Let's break down the problem step by step: ### **Given Data** | Pipe Type | Length (feet) | Required Quantity (units) | |-----------|--------------|---------------------------| | 1 | 3 | 10 | | 2 | 5 | 20 | | 3 | 8 | 15 | - **Standard pipe length:** 15 feet --- ### **(a) Generate All Feasible Cutting Patterns** A cutting pattern is a way to cut a 15-feet pipe into pieces of 3, 5, and 8 feet such that the sum does not exceed 15. Let: - \( x_1 \) = number of 3-foot pipes cut in a pattern - \( x_2 \) = number of 5-foot pipes cut in a pattern - \( x_3 \) = number of 8-foot pipes cut in a pattern Subject to: \( 3x_1 + 5x_2 + 8x_3 \leq 15 \) \( x_1, x_2, x_3 \) = integers \( \geq 0 \) Let’s enumerate all non-negative integer solutions: #### **Case 1: \( x_3 = 0 \) (no 8-ft pipes)** \( 3x_1 + 5x_2 \leq 15 \) - \( x_2 = 0 \): \( 3x_1 \leq 15 \) → \( x_1 = 0, 1, 2, 3, 4, 5 \) - \( x_2 = 1 \): \( 3x_1 + 5 \leq 15 \to 3x_1 \leq 10 \to x_1 = 0,1,2,3 \) - \( x_2 = 2 \): \( 3x_1 + 10 \leq 15 \to 3x_1 \leq 5 \to x_1 = 0,1 \) - \( x_2 = 3 \): \( 3x_1 + 15 \leq 15 \to 3x_1 \leq 0 \to x_1 = 0 \) - \( x_2 \geq 4 \): Not possible (exceeds 15) So, patterns with \( x_3 = 0 \): | Pattern | (x1, x2, x3) | |---------|--------------| | 1 | (5,0,0) | | 2 | (4,0,0) | | 3 | (3,0,0) | | 4 | (2,0,0) | | 5 | (1,0,0) | | 6 | (0,0,0) | (trivial, not useful) | 7 | (3,1,0) | | 8 | (2,1,0) | | 9 | (1,1,0) | | 10 | (0,1,0) | | 11 | (1,2,0) | | 12 | (0,2,0) | | 13 | (0,3,0) | But only retain those with a total length > 0 and ≤ 15. #### **Case 2: \( x_3 = 1 \) (one 8-ft pipe)** \( 3x_1 + 5x_2 + 8 \leq 15 \to 3x_1 + 5x_2 \leq 7 \) - \( x_2 = 0 \): \( 3x_1 \leq 7 \to x_1 = 0,1,2 \) - \( x_2 = 1 \): \( 3x_1 + 5 \leq 7 \to 3x_1 \leq 2 \to x_1 = 0 \) - \( x_2 \geq 2 \): Not possible Patterns with \( x_3 = 1 \): | Pattern | (x1, x2, x3) | |---------|--------------| | 14 | (2,0,1) | | 15 | (1,0,1) | | 16 | (0,0,1) | | 17 | (0,1,1) | #### **Case 3: \( x_3 = 2 \) (two 8-ft pipes)** \( 3x_1 + 5x_2 + 16 \leq 15 \) → Not possible. --- #### **Case 4: Mixed patterns** Check for other combinations: - \( x_3 = 1, x_2 = 1 \): \( 3x_1 + 5 + 8 \leq 15 \to 3x_1 \leq 2 \to x_1 = 0 \) - \( x_3 = 1, x_2 = 0 \): Already listed. --- ### **Summary of Feasible Patterns** Let's list only unique, non-trivial, useful patterns (that make use of pipe length efficiently): | Pattern | (x1, x2, x3) | Pipe composition (3', 5', 8') | Total length used | |---------|--------------|-------------------------------|------------------| | P1 | (5,0,0) | 5 × 3' | 15 | | P2 | (0,3,0) | 3 × 5' | 15 | | P3 | (1,0,1) | 1 × 3', 1 × 8' | 3+8=11 | | P4 | (2,1,0) | 2 × 3', 1 × 5' | 6+5=11 | | P5 | (0,1,1) | 1 × 5', 1 × 8' | 5+8=13 | | P6 | (0,0,1) | 1 × 8' | 8 | | P7 | (2,0,1) | 2 × 3', 1 × 8' | 6+8=14 | | P8 | (3,0,0) | 3 × 3' | 9 | | P9 | (0,2,0) | 2 × 5' | 10 | | P10 | (1,2,0) | 1 × 3', 2 × 5' | 3+10=13 | But for efficiency, generally, we use patterns that come close to or exactly 15 ft. Main patterns: - P1: (5,0,0) → 15' - P2: (0,3,0) → 15' - P3: (1,1,1) → 3+5+8=16 (exceeds 15, not allowed) - P4: (2,0,1) → 6+8=14' - P5: (0,1,1) → 5+8=13' So, **efficient patterns**: - (5,0,0), (0,3,0), (2,0,1), (0,1,1) --- ### **(b) Integer Linear Programming (ILP) Formulation** #### **Decision Variables** Let: - \( y_j \) = number of times pattern \( j \) is used, where \( j \) indexes feasible patterns. Let’s number the main patterns: 1. \( (5,0,0) \): Pattern 1 2. \( (0,3,0) \): Pattern 2 3. \( (2,0,1) \): Pattern 3 4. \( (0,1,1) \): Pattern 4 So, \( y_1, y_2, y_3, y_4 \) are the number of times we use each pattern. #### **Pattern Matrix** | Pattern | \( x_1 \) (3') | \( x_2 \) (5') | \( x_3 \) (8') | |---------|--------------|--------------|--------------| | 1 | 5 | 0 | 0 | | 2 | 0 | 3 | 0 | | 3 | 2 | 0 | 1 | | 4 | 0 | 1 | 1 | #### **Objective Function** Minimize the total number of standard pipes used: \[ \text{Minimize } Z = y_1 + y_2 + y_3 + y_4 \] #### **Constraints** Each small pipe demand must be met (possibly exceeded): - For 3' pipes: \[ 5y_1 + 2y_3 \geq 10 \] - For 5' pipes: \[ 3y_2 + y_4 \geq 20 \] - For 8' pipes: \[ y_3 + y_4 \geq 15 \] - Non-negativity and integer constraints: \[ y_1, y_2, y_3, y_4 \geq 0, \quad \text{integer} \] --- ### **Explanation** - **Decision Variables:** \( y_j \) = number of times pattern \( j \) is used. - **Objective:** Minimize the total number of standard pipes (15-foot) used. - **Constraints:** Ensure at least the required number of each type of small pipe is produced. --- ### **Final ILP Model (in OPEN Mathematical Form)** \[ \begin{align*} \text{Minimize} \quad & y_1 + y_2 + y_3 + y_4 \\ \text{subject to:} \\ & 5y_1 + 2y_3 \geq 10 \qquad \text{(3' pipes demand)} \\ & 3y_2 + y_4 \geq 20 \qquad \text{(5' pipes demand)} \\ & y_3 + y_4 \geq 15 \qquad \text{(8' pipes demand)} \\ & y_1, y_2, y_3, y_4 \geq 0 \text{ and integer} \end{align*} \] --- **All variables and constraints are defined and explained above.** If you want to solve this ILP for the exact minimum, you can use an LP solver or do it by hand with some trial and error. --- **Let me know if you'd like to see the optimal solution!**