Let's solve the ePaint Lustre line inventory problem step by step. #### Given Data - **Annual Demand (D):** 12,000 litres/year - **Holding Cost (H):** $0.75/litre/year - **Setup Cost (S):** $250 - **Production Rate (P):** 200 litres/day - **Number of Working Days:** 311 days/year - **Daily Demand (d):** 12,000 / 311 ≈ 38.59 litres/day --- ## (a) **Economic Production Quantity (EPQ), Number of Orders, Time Between Orders, Costs, and Setup Time** ### **1. Economic Production Quantity (EPQ) Formula** For production (not purchase), the formula is: \[ Q^* = \sqrt{ \frac{2DS}{H} \cdot \frac{P}{P - d} } \] #### **Step 1: Calculate \( P - d \)** \[ P - d = 200 - 38.59 = 161.41 \text{ litres/day} \] #### **Step 2: Calculate the fraction \( \frac{P}{P-d} \)** \[ \frac{P}{P-d} = \frac{200}{161.41} = 1.239 \] #### **Step 3: Calculate the classic EOQ part** \[ \frac{2DS}{H} = \frac{2 \times 12,000 \times 250}{0.75} = \frac{6,000,000}{0.75} = 8,000,000 \] #### **Step 4: Plug into EQP formula** \[ Q^* = \sqrt{8,000,000 \times 1.239} = \sqrt{9,912,000} \approx 3,150 \text{ litres} \] --- ### **2. Number of Production Runs per Year** \[ N = \frac{D}{Q^*} = \frac{12,000}{3,150} \approx 3.81 \text{ orders/year} \] --- ### **3. Time Between Orders** \[ \text{Cycle Time} = \frac{Q^*}{D} \times 1 \text{ year} = \frac{3,150}{12,000} \times 1 \text{ year} = 0.2625 \text{ years} \approx 95.8 \text{ days} \] --- ### **4. Total Annual Setup Cost** \[ \text{Annual Setup Cost} = N \times S = 3.81 \times \$250 = \$952.38 \] --- ### **5. Total Annual Holding Cost** For EPQ, max inventory is not Q but \[ \text{Max Inventory} = Q^* \left(1 - \frac{d}{P}\right) \] \[ \frac{d}{P} = \frac{38.59}{200} = 0.19295 \] \[ \text{Max Inventory} = 3,150 \times (1 - 0.19295) = 3,150 \times 0.80705 = 2,542 \text{ litres} \] Average inventory is half the max: \[ \text{Avg Inventory} = \frac{3,150 \times (1 - 0.19295)}{2} = \frac{2,542}{2} = 1,271 \text{ litres} \] \[ \text{Annual Holding Cost} = 1,271 \times \$0.75 = \$953.25 \] --- ### **6. Setup Time in Hours** \[ \text{Setup Time (hours)} = \frac{S}{\text{Setup Labour Cost/hr}} = \frac{250}{25} = 10 \text{ hours} \] --- ### **7. Diagram** (Since I can't draw, here’s how to label it): - **X-axis:** Order Quantity (Q) - **Y-axis:** Annual Cost (\$) - **Curves:** - **Setup Cost Curve:** Decreases as Q increases (hyperbola) - **Holding Cost Curve:** Increases as Q increases (straight line) - **Total Cost Curve:** U-shaped, minimum at optimal Q* - **Label:** Show the minimum at Q* = 3,150 litres, and the minimum total cost ≈ \$952 (setup) + \$953 (holding) = **\$1,905**. --- ## (b) **Duration of Production Run (L), Maximum Inventory, Demand During Production** ### **1. Duration of Production Run (L)** \[ L = \frac{Q^*}{P} = \frac{3,150}{200} = 15.75 \text{ days} \] ### **2. Maximum Inventory (already found):** \[ \text{Max Inventory} = 2,542 \text{ litres} \] ### **3. Demand During Production Interval** During production (L days), demand is: \[ \text{Demand During Run} = d \times L = 38.59 \times 15.75 = 608 \text{ litres} \] --- ## (c) **New Policy: Order Quantity 2,800 litres; Setup Cost per Order; Number of Orders; Costs; Setup Time** ### **1. New Order Quantity (Q):** 2,800 litres ### **2. Number of Orders** \[ N' = \frac{12,000}{2,800} = 4.29 \text{ orders/year} \] ### **3. Setup Cost per Order** \[ \text{Annual Setup Cost} = 4.29 \times \$250 = \$1,072.50 \] ### **4. Holding Cost** - Max Inventory: \[ Q' \left(1 - \frac{d}{P}\right) = 2,800 \times 0.80705 = 2,260 \text{ litres} \] - Average Inventory: \[ \frac{2,260}{2} = 1,130 \text{ litres} \] - Annual Holding Cost: \[ 1,130 \times \$0.75 = \$847.50 \] ### **5. Total Annual Inventory Cost** \[ \text{Total Cost} = \$1,072.50 + \$847.50 = \$1,920 \] ### **6. Setup Time at \$75/hr** \[ \text{Setup Time (hours)} = \frac{250}{75} = 3.33 \text{ hours} \] --- ## **Summary Table** | Item | Original Policy | New Policy | |------------------------------------|---------------------|-------------------| | Order Quantity (Q) | 3,150 litres | 2,800 litres | | Number of Orders | 3.81 | 4.29 | | Time Between Orders (days) | 95.8 | 72.6 | | Setup Cost per Order | $250 | $250 | | Annual Setup Cost | $952 | $1,073 | | Max Inventory | 2,542 litres | 2,260 litres | | Average Inventory | 1,271 litres | 1,130 litres | | Annual Holding Cost | $953 | $848 | | Total Annual Inventory Cost | $1,905 | $1,920 | | Setup Time (hours) | 10 | 3.33 | --- ## **Step-by-step: How to Solve** **a.** - Use EPQ formula for Q* - Calculate number of orders: D/Q* - Find time between orders: 311 / number of orders OR Q*/d - Setup & holding costs: N*S, Avg Inv * H - Setup time: S / setup rate - Draw the curves as described **b.** - Production run duration: Q*/P - Max inventory: Q*(1-d/P) - Demand during production: d * L **c.** - Use new Q = 2,800 - Number of orders: D/Q - Calculate new setup/holding/total cost as above - Setup time: S / 75 --- If you want detailed workings for any step or a sample diagram, let me know!