--- title: "Big-M method max:Z=2x₁+x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-z-2x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-z-2x site: BCT Engineering --- # Big-M method max:Z=2x₁+x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2082 Baishakh, 8 marks; 2081 Baishakh, 7 marks; 2078 Kartik, 8 marks; 2075 Chaitra, 10 marks; 2070 Asar, 8 marks (asked 5 times) - Video: [Artificial Variable in Big-M Method || Concept of non-zero Artificial Variable in LPP](https://www.youtube.com/watch?v=n3T-k19tCEo) — Swati Theng Mathematics - 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-2x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following linear programming problem by the simplex method using the Big-M method: Maximize Z = 2x₁ + x₂ subject to x₁ + x₂ ≤ 10, −x₁ + x₂ ≥ 2, x₁, x₂ ≥ 0. ## Short answer The problem is solved by Big-M method: introduce artificial variable A1 for the ≥ constraint with cost −M. After simplex iterations, optimal solution is x1 = 4, x2 = 6, max Z = 14. ## 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 = 2x₁ + x₂ − M·A₁ subject to x₁ + x₂ + s₁ = 10 −x₁ + x₂ − s₂ + A₁ = 2 ``` **Step 2: Initial solution.** Put x₁ = x₂ = s₂ = 0. The basic variables are the slack s₁ = 10 (from the ≤ row) and the artificial A₁ = 2 (from the ≥ row). **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 0 -M Basis CB x₁ x₂ s₁ s₂ A₁ b Ratio --------------------------------------- s₁ 0 1 1 1 0 0 10 10 A₁ -M -1 1 0 -1 1 2 2 --------------------------------------- Zj M -M 0 M -M -2M Cj-Zj 2-M 1+M 0 -M 0 Entering: x₂ Leaving: A₁ Pivot = 1 ``` **Table 2** ```text Cj 2 1 0 0 Basis CB x₁ x₂ s₁ s₂ b Ratio -------------------------------- s₁ 0 2 0 1 1 8 4 x₂ 1 -1 1 0 -1 2 - -------------------------------- Zj -1 1 0 -1 2 Cj-Zj 3 0 0 1 Entering: x₁ Leaving: s₁ Pivot = 2 ``` **Table 3** ```text Cj 2 1 0 0 Basis CB x₁ x₂ s₁ s₂ b ------------------------------- x₁ 2 1 0 1/2 1/2 4 x₂ 1 0 1 1/2 -1/2 6 ------------------------------- Zj 2 1 3/2 1/2 14 Cj-Zj 0 0 -3/2 -1/2 ``` **Step 4: Optimality.** All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible. **Answer:** x₁ = 4, x₂ = 6, and Maximum Z = 14. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the constraint -x1 + x2 >= 2, which variables are added? A. Subtract an artificial variable B. Add a surplus only C. Add a slack only D. Subtract a surplus and add an artificial variable Answer: D. A >= row needs a surplus, and the artificial supplies the starting basis. (Key point: Surplus and artificial) 2. What cost does an artificial variable carry in a maximisation? A. 1 B. +M C. -M D. 0 Answer: C. -M with M large drives it out of the basis. (Key point: Artificial cost) 3. In Table 1, which variable enters and which leaves? A. x1 enters, A1 leaves B. x2 enters, s1 leaves C. x1 enters, s1 leaves D. x2 enters, A1 leaves Answer: D. Cj - Zj is 1+M for x2, the largest, and ratio 2 is smallest from A1. (Key point: First pivot) 4. What is the optimal solution? A. x1 = 4, x2 = 6, Z = 14 B. x1 = 4, x2 = 6, Z = 16 C. x1 = 6, x2 = 4, Z = 16 D. x1 = 0, x2 = 10, Z = 10 Answer: A. 2(4) + 6 = 14; both constraints are satisfied. (Key point: Result) ## Close variants set in other exams - [Simplex method max:P=15x+10y](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-p-15x-10y) — Mathematics III, chapter 5, asked 17× - [Big-M method max:Z=4x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-z-4x) — Mathematics III, chapter 5, asked 15× - [Big-M method max:F=2x₁+x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-f-2x) — Mathematics III, chapter 5, asked 17× - [Big-M method max:P=2x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x-two) — Mathematics III, chapter 5, asked 15× - [Simplex method max:Z=3x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-3x-5x-two) — Mathematics III, chapter 5, asked 17× - [Simplex method max:Z=4x+3y](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-4x-3y) — Mathematics III, chapter 5, asked 17× - [Simplex method max:P=20x₂−5x₁](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-p-20x-5x) — Mathematics III, chapter 5, asked 17× - [Big-M method max:P=2x₁+x₂+3x₃](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x) — Mathematics III, chapter 5, asked 17× - [Simplex method max:Z=10x₁+x₂+2x₃](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-10x-x) — Mathematics III, chapter 5, asked 18× - [Simplex method max:Z=15x₁+10x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-z-15x-10x) — Mathematics III, chapter 5, asked 15× - [Two-phase simplex method max:Z=3x₁−x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Atwo-phase-simplex-method-max-z) — Mathematics III, chapter 5, asked 15× ## Related in Mathematics III - [Big-M method max:P=2x₁+x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x-three) — Mathematics III, chapter 5, asked 3× - [Big-M method max:P=2x₁+5x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Abig-m-method-max-p-2x-four) — Mathematics III, chapter 5, asked 3× - [Simplex method max:P=−x₁+2x₂](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asimplex-method-max-p-x-2x) — Mathematics III, chapter 5, asked 8× ## Related in other subjects - [Short note: Linear programming](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Alinear-programming-note) — Project Engineering, chapter 3, asked 3× - [Basic requirements of linear programming](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Alinear-programming-requirements) — Project Engineering, chapter 3 - [Solve y''+3y'+2y=e^(2x) sin x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-d2-3d-2-e2x-sinx) — Mathematics I, chapter 4, asked 2× - [Solve (D²−2D+5)y=e^(2x) sin x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-d2-2d-5-e2x-sinx) — Mathematics I, chapter 4, asked 2×