Visual Park is considering marketing one of its two television models for the coming Christmas season: Model A or Model B. Model A is a unique featured television and appears to have no competition. Estimated profits (in thousands of dollars) under high, medium, and low demand are given below:
Demand
Model AHigh
Medium
Low
Profit1200
900
500
Probability0.2
0.6
0.2
Visual Park is optimistic about the TV Model B. However, the concern is that profitability will be affected if a competitor launches a TV model which has similar features as Model B. Estimated profits (in thousands of dollars) with and without competition is as follows:
Model BDemand
With competitionHigh
Medium
Low
Profit1,200
900
500
Probability0.2
0.3
0.5
Model BDemand
Without competitionHigh
Medium
Low
Profit1,600
1,100
700
Probability0.6
0.2
0.2
a. Develop a decision tree for the Visual Park problem.b. For planning purposes, Visual Park believes there is a 0.7 probability that its competitor will launch a TV model similar to Model B. Given this probability of competition, the director of planning recommends marketing the Model A. Using expected value, what is your recommended decision?c. Show a risk profile for your recommended decision.d. Use sensitivity analysis to determine what the probability of competition for Model B would have to be for you to change your recommended decision alternative.
What will be an ideal response?
a.
b. EV(node 2) = 0.2(1,200) + 0.6(900) + 0.2(500) = 880
EV(node 4) = 0.2(1,200) + 0.3(900) + 0.5(500) = 760
EV(node 5) = 0.6(1,600) + 0.2(1,100) + 0.2(700) = 1,320
EV(node 3) = 0.7EV(node 4) + 0.3EV(node 5) = 0.7(760) + 0.3(1,320) = 928
Model B is recommended as the expected value of $928,000 is $48,000 better than Model A.
c. Risk profile:
1600 | 0.3 × 0.6 | 0.18 |
1200 | 0.7 × 0.2 | 0.14 |
1100 | 0.3 × 0.2 | 0.06 |
900 | 0.7 × 0.3 | 0.21 |
700 | 0.3 × 0.2 | 0.06 |
500 | 0.7 × 0.5 | 0.35 |
d. Let p = probability of competition
p = 1®EV(node 5) = 1,120
p = 0®EV(node 4) = 460
Setting the Expected Value of both decisions equal to each other gives us:
EV(Model B) = EV(Model A)
1320 - p(1320 - 760) = 880
560p = 440
p = 440/560 = 0.7857
For p> 0.7857, the EV of Model A is greater; for p< 0.7857, the EV of Model B is greater. Therefore, the probability of competition would have to be greater than 0.7857 before we would change to the Model A.
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