Given a decision table with conditional values (payoffs) that are monetary values and probability assessments for all states of nature, it is possible to determine the expected monetary value (EMV) for each alternative. The expected value, or the mean value, is the long-run average value of that decision. The EMV for an alternative is just the sum of possible payoffs of the alternative, each weighted by the probability of that payoff occurring.

This could also be expressed simply as the expected value of X, or E(X), which was discussed in Section 2.6 of Chapter 2.

where

• $X_i=\text{payoff for the alternative in state of nature }i$

• $P\left(X_i\right)=\text{probability of achieving payoff }{X}_i\left(\text{i}.\text{e}.\text{, probability of state of nature }i\right)$

• ${\displaystyle \sum _{}^{}}=\text{summation symbol}$

If this were expanded, it would become

The alternative with the maximum EMV is then chosen.

Suppose that John Thompson now believes that the probability of a favorable market is exactly the same as the probability of an unfavorable market; that is, each state of nature has a 0.50 probability. Which alternative would give the greatest EMV? To determine this, John has expanded the decision table, as shown in Table 3.9. His calculations follow:

	STATE OF NATURE
ALTERNATIVE	FAVORABLE MARKET ($)	UNFAVORABLE MARKET ($)	EMV ($)
Construct a large plant	200,000	–180,000	10,000
Construct a small plant	100,000	–20,000
Do nothing	0	0	0
Probabilities	0.50	0.50

• $\text{EMV} \left(\text{large} \text{plant}\right)=\left(\text{\$}200\text{,}000\right)\left(0.50\right)+\left(\mathrm{-}\text{\$}180\text{,}000\right)\left(0.50\right)=\text{\$}10\text{,}000$

• $\text{EMV} \left(\text{small} \text{plant}\right)=\left(\text{\$}100\text{,}000\right)\left(0.50\right)+\left(\mathrm{-}\text{\$}20\text{,}000\right)\left(0.50\right)=\text{\$}40\text{,}000$

• $\text{EMV} \left(\text{do} \text{nothing}\right)=\left(\text{\$}0\right)\left(0.50\right)+\left(\text{\$}0\right)\left(0.50\right)=\text{\$}0$

The largest expected value ($40,000) results from the second alternative, “construct a small plant.” Thus, Thompson should proceed with the project and put up a small plant to manufacture storage sheds. The EMVs for constructing the large plant and for doing nothing are $10,000 and $0, respectively.

When using the EMV criterion with minimization problems, the calculations are the same, but the alternative with the smallest EMV is selected.

John Thompson has been approached by Scientific Marketing, Inc., a firm that proposes to help John make the decision about whether to build the plant to produce storage sheds. Scientific Marketing claims that its technical analysis will tell John with certainty whether the market is favorable for his proposed product. In other words, it will change his environment from one of decision making under risk to one of decision making under certainty. This information could prevent John from making a very expensive mistake. Scientific Marketing would charge Thompson $65,000 for the information. What would you recommend to John? Should he hire the firm to make the marketing study? Even if the information from the study is perfectly accurate, is it worth $65,000? What would it be worth? Although some of these questions are difficult to answer, determining the value of such perfect information can be very useful. It places an upper bound on what you should be willing to spend on information such as that being sold by Scientific Marketing. In this section, two related terms are investigated: the expected value of perfect information (EVPI) and the expected value with perfect information (EVwPI). These techniques can help John make his decision about hiring the marketing firm.

The expected value with perfect information is the expected or average return, in the long run, if we have perfect information before a decision has to be made. To calculate this value, we choose the best alternative for each state of nature and multiply its payoff times the probability of occurrence of that state of nature.

If this were expanded, it would become

The EVPI is the expected value with perfect information minus the expected value without perfect information (i.e., the best or maximum EMV). Thus, the EVPI is the improvement in EMV that results from having perfect information.

By referring to Table 3.9, Thompson can calculate the maximum that he would pay for information, that is, the EVPI. He follows a three-stage process. First, the best payoff in each state of nature is found. If the perfect information says the market will be favorable, the large plant will be constructed, and the profit will be $200,000. If the perfect information says the market will be unfavorable, the “do nothing” alternative is selected, and the profit will be 0. These values are shown in the “with perfect information” row in Table 3.10. Second, the expected value with perfect information is computed. Then, using this result, EVPI is calculated.

	STATE OF NATURE
ALTERNATIVE	FAVORABLE MARKET ($)	UNFAVORABLE MARKET ($)	EMV ($)
Construct a large plant	200,000	–180,000	10,000
Construct a small plant	100,000	–20,000	40,000
Do nothing	0	0	0
With perfect information	200,000	0
Probabilities	0.50	0.50

The expected value with perfect information is

Thus, if we had perfect information, the payoff would average $100,000.

The maximum EMV without additional information is $40,000 (from Table 3.9 ). Therefore, the increase in EMV is

Thus, the most Thompson would be willing to pay for perfect information is $60,000. This, of course, is again based on the assumption that the probability of each state of nature is 0.50.

This EVPI also tells us that the most we would pay for any information (perfect or imperfect) is $60,000. In a later section, we’ll see how to place a value on imperfect or sample information.

In finding the EVPI for minimization problems, the approach is similar. The best payoff in each state of nature is found, but this is the lowest payoff for that state of nature rather than the highest. The EVwPI is calculated from these lowest payoffs, and this is compared to the best (lowest) EMV without perfect information. The EVPI is the improvement that results, and this is the best $\text{EMV}-\text{EVwPI}.$

An alternative approach to maximizing EMV is to minimize expected opportunity loss (EOL). First, an opportunity loss table is constructed. Then the EOL is computed for each alternative by multiplying the opportunity loss by the probability and adding these together. In Table 3.7, we presented the opportunity loss table for the Thompson Lumber example. Using these opportunity losses, we compute the EOL for each alternative by multiplying the probability of each state of nature times the appropriate opportunity loss value and adding these together:

Table 3.11 gives these results. Using minimum EOL as the decision criterion, the best decision would be the second alternative, “construct a small plant.”

	STATE OF NATURE
ALTERNATIVE	FAVORABLE MARKET ($)	UNFAVORABLE MARKET ($)	EOL ($)
Construct a large plant	0	180,000	90,000
Construct a small plant	100,000	20,000
Do nothing	200,000	0	100,000
Probabilities	0.50	0.50

It is important to note that minimum EOL will always result in the same decision as maximum EMV and that the EVPI will always equal the minimum EOL. Referring to the Thompson case, we used the payoff table to compute the EVPI to be $60,000. Note that this is the minimum EOL we just computed.

In previous sections, we determined that the best decision (with the probabilities known) for Thompson Lumber was to construct the small plant, with an expected value of $40,000. This conclusion depends on the values of the economic consequences and the two probability values of a favorable and an unfavorable market. Sensitivity analysis investigates how our decision might change given a change in the problem data. In this section, we investigate the impact that a change in the probability values would have on the decision facing Thompson Lumber. We first define the following variable:

Because there are only two states of nature, the probability of an unfavorable market must be $1-P.$

We can now express the EMVs in terms of P, as shown in the following equations. A graph of these EMV values is shown in Figure 3.1.

As you can see in Figure 3.1, the best decision is to do nothing as long as P is between 0 and the probability associated with point 1, where the EMV for doing nothing is equal to the EMV for the small plant. When P is between the probabilities for points 1 and 2, the best decision is to build the small plant. Point 2 is where the EMV for the small plant is equal to the EMV for the large plant. When P is greater than the probability for point 2, the best decision is to construct the large plant. Of course, this is what you would expect as P increases. The value of P at points 1 and 2 can be computed as follows:

• $\text{Point} 1: \text{EMV} \left(\text{do} \text{nothing}\right)=\text{EMV} \left(\text{small} \text{plant}\right)$

• $0=\$120,000P-\$20,000P=\frac{20,000}{120,000}=0.167$

• $\text{Point} 2\text{:} \text{EMV} \left(\text{small} \text{plant}\right)=\text{EMV} \left(\text{large} \text{plant}\right)$

• $\$120,000P-\$20,000=\$380,000P-\$180,000$

• $260,000P=160,000P=\frac{160,000}{260,000}=0.615$

The results of this sensitivity analysis are displayed in the following table:

The previous examples have illustrated how to apply the decision-making criterion when the payoffs are to be maximized. The following example illustrates how the criteria are applied to problems in which the payoffs are costs that should be minimized.

The Business Analytics department at State University will be signing a 3-year lease for a new copy machine, and three different machines are being considered. For each of these, there is a monthly fee, which includes service on the machine, plus a charge for each copy. The number of copies that would be made each month is uncertain, but the department has estimated that the number of copies per month could be 10,000 or 20,000 or 30,000 per month. The monthly cost for each machine based on each of the three levels of activity is shown in Table 3.12.

Which machine should be selected? To determine the best alternative, a specific criterion must be chosen. If the decision maker is optimistic, only the best (minimum) payoff for each decision is considered. These are shown in Table 3.13, and the best (minimum) of these is 700. Thus, Machine C would be selected, allowing the possibility of achieving this best cost of $700.

If the decision maker is pessimistic, only the worst (maximum) payoff for each decision is considered. These are also shown in Table 3.13, and the best (minimum) of these is 1,150. Thus, Machine A would be selected based on the pessimistic criterion. This would guarantee that the cost would be no more than 1,150, regardless of which state of nature occurred.

Using the Hurwicz criterion, if we assume that the decision maker is 70% optimistic (the coefficient of realism is 0.7), the weighted average of the best and the worst payoff for each alternative would be calculated using the formula

For each of the three copy machines, we would get the following values:

The decision would be to select Machine C based on this criterion because it has the lowest weighted average cost.

For the equally likely criterion, the average payoff for each machine would be calculated.

Based on the equally likely criterion, Machine C would be selected because it has the lowest average cost.

To apply the EMV criterion, probabilities must be known for each state of nature. Past records indicate that 40% of the time the number of copies made in a month was 10,000, while 30% of the time it was 20,000 and 30% of the time it was 30,000. The probabilities for the three states of nature would be 0.4, 0.3, and 0.3. We can use these to calculate the EMVs, and the results are shown in Table 3.14. Machine C would be selected because it has the lowest EMV. The monthly cost would average $970 with this machine, while the other machines would average a higher cost.

To find the EVPI, we first find the payoffs (costs) that would be experienced with perfect information. The best payoff in each state of nature is the lowest value (cost) in that state of nature, as shown in the bottom row of Table 3.14. These values are used to calculate the EVwPI. With these costs, we find

Perfect information would lower the expected value by $45.

To apply criteria based on opportunity loss, we must first develop the opportunity loss table. In each state of nature, the opportunity loss indicates how much worse each payoff is than the best possible payoff in that state of nature. The best payoff (cost) would be the lowest cost. Thus, to get the opportunity loss in this case, we subtract the lowest value in each column from all the values in that column, and we obtain the opportunity loss table.

Once the opportunity loss table has been developed, the minimax regret criterion is applied exactly as it was for the Thompson Lumber example. The maximum regret for each alternative is found, and the alternative with the minimum of these maximums is selected. As seen in Table 3.15, the minimum of these maximums is 150, so Machine C would be selected based on this criterion.

The probabilities are used to compute the expected opportunity losses as shown in Table 3.15. Machine C has the lowest EOL of $45, so it would be selected based on the minimum EOL criterion. As previously noted, the minimum EOL is equal to the expected value of perfect information.