--- title: "Big-M method max:P=2x₁+x₂+3x₃" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-p-2x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-p-2x site: BCT Engineering --- # Big-M method max:P=2x₁+x₂+3x₃ Mathematics III · Chapter 5: Linear Programming - Asked in: 2080 Baishakh, 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-p-2x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the Linear Programming Problem by the Big-M method: Maximize P = 2x₁ + x₂ + 3x₃ subject to x₁ + x₂ + 2x₃ ≤ 5, 2x₁ + 3x₂ + 4x₃ = 12 and x₁, x₂, x₃ ≥ 0. ## Short answer The Big-M method, applied with an artificial variable for the equality constraint, gives optimal values x1 = 3, x2 = 2, x3 = 0 and maximum P = 8. ## Answer **Step 1: Make the problem ready for the simplex method.** - Constraint 1 is ≤: add slack variable s₁. - Constraint 2 is an equation: 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 + x₂ + 3x₃ − M·A₁ subject to x₁ + x₂ + 2x₃ + s₁ = 5 2x₁ + 3x₂ + 4x₃ + A₁ = 12 ``` **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 3 0 -M Basis CB x₁ x₂ x₃ s₁ A₁ b Ratio -------------------------------------------- s₁ 0 1 1 2 1 0 5 5/2 A₁ -M 2 3 4 0 1 12 3 -------------------------------------------- Zj -2M -3M -4M 0 -M -12M Cj-Zj 2+2M 1+3M 3+4M 0 0 Entering: x₃ Leaving: s₁ Pivot = 2 ``` **Table 2** ```text Cj 2 1 3 0 -M Basis CB x₁ x₂ x₃ s₁ A₁ b Ratio --------------------------------------------------- x₃ 3 1/2 1/2 1 1/2 0 5/2 5 A₁ -M 0 1 0 -2 1 2 2 --------------------------------------------------- Zj 3/2 3/2-M 3 3/2+2M -M 15/2-2M Cj-Zj 1/2 -1/2+M 0 -3/2-2M 0 Entering: x₂ Leaving: A₁ Pivot = 1 ``` **Table 3** ```text Cj 2 1 3 0 Basis CB x₁ x₂ x₃ s₁ b Ratio -------------------------------------- x₃ 3 1/2 0 1 3/2 3/2 3 x₂ 1 0 1 0 -2 2 - -------------------------------------- Zj 3/2 1 3 5/2 13/2 Cj-Zj 1/2 0 0 -5/2 Entering: x₁ Leaving: x₃ Pivot = 1/2 ``` **Table 4** ```text Cj 2 1 3 0 Basis CB x₁ x₂ x₃ s₁ b -------------------------- x₁ 2 1 0 2 3 3 x₂ 1 0 1 0 -2 2 -------------------------- Zj 2 1 4 4 8 Cj-Zj 0 0 -1 -4 ``` **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 = 2, x₃ = 0, and Maximum P = 8. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the equality 2x1 + 3x2 + 4x3 = 12 handled in the Big-M method? A. Add a slack variable B. Subtract a surplus variable C. Add only an artificial variable D. Convert it into two <= constraints and add two slacks Answer: C. An equation needs no slack or surplus; an artificial variable A1 is added to provide the starting basic variable. (Key point: Equality constraint) 2. In Table 1, x3 enters. The ratios are 5/2 for s1 and 12/4 = 3 for A1. Which leaves? A. A1 B. x1 C. s1 D. x2 Answer: C. The minimum ratio is 5/2, in the s1 row, so s1 leaves with pivot 2. (Key point: Leaving variable) 3. What happens to the A1 column once A1 leaves the basis? A. It is dropped from later tables B. It stays and keeps cost -M C. It re-enters in the next table D. It is replaced by a surplus column Answer: A. An artificial variable has no meaning in the original problem, so its column is dropped once it leaves. (Key point: Dropping the artificial column) 4. What is the optimal solution? A. x1 = 3, x2 = 2, x3 = 0, P = 8 B. x1 = 0, x2 = 4, x3 = 0, P = 4 C. x1 = 2, x2 = 0, x3 = 2, P = 10 D. x1 = 5, x2 = 0, x3 = 0, P = 10 Answer: A. Check: 3 + 2 = 5 and 6 + 6 = 12, giving P = 6 + 2 = 8; Cj - Zj are 0, 0, -1, -4 so it is optimal. 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