--- title: "Big-M method max:P=2x₁+5x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-max-p-2x-four" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-max-p-2x-four site: BCT Engineering --- # Big-M method max:P=2x₁+5x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2079 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-four - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Solve the following LPP by the Big-M method: Maximize P = 2x₁ + 5x₂ subject to x₁ + 2x₂ ≤ 18, 2x₁ + x₂ ≥ 21, x₁ + x₂ ≥ 10, x₁, x₂ ≥ 0. ## Short answer Introduce slack s1 for ≤, surplus s2,s3 and artificial A1,A2 for ≥ constraints. The objective becomes P=2x1+5x2−MA1−MA2. After pivot operations entering x1, s3, then x2, optimal feasible solution: x1=8, x2=5, Maximum P=41. ## 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. - Constraint 3 is ≥: subtract surplus variable s₃ and add artificial variable A₂. 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 + 5x₂ − M·A₁ − M·A₂ subject to x₁ + 2x₂ + s₁ = 18 2x₁ + x₂ − s₂ + A₁ = 21 x₁ + x₂ − s₃ + A₂ = 10 ``` **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 5 0 0 -M 0 -M Basis CB x₁ x₂ s₁ s₂ A₁ s₃ A₂ b Ratio ------------------------------------------------ s₁ 0 1 2 1 0 0 0 0 18 18 A₁ -M 2 1 0 -1 1 0 0 21 21/2 A₂ -M 1 1 0 0 0 -1 1 10 10 ------------------------------------------------ Zj -3M -2M 0 M -M M -M -31M Cj-Zj 2+3M 5+2M 0 -M 0 -M 0 Entering: x₁ Leaving: A₂ Pivot = 1 ``` **Table 2** ```text Cj 2 5 0 0 -M 0 Basis CB x₁ x₂ s₁ s₂ A₁ s₃ b Ratio --------------------------------------------- s₁ 0 0 1 1 0 0 1 8 8 A₁ -M 0 -1 0 -1 1 2 1 1/2 x₁ 2 1 1 0 0 0 -1 10 - --------------------------------------------- Zj 2 2+M 0 M -M -2-2M 20-M Cj-Zj 0 3-M 0 -M 0 2+2M Entering: s₃ Leaving: A₁ Pivot = 2 ``` **Table 3** ```text Cj 2 5 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b Ratio ------------------------------------------ s₁ 0 0 3/2 1 1/2 0 15/2 5 s₃ 0 0 -1/2 0 -1/2 1 1/2 - x₁ 2 1 1/2 0 -1/2 0 21/2 21 ------------------------------------------ Zj 2 1 0 -1 0 21 Cj-Zj 0 4 0 1 0 Entering: x₂ Leaving: s₁ Pivot = 3/2 ``` **Table 4** ```text Cj 2 5 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b ---------------------------------- x₂ 5 0 1 2/3 1/3 0 5 s₃ 0 0 0 1/3 -1/3 1 3 x₁ 2 1 0 -1/3 -2/3 0 8 ---------------------------------- Zj 2 5 8/3 1/3 0 41 Cj-Zj 0 0 -8/3 -1/3 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 = 8, x₂ = 5, and Maximum P = 41. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many artificial variables are needed, and why? A. Two (A1, A2), one for each >= constraint B. Three, one for each constraint C. None, surplus variables are enough D. One, for the <= constraint Answer: A. Each >= row needs a surplus variable subtracted and an artificial variable added to give a starting basis; the <= row uses its slack. (Key point: Setting up the constraints) 2. In Table 1, Cj - Zj is 2 + 3M for x1 and 5 + 2M for x2. Which enters? A. x2, since 5 > 2 B. x1, since 3M is the larger M part C. s1, since its Cj - Zj is 0 D. A1, since it has cost -M Answer: B. M is very large, so the M part decides: 3M > 2M and x1 enters. (Key point: Entering variable with M) 3. In Table 1 the ratios are 18 (s1), 21/2 (A1), 10 (A2). Which variable leaves, and what happens to its column? A. s1 leaves; its column is kept B. A1 leaves; the column is dropped C. A2 leaves; its column is dropped as it never re-enters D. A2 leaves; its column is kept Answer: C. The smallest ratio is 10 (A2), so A2 leaves, and an artificial variable's column is dropped once it leaves the basis. (Key point: Artificial variables leaving) 4. What is the final answer? A. x1 = 5, x2 = 8, Max P = 50 B. x1 = 8, x2 = 5, Max P = 41 C. x1 = 8, x2 = 5, Max P = 26 D. x1 = 10, x2 = 4, Max P = 40 Answer: B. The final table gives x2 = 5, x1 = 8, so P = 2(8) + 5(5) = 41, and no artificial variable remains. 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