# 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%.

## Course section 11.7: Subsections and their intended learning outcomes

• ##### 11.7.1 Exercise: ABC Category and ABC Classification

Intended learning outcomes: Determine meaningful ABC categories of some items and perform the ABC classification.

• ##### 11.7.2 Scenario: The ABC-XYZ Analysis and its Use in Materials Management

Intended learning outcomes: Explain the combined ABC-XYZ classification. Describe its use in materials management. Identify the appropriateness of Kanban control as the result of an ABC-XYZ analysis.

• ##### 11.7.3 Exercise: Safety Stock Variation versus Demand Variation

Intended learning outcomes: Find out whether the safety stock level increases with increasing demand.

• ##### 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.

• ##### 11.7.5 Scenario: Effectiveness of the Order Point Technique

Intended learning outcomes: Explore the changing shape of the inventory curve for continuous and less continuous demand.

• ##### 11.7 Scenarios and Exercises

Intended learning outcomes: Calculate examples for the ABC Classification. Disclose the ABC-XYZ analysis in materials management. Differentiate between safety stock variation and demand variation. Determine batch size depending on stockout costs. Assess the effectiveness of the order point technique.