Finding the inventory level with the lowest cost is not difficult when we follow the marginal analysis procedure. This approach says that we would stock an additional unit only if the expected marginal profit for that unit equals or exceeds the expected marginal loss. This relationship is expressed symbolically as follows:

The expected marginal profit is found by multiplying the probability that a given unit will be sold by the marginal profit, P(MP). Similarly, the expected marginal loss is the probability of not selling the unit multiplied by the marginal loss, or $\left(1-\text{P}\right)\left(\text{ML}\right).$

The optimal decision rule is to stock the additional unit if

With some basic mathematical manipulations, we can determine the level of P for which this relationship holds:

or

In other words, as long as the probability of selling one more unit (P) is greater than or equal to $\text{ML/}\left(\text{MP}+\text{ML}\right),$ we would stock the additional unit.

Café du Donut is a popular New Orleans dining spot on the edge of the French Quarter. Its specialty is coffee and doughnuts; it buys the doughnuts fresh daily from a large industrial bakery. The café pays $4 for each carton (containing two dozen doughnuts) delivered each morning. Any cartons not sold at the end of the day are thrown away, for they would not be fresh enough to meet the café’s standards. If a carton of doughnuts is sold, the total revenue is $6. Hence, the marginal profit per carton of doughnuts is

The marginal loss is $\text{ML}=\text{\$}4$, since the doughnuts cannot be returned or salvaged at day’s end.

From past sales, the café’s manager estimates that the daily demand will follow the probability distribution shown in Table 6.6. The manager then follows the three steps to find the optimal number of cartons of doughnuts to order each day.

1. Step 1. Determine the value of $\frac{\text{ML}}{\text{ML}+\text{MP}}$ for the decision rule

   $$\begin{array}{lll}P\hfill & \ge \hfill & \frac{\text{ML}}{\text{ML}+\text{MP}}=\frac{\$4}{\$4+\$2}=\frac{4}{6}=0.67\hfill \\ P\hfill & \ge \hfill & 0.67\hfill \end{array}$$

   So the inventory stocking decision rule is to stock an additional unit if $P\ge \mathrm{0.67.}$

2. Step 2. Add a new column to the table to reflect the probability that doughnut sales will be at each level or greater. This is shown in the right-hand column of Table 6.7. For example, the probability that demand will be 4 cartons or greater is 1.00 (= 0.05 + 0.15 + 0.15 + 0.20 + 0.25 + 0.10 + 0.10). Similarly, the probability that sales will be 8 cartons or greater is $0.45\left(=0.25+0.10+0.10\right)$—that is, the sum of the probabilities for sales of 8, 9, and 10 cartons.

3. Step 3. Keep ordering additional cartons as long as the probability of selling at least one additional carton is greater than P, which is the indifference probability. If Café du Donut orders 6 cartons, marginal profits will still be greater than marginal loss, since

   $$P\text{ at }6\text{ cartons}=0.80>0.67$$

When product demand or sales follow a normal distribution, which is a common business situation, marginal analysis with the normal distribution can be applied. First, we need to find four values:

1. The average or mean sales for the product, $\mu$

2. The standard deviation of sales, $\sigma$

3. The marginal profit for the product, MP

4. The marginal loss for the product, ML

Once these quantities are known, the process of finding the best stocking policy is somewhat similar to marginal analysis with discrete distributions. We let $X^{*}=\text{optimal}$ stocking level.

Demand for copies of the Chicago Tribune newspaper at Joe’s Newsstand is normally distributed and has averaged 60 papers per day, with a standard deviation of 10 papers. With a marginal loss of 20 cents and a marginal profit of 30 cents, what daily stocking policy should Joe follow?

1. Step 1. Joe should stock the Chicago Tribune as long as the probability of selling the last unit is at least $\text{ML/}\left(\text{ML}+\text{MP}\right):$

   $$\frac{\text{ML}}{\text{ML}+\text{MP}}=\frac{20\text{ cents}}{20\text{ cents}+30\text{ cents}}=\frac{20}{50}=0.40$$

   Let $P=\mathrm{0.40.}$

2. Step 2. Figure 6.10 shows the normal distribution. Since the normal table has cumulative areas under the curve between the left side and any point, we look for $0.60\left(=1.0-0.40\right)$ in order to get the corresponding Z value:

   $$Z=0.25\text{ standard deviations from the mean}$$

3. Step 3. In this problem, $\mu =60$ and $\sigma =10,$ so

   $$0.25=\frac{X^{*}-60}{10}$$

   or

   $$X^{*}=60+0.25\left(10\right)=62.5,\text{ or }62\text{ newspapers}.$$

   Thus, Joe should order 62 copies of the Chicago Tribune daily, since the probability of selling 63 is slightly less than 0.40.

When P is greater than 0.50, the same basic procedure is used, although caution should be used when looking up the Z value. Let’s say that Joe’s Newsstand also stocks the Chicago Sun-Times, which has a marginal loss of 40 cents and a marginal profit of 10 cents. The daily sales have averaged 100 copies of the Chicago Sun-Times, with a standard deviation of 10 papers. The optimal stocking policy is as follows:

The normal curve is shown in Figure 6.11. Since the normal curve is symmetrical, we find Z for an area under the curve of 0.80 and multiply this number by $\mathrm{-}1$ because all values below the mean are associated with a negative Z value:

With $\mu =100$ and $\sigma =10,$

or

So Joe should order 91 copies of the Chicago Sun-Times every day.

The optimal stocking policies in these two examples are intuitively consistent. When marginal profit is greater than marginal loss, we would expect $X^{*}$ to be greater than the average demand, $\mu$, and when marginal profit is less than marginal loss, we would expect the optimal stocking policy, $X^{*},$ to be less than $\mu .$