Question

PLEASE READ: This is one question with 3 parts to it, please answer the full question.

Mark M. Upp has just been fired as the university bookstore manager for setting prices too low (only 20 percent above suggest retail). He is considering opening a competing bookstore near the campus, and he has begun an analysis of the situation. There are two possible sites under consideration. One is relatively small, while the other is large. If he opens at Site 1 and demand is good, he will generate a profit of $ 200,000. If demand is low, he will lose $180,000. If he opens at Site 2 and demand is high, he will generate a profit of $100,000, but he will lose $20,000 if demand is low. He also has the option of not opening either. He believes that there is a 50 percent chance that demand will be high. Mark can purchase a market research study from Brooklyn College. The survey costs $10,000. The probability of a good demand given a favorable study is 0.8. The probability of a good demand given an unfavorable study is 0.3. There is a 45 percent chance that the study will be favorable. Draw a decision tree to determine the following:

a)What should Mark’s decisions be?

b)What is the maximum amount Mark should be willing to pay for this study?

c)What is the efficiency of the study?

Hint: The revised probabilities have already been calculated for you.

Answer 1

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ANS;

(a) Decision without help of study

EPV of site 1 = 0.5 x200000-0.5x180000 = 10000

EPV of site 2 = 0.5x100000-0.5x20000 = 40000

The decision whould be in favour of site 2 as its EPV is higher.

(b) If study is taken

EPV = 0.45 [ better of EPV of site I or II under favourable demand] +0.55 [ better of EPV of site I or II under unfavorable demand]

= 0.45 [ better of ( 0.8x200000-0.2x180000), ( 0.8x100000-0.2x20000)] + 0.55 [ better of (0.3x200000-0.7x180000), (0.3x100000-0.7x20000)] -10000

= 0.45 [ 124000]+0.55[ 16000]-10000

= 54600

Hence the study is beneficial.

Max value that can be paid for the study = EV with study - EV without study

= 54600-40000 =14600

(c) Expected value with perfect information, ( choice of best alternative)

EVwPI = 0.5x200000 =100000

EV of PI = 100000-40000 =60000

Efficiency of study = value of information with study / value of PI

= 14600/60000 = 24.33%

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