Suppose there is a risky prospect associated with the amount of code that can be

Question

Suppose there is a risky prospect associated with the amount of code that can be reused from an existing application for the development of a new technology program's software. The higher the percentage of reuse the better for the program's cost and schedule: for example, 90 percent reuse of the existing application's code has higher utility to the program than 50 percent reuse and so forth. Suppose the uncertainty in the engineering assessments on the reuse percentage are as follows: An outcome of 90 percent reuse has probability 0.70, an outcome of 50 percent reuse has probability 0.20, and an outcome of 25 percent reuse has probability 0.10. Suppose the utility function for this outcome is U(x) = k(1 -e^-x/p), where 0 lessthanorequalto x lessthanorequalto 100 is the reuse percent, k= 1/(l-e^-100/p), and p is the risk tolerance of the technology program's manager. Given this, answer the following questions. (A) Determine the expected value of the reuse outcome. (B) Compute the expected utility of the reuse outcome with p = 20. Find a value of p such that the expected value and the expected utility of this outcome approach being equal. Characterize the risk attitude of the technology program manager (risk neutral, risk seeking, or risk averse) and provide a mathematical reason for your conclusion. (C) Compute the certainty equivalent of the reuse outcome, as given by the risky prospect presented above. Explain in non-mathematical terms what the computed value of the certainty equivalent means to the program manager.

Explanation / Answer

(A) According to the question we are provided with tProhe following information:

Thus, the expected value of the reuse outcome is given by E(X)=x.p(x)

Putting the values from the above table we have E(X)=90x0.70 + 50x0.20 + 25x0.10=75.5%

Thus, the expected value of the reuse outcome is 75.5%.

(B) It is given in the question that the utility function of X is U(x) =k(1-e-x/) where 0<=X<=100 is the reuse percentage.and k=1/(1-e-100/) Now expected utility when =20 is given by intergrating xU(x) over 0<=X<=100. and putting =20 in the function.

For =20 our k becomes 1.0068(This is simply derived by putting the value of in the expression for k.

Integrating the expression we get the value as E[U(x)]= 1.006801oox(1-e-x/20)dx = 4647.561027.

This is the expected value of the utlity when =20.

Now, to find a value of such that the expected value and the expected utility of this outcome becomes zero we set the expectation of the utility function as 75.5(i.e the expected value of the outcome) and solve for .

So, we have E[U(x)] =75.5

This is a very complicated integral and by trying to solve it by the method of trial and error I noticed that the value of becomes indefinitely small.

Since, denotes the risk attitude of the technology program manager so we can say that the program manager is strongly risk averse.

(C) Certainty equivalent gives the certain amount of utility that the technology program manager will be happy to accept to avoid the risks or not have any risk.

The ertainty eqivalent can be calculated by finding the value of the utility finction at each reuse outcome i.e for reuse outcome 90 50 and 25. So, we have the following table.

*we take =20 for the above calculations.

The certainty equivalent is given by taking the sum product of the utility for each reuse percetages multiplied by their respective probabilities.

So, Certainty Equivalent = 0.9956x0.70 +0.9241x0.20 + 0.7135x0.10 =0.69692 + 0.18482 + 0.07135 =0.95309

Thus, from the above value of the certainty equivalent we can say that the technology program manager will be able to accept only 0.95309 utility to avoid the risks. This means that the manager is highly risk averse and will be willing to sacrifice all utility to avoid risks for a risky prospects of level =20.

Reuse Percentages(X) 90% 50% 25% ProbabilitiesP(X=x) 0.70 0.20 0.10

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