--- title: "Big-M method max:F=2x₁+x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-f-2x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-f-2x site: BCT Engineering --- # Big-M method max:F=2x₁+x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2080 Bhadra, 8 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-f-2x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the linear programming problem by simplex method using the Big-M method: Maximize F = 2x₁ + x₂ subject to 2x₁ − x₂ ≥ 2, x₁ − x₂ ≤ 2, x₁ + x₂ ≤ 4 and x₁, x₂ ≥ 0. ## Short answer Introducing an artificial variable for the ≥ constraint, the Big-M method yields an optimal feasible solution with x1 = 3, x2 = 1 and maximum F = 7. ## Answer **Step 1: Make the problem ready for the simplex method.** - Constraint 1 is ≥: subtract surplus variable s₁ and add artificial variable A₁. - Constraint 2 is ≤: add slack variable s₂. - Constraint 3 is ≤: add slack variable s₃. 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 F = 2x₁ + x₂ − M·A₁ subject to 2x₁ − x₂ − s₁ + A₁ = 2 x₁ − x₂ + s₂ = 2 x₁ + x₂ + s₃ = 4 ``` **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 1 0 -M 0 0 Basis CB x₁ x₂ s₁ A₁ s₂ s₃ b Ratio ------------------------------------------- A₁ -M 2 -1 -1 1 0 0 2 1 s₂ 0 1 -1 0 0 1 0 2 2 s₃ 0 1 1 0 0 0 1 4 4 ------------------------------------------- Zj -2M M M -M 0 0 -2M Cj-Zj 2+2M 1-M -M 0 0 0 Entering: x₁ Leaving: A₁ Pivot = 2 ``` **Table 2** ```text Cj 2 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio --------------------------------------- x₁ 2 1 -1/2 -1/2 0 0 1 - s₂ 0 0 -1/2 1/2 1 0 1 - s₃ 0 0 3/2 1/2 0 1 3 2 --------------------------------------- Zj 2 -1 -1 0 0 2 Cj-Zj 0 2 1 0 0 Entering: x₂ Leaving: s₃ Pivot = 3/2 ``` **Table 3** ```text Cj 2 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio --------------------------------------- x₁ 2 1 0 -1/3 0 1/3 2 - s₂ 0 0 0 2/3 1 1/3 2 3 x₂ 1 0 1 1/3 0 2/3 2 6 --------------------------------------- Zj 2 1 -1/3 0 4/3 6 Cj-Zj 0 0 1/3 0 -4/3 Entering: s₁ Leaving: s₂ Pivot = 2/3 ``` **Table 4** ```text Cj 2 1 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b --------------------------------- x₁ 2 1 0 0 1/2 1/2 3 s₁ 0 0 0 1 3/2 1/2 3 x₂ 1 0 1 0 -1/2 1/2 1 --------------------------------- Zj 2 1 0 1/2 3/2 7 Cj-Zj 0 0 0 -1/2 -3/2 ``` **Step 4: Optimality.** All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible. **Answer:** x₁ = 3, x₂ = 1, and Maximum F = 7. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this problem, which constraint needs an artificial variable? A. x1 - x2 <= 2 B. x1 + x2 <= 4 C. 2x1 - x2 >= 2 only D. All three constraints Answer: C. Only the >= constraint needs a surplus and an artificial variable; the <= rows get slack variables that serve as basic variables. (Key point: Which rows need artificial variables) 2. What is the initial basis in Table 1? A. s1, s2, s3 B. A1, s2, s3 C. x1, x2, A1 D. A1, s1, s3 Answer: B. A1 is basic in row 1, while slack variables s2 and s3 are basic in rows 2 and 3. (Key point: Starting basis) 3. With x1 entering in Table 1, ratios are 2/2 = 1, 2/1 = 2, 4/1 = 4. Which variable leaves? A. A1 B. s2 C. s3 D. x2 Answer: A. The smallest ratio is 1, in the A1 row, so A1 leaves with pivot 2. (Key point: Leaving variable) 4. What is the optimal solution? A. x1 = 2, x2 = 2, F = 6 B. x1 = 4, x2 = 0, F = 8 C. x1 = 1, x2 = 3, F = 5 D. x1 = 3, x2 = 1, F = 7 Answer: D. In the last table all Cj - Zj <= 0 and x1 = 3, x2 = 1 give F = 6 + 1 = 7. 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