Let's break down the **blending problem** step by step: ### **Step 1: Define Decision Variables** Let \( x_{ij} \) be the amount of rock type \( i \) used in order \( j \). - \( i \) = rock type (A, B, C, D, E) - \( j \) = order (1, 2, 3, 4) So, \( x_{ij} \) = pounds of rock \( i \) in order \( j \). ### **Step 2: Objective Function (Minimize Cost)** Minimize the total cost: \[ \text{Minimize} \quad Z = \sum_{i=1}^5 \sum_{j=1}^4 (\text{Cost}_i \times x_{ij}) \] Where Cost: - A: \$1.00 - B: \$5.00 - C: \$5.50 - D: \$2.00 - E: \$1.20 ### **Step 3: Constraints** #### **a) Availability of Each Rock** \[ \sum_{j=1}^4 x_{ij} \leq \text{Amount}_i \quad \forall i \] Where Amount: - A: 500 - B: 600 - C: 700 - D: 400 - E: 450 #### **b) Each Order Must Be Filled** \[ \sum_{i=1}^5 x_{ij} = \text{Size}_j \quad \forall j \] Where Size: - Order 1: 500 - Order 2: 600 - Order 3: 500 - Order 4: 350 #### **c) Phosphate Percentage Constraints** Let \( p_i \) be the % phosphate in rock \( i \). - A: 2.0% - B: 5.0% - C: 4.5% - D: 3.0% - E: 6.0% The weighted average phosphate % in each order \( j \): \[ \frac{\sum_{i=1}^5 p_i x_{ij}}{\sum_{i=1}^5 x_{ij}} \quad \text{(for order \( j \))} \] This must be within the min and max bounds for each order: For each order \( j \): \[ \text{Min}_j \leq \frac{\sum_{i=1}^5 p_i x_{ij}}{\text{Size}_j} \leq \text{Max}_j \] Multiply both sides by \(\text{Size}_j\): \[ \text{Min}_j \cdot \text{Size}_j \leq \sum_{i=1}^5 p_i x_{ij} \leq \text{Max}_j \cdot \text{Size}_j \] Where Min and Max for each order are given in the table. ### **Step 4: Write the Explicit Linear Program** **Variables:** \( x_{ij} \geq 0 \) for \( i = A-E \), \( j = 1-4 \) #### **Objective:** \[ \text{Minimize}\quad Z = 1.00\sum_{j=1}^4 x_{Aj} + 5.00\sum_{j=1}^4 x_{Bj} + 5.50\sum_{j=1}^4 x_{Cj} + 2.00\sum_{j=1}^4 x_{Dj} + 1.20\sum_{j=1}^4 x_{Ej} \] #### **Subject to:** **Availability:** \[ \sum_{j=1}^4 x_{Aj} \leq 500 \\ \sum_{j=1}^4 x_{Bj} \leq 600 \\ \sum_{j=1}^4 x_{Cj} \leq 700 \\ \sum_{j=1}^4 x_{Dj} \leq 400 \\ \sum_{j=1}^4 x_{Ej} \leq 450 \\ \] **Order size:** \[ x_{A1}+x_{B1}+x_{C1}+x_{D1}+x_{E1} = 500 \\ x_{A2}+x_{B2}+x_{C2}+x_{D2}+x_{E2} = 600 \\ x_{A3}+x_{B3}+x_{C3}+x_{D3}+x_{E3} = 500 \\ x_{A4}+x_{B4}+x_{C4}+x_{D4}+x_{E4} = 350 \\ \] **Phosphate content:** For order 1: \[ 0.035 \times 500 \leq 0.02x_{A1} + 0.05x_{B1} + 0.045x_{C1} + 0.03x_{D1} + 0.06x_{E1} \leq 0.044 \times 500 \] Order 2: \[ 0.038 \times 600 \leq 0.02x_{A2} + 0.05x_{B2} + 0.045x_{C2} + 0.03x_{D2} + 0.06x_{E2} \leq 0.046 \times 600 \] Order 3: \[ 0.04 \times 500 \leq 0.02x_{A3} + 0.05x_{B3} + 0.045x_{C3} + 0.03x_{D3} + 0.06x_{E3} \leq 0.047 \times 500 \] Order 4: \[ 0.036 \times 350 \leq 0.02x_{A4} + 0.05x_{B4} + 0.045x_{C4} + 0.03x_{D4} + 0.06x_{E4} \leq 0.048 \times 350 \] ### **Step 5: Solve the Linear Program** This is a standard **Linear Programming** (LP) problem and can be solved using tools like Excel Solver, Python (PuLP), or any LP solver. --- #### **Summary of Steps:** 1. **Define variables**: \( x_{ij} \) = pounds of rock \( i \) in order \( j \) 2. **Set up cost minimization**: Objective function as shown. 3. **Write supply constraints**: Each rock type's total use ≤ available amount. 4. **Write demand constraints**: Each order's total = required size. 5. **Write quality constraints**: Weighted average phosphate % in each order within bounds. 6. **Solve the LP**: Use software for the solution. --- #### **Final Answer:** **The cheapest way to fill the orders is found by solving the above linear program.** If you want the actual values for each \( x_{ij} \), you need to use an LP solver. If you want, I can show you how to set this up in Excel or Python, or solve a simplified version here! Let me know how you'd like to proceed.

Certainly! Let's proceed with a detailed, step-by-step solution, including calculations, to determine the cheapest way to fill the four orders using the given rocks. --- ## **Step 1: Gather and organize the data** ### Rock Types and their properties: | Rock | Cost ($/lb) | Phosphate % | Available (lbs) | |--------|--------------|--------------|-----------------| | A | 1.00 | 20% (0.20) | 500 | | B | 5.00 | 5% (0.05) | 600 | | C | 5.50 | 45% (0.45) | 700 | | D | 2.00 | 3% (0.03) | 400 | | E | 1.20 | 60% (0.60) | 250 | ### Orders: | Order | Size (lbs) | Min Phosphate % | Max Phosphate % | |--------|--------------|------------------|------------------| | 1 | 500 | 35% (0.35) | 44% (0.44) | | 2 | 600 | 38% (0.38) | 46% (0.46) | | 3 | 500 | 40% (0.40) | 47% (0.47) | | 4 | 350 | 38% (0.38) | 48% (0.48) | --- ## **Step 2: Define decision variables** Let: - \( x_{ij} \) = pounds of rock \( i \) used in order \( j \) where \( i \in \{A, B, C, D, E\} \), \( j \in \{1,2,3,4\} \). --- ## **Step 3: Write the objective function** Minimize total cost: \[ Z = \sum_{i} \sum_{j} \text{Cost}_i \times x_{ij} \] Explicitly: \[ Z = 1.00(x_{A1} + x_{A2} + x_{A3} + x_{A4}) + 5.00(x_{B1} + x_{B2} + x_{B3} + x_{B4}) + 5.50(x_{C1} + x_{C2} + x_{C3} + x_{C4}) + 2.00(x_{D1} + x_{D2} + x_{D3} + x_{D4}) + 1.20(x_{E1} + x_{E2} + x_{E3} + x_{E4}) \] --- ## **Step 4: Write the constraints** ### 4.1. Availability constraints: \[ \begin{cases} x_{A1} + x_{A2} + x_{A3} + x_{A4} \leq 500 \\ x_{B1} + x_{B2} + x_{B3} + x_{B4} \leq 600 \\ x_{C1} + x_{C2} + x_{C3} + x_{C4} \leq 700 \\ x_{D1} + x_{D2} + x_{D3} + x_{D4} \leq 400 \\ x_{E1} + x_{E2} + x_{E3} + x_{E4} \leq 250 \end{cases} \] ### 4.2. Volume constraints for each order: \[ \begin{cases} x_{A1} + x_{B1} + x_{C1} + x_{D1} + x_{E1} = 500 \\ x_{A2} + x_{B2} + x_{C2} + x_{D2} + x_{E2} = 600 \\ x_{A3} + x_{B3} + x_{C3} + x_{D3} + x_{E3} = 500 \\ x_{A4} + x_{B4} + x_{C4} + x_{D4} + x_{E4} = 350 \end{cases} \] ### 4.3. Phosphate percentage constraints: For each order, the weighted average phosphate content must be within the given bounds. Calculate the total phosphate in pounds for each order: \[ \sum_{i} p_i \times x_{ij} \] where \( p_i \) are the phosphate percentages: | Rock | \( p_i \) | |---------|------------| | A | 0.20 | | B | 0.05 | | C | 0.45 | | D | 0.03 | | E | 0.60 | Order 1 (size 500): \[ 0.35 \leq \frac{0.20 x_{A1} + 0.05 x_{B1} + 0.45 x_{C1} + 0.03 x_{D1} + 0.60 x_{E1}}{500} \leq 0.44 \] Multiply through by 500: \[ 175 \leq 0.20 x_{A1} + 0.05 x_{B1} + 0.45 x_{C1} + 0.03 x_{D1} + 0.60 x_{E1} \leq 220 \] Similarly for other orders: **Order 2 (size 600):** \[ 0.38 \leq \frac{0.20 x_{A2} + 0.05 x_{B2} + 0.45 x_{C2} + 0.03 x_{D2} + 0.60 x_{E2}}{600} \leq 0.46 \] Multiply by 600: \[ 228 \leq 0.20 x_{A2} + 0.05 x_{B2} + 0.45 x_{C2} + 0.03 x_{D2} + 0.60 x_{E2} \leq 276 \] **Order 3 (size 500):** \[ 0.40 \leq \frac{0.20 x_{A3} + 0.05 x_{B3} + 0.45 x_{C3} + 0.03 x_{D3} + 0.60 x_{E3}}{500} \leq 0.47 \] Multiply by 500: \[ 200 \leq 0.20 x_{A3} + 0.05 x_{B3} + 0.45 x_{C3} + 0.03 x_{D3} + 0.60 x_{E3} \leq 235 \] **Order 4 (size 350):** \[ 0.38 \leq \frac{0.20 x_{A4} + 0.05 x_{B4} + 0.45 x_{C4} + 0.03 x_{D4} + 0.60 x_{E4}}{350} \leq 0.48 \] Multiply by 350: \[ 133 \leq 0.20 x_{A4} + 0.05 x_{B4} + 0.45 x_{C4} + 0.03 x_{D4} + 0.60 x_{E4} \leq 168 \] --- ## **Step 5: Summary of the LP model** **Minimize:** \[ Z = 1.00 \sum_{j=1}^{4} x_{A j} + 5.00 \sum_{j=1}^{4} x_{B j} + 5.50 \sum_{j=1}^{4} x_{C j} + 2.00 \sum_{j=1}^{4} x_{D j} + 1.20 \sum_{j=1}^{4} x_{E j} \] **Subject to:** - Availability constraints - Order volume constraints - Phosphate constraints (as above) - Non-negativity constraints: \( x_{ij} \ge 0 \) --- ## **Step 6: Numerical solution (approximate)** Due to the complexity, solving this LP manually is impractical. Typically, we'd use solver software (Excel Solver, Python's PuLP, or similar). However, to give an **approximate solution**, let's make some logical assumptions: ### Simplified approach: - Use the **cheapest rock (A)** as much as possible, respecting constraints. - Fill the orders with low-cost rocks first, then adjust to meet phosphate constraints. --- ## **Step 7: Step-by-step approximate solution** ### **Order 1 (500 lbs, 35-44%)** - Min phosphate: 175 lbs - Max phosphate: 220 lbs **Strategy:** - Use mostly low-cost rocks with moderate phosphate content—preferably rock A (20%) and D (3%) are too low, so focus on A, B, and C. - To meet minimum phosphate (175 lbs), use some C (45%) and E (60%). **Example:** - Maximize use of A (20%) for cost savings: Suppose: - \( x_{A1} = 250 \) lbs (cost-effective) Phosphate from A: \[ 0.20 \times 250 = 50 \text{ lbs phosphate} \] Remaining: \[ 175 - 50 = 125 \text{ lbs} \] Use C (45%): \[ x_{C1} = \frac{125}{0.45} \approx 278 \text{ lbs} \] Check total: \[ x_{A1} + x_{C1} = 250 + 278 = 528 \text{ lbs} > 500 \] So, reduce \( x_{C1} \) to fit total: Total volume constraint: \[ x_{A1} + x_{C1} \le 500 \] Try: - \( x_{A1} = 250 \) - \( x_{C1} = 250 - 250 = 250 \) Phosphate: \[ 0.20 \times 250 + 0.45 \times 250 = 50 + 112.5 = 162.5 \text{ lbs} \approx 163 \text{ lbs} \] Below the minimum (175). So increase \( x_{A1} \): - \( x_{A1} = 300 \): Phosphate: \[ 0.20 \times 300 + 0.45 \times (500 - 300) = 60 + 0.45 \times 200 = 60 + 90 = 150 \] Still below 175. Increase \( x_{A1} \): - \( x_{A1} = 350 \): Phosphate: \[ 70 + 0.45 \times 150 = 70 + 67.5 = 137.5 \] No, too low. Alternatively, use E (60%) for higher phosphate: Suppose: - \( x_{A1} = 200 \) Phosphate: \[ 0.20 \times 200 = 40 \] Remaining: \[ 175 - 40 = 135 \text{ lbs} \] Use E: \[ x_{E1} = \frac{135}{0.60} = 225 \text{ lbs} \] Total: \[ 200 + 225 = 425 \text{ lbs} < 500 \] Remaining: \[ 75 \text{ lbs} \] Fill with rock B (5%): \[ x_{B1} = 75 \] Total: \[ 200 + 225 + 75 = 500 \text{ lbs} \quad \text{(fits order size)} Phosphate: \[ 0.20 \times 200 + 0.60 \times 225 + 0.05 \times 75 = 40 + 135 + 3.75 = 178.75 \text{ lbs} \] Phosphate percentage: \[ 178.75 / 500 = 0.3575 = 35.75\% \] Within bounds (35%-44%). Excellent. --- ## **Similarly, approximate allocations for other orders:** - **Order 2 (600 lbs, 38-46%)** Use similar approach: - Maximize low-cost rock A and B, add some E for phosphate: Suppose: - \( x_{A2} = 250 \) - \( x_{E2} = 200 \) (for high phosphate) - Remaining: \[ 600 - 250 - 200 = 150 \] Fill with B: \[ x_{B2} = 150 \] Phosphate: \[ 0.20 \times 250 + 0.60 \times 200 + 0.05 \times 150 = 50 + 120 + 7.5 = 177.5 \] Phosphate percentage: \[ 177.5/600 \approx 29.58\% \] Too low; need more phosphate. Add C: - \( x_{C2} = 150 \) Total: \[ 250 + 200 + 150 = 550 \text{ lbs} < 600 \] Remaining: \[ 50 \text{ lbs} \] Add to E: \[ x_{E2} = 200 + 50 = 250 \] Now total: \[