Question 18.14: Calculating Risk/Return Trade-offs for MedEQuipt (Examples 1...
Calculating Risk/Return Trade-offs for MedEQuipt (Examples 18.9 and 18.10 continued)
If MedEQuipt (in Examples 18.9 and 18.10) maximizes PW and minimizes risk, which alternative should be adopted?
The decision tree with expected values can be used here as well (Exhibit 18.7 repeats Exhibit 18.6, but with standard deviations also shown).
Stop alternative. This alternative’s expected value and standard deviation are both $0. The P(loss) is also $0.
Normal development. The expected value of the PW for normal development was calculated as follows:
E_{normal returns} = .05(0) + .95(100,000)(P/A,.12,10)(P/F,.12,1)
= 0 + .95(504,484) = $479,260
E_{PW normal} = $479,260 − $450,000 = $29,260
To calculate the standard deviation, the first step is to identify the possible outcomes and their probabilities. That is the probability distribution for the PW under normal development.
These values are found by subtracting the $450K for normal development from the PW of each possible outcome. This also defines the 5% probability of losing $450K.
| Normal PW | Probabilities |
| −$450,000 | .05 |
| 54,484 | .95 |
This probability distribution still has an expected value of $29,260. The second step is to calculate the expected value of the squares and the standard deviation.
E(PW^{2}) = .05 · (−450,000)^{2} + .95 · 54,484^{2} = 1.2945 · 10^{10}
σ_{PW} = \sqrt{1.2945 · 10^{10} − 29,260^{2}} = \$109,949
Accelerated Development. The expected PW under accelerated development was calculated as follows:
PW_{accelerated returns} = (P/F,.12,1)[50,000 + 125,000(P/A,.12,10)]
= (50,000 + 706,278)/1.12 = $675,248
E_{accelerated returns} = .4(0) + .6(675,248) = $405,149
E_{PW} = $405,149 − $250,000, or $155,149
To calculate the standard deviation, the first step is to identify the possible outcomes and their probabilities. That is the probability distribution for the PW under accelerated development.
This has a much larger probability (.4) of a loss, but the loss is smaller.
| Accelerated PW | Probabilities |
| −$250,000 | .4 |
| 425,248 | .6 |
This probability distribution still has an expected value of $155,149. The second step is to calculate the expected value of the squares and the standard deviation.
E(PW^{2}) = .4 · (−250,000)^{2} + .6 · 425,248^{2} = 1.3350 · 10^{11}
σ_{PW} = \sqrt{1.3350 · 10^{11} − 155,149^{2}} = \$330,803
Comparing the Alternatives. As we can see in the following table, none of the three alternatives is dominant. When the three are compared, an increasing level of return is associated with more risk.
| Loss | P(loss) | σ_{PW} | EV_{PW} | Alternative |
| $0 | 0 | $0 | $0 | Stop |
| 450K | .05 | 109,949 | 29,260 | Normal development |
| 250K | .4 | 330,803 | 155,149 | Accelerated development |
In this case, tripling the risk (measured by standard deviation) increases the return more than fivefold, so accelerated development likely is still the best alternative. The P(loss) is harder to evaluate, because accelerating development dramatically increases failure probabilities while about halving the consequences. A rigorous treatment of the trade-off question is beyond the scope of this text.
