BDM Sample Final Exam + Solutions

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NCCB501/MBQC 862: BUSINESS DECISION MODELS

PRACTICE FINAL EXAMINATION

Please sign the following confidentiality statement.

I will not convey nor have I conveyed any information relating to the contents of this examination to any other person. Furthermore, I have received no unauthorized information relating to the contents of this exam.

Signature: __________________________________________________

The next page contains instructions for the exam. Read them carefully.Solutions to the exam are entered in blue text. Numbers below may not match values shown on exam.

Question Value

1 16%2 15%

3 10%

4 12%

5 18%

6 18%

7 11%

Total 100%

Participant Name: Student Number

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Instructions

1. This is an OPEN BOOK and NOTES examination. Hand-held calculators and personal computers are permitted, but personal computers must not be connected to a communications port.

2. Be sure to put your name and/or student number on the cover page in the space provided.

3. Be sure to sign the confidentiality statement.

4. This examination consists of 7 questions on 14 pages, including the cover page, the instructions page and a datasheet. Be sure that all pages are present before starting.

5. Error: Reference source not foundError: Reference source not foundALL 14 PAGES MUST BE RETURNED WITH YOUR COMPLETED EXAMINATION. If you separate the pages, be sure to staple them in the correct order before handing in the exam.

6. If you have difficulty understanding a question or feel that some information is missing, describe your problem, and make and state an assumption that will allow you continue with your analysis, and proceed with the question.

7. Answer all questions in the spaces provided on this examination. The back side of the preceding page may be used for your rough work and calculations if necessary.

8. Time: 3 hours.

9. The relative values of the questions are shown on each page and in the table on the cover page. Budget your time accordingly.

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Question 1 (16%)

Suppose you have set up a local firm to manufacture anchors and related equipment for sailing yachts. Your principal product lines are: 2 sizes of plow anchor (A and B) and 2 sizes of folding anchor (C and D). Your scarce resources in production are the following:

• you have 150 hours of manual labour per month (your own labour; cost is zero)• you have accepted a contract committing you to buy 120 hours of welding per month @

$8.00/hr. Additional welding time can be purchased @ $12.00/hr• you have a storage space limitation of 55 cu.meters. (useless for any other purpose)

Anticipating a rise in steel prices, you have laid in a large inventory of steel which is the only major raw material requirement. You anticipate product demands, on a monthly basis, as follows:

Product A B C DDemand 20 30 40 50

Due to the fixed nature of the resource base, you have chosen to maximize the contribution to gross revenue. The coefficients below, in dollars, therefore do not include the cost of labour, welding, steel or overheads. An LP formulation and solution for your problem is:MAX 60 A + 35 B + 24 C + 25 D SUBJECT TO 2) 3 A + 3 B + 1.5 C + 1.5 D <= 150 (manual labour) 3) 1.2 A + B + .5 C + .5 D <= 120 (welding hours) 4) A + .9 B + .6 C + .5 D <= 55 (storage space) 5) A <= 20 ("A" demand) 6) B <= 30 ("B" demand) 7) C <= 40 ("C" demand) 8) D <= 30 ("D" demand) 9) A,B,C,D >= 0

Question 1. (continued)Consider each part below to be independent of the other parts.

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(a) What would you pay for an additional hour of manual labour?

Since the cost of labour in the objective function coefficients = 0 you’d pay $16 for one more hour of labour (the shadow price of labour).

What would you pay for an additional hour of welding time?The shadow price of welding is $0, so you’d pay nothing for more welding time (you have

too much already).

(b) Suppose you are requested to sell some welding time to another firm. What price would you charge, and how much would you sell at this price?You have an excess of 66 hours of welding time at the moment, so you’d sell this amount for anything you can get (you’ve already paid $8/hour by the contract, so you’d like to get$8/hr, but anything would be better than leaving it idle.

(c) An acquaintance offers to become a manual labourer for you at a wage of $7/hr, but insists on a minimum work month of 10 hours. Explain why you should hire him or not.Since the offer is only $7 and your shadow price for labour is $16, you’d like to hire this person for at least one hour. We know that the shadow price of labour is $16 for an allowable increase of 5 hours, so the net profit contribution for the first 5 hours is 5 x 16 = $80. If you hire the acquaintance for the full 10 hours, the cost will be $70. So you should hire him, even if the second 5 hours of his time are unproductive.

(c) A dealer offers to sell some more of your A type anchors, under the following conditions - you must reduce your price to him by $10 per anchor but guarantee that you will provide him 10 "A" anchors/month. Explain why you should or should not take the offer.We currently sell 20 A anchors. We know the shadow price on the “A Demand” constraint is $12, so each additional “A” we can sell at the current price would increase profits by $12. Even if we reduce the price by $10 on these extra A’s, we’ll still be ahead$2 for each one, so at the margin it’s a good deal. The allowable increase in the RHS of the “A Demand” constraint is 15, so we should take the offer for the full 10 new A anchors.

(d) Finally, suppose that your data on the profit contribution of A Anchors is subject to uncertainty, in that you have not negotiated the contract yet. You feel that the actual contribution could be as low as $50/anchor, or as high as $80/anchor. Describe as best you can what the objective function value will be at these two extremes (giving either a specific value, or a range of values).We see that the allowable increase of A price for which the current solution is optimal is +infinity, so if the price is anything more that $60, we’ll still make 20 of them. The allowable decrease is $12, so we will continue to make 20 A’s until the price drops to $48 or less. Since $50 is within the range, we’ll make 20 A anchors anyway. If the price was $50 instead of $60, we’d lose $10 on each of the 20 A’s, so we’d lose $200. If the price was $80, we’d get an additional $20 per anchor, for a net gain of $400.

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Question 2 (15%)

Nanton Biometrics is a small firm that uses the techniques of genetic engineering to produce pharmaceuticals. Its primary product is Setorin, a drug that has shown good results in treating a common form of bowel cancer. Setorin is now beginning to show its age and the company must decide whether to stay with Setorin or to take the risk of developing a new drug to replace it. If the company stays with Setorin and the competition does not develop a competitor drug, the company can expect to earn $20 million over the next few years. The company estimates only a 0.40 chance of a new drug being developed by the competition but if that happens its earnings will be reduced from $20 million to $8 million. The research program to develop the new drug will cost $10 million and has a projected probability for a successful outcome of 0.80. If the development of the new drug is unsuccessful, the company can continue to market Setorin with the same possible exposure to competition described above. The new drug, if the development is successful, will earn the company $40 million over the next few years. If the development is successful, the company has the opportunity to spend an additional $8 million on a research project that will refine the new drug and result in a “great leap forward” in anti-cancer drug therapies. The probability of success of this follow-on project is only 0.20, but if successful, it will result in extra earnings of $35 million for a gross total of $75 million. If the follow-on project is unsuccessful, the company will still have the “unrefined” version of the new drug to fall back on. In the space provided below, represent the company’s decision problem as a tree and state the course of action that would be dictated by an EMV based decision.Below is the treeplan version of the problem… you could also have done a pencil and paper version.The course of action that yield the highest expected monetary value is to Develop the new drug, and if it is successful, do not attempt to refine it.

Question 3 (10%)

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You are the production manager for a firm that produces computer disk drives. You believe that, in a re-cent batch of drives, 10% are defective. A good drive installed nets the firm a profit of $300. If a bad drive is installed, it is reworked and replaced with a net cost of $200.

The cost of reworking a drive before installation is $100, but rework yields a good drive for sure. A drive can also be tested for $25, and the test indicates “Pass” or “Fail”.

The test is not perfect, and has a tendency to fail drives that are good, but sometimes passes bad drives. Experience with the test indicates that, with a 10% actual defective proportion, 2/3 of the drives will pass the test and 1/3 will fail. If a drive passes the test, there is only a 5% chance it will be a bad drive in fact.

The decision tree below describes the problem, showing the possibilities, probabilities and outcomes from each choice.

0.9Drive OK

$300.00Install the drive $300

0.1Drive fails

-$200.00-$200

Re-work the drive$200.00

0.95Drive OK

0.67 $275.00Drive Passes: install $300

0.05Drive fails

Test the drive -$225.00-$200

-250.33

Drive fails: rework$175.00

$200

a) [3 marks] From this tree, using expected value as a criterion, what should the company do?Expected values are inserted in the tree above. The decision that maximizes the EMV is to Install the Drive, for an EV of $250.

EV = $250

EV = $200

EV = $250EV = $225.25

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Question 3 Continued

The company has also developed a utility function that is appropriate for making decisions on one drive, shown below. Note that on this curve, the utility of -$250 is zero, and of +$300 is 1.

0

0.1

0.20.3

0.4

0.5

0.6

0.70.8

0.9

1

-250 -200 -150 -100 -50 0 50 100 150 200 250 300

dollar value

uti

lity

b) [5 marks]Using this utility function and the tree on the preceding page, what decision should the company make? The tree is repeated below – you could have used the same tree, and superimposed the utility values on the same sketch. The utility for each outcome, in Red, is found from the above graph. The maximum expected utility occurs with the decision to re-work the drive, unsurprisingly since the curve shows risk aversion.

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0.9Drive OK

$300.00Install the drive $300

0.1Drive fails

-$200.00-$200

Re-work the drive$200.00

0.95Drive OK

0.67 $275.00Drive Passes: install $300

0.05Drive fails

Test the drive -$225.00-$200

-250.33

Drive fails: rework$175.00

$200

c) [2 marks] What is the certainty equivalent of the decision on one drive? What is the risk premium?From the graph, since the EU = 0.95, the Certainty Equivalent sum of money that yields this utility, or the CE, = $200. The risk premium is the difference between the EV of $250 under the best EMV criterion and the CE, or RP = 250 – 200 = $50 per drive.

Util = 1

Util = 0.2

Util = 0.95

Util = 0.98

Util = 0.1

Util = 0.92

Exp U = .92

Exp U = .95

Exp U = .936

Exp U = .931

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Question 4 (12%)REW advertising in Toronto helps small and new companies identify target markets. A new client, OQ Cycle, carries 3 lines of racing and mountain bikes, currently sold in six retail outlets in Eastern Canada. OQ wants to market nationally, and to focus its advertising by identifying customer characteristics. REW has conducted a survey of a sample of OQ customers, as follows:Value: price of the package purchased by the customer (found from product registration check).Client Age: on last birthdaySex: 0 = male, 1 = femaleEducation: 1 = didn’t finish high school; 2 = high school; 3 = some college; 4 = college degree;

5 = started or completed graduate workMarital status: 0 = single or equivalent; 1 = married or equivalentIncome: Annual family income, in dollars.Times/wk: average number of times used, or planned to use, the bike each week in season.Km/wk: kilometers completed or planned per weekFitness: Self-rated fitness level, where 1 = poor shape to 5 = excellent shape.

OQ wants to know whether it can predict the Value of the package a potential customer might purchase based on the customer’s characteristics. A sample of 180 customers yielded the correlation matrix:

Value Age Sex Education Married Income Times/wk Km/wk FitnessValue 1Age 0.040 1Sex -0.161 -0.013 1Education 0.442 0.247 -0.064 1Married 0.006 0.179 0.078 0.051 1Income 0.552 0.576 -0.142 0.552 0.124 1Times/wk 0.454 -0.015 -0.234 0.192 -0.013 0.375 1Km/wk 0.523 0.018 -0.246 0.214 0.044 0.465 0.805 1Fitness 0.580 0.037 -0.281 0.329 -0.042 0.461 0.639 0.779 1

Circle two values from the first column of this matrix, and circle two other numbers (not = 1) from the other columns, that are meaningful for identifying a potential explanatory model. Using the space below and the next page as necessary, briefly explain what each of your selected values indicates about the potential model..040 : There is little correlation between the variable of interest, Value, and the Age of the customer. We should not likely consider including Age as an explanatory factor in modeling the Package Value.

.552 There is a reasonable correlation between Value of Package and the Income of the customer (this makes sense – the better off tend to purchase more expensive packages). Income is a good candidate to include in an explanatory model of Package Value.

0.375 There is relatively little correlation between Income and Times/wk. Both of these variables could reasonably be included in a multiple regression model to explain Value.

0.805 There is a high correlation between Times/wk and Km/wk. This makes sense – if a customer cycles more often during the week, they will also tend to put more kilometers on their cycle during the week. Both of these variables should not be included in the same regression, as they are co-linear. We could drop one of them (likely Times/wk, since it’s correlation with Value is lower), or we could think of creating a new variable, like Km/use by dividing Km/wk by Times/wk, maybe.

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Question 5 (18%) Continuing with the case introduced in Question 4, REW is pondering what to tell OQ about its customer profile. Suppose (despite your answer to the previous question) REW has run a multiple regression of Value versus customer Fitness, Sex, and Income. In this output (shown below), 6 values are shown in bold. Briefly summarize, in an English sentence, what each value tells you about this regression model. (Each of the six parts is worth 3 marks.)

Regression StatisticsMultiple R 0.66R Square 0.44

Adjusted R Square 0.43Standard Error 591.174222Observations 180

ANOVAdf SS MS F Significance F

Regression 3 48151405.99 16050469 45.93 5.67E-22Residual 176 61509705.12 349487Total 179 109661111.11

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept -131.63 187.09 -0.70 0.48 -500.86 237.61Fitness 330.42 52.24 6.33 0.00 227.33 433.51

Sex 10.54 93.32 0.11 0.91 -173.63 194.70

Income 0.013 0.004 5.69 5.34E-08 0.01 0.03

a) 0.43 This indicates that 43% of the variability in Value of Package purchased is explained by the variables Fitness, Sex, and Income. The remaining 57% is due to some other factors that are not in the model.

b) 5.67E-22 This is a test of the null hypothesis that all of the slopes in the model are simultaneously = 0. The probability of observing the values we did for the slopes (or larger) if the null hypothesis is true is very small, at 5.67 10 -22. This means we can reject the null hypothesis and conclude we have a significant overall explanatory model here.

c) 10.54 The estimated slope coefficient for the Sex variable. Since Sex = 1 corresponds to Females, this indicates that on average Females spend $10.54 more on their package than do Males.

d) 0.11 The t-Stat for the hypothesis test that the population slope coefficient for the Sex variable = 0. Since this t-Stat is so small, we can conclude that the Sex difference (ie the slope coefficient for Sex) could well be zero. We cannot reject the null hypothesis that the slope = 0.

e) 0.013 The slope coefficient for Income = .013. Since income is measured in dollars, this indicates that for each additional dollar of income, customers on average will spend an additional $.013 in cycle package.

f) 5.69 The t-Stat for the slope coefficient hypothesis test is 5.69. This indicates that we can reject the null hypothesis that the slope coefficient = 0, and conclude there is a strong linear relationship between Income and Package Value.

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Question 6 (18%) The City of Kingston recently hosted a Tall Ships Weekend, and part of the budgeting for the costs of the visit involved determining the economic benefit to the community from additional visitors during the weekend. Suppose you have been retained by the City to estimate this benefit. You have circulated a survey of visitors to the City on a typical holiday weekend and asked the respondents to estimate their daily expenditure during their stay, including lodging, food, and miscellaneous purchases. Suppose you received 120 completed questionnaires, and found the following summary statistics from your sample:

a) [6 marks] Based on this data, develop a confidence interval for the average daily expenditure per visitor to Kingston. Interpret the meaning of your answer to City Council (none of whom speak “statistics”).

We can calculate the standard error of the mean, as

Then a 95% confidence interval for the mean is 93.75 1.96 * 1.69 = 93.75 3.31, or [90.44, 97.06].

This can be interpreted as saying “We are quite certain that the average daily expenditure per visitor is within the range from $90.44 to $97.04.”

b) [6 marks] A report commissioned by the Canadian Tourist Booster Association makes the claim that visitors to a city in Canada spend an average of at least $100 per day. Set up and conduct an appropriate hypothesis test to examine this claim.

We can set up hypothesis assuming that the Boosters are right, and testing this assumption. Letting be the actual average daily expenditure of visitors to Canada, we can set up the hypothesis asH0 : 100H1: < 100

The observed z score for the hypothesis test is z = (93.75 – 100)/1.69 = -3.70. Since this is less than about 2, we can confidently reject the null hypothesis and conclude the Boosters are not correct in their assertion, given this evidence. (Of course our sample only includes visitors to Kingston, and we must assume that these are representative of the population of all visitors to Canada for this test to be valid.)You could, if you wish, also calculate the P-value of the test as = 2*NORMSDIST(-3.70) = .000216, and say that with such small P-value, we can reject the null hypothesis.

c) [6 marks] A busload of 49 tourists visited Kingston during this weekend. What is a 95% confidence interval for the average expenditure of the people on this bus, based on the above data?

Since we know in this case n = 49, and we have estimated the standard deviation of the population as

$18.50, the standard error for the mean of the busload will be Then a

95% confidence interval for the average expenditure of people on the bus will be93.75 1.96 * 2.64 = 93.75 5.18 = [88.57, 98.93].

Question 7 (11%)

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The Kingston Brewing Co. was the second pub to open in Ontario which was allowed to brew beer and/or ale on the premises. The quarterly sales of one of its products, Dragon's Breath Ale, for 9 quarters are shown below (the fourth quarter of 1994 is designated as 94.4, for example):

One of the owners, a professor of biology at Queen's, had studied some simple forecasting methods during his undergraduate course in statistics.

[3] a. Suppose he had used a simple exponential forecasting method with a weighting constant of 0.7. Develop the first four values he would have computed (that is, for quarters 94.4, 95.1, 95.2 and 95.3). Show any calculations clearly.

The computations are as follows:Quarter Observed Sales Smoothed values94.4 32 32 (choose the first observation as the anchor for the smoothed series)95.1 38 36.2 ( = .7 × 38 + .3 × 32)95.2 37 36.76 ( = .7 × 37 + .3 × 36.2)95.3 33 34.13 ( = .7 × 33 + .3 × 36.76)

ii) What do you think of his choice of weighting constant? Why?The high value of , at 0.7, means that the smoothed series will be very responsive to changes in the next observation. The resulting series will not be very smooth, but will be almost as variable as the original data. This value is likely too high to be useful. A number in the range from 0.2 to 0.4 is much more reasonable, striking a balance between responsiveness and degree of smoothing.

[4] b. Recently, one of his customers, a professor from the School of Business had suggested (after a pint of Dragon’s Breath) that the Brewing Company consider using a classical (multiplicative) decomposition time-series method of forecasting. After some minutes of struggling he came up with the following values for the seasonal adjustments: Quarter 1: 1.10; Quarter 2, 1.05; Quarter 3, 0.90; and Quarter 4, 0.95. How did the professor calculate the Quarter 1 value of 1.10?

The computations are done as follows:

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From this, you can see how the highlighted multiplicative seasonal adjustment was calculated. Since there is only one “first quarter” in this data, this is the best estimate of the seasonal adjustment for the first quarter.

[4] c. Given the seasonal adjustments in part b., what steps would the professor follow next (after another pint, of course)? Describe the steps and results below

Step 1. Divide each observed sales figure by the average seasonal adjustment, to deseasonalize the series.Step 2. Create a time index, and regress the deseasonalized sales against “time”.Step 3. Extend the regression results to cover the desired forecast period.Step 4. Reseasonalize the regression results to form the forecasted series.

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SELECTED STATISTICAL TABLE VALUES

Selected Critical Critical Values from theValues from the t-distribution for 95% Normal Distribution confidence.

Probability Z Degrees of80% 1.28 Freedom t90% 1.65 3 3.18295% 1.96 4 2.77699% 2.58 5 2.571

99.5% 2.81 6 2.4477 2.3658 2.3069 2.262

10 2.22811 2.20112 2.17913 2.16014 2.14515 2.13116 2.12017 2.11018 2.10119 2.09320 2.08625 2.06030 2.04235 2.03040 2.02141 2.02042 2.01843 2.01744 2.01545 2.01446 2.01347 2.01248 2.01149 2.01050 2.009

BDM Sample Final Exam + Solutions

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