## Continuous-Review Inventory: Reorder Point and Backorders ### Given Data - **Daily Demand:** \( D \sim N(\mu = 100, \sigma = 40) \) - **Lead Time:** \( L = 7 \) days - **Service Target:** Cycle Service Level \( \text{CSL} = 95\% \) - **Z-Value for CSL 95%:** \( z_{.95} \approx 1.645 \) ### Calculations #### 1. Lead-Time Demand - **Mean:** \( \mu_L = \mu \times L = 100 \times 7 = 700 \) - **Standard Deviation:** \( \sigma_L = \sigma \times \sqrt{L} = 40 \times \sqrt{7} \approx 105.83 \) #### 2. Reorder Point \( r \) \[ r = \mu_L + z_{CSL} \times \sigma_L \] \[ r = 700 + 1.645 \times 105.83 \approx 873.92 \] #### 3. Expected Backorders per Cycle - **Standard Normal Variable \( z \):** \[ z = \frac{r - \mu_L}{\sigma_L} = \frac{873.92 - 700}{105.83} \approx 1.645 \] - **Expected Backorders:** \[ E[B] = \sigma_L \times \phi(z) - z \times \sigma_L \times (1 - \Phi(z)) \] - \( \phi(z) \) is the standard normal probability density function. - \( \Phi(z) \) is the standard normal cumulative distribution function. Using \( z = 1.645 \): - \( \phi(1.645) \approx .103 \) - \( \Phi(1.645) \approx .95 \) \[ E[B] = 105.83 \times .103 - 1.645 \times 105.83 \times (1 - .95) \] \[ E[B] = 10.90 - 8.70 \approx 2.20 \] #### 4. Total Expected Backorders over 90 Days - **Number of Cycles in 90 Days:** \( \frac{90}{7} \approx 12.86 \) - **Total Expected Backorders:** \[ \text{Total Backorders} \approx 12.86 \times 2.20 \approx 28.29 \] ### Summary - **Reorder Point \( r \):** Approximately 874 units. - **Expected Total Backorders over 90 Days:** Approximately 28 units.