6 0 0 0

b. Yes, it is important to establish the most appropriate approach before analyzing, because in that way you can minimize the loses and maximize the profits. I prefer the minimax regret approach because like this the possibility to lose more is the smallest.

c. Optimistic approach Select D1, 5Conservative approach Select D2, D3, 11Minimax regret approach Select D2

Problem 3:

Southland Corporation’s decision to produce a new line of recreational products has resulted in the need to construct either a small plant or a large plant. The selection of plant size depends on how the marketplace reacts to the new product line. To conduct an analysis, marketing management has decided to view the possible long-run demand as either low, medium, or high. The following payoff table shows the projected profit in millions of dollars:

Decision Alternative Long-run demandLow Medium High

Small plant 150 200 200Large plant 50 200 500

a. Construct a decision tree for this problemb. Recommend a decision based on the use of the optimistic, conservative, and minimax regret

approaches.

Low 150

a.

Medium 200

1

2

Small plant

High 200

Low 50

Large plant

Medium 200

High 500

b. Optimistic Approach Choose High long-run demand and Large plant 500Conservative approach Choose Low long-run demand and Small plant 150Minimax regret approach Choose Low long-run demand and Large plant 100

Decision Alternative Long-run demandLow Medium High

Small plant 0 0 300Large plant 100 0 0

Problem 4:Investment Advisors, Inc. , has three investment strategies under consideration. Profits from the strategies will depend on what happens to the prime interest rate over the next three months. Payoffs (in thousands of dollars) are shown.

Decision Alternative State of natureRate decrease S1 No Change S2 Rate Increase S3

Strategy D1 50 70 40Strategy D2 55 35 80Strategy D3 15 60 70

What investment strategy would you recommend based on the use of optimistic, conservative and minimax regret approaches?

Optimistic Strategy D2, 80Conservative Strategy D1, 40Minimax regret Strategy D2, 35

Decision Alternative State of natureRate decrease S1 No Change S2 Rate Increase S3

Strategy D1 5 0 40Strategy D2 0 35 0Strategy D3 40 10 10

Problem 5:The following profit payoff table was presented in Problem 1. Suppose that the decision maker has obtained the probability estimates: P(s1) = 0.65, P(s2) = 0.15, and P(s3) = 0.20. Use the expected value approach to determine the optimal decision.

Decision Alternative State of natureS1 S2 S3

D1 250 100 25D2 100 100 75

EV (d1) = 0.65(250) +0.15(100) +0.20(25) == 162.5 +15 +5 == 182.5

EV (d2) = 0.65(100) +0.15(100) +0.20(75) == 65+15+15 == 95

The optimal decision is D1 with an expected value of 182.5!

Chapter 17: Markov Processes

Problem 1:

In the market share analysis of section 17.1, suppose that we are considering the Markov process associated with the shopping trips of one customer, but we don’t know where the

customer shopped during the last week. Thus, we might assume a 0.5 probability that the customer shopped at Murphy’s and 0.5 probability that the customer shopped at Ashley’s at period 0; that is π1(0) = 0.5 and π2(0) = 0.5. Given these initial state probabilities, develop a table similar to table 17.2 showing the probability of each state in future periods. What do you observe about the long-run probabilities of each state?

Current weekly shopping period

Next weekly shopping periodMurphy’s Food liner Ashley’s Supermarket

Murphy’s Food liner 0.9 0.1Ashley’s Supermarket 0.2 0.8

[π1(0) π2(0)] = [0.5 0.5]

[π1(1) π2(1)] = [π1(0) π2(0)] [p11 p12p21 p22

]

= [0.5 0.5] [0.9 0.10.2 0.8

]

= [0.55 0.45]

[π1(2) π2(2)] = [0.55 0.45] [0.9 0.10.2 0.8

]

= [0.585 0.415]

[π1(3) π2(3)] = [0.585 0.415] [0.9 0.10.2 0.8

]

= [0.609 0.39]

[π1(4) π2(4)] = [0.609 0.39] [0.9 0.10.2 0.8

]

= [0.626 0.373]

[π1(5) π2(5)] = [0.626 0.373] [0.9 0.10.2 0.8

]

= [0.638 0.361]

[π1(6) π2(6)] = [0.638 0.361] [0.9 0.10.2 0.8

]

= [0.646 0.353]

[π1(7) π2(7)] = [0.646 0.353] [0.9 0.10.2 0.8

]

= [0.652 0.347]

State probability

Period (n)0 1 2 3 4 5 6 7

π1(n) 0.5 0.55 0.585 0.609 0.626 0.638 0.646 0.652π2(n) 0.5 0.45 0.415 0.39 0.373 0.361 0.353 0.347

Problem 2:Management of the New Fangled Softdrink Company believes that the probability of a customer purchasing Red Pop or the company’s major competition, Super Cola, is based on the customer’s most recent purchase. Suppose that the following transition probabilities are appropriate:

From ToRed Pop Super Cola

Red Pop 0.9 0.1Super Cola 0.1 0.9

a. Show the two-period tree diagram for a customer who last purchased Red Pop. What is the probability that this customer purchases Red Pop on the second purchase?

b. What is the long-run market share for each of these two products?c. A RP advertising campaign is being planned to increase the probability of attracting SC

customers. Management believes that the new campaign will increase to 0.15 the probability of a customer switching from SC to RP. What’s the projected effect of the advertising campaign on the market shares?

Period 1 Period 2

a.

Red Pop (0.9)(0.9) = 0.81

0.9

0.9 0.1 Super Cola (0.9)(0.1) = 0.09

Shops at Red Pop

0.1 0.1 Red Pop (0.1)(0.1) = 0.01

0.9 Super Cola (0.1)(0.9) = 0.09

Super Cola = 0.18

Red Pop = 0.81 + 0.01 = 0.82 is the probability the customer purchases Red Pop on the second Purchase!

b. [π1(0) π2(0)] = [0.5 0.5]

[π1(1) π2(1)] = [π1(0) π2(0)] [p11 p12p21 p22

]

= [0.5 0.5] [0.9 0.10.1 0.9

]

= [0.5 0.5] [π1(2) π2(2)] = [0.5 0.5]

So, the long-run market share is π1 = 0.5, π2 = 0.5!

c. π1 = 0.9π1 + 0.15π2

π2 = 0.1π1 + 0.9 π2

π1+π2 = 1

π1 = 0.9π1 + 0.15(1−π 1)π1 = 0.9 π1 + 0.15 – 0.15π1

π1 = 0.75π1 +0.150.25 π1 = 0.15π1 = 0.6

π2 = 1 – π1

π2 = 1 - 0.6π2 = 0.4

Customer who last purchased Red Pop

Shops at Super Cola

Problem 3:

a. [π1(0) π2(0)] = [1 0]

[π1(1) π2(1)] = [π1(0) π2(0)] [p11 p12p21 p22

]

= [1 0] [0.90 0.100.30 0.70

]

= [0.9 0.10]

The probability of the system being down in the next our of operation: π2(1) = 0.10

b.

π1 = 0.90 π1 + 0.30 π2

π2 = 0.10π1 + 0.70 π2

π1+π2 = 1

π1 = 0.90π1 + 0.30(1−π 1)π1 = 0.90 π1 + 0.30 – 0.30π1

π1 = 0.60π1 +0.300.40 π1 = 0.30π1 = 0.75

π2 = 1 – π1

π2 = 1 - 0.75π2 = 0.25

Problem 4:

a. π1 = 0.95π1 + 0.60π2

π2 = 0.05π1 + 0.40 π2

π1+π2 = 1

π1 = 0.95π1 + 0.60(1−π 1)π1 = 0.95 π1 + 0.60 – 0.60π1

π1 = 0.35π1 +0.600.65 π1 = 0.60π1 = 0.92 is the steady-state probability of the system being in the running state.

π2 = 1 – π1

π2 = 1 - 0.92π2 = 0.08 is the steady-state probability the system being in the down state.

Problem 5:

a.

From ToNo traffic delay Traffic delay

No traffic delay 0.85 0.15Traffic delay 0.25 0.75

[π1(0) π2(0)] = [0.5 0.5]

[π1(1) π2(1)] = [π1(0) π2(0)] [p11 p12p21 p22

]

= [0.5 0.5] [0.85 0.150.25 0.75

]

= [0.55 0.45]

[π1(2) π2(2)] = [0.55 0.45] [0.85 0.150.25 0.75

]

= [0.58 0.42]

π2(2) = 0.42The probability for the next 60 minutes the system will be in the delay state is 0.42!

b. π1 = 0.85π1 + 0.25π2

π2 = 0.15π1 + 0.75 π2

π1+π2 = 1

π1 = 0.85π1 + 0.25(1−π 1)π1 = 0.85 π1 + 0.25 – 0.25π1

π1 = 0.6π1 +0.250.4 π 1 = 0.25π1 = 0.625 – That is the probability that in the long run the traffic will not be in the delay state!

π2 = 1 – π1

π2 = 1 – 0.625π2 = 0.375

Chapter 18: Dynamic Programming

Problem 1:

Route Value1-4-5-8-10 221-4-5-9-10 251-4-6-8-10 271-4-6-9-10 231-3-5-8-10 191-3-5-9-10 221-3-6-8-10 261-3-6-9-10 221-3-7-8-10 22

1-3-7-9-10 211-2-5-8-10 221-2-5-9-10 251-2-6-8-10 241-2-6-9-10 201-2-7-8-10 251-2-7-9-10 24

The dynamic programming reduces the number of computation by decomposing the large problem into a number of interrelated smaller problems, which are easier to solve.

Problem 2:

a.

Stage 1:

Input Node Arc (decision) Shortest distance to Node 107 7-10 88 8-10 109 9-10 6

Stage 2:

Input Node Arc (decision) Shortest distance to Node 105 5-9 116 6-9 17

Stage 3:

Input Node Arc (decision) Shortest distance to Node 102 2-5 183 3-5 194 4-6 21

Stage 4:

1-2 10

1-3 8

1-4 5

So the shortest route from node 1 to node 10 is 1-4-6-9-10 with a distance of (5+4+11+6) = 26.

b. In the same way it is: 4-6-9-10 with a distance of (4+11+6) = 21.

c.

Route Value1-2-5-7-10 321-2-5-8-10 361-2-5-9-10 281-3-5-7-10 311-3-5-8-10 351-3-5-9-10 271-3-6-8-10 341-3-6-9-10 311-4-6-8-10 291-4-6-9-10 26

Dynamic programming reduces the number of computation by decomposing the large problem into a number of interrelated smaller problems, which are easier to solve.

Problem 3:

Problem 4:

Problem 5: