1. **Problem Statement:** Calculate the safety stock, reorder point, and economic order quantity (EOQ) for a part with given demand, lead time, and cost parameters. 2. **Given Data:** - Cost per case = 600 - Parts per case = 72 - Daily demand ($D$) = 37 parts/day - Standard deviation of daily demand ($\sigma_d$) = 13 parts/day - Lead time ($L$) = 16 days - Desired service level = 95% (corresponds to $z$-value) - Ordering cost ($S$) = 410 - Annual interest rate ($i$) = 6% (used as holding cost rate) - Operating days per year = 365 3. **Formulas and Important Rules:** - Safety stock ($SS$) accounts for demand variability during lead time: $$SS = z \times \sigma_L$$ where $\sigma_L = \sigma_d \times \sqrt{L}$ is the standard deviation of demand during lead time. - Reorder point ($ROP$) is the inventory level to trigger a new order: $$ROP = D \times L + SS$$ - Economic Order Quantity (EOQ) minimizes total inventory cost: $$EOQ = \sqrt{\frac{2DS}{H}}$$ where $D$ is annual demand, $S$ is ordering cost, and $H$ is holding cost per unit per year. - Holding cost per unit per year ($H$) is calculated as: $$H = i \times C_u$$ where $C_u$ is cost per unit (part). - $z$-value for 95% service level (from standard normal distribution) is approximately 1.645. 4. **Calculations:** - Cost per part: $$C_u = \frac{600}{72} = 8.3333$$ - Annual demand: $$D = 37 \times 365 = 13505$$ - Holding cost per unit per year: $$H = 0.06 \times 8.3333 = 0.5$$ - Standard deviation of demand during lead time: $$\sigma_L = 13 \times \sqrt{16} = 13 \times 4 = 52$$ - Safety stock: $$SS = 1.645 \times 52 = 85.54$$ - Reorder point: $$ROP = 37 \times 16 + 85.54 = 592 + 85.54 = 677.54$$ - Economic order quantity: $$EOQ = \sqrt{\frac{2 \times 13505 \times 410}{0.5}} = \sqrt{22108100} = 4702$$ 5. **Final Answers:** - Safety stock = 85.54 parts - Reorder point = 677.54 parts - Economic order quantity = 4702 parts These values help Boeing maintain inventory to meet demand with a 95% service level while minimizing costs.