# 11.7.4 Exercise: Determining Batch Size Depending on Stockout Costs

### Intended learning outcomes: Determine the batch size in dependence of carrying costs and stockout costs, average annual consumption, and the standard deviation of demand during lead time.

This exercise refers to Sections 11.3.3 and 11.3.4.

The carrying costs for a certain article are 2 per unit and year. Stockout costs are 5 per unit. The average annual consumption amounts to 1000, and the standard deviation of demand during lead time is 10. No safety stock is intended. Normal distribution is assumed.

a.    How large should the batch size be, considering the optimum stockout probability? Can the fill rate target of 99% be met? What are the carrying costs per year?

b.    Assume a batch size of only 250. What are the values for safety stock and fill rate corresponding to the optimum probability of stockout per order cycle?

c.    Now assume a safety stock of 20 units. Again, the batch size is 250. What are the values for service level and fill rate?

Solution:

a.    Zero safety stock entails a service level of 50% (see Figure 11.3.3.6, for example) and — by Figure 11.3.3.3 — a probability of stockout per order cycle of 50%. Because stockout can be expressed as cost per unit, the formulas in Figures 11.3.4.2, 11.3.4.4, and 11.3.4.5 apply. Therefore,

·     Batch size = 1000 * 50% * (5/2) = 1250.

·     Stockout quantity coefficient P(s) = 0.399.

·     → Fill rate = 1 – ((10/1250) * 0.399) = 99.68% > 99%.

·     Average inventory = 1250/2 = 625.

·     → Carrying costs per year = 625 * 2 = 1250.

b.    Again, the formulas in Figures 11.3.4.2, 11.3.4.4, and 11.3.4.5 apply:

·     Optimum probability of stockout = (2/5) * (250/1000) = 10%.

·     → Optimum service level = 1 – 10 % = 90%.

·     → Safety stock = 1.282 * 10 {note: the standard deviation} ≈ 13.

·     → Stockout quantity coefficient P(s) = 0.048.

·     → Fill rate = 1 – ((10/250) * 0.048) = 99.81%.

c.    Applying the formulas in Figures 11.3.4.4, 11.3.4.2, and 11.3.4.5:

·     Standard deviation = 10; => safety factor = 20/10 = 2.

·     → Service level ≈ 98%.

·     → Stockout quantity coefficient P(s) = 0.008.

·     → Fill rate = 1 – ((10/250) * 0.008) = 99.97%.