--- title: "Big-M method max:P=2x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-p-2x-two" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-p-2x-two site: BCT Engineering --- # Big-M method max:P=2x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2076 Asoj, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-p-2x-two - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following linear programming problem using the Big-M method: Maximize P = 2x₁ + 5x₂ subject to x₁ + 2x₂ ≤ 18, 2x₁ + x₂ ≥ 21, x₁, x₂ ≥ 0. ## Short answer Add slack, surplus, and artificial variables, then apply the Big-M simplex method. The optimal solution is Maximum P = 41 at x₁ = 8, x₂ = 5. ## Answer **Step 1: Make the problem ready for the simplex method.** - Constraint 1 is ≤: add slack variable s₁. - Constraint 2 is ≥: subtract surplus variable s₂ and add artificial variable A₁. Artificial variables carry a cost of −M in the objective, where M is a very large positive number, so that the simplex method drives them out of the basis. ```text Maximize P = 2x₁ + 5x₂ − M·A₁ subject to x₁ + 2x₂ + s₁ = 18 2x₁ + x₂ − s₂ + A₁ = 21 ``` **Step 2: Initial solution.** Basic variables: the slack variables (for ≤ rows) and the artificial variables (for ≥ and = rows). **Step 3: Simplex tables.** An artificial variable's column is dropped as soon as it leaves the basis. Cj − Zj entries contain M, and the large-M part decides which variable enters. **Table 1** ```text Cj 2 5 0 0 -M Basis CB x₁ x₂ s₁ s₂ A₁ b Ratio ----------------------------------------- s₁ 0 1 2 1 0 0 18 18 A₁ -M 2 1 0 -1 1 21 21/2 ----------------------------------------- Zj -2M -M 0 M -M -21M Cj-Zj 2+2M 5+M 0 -M 0 Entering: x₁ Leaving: A₁ Pivot = 2 ``` **Table 2** ```text Cj 2 5 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio -------------------------------------- s₁ 0 0 3/2 1 1/2 15/2 5 x₁ 2 1 1/2 0 -1/2 21/2 21 -------------------------------------- Zj 2 1 0 -1 21 Cj-Zj 0 4 0 1 Entering: x₂ Leaving: s₁ Pivot = 3/2 ``` **Table 3** ```text Cj 2 5 0 0 Basis CB x₁ x₂ s₁ s₂ b ------------------------------- x₂ 5 0 1 2/3 1/3 5 x₁ 2 1 0 -1/3 -2/3 8 ------------------------------- Zj 2 5 8/3 1/3 41 Cj-Zj 0 0 -8/3 -1/3 ``` **Step 4: Optimality.** All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible. **Answer:** x₁ = 8, x₂ = 5, and Maximum P = 41. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is 2x1 + x2 >= 21 treated in the Big-M method? A. Add a slack variable B. Add an artificial variable only C. Subtract a slack variable D. Subtract a surplus variable and add an artificial variable Answer: D. A >= row needs both a surplus (subtracted) and an artificial variable (added) to start. (Key point: Setup) 2. In Table 1, Cj - Zj is 2+2M for x1 and 5+M for x2. Which enters? A. x2, because 5 is larger than 2 B. x1, because 2M is larger than M C. A1 D. s1 Answer: B. With M very large, the M-part decides: 2+2M > 5+M, so x1 enters. (Key point: Entering variable) 3. With x1 entering, ratios are 18/1 = 18 and 21/2 = 21/2. Which leaves? A. A1 B. s1 C. x2 D. s2 Answer: A. The smallest ratio is 21/2, in the A1 row, so A1 leaves. (Key point: Leaving variable) 4. What is the optimal solution? A. x1 = 21/2, x2 = 0, P = 21 B. x1 = 0, x2 = 9, P = 45 C. x1 = 8, x2 = 5, P = 41 D. x1 = 5, x2 = 8, P = 50 Answer: C. Check: 8 + 10 = 18 and 16 + 5 = 21, giving P = 16 + 25 = 41. 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