--- title: "Big-M method max:Z=4x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-z-4x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-z-4x site: BCT Engineering --- # Big-M method max:Z=4x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2081 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-z-4x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following linear programming problem by using the Big-M method: Maximize Z = 4x₁ + 5x₂ subject to x₁ + x₂ ≤ 5, 3x₁ + 10x₂ ≥ 30, x₁, x₂ ≥ 0. ## Short answer Using the Big-M method, the optimal solution is obtained after introducing artificial variable A1 for the ≥ constraint. The maximum Z is 25, achieved at x1 = 0, 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 Z = 4x₁ + 5x₂ − M·A₁ subject to x₁ + x₂ + s₁ = 5 3x₁ + 10x₂ − s₂ + A₁ = 30 ``` **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 4 5 0 0 -M Basis CB x₁ x₂ s₁ s₂ A₁ b Ratio ------------------------------------------- s₁ 0 1 1 1 0 0 5 5 A₁ -M 3 10 0 -1 1 30 3 ------------------------------------------- Zj -3M -10M 0 M -M -30M Cj-Zj 4+3M 5+10M 0 -M 0 Entering: x₂ Leaving: A₁ Pivot = 10 ``` **Table 2** ```text Cj 4 5 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio -------------------------------------- s₁ 0 7/10 0 1 1/10 2 20/7 x₂ 5 3/10 1 0 -1/10 3 10 -------------------------------------- Zj 3/2 5 0 -1/2 15 Cj-Zj 5/2 0 0 1/2 Entering: x₁ Leaving: s₁ Pivot = 7/10 ``` **Table 3** ```text Cj 4 5 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio ----------------------------------------- x₁ 4 1 0 10/7 1/7 20/7 20 x₂ 5 0 1 -3/7 -1/7 15/7 - ----------------------------------------- Zj 4 5 25/7 -1/7 155/7 Cj-Zj 0 0 -25/7 1/7 Entering: s₂ Leaving: x₁ Pivot = 1/7 ``` **Table 4** ```text Cj 4 5 0 0 Basis CB x₁ x₂ s₁ s₂ b --------------------------- s₂ 0 7 0 10 1 20 x₂ 5 1 1 1 0 5 --------------------------- Zj 5 5 5 0 25 Cj-Zj -1 0 -5 0 ``` **Step 4: Optimality.** All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible. **Answer:** x₁ = 0, x₂ = 5, and Maximum Z = 25. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the Big-M method, how is the constraint 3x1 + 10x2 >= 30 converted to an equation? A. Add a slack variable only B. Subtract a surplus variable and add an artificial variable C. Add an artificial variable only D. Subtract a slack variable and leave the rest unchanged Answer: B. A >= row needs a surplus variable subtracted, and since that gives a negative basic value, an artificial variable is added to get a starting basis. (Key point: Handling a >= constraint) 2. Why does the artificial variable carry a cost of -M in the maximisation objective? A. To make the objective value larger B. To keep the constraint as an inequality C. To penalise it so heavily that the simplex method drives it out of the basis D. To make the slack variable enter first Answer: C. With M very large and positive, -M makes any solution containing A1 very poor, so the simplex iterations remove it. (Key point: Role of -M) 3. In Table 1, Cj - Zj is 4+3M for x1 and 5+10M for x2. Which variable enters? A. x2, because 10M is the largest M-part B. x1, because 4 is smaller C. s2, because it is a surplus variable D. A1, because it has cost -M Answer: A. The M-part dominates, and 5+10M is the largest positive value, so x2 enters (pivot 10, A1 leaves). (Key point: Entering variable) 4. What is the optimal solution? A. x1 = 20/7, x2 = 15/7, Z = 155/7 B. x1 = 5, x2 = 0, Z = 20 C. x1 = 0, x2 = 3, Z = 15 D. x1 = 0, x2 = 5, Z = 25 Answer: D. In the last table all Cj - Zj <= 0 and no artificial variable is in the basis, giving x2 = 5 and Z = 5(5) = 25. 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