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| 1 | + |
| 2 | +# 📊 Linear Programming Model — Production Optimization |
| 3 | + |
| 4 | +## ✅ Problem Statement |
| 5 | + |
| 6 | +A company, after going through a production streamlining process, ended up with the availability of 3 productive resources: **R1**, **R2**, and **R3**. |
| 7 | + |
| 8 | +A study on resource usage showed the possibility of producing two products: **P1** and **P2**. After evaluating costs and consulting the sales department, it was found that: |
| 9 | + |
| 10 | +- **P1 yields a profit of 120 monetary units per unit** |
| 11 | +- **P2 yields a profit of 150 monetary units per unit** |
| 12 | + |
| 13 | +The production department provided the following **resource usage** table: |
| 14 | + |
| 15 | +| Product | R1/unit | R2/unit | R3/unit | |
| 16 | +|---------|---------|---------|---------| |
| 17 | +| **P1** | 2 | 3 | 5 | |
| 18 | +| **P2** | 4 | 0 | 3 | |
| 19 | + |
| 20 | +And the **monthly resource availability**: |
| 21 | + |
| 22 | +| Resource | Monthly Availability | |
| 23 | +|----------|----------------------| |
| 24 | +| **R1** | 100 | |
| 25 | +| **R2** | 90 | |
| 26 | +| **R3** | 120 | |
| 27 | + |
| 28 | +--- |
| 29 | + |
| 30 | +## 🎯 Objective |
| 31 | + |
| 32 | +Mathematically model the **Linear Programming (LP)** problem to **maximize profit** under resource constraints. |
| 33 | + |
| 34 | +--- |
| 35 | + |
| 36 | +## 🧠 Step-by-Step Modeling |
| 37 | + |
| 38 | +### 1. 🧮 Decision Variables |
| 39 | + |
| 40 | +Let: |
| 41 | + |
| 42 | +x₁ = quantity produced of product P1 |
| 43 | +x₂ = quantity produced of product P2 |
| 44 | + |
| 45 | +Or in LaTeX (for use in documents): |
| 46 | + |
| 47 | +```latex |
| 48 | +x_1 = \text{quantity produced of product P1} \\ |
| 49 | +x_2 = \text{quantity produced of product P2} |
| 50 | +
|
| 51 | +
|
| 52 | +
|
| 53 | +⸻ |
| 54 | +
|
| 55 | +2. 📈 Objective Function |
| 56 | +
|
| 57 | +Maximize total profit: |
| 58 | +
|
| 59 | +\text{Maximize } Z = 120x_1 + 150x_2 |
| 60 | +
|
| 61 | +
|
| 62 | +
|
| 63 | +⸻ |
| 64 | +
|
| 65 | +3. 📏 Resource Constraints |
| 66 | +
|
| 67 | +Each resource has limited availability: |
| 68 | + • R1 constraint: |
| 69 | +
|
| 70 | +2x_1 + 4x_2 \leq 100 |
| 71 | +
|
| 72 | + • R2 constraint: |
| 73 | +
|
| 74 | +3x_1 \leq 90 |
| 75 | +
|
| 76 | + • R3 constraint: |
| 77 | +
|
| 78 | +5x_1 + 3x_2 \leq 120 |
| 79 | +
|
| 80 | +
|
| 81 | +
|
| 82 | +⸻ |
| 83 | +
|
| 84 | +4. 🚫 Non-Negativity Constraints |
| 85 | +
|
| 86 | +We cannot produce a negative quantity of products: |
| 87 | +
|
| 88 | +x_1 \geq 0, \quad x_2 \geq 0 |
| 89 | +
|
| 90 | +
|
| 91 | +
|
| 92 | +⸻ |
| 93 | +
|
| 94 | +🧾 Complete Mathematical Model |
| 95 | +
|
| 96 | +\boxed{ |
| 97 | +\begin{cases} |
| 98 | +\text{Maximize } Z = 120x_1 + 150x_2 \\ |
| 99 | +2x_1 + 4x_2 \leq 100 \\ |
| 100 | +3x_1 \leq 90 \\ |
| 101 | +5x_1 + 3x_2 \leq 120 \\ |
| 102 | +x_1 \geq 0, \quad x_2 \geq 0 |
| 103 | +\end{cases} |
| 104 | +} |
| 105 | +
|
| 106 | +
|
| 107 | +
|
| 108 | +⸻ |
| 109 | +
|
| 110 | +## 📌 Summary Tables |
| 111 | +
|
| 112 | +### 🔢 Profit per Product |
| 113 | +
|
| 114 | +| Product | Profit per Unit (u.m.) | |
| 115 | +|:--------|:----------------------:| |
| 116 | +| **P1** | 120 | |
| 117 | +| **P2** | 150 | |
| 118 | +
|
| 119 | +--- |
| 120 | +
|
| 121 | +### 🧰 Resource Usage per Unit |
| 122 | +
|
| 123 | +| Product | R1/unit | R2/unit | R3/unit | |
| 124 | +|:--------|:-------:|:-------:|:-------:| |
| 125 | +| **P1** | 2 | 3 | 5 | |
| 126 | +| **P2** | 4 | 0 | 3 | |
| 127 | +
|
| 128 | +--- |
| 129 | +
|
| 130 | +### 📦 Monthly Resource Availability |
| 131 | +
|
| 132 | +| Resource | Available Units | |
| 133 | +|:---------|:----------------:| |
| 134 | +| **R1** | 100 | |
| 135 | +| **R2** | 90 | |
| 136 | +| **R3** | 120 | |
| 137 | +
|
| 138 | +
|
| 139 | +⸻ |
| 140 | +
|
| 141 | +🧠 Notes |
| 142 | + • This LP model can be solved using methods such as the Simplex Algorithm. |
| 143 | + • Can also be implemented in software such as Python (PuLP), MATLAB, or Excel Solver. |
| 144 | +
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