Let's break down the problem step by step and solve it using the **transportation method** as required. ### **Step 1: Problem Summary** - **5 months** planning period. - **Demand:** 150, 160, 130, 200, 210 units per month. - **Production capacity (per month):** - Regular: 150 units - Overtime: 20 units (except months 3, 4, 5 where it's 10 units) - **Subcontracting:** Max 100 units over 5 months. - **Inventory:** Start = 0 units, End = 20 units minimum. - **Costs:** - Regular: $100/unit - Overtime: $125/unit - Subcontract: $135/unit - Inventory: $3/unit/month --- ### **Step 2: Structure the Transportation Table** **Sources (rows):** - Regular (for each month, up to 150) - Overtime (for each month, up to 20 or 10 as listed) - Subcontracting (max 100 units over 5 months) - Inventory (carryover from previous month) **Destinations (columns):** - Demand for each month **Inventory flow:** Excess produced in one month can be used in the next month, with a $3/unit/month holding cost. --- ### **Step 3: Set Up the Transportation Table** | Source | Supply | Month 1 | Month 2 | Month 3 | Month 4 | Month 5 | |-------------|--------|---------|---------|---------|---------|---------| | Regular 1 | 150 | 100 | 103 | 106 | 109 | 112 | | Overtime 1 | 20 | 125 | 128 | 131 | 134 | 137 | | Regular 2 | 150 | | 100 | 103 | 106 | 109 | | Overtime 2 | 20 | | 125 | 128 | 131 | 134 | | ... | ... | ... | ... | ... | ... | ... | | Subcontract | 100 | 135 | 135 | 135 | 135 | 135 | | Inventory | | 3 | 6 | 9 | 12 | | - The cost in each cell is the **cost to produce (or carry from inventory) to meet demand in that month**. --- #### **Let’s define variables:** - \( R_i \): Regular production in month \( i \) - \( O_i \): Overtime production in month \( i \) - \( S_i \): Subcontracted in month \( i \) - \( I_i \): Inventory at end of month \( i \) #### **Constraints:** 1. **Demand satisfaction:** \( R_1 + O_1 + S_1 + I_0 = D_1 + I_1 \) \( R_2 + O_2 + S_2 + I_1 = D_2 + I_2 \) ... \( R_5 + O_5 + S_5 + I_4 = D_5 + I_5 \) Where \( I_0 = 0 \), \( I_5 = 20 \) 2. **Capacities:** - \( R_i \leq 150 \) - \( O_1, O_2 \leq 20 \), \( O_3, O_4, O_5 \leq 10 \) - \( \sum_{i=1}^5 S_i \leq 100 \) - No backorders: \( I_i \geq 0 \) --- ### **Step 4: Initial Feasible Solution (Greedy Approach)** Let's try to meet as much demand as possible with regular time, then overtime, then subcontracting, and carry excess as inventory: #### **Month 1: Demand = 150** - Regular: 150 (max capacity) - Overtime: 0 - Subcontract: 0 - Inventory carried to month 2: \(150 - 150 = 0\) #### **Month 2: Demand = 160** - Regular: 150 - Overtime: 10 (to make up the 10-unit shortfall, within 20-unit capacity) - Subcontract: 0 - Inventory: \(150 + 10 = 160\), so inventory at end = 0 #### **Month 3: Demand = 130** - Regular: 130 (within 150 cap) - Overtime: 0 - Subcontract: 0 - Inventory: 0 #### **Month 4: Demand = 200** - Regular: 150 (max) - Overtime: 10 (max for month 4) - Subcontract: 40 (needed to meet demand) - Inventory: \(150 + 10 + 40 = 200\), so inventory at end = 0 #### **Month 5: Demand = 210** - Regular: 150 - Overtime: 10 (max) - Subcontract: 50 (to meet demand and have 20 units inventory at end) - Inventory: \(150 + 10 + 50 = 210\), so inventory at end = 20 Total subcontract used = 40 + 50 = 90 units (within 100 allowed). --- ### **Step 5: Compute the Total Cost** #### **Month 1** - Regular: 150 × $100 = $15,000 - Overtime: 0 - Subcontract: 0 - Inventory: 0 #### **Month 2** - Regular: 150 × $100 = $15,000 - Overtime: 10 × $125 = $1,250 - Subcontract: 0 - Inventory: 0 #### **Month 3** - Regular: 130 × $100 = $13,000 - Overtime: 0 - Subcontract: 0 - Inventory: 0 #### **Month 4** - Regular: 150 × $100 = $15,000 - Overtime: 10 × $125 = $1,250 - Subcontract: 40 × $135 = $5,400 - Inventory: 0 #### **Month 5** - Regular: 150 × $100 = $15,000 - Overtime: 10 × $125 = $1,250 - Subcontract: 50 × $135 = $6,750 - Inventory: 20 × $3 (carried over to end) = $60 --- #### **Total Cost:** - Regular: $15,000 + $15,000 + $13,000 + $15,000 + $15,000 = **$73,000** - Overtime: $0 + $1,250 + $0 + $1,250 + $1,250 = **$3,750** - Subcontract: $0 + $0 + $0 + $5,400 + $6,750 = **$12,150** - Inventory: $60 **Grand Total = $73,000 + $3,750 + $12,150 + $60 = $88,960** --- ## **Final Answer** ### **The total cost of the optimal plan is:** \[ \boxed{\$88,960} \] --- **Note:** This solution uses a greedy approach with the transportation method's logic (prioritizing regular, then overtime, then subcontracting, and ensuring inventory meets the final requirement). For fine-tuning, a full transportation tableau or LP solver could be used, but this will be very close to the optimal, if not exactly optimal, due to the problem's structure. If you want a full transportation tableau setup, let me know!

Certainly! Let's go through the detailed step-by-step solution, including calculations for each part, to find the total cost of the optimal aggregate plan. --- ### **Problem Recap** - **Planning period:** 5 months - **Demand per month:** 150, 160, 130, 200, 210 units - **Production capacities:** - Regular: 150 units/month - Overtime: 20 units/month (except months 3, 4, 5 where it's 10 units) - **Subcontracting:** Max 100 units over 5 months - **Initial inventory:** 0 units - **Ending inventory:** 20 units - **Costs:** - Regular: $100/unit - Overtime: $125/unit - Subcontract: $135/unit - Inventory holding: $8/unit/month - **Backorders:** Not permitted --- ### **Step 1: Determine total demand and supply constraints** | Month | Demand | Regular capacity | Overtime capacity | Max subcontract | End inventory goal | |---------|--------|-------------------|-------------------|-----------------|---------------------| | 1 | 150 | 150 | 20 | 100 | 20 | | 2 | 160 | 150 | 20 | 100 | | | 3 | 130 | 150 | 10 | 100 | | | 4 | 200 | 150 | 10 | 100 | | | 5 | 210 | 150 | 10 | 100 | 20 (final inventory) | --- ### **Step 2: Initial feasible plan assumptions** - Use **full regular capacity** first each month - Use **overtime** up to capacity when necessary - Use **subcontracting** to fill any gaps - Carry over inventory from month to month to meet demand - End with **20 units** inventory at month 5 --- ### **Step 3: Month-by-month production plan** **Month 1:** - Demand = 150 units - Use full regular capacity: **150 units** - Overtime = 0 - Subcontract = 0 - Inventory at end of month: \( 0 + 150 - 150 = 0 \) **Remaining capacity for month 1:** - Regular: 150 (full capacity used) - Overtime: 0 - Subcontract: 0 --- **Month 2:** - Demand = 160 units - Use full regular capacity: **150 units** - Remaining shortfall: \(160 - 150 = 10\) units - Use overtime capacity: **10 units** (within 20-unit limit) - Subcontract: 0 - Inventory at end of month: \( 0 + 150 + 10 - 160 = 0 \) --- **Month 3:** - Demand = 130 units - Use full regular capacity: **130 units** - Remaining capacity: - Regular: 20 units (since total is 150) - Overtime: 10 units (max for months 3,4,5) - Use 130 units for demand, remaining capacity: \( 20 + 10 = 30 \) units - Inventory at end of month: \( 0 + 130 - 130 = 0 \) --- **Month 4:** - Demand = 200 units - Use full regular capacity: **150 units** - Remaining shortfall: \(200 - 150 = 50\) - Use overtime: **10 units** (max for month 4) - Remaining shortfall after regular + overtime: \(50 - 10 = 40\) - Use subcontracting: **40 units** (within 100 limit) - Inventory at end of month: \(0 + 150 + 10 + 40 - 200 = 0\) --- **Month 5:** - Demand = 210 units - Use full regular capacity: **150 units** - Remaining shortfall: \(210 - 150 = 60\) - Use overtime: **10 units** (max for month 5) - Remaining shortfall after regular + overtime: \(60 - 10 = 50\) - Use subcontracting: **50 units** (within 100 limit) - Inventory at end of month: \(0 + 150 + 10 + 50 - 210 = 0\) - **Final inventory** must be 20 units; to achieve that, produce **additional units** or adjust. **However,** since the last month needs to finish with 20 units, we can produce **20 more units** in month 5 and store as inventory, or produce an extra 20 units in month 5 for inventory. **Adjustment for final inventory:** - Produce **20 units** extra in month 5 (beyond demand) and not use for demand - Total production in month 5 becomes: \(150 + 10 + 50 + 20 = 230\) - **Demand covered:** \(\text{Regular}: 150\) \(\text{Overtime}: 10\) \(\text{subcontract}: 50\) \(\text{Extra units for inventory}\): 20 - **Total units produced in Month 5:** 230 units - Inventory at end of month 5: \(20\) units (as desired) --- ### **Step 4: Compute the total units produced** | Month | Regular | Overtime | Subcontract | Extra for inventory | Total production | |---------|---------|------------|--------------|----------------------|------------------| | 1 | 150 | 0 | 0 | 0 | 150 | | 2 | 150 | 10 | 0 | 0 | 160 | | 3 | 130 | 10 | 0 | 0 | 140 | | 4 | 150 | 10 | 40 | 0 | 200 | | 5 | 150 | 10 | 50 | 20 | 230 | --- ### **Step 5: Calculate costs** **(a) Regular production cost:** \[ 150 \times \$100 + 150 \times \$100 + 130 \times \$100 + 150 \times \$100 + 150 \times \$100 \] \[ = (150+150+130+150+150) \times \$100 = 730 \times \$100 = \$73,000 \] --- **(b) Overtime cost:** \[ 0 + 10 + 10 + 10 + 10 = 40 \text{ units} \] \[ 40 \times \$125 = \$5,000 \] --- **(c) Subcontracting cost:** \[ 0 + 0 + 0 + 40 + 50 = 90 \text{ units} \] \[ 90 \times \$135 = \$12,150 \] --- **(d) Inventory holding cost:** - Extra units produced for month 5: 20 units held for 1 month (from month 5 end inventory) - Inventory at end of month 5 = 20 units Inventory at end of month 4: 0 units Inventory at end of month 3: 0 units Inventory at end of month 2: 0 units Inventory at end of month 1: 0 units - Inventory cost: \[ 20 \times \$8 = \$160 \] --- ### **Step 6: Total cost** \[ \boxed{ \text{Total Cost} = \text{Regular} + \text{Overtime} + \text{Subcontract} + \text{Inventory} } \] \[ = \$73,000 + \$5,000 + \$12,150 + \$160 = \boxed{\$90,310} \] --- ### **Final answer:** ### **The total cost of the optimal plan is \(\boxed{\$90,310}\).** --- **Note:** This detailed plan assumes maximum utilization of regular time, with overtime and subcontracting filling the shortfalls, and produces an extra 20 units in the last month to meet the inventory requirement. Adjustments can be made based on different production sequencing, but this plan minimizes costs given the constraints.