Let's solve part (a) **step-by-step** and formulate the linear programming (LP) model. --- ### **Part (a): Formulating the LP Model** #### **Step 1: Define variables** Let: - \( x_t \): Number of lamps produced by Company A in month \( t \) (at $35 each) - \( y_t \): Number of lamps purchased from Company C in month \( t \) (at $50 each) - \( I_t \): Inventory at the end of month \( t \) (each unit costs $5/month to hold) where \( t = 1 \) (January), \( t = 2 \) (February), \( t = 3 \) (March), \( t = 4 \) (April). #### **Step 2: Objective Function** Minimize total cost: \[ \text{Minimize:} \quad Z = 35 \sum_{t=1}^4 x_t + 50 \sum_{t=1}^4 y_t + 5 \sum_{t=1}^4 I_t \] #### **Step 3: Constraints** ##### **Production limits** \[ x_t \leq 160 \qquad \forall t \] \[ x_t \geq 0, \quad y_t \geq 0, \quad I_t \geq 0 \qquad \forall t \] ##### **Demand Satisfaction and Inventory Balance** Let demand in each month be \( d_t \): - \( d_1 = 150 \) - \( d_2 = 160 \) - \( d_3 = 225 \) - \( d_4 = 180 \) Inventory balance for each month: - **January:** \( x_1 + y_1 = d_1 + I_1 \) - **February:** \( x_2 + y_2 + I_1 = d_2 + I_2 \) - **March:** \( x_3 + y_3 + I_2 = d_3 + I_3 \) - **April:** \( x_4 + y_4 + I_3 = d_4 + I_4 \) - **Final inventory:** Assume ending inventory is zero: \( I_4 = 0 \) Or, more generally: \[ x_t + y_t + I_{t-1} = d_t + I_t \quad \forall t = 1,2,3,4 \] Where \( I_0 = 0 \) (no initial inventory). ##### **Explicitly:** - \( x_1 + y_1 = d_1 + I_1 \) - \( x_2 + y_2 + I_1 = d_2 + I_2 \) - \( x_3 + y_3 + I_2 = d_3 + I_3 \) - \( x_4 + y_4 + I_3 = d_4 + I_4 \) - \( I_0 = 0, \quad I_4 = 0 \) --- ### **Summary of the LP Model** #### **Variables:** - \( x_t \): Produced by Company A in month \( t \) - \( y_t \): Purchased from Company C in month \( t \) - \( I_t \): Inventory at end of month \( t \) #### **Objective:** \[ \text{Minimize} \quad Z = 35 \sum_{t=1}^4 x_t + 50 \sum_{t=1}^4 y_t + 5 \sum_{t=1}^4 I_t \] #### **Subject to:** \[ x_1 + y_1 = 150 + I_1 \] \[ x_2 + y_2 + I_1 = 160 + I_2 \] \[ x_3 + y_3 + I_2 = 225 + I_3 \] \[ x_4 + y_4 + I_3 = 180 + I_4 \] \[ x_t \leq 160 \quad \forall t \] \[ I_0 = 0, \quad I_4 = 0 \] \[ x_t, y_t, I_t \geq 0 \quad \forall t \] --- ## **Final LP formulation (answer for (a)):** \[ \begin{align*} \text{Minimize} \quad & Z = 35(x_1 + x_2 + x_3 + x_4) + 50(y_1 + y_2 + y_3 + y_4) + 5(I_1 + I_2 + I_3 + I_4) \\ \text{Subject to:} \\ & x_1 + y_1 = 150 + I_1 \\ & x_2 + y_2 + I_1 = 160 + I_2 \\ & x_3 + y_3 + I_2 = 225 + I_3 \\ & x_4 + y_4 + I_3 = 180 + I_4 \\ & x_t \leq 160, \quad x_t \geq 0, \quad y_t \geq 0, \quad I_t \geq 0 \quad (t = 1,2,3,4) \\ & I_0 = 0, \quad I_4 = 0 \\ \end{align*} \] --- If you'd like to see the solution for (b) or further parts, let me know!

Certainly! Let's proceed step-by-step through each part of the problem, providing detailed calculations and explanations. --- ## **Part (a): Formulating the LP Model** **Already done in previous response.** **Summary:** - Variables: \(x_t, y_t, I_t\) - Objective: Minimize total costs (production, purchase, inventory holding) - Constraints: Production limits, inventory balance, demand satisfaction --- ## **Part (b): Solve the problem using a linear programming package** **Note:** Since this involves multiple variables and constraints, solving manually is complex. Typically, LP solvers like Excel Solver, LINDO, or any LP software can be used. **However,** I will outline the **step-by-step** approach to obtain the optimal solution manually, focusing on key insights. --- ### **Step 1: Understand the cost structure** - Production cost per unit: **\$35** - Purchase cost per unit: **\$50** - Inventory holding cost per unit per month: **\$5** Since **production is cheaper** (\$35) than **buying** (\$50), we prefer to produce as much as possible, subject to capacity, and buy only when necessary. --- ### **Step 2: Demand and capacity** | Month | Demand | Max production | Cost per unit | |---------|----------|------------------|--------------| | January | 150 | 160 | \$35 | | February| 160 | 160 | \$35 | | March | 225 | 160 (normal), 153 (maintenance) | \$35 | | April | 180 | 160 | \$35 | **Note:** Maintenance schedule affects maximum production in months 1-3. --- ### **Step 3: Initial plan without maintenance** Assuming **no maintenance**, produce the maximum in months 1-3: - January: produce 150 units (to meet demand exactly) - February: produce 160 units - March: produce 160 units - April: produce 180 units (but max is 160, so buy the remaining 20 units from Company C) But inventory costs and actual demand may differ, so let's **simulate** the production and inventory flow. --- ### **Step 4: Calculate the total cost** **Option 1:** Fully produce to meet demand in each month, minimizing purchases. - **Month 1:** - Produce 150 units (cost: 150 * \$35 = \$5,250) - Inventory at end of January: 0 units (since production exactly meets demand) - **Month 2:** - Produce 160 units (cost: 160 * \$35 = \$5,600) - Inventory at end of February: 0 units - **Month 3:** - Produce 160 units (cost: \$5,600) - Demand is 225, but only 160 produced, so need to buy 65 units: - Purchase 65 units (cost: 65 * \$50 = \$3,250) - Inventory at end of March: 0 units (since demand is met) - **Month 4:** - Produce 160 units (cost: \$5,600) - Need total demand of 180, so buy 20 units: - Purchase 20 units (cost: 20 * \$50 = \$1,000) - Inventory at end of April: 0 units **Total costs:** - Production: \( (150 + 160 + 160 + 160) \times \$35 = 630 \times \$35 = \$22,050 \) - Purchases: \( (65 + 20) \times \$50 = 85 \times \$50 = \$4,250 \) - Inventory holding: 0 (since exact meeting of demand) **Total Cost:** \( \$22,050 + \$4,250 = \$26,300 \) --- ### **Step 5: Considering inventory costs** If we produce more than demand in some months, inventory costs accrue, but may reduce purchase costs later. Without explicit calculations, the **initial estimate** suggests that producing exactly the demand and buying only when necessary is close to optimal. --- ### **Summary of (b):** **Optimal solution** involves: - Producing the maximum allowed (160 units) in months 1-3, adjusting for maintenance constraints. - Purchasing from Company C only when production capacity is insufficient. **Due to the complexity and multiple constraints,** the **best way to get the precise optimal values** is to use LP solver software with the formulation from part (a). --- ## **Part (c): Maintenance impact** **Maintenance schedule options:** | Month | Max production after maintenance | |--------|----------------------------------| | January | 151 units | | February | 153 units | | March | 155 units | ### **Step 1: Adjust the LP constraints** For the month scheduled for maintenance, update: \[ x_t \leq \text{max units after maintenance} \] ### **Step 2: Recalculate the production plan** - **January:** - Max production: 151 units - Demand: 150 units - Produce 151 units (cost: \$5,285) - Inventory at end of January: 1 unit (cost: \$5) - **February:** - Max production: 153 units - Demand: 160 units - Produce 153 units (cost: \$5,355) - Remaining demand: 7 units - Purchase 7 units (cost: \$350) - Inventory at end of February: 0 - **March:** - Max production: 155 units - Demand: 225 units - Produce 155 units (cost: \$5,425) - Remaining demand: 70 units - Purchase 70 units (cost: \$3,500) - Inventory at end of March: 0 - **April:** - Produce 160 units (cost: \$5,600) - Demand: 180 units - Purchase 20 units (cost: \$1,000) **Total costs:** - Production: \(151 + 153 + 155 + 160 = 619 \times \$35 = \$21,665\) - Purchases: \(7 + 70 + 20 = 97 \times \$50 = \$4,850\) - Inventory holding: \$5 (from January) **Total approximate cost:** \$21,665 + \$4,850 + \$5 = **\$27,525** **Observation:** Maintaining in January reduces inventory costs but limits production, increasing purchase costs. The optimal schedule requires balancing these factors. **Recommendation:** Schedule maintenance in the month that results in the lowest total cost after comparing options. --- ## **Part (d): Buying from Company D** - Company D offers **up to 50 lamps** during Jan, Feb, or March at **\$45 per lamp**. ### **Step 1: Is it cost-effective?** - Cost of buying from D: \$45 per lamp - Cost of producing: \$35 - Cost of purchasing from C: \$50 Since \$45 < \$50, **buying from D is cheaper than from C**. Compare costs: - **From D:** \$45 per lamp - **From C:** \$50 per lamp - **From A:** \$35 per lamp (production cost) But **producing internally** remains cheapest (\$35), unless capacity or demand constraints prevent it. ### **Step 2: When to buy from D?** Suppose there's a **demand for additional lamps** or **capacity limitations**. - For example, in month 2, demand exceeds production capacity (160 units), so buying from D can be beneficial. ### **Step 3: Optimal purchase decision** - **Identify months where demand exceeds capacity or where production is limited.** - Assuming we want to reduce overall costs, **buy up to 50 units from D in the month with the highest extra demand or lowest production capacity**. **If we buy 50 units in February:** - Cost: 50 * \$45 = \$2,250 - This reduces purchases from C or internal production. **Impact:** - Total cost reduces by saving \$5 per unit compared to buying from C. **Conclusion:** - Yes, **buying from Company D** is advantageous when demand exceeds internal capacity. **Final decision:** - Purchase **50 units in February** from D at \$45 each. **Impact:** - Reduce purchases from C by 50 units at \$50 each (saving \$250). - Total cost decreases by \$250. --- ## **Part (e): Price reduction from Company C in February** - Current price from C in Feb: \$50 - To make buying from C attractive, the **price must be less than \$35** (production cost), which is impossible. - Alternatively, compare with D's price (\$45): - For **buying from C to be attractive over D**: \[ \text{New price} < \$45 \] - **Maximum decrease:** \[ \$50 - \text{new price} < \$45 \Rightarrow \text{new price} > \$45 \] - Since the current price is \$50, the maximum decrease: \[ \$50 - \$45 = \$5 \] **Therefore,** if Company C offers **less than \$45 per unit** in February, it becomes more attractive than Company D. **Answer:** The **maximum decrease** in price for C in February that would make it attractive: \[ \$50 - \$45 = \boxed{\$5} \] --- ## **Part (f): Increase in holding cost to \$8 in February** - **Effect:** Inventory holding cost increases, making holding inventory more expensive. ### **Step 1: Recalculate inventory costs** - Original inventory cost per unit: \$5 - New inventory cost per unit in February: \$8 ### **Step 2: Impact on the solution** - **Higher inventory costs** discourage holding inventory. - The LP will favor **reducing inventory** and **matching demand more closely** each month, possibly increasing production or purchases. ### **Step 3: Effect on total cost** - The **total inventory holding cost** increases, likely increasing overall total cost. - The **optimal production plan** will be adjusted to produce and use inventory within the same month to avoid high holding costs. **Overall:** - The total cost **increases** due to higher inventory costs. - The **solution** will shift towards **more just-in-time production**. --- ## **Part (g): Impact of revised January demand (90 units)** - Original January demand: 150 units - New demand: 90 units ### **Step 1: Recalculate bounds** - **Lower bound:** The **minimum** total cost will **decrease** since demand is lower, reducing production and purchase costs. - **Upper bound:** The **maximum** increase in total cost is limited; the total cost cannot be less than the cost of producing 90 units, nor can it increase significantly. ### **Step 2: Using LP optimal solution** - The **original LP solution** provides **shadow prices** (dual variables), indicating how much the total cost would change with a change in demand. - **Lower bound:** \[ \text{Change in cost} \geq \text{Shadow price of January demand} \times (150 - 90) \] - **Upper bound:** \[ \text{Change in cost} \leq \text{Shadow price of January demand} \times (150 - 90) \] *Without explicit dual prices,** we can estimate that **cost reductions** are proportional to the decrease in demand. ### **Final notes:** - The **bounds** on the change in total cost depend on the **dual variables** from the LP solution. - **In practice**, solving the LP with demand of 90 units in January yields the exact change. --- # **Summary of Final Answers:** | Part | Summary | |---------|--------------------------------------------------------------| | (a) | LP formulation with variables \(x_t, y_t, I_t\), objective, and constraints | | (b) | Approximate total cost: **\$26,300** (exact solution via LP software recommended) | | (c) | Maintenance in Jan (151 units): approx **\$27,525**; in Feb (153 units): similar; choose the month with minimal total cost | | (d) | Buy 50 lamps from D in February: reduces total cost by **\$2,250** | | (e) | Max price decrease for C in Feb to \$45: **\$5** | | (f) | Higher holding cost (\$8): increases total cost, promotes just-in-time production | | (g) | Lower demand in January (90 units): total cost decreases; bounds estimated via LP dual prices | --- **If you need detailed LP solutions with exact numbers or software outputs, please specify, and I can guide you through setting up and solving in a specific LP solver.**