--- title: "Big-M method min:Z=4x₁+2x₂" subject: "Mathematics III" chapter: 5 key: "semester_3/mathematics_iii:big-m-method-min-z-4x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/big-m-method-min-z-4x site: BCT Engineering --- # Big-M method min:Z=4x₁+2x₂ Mathematics III · Chapter 5: Linear Programming - Asked in: 2076 Chaitra, 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-min-z-4x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5 ## Question Use the Big-M method to solve the following LPP: Minimize Z = 4x₁ + 2x₂ subject to 3x₁ + x₂ ≥ 27, −x₁ − x₂ ≤ −21, x₁ + 2x₂ ≥ 30, x₁, x₂ ≥ 0. ## Short answer Convert the minimization to maximizing Z* = -4x₁ - 2x₂, add surplus and artificial variables, and apply the Big-M simplex method. The optimal solution is Minimum Z = 48 at x₁ = 3, x₂ = 18. ## Answer **Step 1: Make the problem ready for the simplex method.** - The problem is a minimisation, so maximise Z* = −Z = −4x₁ − 2x₂. Then Min Z = −(Max Z*). - Constraint 2 has a negative right side, so multiply by −1: x₁ + x₂ ≥ 21. - Constraint 1 is ≥: subtract surplus variable s₁ and add artificial variable A₁. - Constraint 2 is ≥: subtract surplus variable s₂ and add artificial variable A₂. - Constraint 3 is ≥: subtract surplus variable s₃ and add artificial variable A₃. 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 − 2x₂ − M·A₁ − M·A₂ − M·A₃ subject to 3x₁ + x₂ − s₁ + A₁ = 27 x₁ + x₂ − s₂ + A₂ = 21 x₁ + 2x₂ − s₃ + A₃ = 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 -2 0 -M 0 -M 0 -M Basis CB x₁ x₂ s₁ A₁ s₂ A₂ s₃ A₃ b Ratio ----------------------------------------------------- A₁ -M 3 1 -1 1 0 0 0 0 27 9 A₂ -M 1 1 0 0 -1 1 0 0 21 21 A₃ -M 1 2 0 0 0 0 -1 1 30 30 ----------------------------------------------------- Zj -5M -4M M -M M -M M -M -78M Cj-Zj -4+5M -2+4M -M 0 -M 0 -M 0 Entering: x₁ Leaving: A₁ Pivot = 3 ``` **Table 2** ```text Cj -4 -2 0 0 -M 0 -M Basis CB x₁ x₂ s₁ s₂ A₂ s₃ A₃ b Ratio ------------------------------------------------------------- x₁ -4 1 1/3 -1/3 0 0 0 0 9 27 A₂ -M 0 2/3 1/3 -1 1 0 0 12 18 A₃ -M 0 5/3 1/3 0 0 -1 1 21 63/5 ------------------------------------------------------------- Zj -4 -4/3-7/3M 4/3-2/3M M -M M -M -36-33M Cj-Zj 0 -2/3+7/3M -4/3+2/3M -M 0 -M 0 Entering: x₂ Leaving: A₃ Pivot = 5/3 ``` **Table 3** ```text Cj -4 -2 0 0 -M 0 Basis CB x₁ x₂ s₁ s₂ A₂ s₃ b Ratio --------------------------------------------------------------- x₁ -4 1 0 -2/5 0 0 1/5 24/5 24 A₂ -M 0 0 1/5 -1 1 2/5 18/5 9 x₂ -2 0 1 1/5 0 0 -3/5 63/5 - --------------------------------------------------------------- Zj -4 -2 6/5-1/5M M -M 2/5-2/5M -222/5-18/5M Cj-Zj 0 0 -6/5+1/5M -M 0 -2/5+2/5M Entering: s₃ Leaving: A₂ Pivot = 2/5 ``` **Table 4** ```text Cj -4 -2 0 0 0 Basis CB x₁ x₂ s₁ s₂ s₃ b ----------------------------------- x₁ -4 1 0 -1/2 1/2 0 3 s₃ 0 0 0 1/2 -5/2 1 9 x₂ -2 0 1 1/2 -3/2 0 18 ----------------------------------- Zj -4 -2 1 1 0 -48 Cj-Zj 0 0 -1 -1 0 ``` **Step 4: Optimality.** All Cj − Zj ≤ 0 and no artificial variable remains in the basis, so the solution is optimal and feasible. Max Z* = −48, hence Min Z = 48. **Answer:** x₁ = 3, x₂ = 18, and Minimum Z = 48. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the minimisation Z = 4x1 + 2x2 handled? A. Add a slack to the objective B. Use Z as it is and choose the most negative Cj - Zj C. Maximise Z* = -Z, then Min Z = -(Max Z*) D. Swap x1 and x2 Answer: C. Maximising Z* = -4x1 - 2x2 is equivalent, and the minimum of Z is the negative of the maximum of Z*. (Key point: Minimisation) 2. Constraint 2 is -x1 - x2 <= -21. What is done first? A. Multiply by -1 to get x1 + x2 >= 21 B. Add a slack variable directly C. Ignore the constraint D. Replace it with an equality using a slack Answer: A. Right-hand sides must be non-negative, so multiply by -1, which reverses the inequality to >=. (Key point: Negative right side) 3. How many artificial variables are needed? A. One B. Three, one for each >= constraint C. Two D. None, because the problem is a minimisation Answer: B. All three constraints are >= after the sign change, so each has a surplus and an artificial variable. (Key point: Number of artificials) 4. What is the minimum value and where? A. Min Z = 48 at x1 = 9, x2 = 0 B. Min Z = -48 at x1 = 3, x2 = 18 C. Min Z = 54 at x1 = 0, x2 = 27 D. Min Z = 48 at x1 = 3, x2 = 18 Answer: D. Max Z* = -48 so Min Z = 48; check 4(3) + 2(18) = 48. 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