The manager of a grocery store wants to anticipate the sales

The manager of a grocery store wants to anticipate the sales of frozen pizza for the following month, so that he can prepare enough inventories to meet the demand. He predicts that the demand for a particular brand of frozen pizza will follow a normal distribution, and the average monthly demand will be 800 pieces with standard deviation of 120.

(a) How many frozen pizza from this brand should be prepared to satisfy a 95% expected fill rate?

(b) How many frozen pizza from this brand should be prepared so that the manager can keep the stock-out probability within 5%?

(c) Even though the percentages for the expected fill rate and in-stock probability are the same, the order quantities differ. In a few sentences, explain the reason behind this difference (i.e., do your answers make sense? which one is greater and why? etc.).

Solution

(a) Fill rate = Expected sales/ Mean demand = (mean demand - expected lost sales)/mean demand

Fill rate is given to be 95%

Therefore, target maximum expected lost sales = (1-95%)*mean demand = 5%*800 = 40

Target L(z) = 40 / Std dev = 40/120 = 0.3333

Corresponding z value = 0.1392 (taken from standard loss table corresponding to L(z) = 0.3333)

Therefore, Quantity of frozen pizza should be prepared = m+zs = 800 + 0.1392*120 = 817

(b) For stockout probability within 5%, stock-in probability = 95%

Corresponding z value = NORMSINV(0.95) = 1.645

Quantity of frozen pizza should be prepared = m + zs = 800 + 1.645*120 = 997

(c) The quantities differ, because fill rate is different from in-stock probability. Fill rate is measured as the ratio of expected sales versus mean demand, whereas in-stock probability is measured as probability of demand being less than or equal to stock available. Therefore, it makes sense that order quantity corresponding to in-stock probability would be higher.