YORK UNIVERSITY Faculty of Liberal Arts and Professional Studies School of Administrative Studies

AP/ADMS 3300 Section “A” - Decision Analysis Fall 2014

Assignment #2 Submitted to

Course Director: S. Abdullah

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|Last Name, First Name |Student # |Signature |
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|Vysotskiy Alexander |211991288 | |
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Following questions are submitted, please check the respective boxes: (Note: For full marks you need to solve all questions.)

|Case # 1 |27 marks | | |
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|Case # 2 |25 marks | | |
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|Case # 3 |18 marks | | |
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|Following Rule 2,3&4 |5 marks | | |
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|Total : | | | |
| |75 marks | | |

Case 1

Part A

a)

EMV(1)=0.42*238’000’000+0.26*141’000’000-0.32*79’000’000=111’340’000$
EMV(1) is higher than 30 000 000 so the product should be launched.

b)

EMV=0.42*238’000’000+0.26*141’000’000+0.32*30’000’000=146’220’000$
EVPI=EMV-EMV(1) = 146’220’000-111’340’000 = 34’880’000$

Part B

c)

| |BRIGHT Consultant prediction |
|Actual Market |High Market |Medium Market |Low market |
|High |0.67 |0.15 |0.18 |
|Mid |0.11 |0.82 |0.07 |
|Low |0.13 |0.2 |0.67 |

P(BRIGHT consultant predict about High market | When actual market is Medium) = 1-0.67-0.15=0.18
P(BRIGHT consultant predict about Medium market | When actual market is Low) = 1-0.82-0.7=0.11
P(BRIGHT consultant predict about Low market | When actual market is High) = 1-0.67-0.13=0.2

| |Posterior probabilities |
|Actual Market |‘High’ |‘Medium’ |‘Low’ |
|High |0.8003 |0.1852 |0.2453 |
|Mid |0.0813 |0.6267 |0.0591 |
|Low |0.1183 |0.1881 |0.6957 |

H means High market
M means Medium market
L means Low market

P(H|’H’) = P(‘H’|H)*P(H) / (P(‘H’|H)*P(H)+P(‘H’|M)*P(M)+P(‘H’|L)*P(L) = 0.67*0.42 / (0.67*0.42+0.11*0.26+0.13*0.32) = 0.8003

P(‘H’) is given by the denominator and equals 0.3516

P(M|’H’) = P(‘H’|M)*P(M) / (P(‘H’|H)*P(H)+P(‘H’|M)*P(M)+P(‘H’|L)*P(L) = 0.11*0.26 / 0.3516 = 0.0813

P(L|’H’) = P(‘H’|L)*P(L) / (P(‘H’|H)*P(H)+P(‘H’|M)*P(M)+P(‘H’|L)*P(L) = 0.32*0.13 / 0.3516 = 0.1183

P(H|’M’) = P(‘M’|H)*P(H) / (P(‘M’|H)*P(H)+P(‘M’|M)*P(M)+P(‘M’|L)*P(L) = 0.15*0.42 / (0.15*0.42+0.82*0.26+0.2*0.32) = 0.1852

P(‘M’) is given by the denominator and equals 0.3402

P(M|’M’) = P(‘M’|M)*P(M) / (P(‘M’|H)*P(H)+P(‘M’|M)*P(M)+P(‘M’|L)*P(L) = 0.82*0.26 / 0.3402 =0.6267

P(L|’M’) = P(‘M’|L)*P(L) / (P(‘M’|H)*P(H)+P(‘M’|M)*P(M)+P(‘M’|L)*P(L) = 0.2*0.32 / 0.3402 = 0.1881

P(H|’L’) = P(‘L’|H)*P(H) / (P(‘L’|H)*P(H)+P(‘L’|M)*P(M)+P(‘L’|L)*P(L) = 0.18*0.42 / (0.18*0.42+0.07*0.26+0.67*0.32) = 0.2453

P(‘L’) is given by the denominator and equals 0.3082

P(M|’L’) = P(‘L’|M)*P(M) / (P(‘L’|H)*P(H)+P(‘L’|M)*P(M)+P(‘L’|L)*P(L) = 0.07*0.26 / 0.3082 = 0.0591

P(L|’L’) = P(‘L’|L)*P(L) / (P(‘L’|H)*P(H)+P(‘L’|M)*P(M)+P(‘L’|L)*P(L) = 0.67*0.32 / 0.3082 = 0.6957

EMV(1) = (0.8003*238+0.0813*141-0.1183*79)*1000000= 192’589’000 $
EMV(2) = (0.1852*238+0.6267*141-0.1881*79)*1000000= 117’582’400 $
EMV(3) = (0.2453*238+0.0591*141-0.6957*79)*1000000= 11’754’200 $
EMV(4) = 0.3516*192’589’000+0.3402*117’582’400+0.3082*30’000’000 = 116’961’524.88 $
EMV(5) = 111’340’000 $ (calculated in part A)

EVof imperfect information = EMV(4)-EMV(5) = 116’961’524.88 - 111’340’000 = 5’621’524.88$

d)

EV of imperfect information is greater than $5 million (consultant fee), so the company should consult with BRIGHT.

Case 2

a) Without doing calculations I would advise Shirley to buy laptop F because it is one of the most reliable laptops (there is only 1 more reliable). And at the same time laptop F is just a little bit more expensive than the cheaper models.

b)

|Laptops |C |D |
|Up |1 |0 |
| | | |
|Ud |0 |1 |

Shirley estimates that the expenses for every extra day in the shop will be 150$. If we take laptop C as a base, Shirley will be indifferent between laptop C (price 900, number of days 10) and laptop X (price 1050, number of days 9).

We can derive the corresponding scores for laptop X by using a formula:
Ui (X) = (x-x-)/(x+-x-)

Up(X) = (1050-1625)/(900-1625)=0.7931
Ud(X) = (9-10)/(3-10)=0.1429

U(C)=U(X)
Kp(1)+Kd(0) = Kp(0.7931)+Kd(0.1429)
Kp(0.2069)=Kd(0.1429)
Kd=1.4479Kp

Including the condition that the weights must sum to 1:
Kd=1.4479(1-Kd)
Kd=1.4479-1.4479Kd
2.4479Kd=1.4479
Kd=0.5915 ; Kp=0.4085

c)

Laptops A and E are both more expensive and require more days in the shop than laptop D, so laptop D dominates laptops A and E. So we can eliminate those two options.

|Laptops |C |D |F |B |
|Up |1 | | |0 |
| | | | | |
|Ud |0 | | |1 |

We can derive the corresponding scores for other laptops by using a formula:
Ui (x) = (x-x-)/(x+-x-)

Up(D) = (1025-1625)/(900-1625) = 0.8276
Up(F) = (1100-1625)/(900-1625) = 0.7241
Ud(D) = (6-10)/(3-10) = 0.5714
Ud(F) = (5-10)/(3-10) = 0.7143

|Laptops |C |D |F |B |
|Up |1 |0.8276 |0.7241 |0 |
| | | | | |
|Ud |0 |0.5714 |0.7143 |1 |

Calculating overall utilities:

U(C) = 1*0.4085=0.4085
U(B) = 1*0.5915=0.5915
U(D) = 0.8276*0.4085+0.5714*0.5915 = 0.6761
U(F) = 0.7241*0.4085+0.7143*0.5915 = 0.7183

Purchasing laptop F comes out to be the best option comparing to other alternatives.

d) There could be some factors (other than money losses) that would be important in demining the tradeoff between cost and the reliability. For example, the amount of time Shirley will have to loose on bringing the laptop to the repair services and taking it back. Also, inability to perform some urgent activities via your computer. And finally, some valuable information can be lost if the laptop will suddenly break, so considering this factor is also important.

Case 3

a) EU(A) = 0.24*U(13208+3500)+0.76*U(-1273+3500) = 0.24*(0.008*16708-3.85*ln(16708))+0.76*(0.008*2227-3.85*ln(2227)) = 14.0801

EU(B) = 0.21*U(13894+3500)+0.79*U(-898+3500) = 0.21*(0.008*17394-3.85*ln(17394))+0.79*(0.008*2602-3.85*ln(2602)) = 13.8540

EU(C) = 0.18*U(6745+3500)+0.82*U(1200+3500) = 0.18*(0.008*10245-3.85*ln(10245))+0.82(0.008*4700-3.85*ln(4700)) = 12.4918

Expected utility of alternative A is higher so Brigg should choose alternative A.

b) EU(A) = 0.24*U(13208+7300)+0.76*U(-1273+7300) = 0.24*(0.008*20508-3.85*ln(20508))+0.76*(0.008*6027-3.85*ln(6027)) = 41.3776

EU(B) = 0.21*U(13894+7300)+0.79*U(-898+7300) = 0.21*(0.008*21194-3.85*ln(21194))+0.79*(0.008*6402-3.85*ln(6402)) = 41.3559

EU(C) = 0.18*U(6745+7300)+0.82*U(1200+7300) = 0.18*(14045*0.008-3.85*ln(14045))+0.82*(8500*0.008-3.85*ln(8500)) = 40.8027

Expected utility of alternative A is higher so Brigg should choose alternative A.

c) EU(A) = 0.24*U(13208+11500)+0.76*U(-1273+11500) = 0.24*(0.008*24708-3.85*ln(24708))+0.76*(0.008*10227-3.85*ln(10227)) = 73.2583

EU(B) = 0.21*U(13894+11500)+0.79*U(-898+11500) = 0.21*(0.008*25394-3.85*ln(25394))+0.79*(0.008*10602-3.85*ln(10602)) = 73.2755

EU(C) = 0.18*U(6745+11500)+0.82*U(1200+11500) = 0.18*(0.008*18245-3.85*ln(18245))+0.82*(0.008*12700-3.85*ln(12700)) = 72.9537

Expected utility of alternative B is higher so Brigg should choose alternative B.

d) EU(A) = 0.24*U(13208+3500)+0.76*U(-1273+3500) = 0.24*(0.0019*16708-1.25*e-16708/1215)+0.76*(0.0019*2227-1.25*e-2227/1215) = 10.6827

EU(B) = 0.21*U(13894+3500)+0.79*U(-898+3500) = 0.21*(0.0019*17394-1.25*e-17394/1215)+0.79*(0.0019*2602-1.25*e-2602/1215) = 10.7298

EU(C) = 0.18*U(6745+3500)+0.82*U(1200+3500) = 0.18*(0.0019*10245-1.25*e-10245/1215)+0.82*(0.0019*4700-1.25*e-4700/1215) = 10.8049

Expected utility of alternative C is higher so Brigg should choose alternative C.

e) EU(A) = 0.24*U(13208+7300)+0.76*U(-1273+7300) = 0.24*(0.0019*20508-1.25*e-20508/1215)+0.76*(0.0019*6027-1.25*e-6027/1215) = 18.0480

EU(B) = 0.21*U(13894+7300)+0.79*U(-898+7300) = 0.21*(0.0019*21194-1.25*e-21194/1215)+0.79*(0.0019*6402-1.25*e-6402/1215) = 18.0607

EU(C) = 0.18*U(6745+7300)+0.82*U(1200+7300) = 0.18*(0.0019*14045-1.25*e-14045/1215)+0.82*(0.0019*8500-1.25*e-8500/1215) = 18.0454

Expected utility of alternative B is higher so Brigg should choose alternative B.

f) EU(A) = 0.24*U(13208+11500)+0.76*U(-1273+11500) = 0.24*(0.0019*24708-1.25*e-24708/1215)+0.76*(0.0019*10227-1.25*e-10227/1215) = 26.0344

EU(B) = 0.21*U(13894+11500)+0.79*U(-898+11500) = 0.21*(0.0019*25394-1.25*e-25394/1215)+0.79*(0.0019*10602-1.25*e-10602/1215) = 26.0456

EU(C) = 0.18*U(6745+11500)+0.82*U(1200+11500) = 0.18*(0.0019*18245-1.25*e-18245/1215)+0.82*(0.0019*12700-1.25*e-12700/1215) = 26.0264

Expected utility of alternative B is higher so Brigg should choose alternative B.

g) Risk neutral person would make his decision according to EMV. EMV(A) = 13208*0.24-1273*0.76 = 2202.44 EMV(B) = 13894*0.21-898*0.79 = 2208.32 EMV(C) = 6745*0.18+1200*0.82 = 2198.1 If Brigg is risk neutral, my recommendation would be – choose alternative B. Because it gives the highest EMV.

The given function represents a risk-averse behavior. U(x)=1.32-0.612*e-x/16054

EU(A) = 0.24*(1.32-0.612*e-13208/16054)+0.76*(1.32-0.612*e1273/16054) = 0.7520 EU(B) = 0.21*(1.32-0.612*e-13894/16054)+0.79*(1.32-0.612*e898/16054) = 0.7546 EU(C) = 0.18*(1.32-0.612*e-6745/16054)+0.82*(1.32-0.612*e-1200/16054) = 0.7819

So if Brigg is not risk neutral and the utility function is U(x)=1.32-0.612*e-x/16054 , I would recommend him to choose alternative C because it has highest expected utility.

To find certainty equivalent, we need to find the certain value of x that U(x)=EU

CE(A): 0.7520=1.32-0.612* e-x/16054 0.568=0.612* e-x/16054 0.9281= e-x/16054 ln(0.9281)= -x/16054 -0.0746*16054=-x x=1197.8819 CE(A)=1197.8819

CE(B): 0.7546=1.32-0.612* e-x/16054 0.5654=0.612* e-x/16054 0.9239= e-x/16054 ln(0.9239)=-x/16054 0.0792*16054=x x=1270.6972 CE(B)=1270.6972

CE(C): 0.7819=1.32-0.612* e-x/16054 0.5381=0.612* e-x/16054 0.8792= e-x/16054 ln(0.8792)=-x/16054 0.1287*16054=x x=2066.8381 CE(C)=2066.8381

The lowest CE has alternative A.

Risk premium = EMV – CE

Risk premium (A) = 2202.44-1197.8819 = 1004.5581 Risk premium (B) = 2208.32-1270.6972 = 937.6228 Risk premium (C) = 2198.1-2066.8381 = 131.2619