--- title: "Big-M method max:P=2x₁+x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-p-2x-three" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-p-2x-three site: BCT Engineering --- # Big-M method max:P=2x₁+x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2068 Chaitra, 7.5 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-three - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following LPP by using the big M-method: Maximize P = 2x₁ + x₂ subject to x₁ + x₂ ≤ 10, −x₁ + x₂ ≥ 2, x₁, x₂ ≥ 0. ## Short answer Introduce slack s₁, surplus s₂, and artificial A₁ with penalty −M. Simplex iterations drive out A₁, yielding optimal solution x₁ = 4, x₂ = 6, and maximum P = 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 P = 2x₁ + x₂ − M·A₁ subject to x₁ + x₂ + s₁ = 10 −x₁ + x₂ − s₂ + A₁ = 2 ``` **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 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 P = 14. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the constraint -x1 + x2 >= 2 handled in the Big-M method? A. Add a slack variable only B. Subtract a surplus variable s2 and add an artificial variable A1 C. Multiply by -1 and add a slack variable D. Add a surplus variable and no artificial variable Answer: B. A >= constraint needs a surplus variable subtracted, and an artificial variable added to give an initial basis. (Key point: Surplus and artificial variables) 2. Why does the artificial variable carry a cost of -M in a maximization problem? A. So it always stays in the basis B. So it increases the objective value C. To make all Cj - Zj negative D. M is very large, so the penalty drives the artificial variable out of the basis Answer: D. A huge penalty makes any solution containing the artificial variable very unattractive, so the simplex method removes it. (Key point: Role of M) 3. In Table 1, Cj - Zj for x1 is 2 - M and for x2 is 1 + M. Which variable enters, and which leaves? A. x1 enters, s1 leaves B. x2 enters, A1 leaves (ratio 2 < 10) C. x2 enters, s1 leaves D. x1 enters, A1 leaves Answer: B. 1 + M is the largest positive entry for large M, so x2 enters; the ratios are 10 and 2, so A1 leaves. (Key point: First pivot) 4. What is the final optimal solution? A. x1 = 6, x2 = 4, Max P = 16 B. x1 = 4, x2 = 6, Max P = 14 C. x1 = 0, x2 = 10, Max P = 10 D. x1 = 4, x2 = 6, Max P = 20 Answer: B. The final table gives x1 = 4, x2 = 6, so P = 2(4) + 6 = 14, with no artificial variable in the basis. 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