The opportunity loss function describes the loss that would be suffered by making the wrong decision. We saw earlier that Rudy’s break-even point is 6,000 sets of the game Strategy. If Rudy produces and markets the new game and sales are greater than 6,000 units, he has made the right decision; in this case, there is no opportunity loss ($0). If, however, he introduces Strategy and sales are less than 6,000 games, he has selected the wrong alternative. The opportunity loss is just the money lost if demand is less than the break-even point; for example, if demand is 5,999 games, Barclay loses $6 (= $10 price/unit - $4 cost/unit). With a $6 loss for each unit of sales less than the break-even point, the total opportunity loss is $6 multiplied by the number of units under 6,000. If only 5,000 games are sold, the opportunity loss will be 1,000 units less than the break-even point times $6 per unit = $6,000. For any level of sales, X, Barclay’s opportunity loss function can be expressed as follows:

In general, the opportunity loss function can be computed by

where

The second step is to find the expected opportunity loss. This is the sum of the opportunity losses multiplied by the appropriate probability values. But in Barclay’s case, there are a very large number of possible sales values. If the break-even point is 6,000 games, there will be 6,000 possible sales values, from 0, 1, 2, 3, up to 6,000 units. Thus, determining the EOL would require setting 6,000 probability values that correspond to the 6,000 possible sales values. These numbers would be multiplied and added together, a very lengthy and tedious task.

When we assume that there are an infinite (or very large) number of possible sales values that follow a normal distribution, the calculations are much easier. Indeed, when the unit normal loss integral is used, the EOL can be computed as follows:

where

$\text{EOL} = \text{expected opportunity loss}$

$K=\text{loss per unit when sales are below the break-even point}$

$\sigma =\text{standard deviation of the distribution}$

$N\left(D\right)=\text{value for the unit normal loss integral in Appendix M}3.2\text{ for a given value of}D$

where

Here is how Rudy can compute the EOL for his situation:

Now refer to the unit normal loss integral table in Appendix M3.2. Look in the “0.6” row and read over to the “0.9” column. This is $N\left(0.69\right),$ which is 0.1453:

Therefore,

Because EVPI and minimum EOL are equivalent, the EVPI is also $2,515.14. This is the maximum amount that Rudy should be willing to spend on additional marketing information.

The relationship between the opportunity loss function and the normal distribution is shown in Figure M3.4. This graph shows both the opportunity loss and the normal distribution with a mean of 8,000 games and a standard deviation of 2,885. To the right of the break-even point, we note that the loss function is 0. To the left of the break-even point, the opportunity loss function increases at a rate of $6 per unit—hence the slope of $\mathrm{-}6.$ The use of Appendix M3.2 and Equation M3-5 allows us to multiply the $6 unit loss times each of the probabilities between 6,000 units and 0 units and to sum these multiplications.