(a) Solve the following linear programming problem: Maximize z = 3x₁ + 5x₂ subject to the constraints 3x₁ + 2x₂ ≤ 18 x₁ ≤ 4 x₂ ≤…

(a) Linear programme and sensitivity to cⱼ

Maximise z = 3x₁ + 5x₂ subject to 3x₁ + 2x₂ ≤ 18, x₁ ≤ 4, x₂ ≤ 6, x₁, x₂ ≥ 0. The corner points are (0,0) with z = 0, (4,0) with 12, (4,3) with 27, (2,6) with 36 and (0,6) with 30. The optimum is x₁ = 2, x₂ = 6, z_max = 36.

Add slacks s₁, s₂, s₃ to the three constraints. At the optimum s₁ = 0, s₂ = 2, s₃ = 0, so the basis is (x₁, x₂, s₂) and the non-basic variables are s₁ and s₃. Solving the constraints gives x₂ = 6 − s₃, x₁ = 2 − s₁/3 + 2s₃/3 and s₂ = 2 + s₁/3 − 2s₃/3.

Let the objective coefficients be c₁ and c₂ for x₁ and x₂, and zero for the slacks. Then z = 2c₁ + 6c₂ + (−c₁/3)s₁ + (2c₁/3 − c₂)s₃. For a maximum, both non-basic coefficients must be ≤ 0, so c₁ ≥ 0 and c₂ ≥ 2c₁/3.

• With c₂ = 5 the solution stays optimal for 0 ≤ c₁ ≤ 15/2. Below 0, s₁ enters the basis; above 15/2, s₃ enters.
• With c₁ = 3 it stays optimal for c₂ ≥ 2. Below 2, s₃ enters; at c₂ = 2 there are alternative optima.

(b) Economic order quantity

D = 600 units a year, ordering cost C₀ = Rs 80 per run and holding cost C_h = 60 paise = Rs 0.60 per unit a year, with no shortages.

(i) EOQ: Q* = √(2DC₀/C_h) = √(2 × 600 × 80 / 0.60) = √160000 = 400 units.

(ii) Optimum period: T* = Q*/D = 400/600 = 2/3 year (8 months).

(iii) Minimum cost: TC(Q*) = √(2DC₀C_h) = 120 + 120 = Rs 240 a year, with TC(Q) = DC₀/Q + C_hQ/2.

• 20% more, Q = 480: TC = 100 + 144 = Rs 244, an increase of Rs 4 a year.
• 40% less, Q = 240: TC = 200 + 72 = Rs 272, an increase of Rs 32 a year.

(c) Mean and range charts

The grand mean is X̄̄ = 442/10 = 44.2 and the average range is R̄ = 58/10 = 5.8. For n = 5, σ̂ = R̄/d₂ = 5.8/2.326 = 2.4936.

Mean chart: 3σ̂/√5 = 3.3455, so UCL = 47.55 and LCL = 40.85. The sample means 49 (sample 2), 37 (3), 37 (6) and 51 (7) lie outside the limits, so the process mean is out of control.

Range chart: UCL = R̄ + 3d₃σ̂ = 5.8 + 3(0.864)(2.4936) = 12.26, and LCL = 5.8 − 6.46 < 0, taken as 0. All ten ranges (4 to 8) lie inside 0 to 12.26, so variability is in control.

The process is not in control overall: the dispersion is stable but the mean is not, so assignable causes affecting the setting of the machine should be traced and removed.