1. Since the decision-making environment is risk (probabilities are known), it is appropriate to use the EMV criterion. The problem can be solved by developing a payoff table that contains all alternatives, states of nature, and probability values. The EMV for each alternative is also computed, as in the following table:

   	STATE OF NATURE
   ALTERNATIVE	GOOD MARKET ($)	AVERAGE MARKET ($)	BAD MARKET ($)	EMV ($)
   Small shop	75,000	25,000	–40,000	15,500
   Medium-sized shop	100,000	35,000	–60,000	19,500
   No shop	0	0	0	0
   Probabilities	0.20	0.50	0.30

   $$\begin{array}{lll}\hfill \text{EMV(small shop)}& =\hfill & \text{(0}\text{.2)(\$75,000) + (0}\text{.5)(\$25,000) + (0}\text{.3)(}-\text{\$40,000) = \$15,500}\hfill \\ \hfill \text{EMV(medium shop)}& =\hfill & \text{(0}\text{.2)(\$100,000) + (0}\text{.5)(\$35,000) + (0}\text{.3)(}-\text{\$60,000) = \$19,500}\hfill \\ \hfill \text{EMV(no shop)}& =\hfill & \text{(0}\text{.2)(\$0) + (0}\text{.5)(\$0) + (0}\text{.3)(\$0) = \$0}\hfill \end{array}$$

   As can be seen, the best decision is to build the medium-sized shop. The EMV for this alternative is $19,500.

2. $\begin{array}{lll}\hfill \text{EVwPI}& =\hfill & \text{(0}\text{.2)\$100,000 + (0}\text{.5)\$35,000 + (0}\text{.3)\$0 = \$37,500}\hfill \\ \hfill \text{EVPI}& =\hfill & \text{\$37,500 - \$19,500 = \$18,000}\hfill \end{array}$

3. The opportunity loss table is shown here.

   	STATE OF NATURE
   ALTERNATIVE	GOOD MARKET ($)	AVERAGE MARKET ($)	BAD MARKET ($)	MAXIMUM ($)	EOL ($)
   Small shop	25,000	10,000	40,000	40,000	22,000
   Medium-sized shop	0	0	60,000	60,000	18,000
   No shop	100,000	35,000	0	100,000	37,500
   Probabilities	0.20	0.50	0.30

   The best payoff in a good market is 100,000, so the opportunity losses in the first column indicate how much worse each payoff is than 100,000. The best payoff in an average market is 35,000, so the opportunity losses in the second column indicate how much worse each payoff is than 35,000. The best payoff in a bad market is 0, so the opportunity losses in the third column indicate how much worse each payoff is than 0.

   The minimax regret criterion considers the maximum regret for each decision, and the decision corresponding to the minimum of these is selected. The decision would be to build a small shop, since the maximum regret for this is 40,000, while the maximum regret for each of the other two alternatives is higher, as shown in the opportunity loss table.

   The decision based on the EOL criterion would be to build the medium shop. Note that the minimum EOL ($18,000) is the same as the EVPI computed in part (b). The calculations are

   $$\begin{array}{lll}\hfill \text{EOL(small)}& =\hfill & \text{(0}\text{.2)25,000 + (0}\text{.5)10,000 + (0}\text{.3)40,000 = 22,000}\hfill \\ \hfill \text{EOL(medium)}& =\hfill & \text{(0}\text{.2)0 + (0}\text{.5)0 + (0}\text{.3)60,000 = 18,000}\hfill \\ \hfill \text{EOL(no shop)}& =\hfill & \text{(0}\text{.2)100,000 + (0}\text{.5)35,000 + (0}\text{.3)0 = 37,500}\hfill \end{array}$$

The problem is one of decision making under uncertainty. Before answering the specific questions, a decision table should be developed showing the alternatives, states of nature, and related consequences.

1. Since Cal is a risk taker, he should use the maximax decision criteria. This approach selects the row that has the highest or maximum value. The $10,000 value, which is the maximum value from the table, is in row 1. Thus, Cal’s decision is to select strategy 1, which is an optimistic decision approach.

2. Becky should use the maximin decision criteria because she wants to avoid risk. The minimum or worst outcome for each row, or strategy, is identified. These outcomes are $\mathrm{-}\$8,000$ for strategy 1, $\mathrm{-}\$4,000$ for strategy 2, and $0 for strategy 3. The maximum of these values is selected. Thus, Becky would select strategy 3, which reflects a pessimistic decision approach.

3. If Cal and Becky are indifferent to risk, they could use the equally likely approach. This approach selects the alternative that maximizes the row averages. The row average for strategy 1 is $\$1,000\left[\text{i}.\text{e}.\text{, }\$1,000=\left(\$10,000-\$8,000\right)/2\right].$ The row average for strategy 2 is $2,000, and the row average for strategy 3 is $0. Thus, using the equally likely approach, the decision is to select strategy 2, which maximizes the row averages.

Before Monica starts to solve this problem, she should develop a decision tree that shows all alternatives, states of nature, probability values, and economic consequences. This decision tree is shown in Figure 3.14.

The EMV at each of the numbered nodes is calculated as follows:

At each of the square nodes with letters, the decisions would be:

Based on the EMV criterion, Monica would select Do Not Conduct Study and then select Large Facility. The EMV of this decision is $42,000. Choosing to conduct the study would result in an EMV of only $32,900. Thus, the expected value of sample information is

This problem requires the use of Bayes’ Theorem. Before we start to solve the problem, we will define the following terms:

$\begin{array}{lll}\hfill P\text{(SF)}& =\hfill & \text{probability of successful driving range facility}\hfill \\ \hfill P\text{(UF)}& =\hfill & \text{probability of unsuccessful driving range facility}\hfill \\ \hfill P\text{(RF | SF)}& =\hfill & \text{probability that the research will be favorable given a successful driving range facility}\hfill \\ \hfill P\text{(RU | SF)}& =\hfill & \text{probability that the research will be unfavorable given a successful driving range facility}\hfill \\ \hfill P\text{(RU | UF)}& =\hfill & \text{probability that the research will be unfavorable given an unsuccessful driving range facility}\hfill \\ \hfill P\text{(RF | UF)}& =\hfill & \text{probability that the research will be favorable given an unsuccessful driving range facility}\hfill \end{array}$

Now, we can summarize what we know:

From this information we can compute the three additional probabilities that we need to solve the problem:

Now we can put these values into Bayes’ Theorem to compute the desired probability:

In addition to using formulas to solve John’s problem, it is possible to perform all calculations in a table:

As you can see from the table, the results are the same. The probability of a successful driving range, given a favorable research result, is $0.36/0.48$, or 0.75.