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EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

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EMGT 501 HW #5 15.3-11 Due Day: Oct. 12. 15.3-11 - PowerPoint PPT Presentation

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Page 1: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

EMGT 501

HW #5

15.3-11

Due Day: Oct. 12

Page 2: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11

Management of the Telemore Company is considering developing and marketing a new product. It is estimated to be twice as likely that the product would prove to be successful as unsuccessful. It is were successful, the expected profit would be $1,500,000. If unsuccessful, the expected loss would be $1,800,000. A marketing survey can be conducted at a cost of $300,000 to predict whether the product would be successful. Past experience with such surveys indicates that successful products have been predicted to be successful 80 percent of the time, whereas unsuccessful products have been predicted to be unsuccessful 70 percent of the time.

Page 3: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

(a) Develop a decision analysis formulation of this problem by identifying the alternative actions, the states of nature, and the payoff table when the market survey is not conducted.

(b) Assuming the market survey is not conducted, use Bayes’ decision rule to determine which decision alternative should be chosen.

(c) Find EVPI. Does this answer indicate that consideration should be given to conducting the market survey?

Page 4: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

(d) Assume now that the market survey is conducted. Find the posterior probabilities of the respective states of nature for each of the two possible predictions from the market survey.

(e) Find the optimal policy regarding whether to conduct the market survey and whether to develop and market the new product.

Page 5: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11a)

b) They should choose to develop the new product because $400,000.

alternative succesfull unsuccesfulldevelop new product $1,500,000 -1,800,000

do not develop new product $0 $0prior probability 2/3 1/3

state of nature

K400$)3/1(K800,1(2/3)K500,1(develop) [payoff E

0$)3/1(0(2/3)0develop)not (do [payoff E

Page 6: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11c)

ationexperiment without payoff expected

ninformatioperfect with payoff Expected EVPI

$1000K(1/3)0(2/3)1,500k=

ninformatioperfect with payoff Expected

K600$K400K1000EVPI

This is larger than $300,000, so that consideration should be given to conducting the market survey.

Page 7: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11d)P(S)=2/3 P(US)=1/3P(+/ S)=0.8 P(-/ S)=0.2P(-/ US)=0.7 P(+/ US)=0.3

842.0

33.03.066.08.0

66.08.0

)()/()()/(

)()/()/(

USPUSPSPSP

SPSPSP

158.066.08.033.03.0

33.03.0

)S(P)S/(P)US(P)US/(P

)US(P)US/(P)/US(P

Page 8: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11

636.066.02.033.07.0

33.07.0

)S(P)S/(P)US(P)US/(P

)US(P)US/(P)/US(P

364.033.07.066.02.0

66.02.0

)US(P)US/(P)S(P)S/(P

)S(P)S/(P)/S(P

633.0

3.03/18.03/2

)US/(P)US(P)S/(P)S(P)(P

367.0633.01)(P1)(P

Page 9: EMGT 501 HW #5 15.3-11 Due Day: Oct. 12

15.3-11e) E(S/+) = 0.842 (1500K)+0.158(-1800K)=$979K E(US/+)=0 E(S/-)=0.364 (1500K)+0.636(-1800K)=-$600K E(US/-)=0 P(+) = 0.633, P(-) = 0.367 Expected Payoff = 0.633($979K)+0.367(0)=$620K EVE=$620K - $400K=$220K Cost=$300K > $220K

Optimal Strategy: Do not conduct survey and market product.