(a) Solve the following Linear Programming problem using Two Phase method : Maximize Z = 3x₁ - x₂ Subject to 2x₁ + x₂ ≥ 2, x₁ +…

(a) Two-phase solution of the linear programme

Maximise Z = 3x₁ − x₂ subject to 2x₁ + x₂ ≥ 2, x₁ + 3x₂ ≤ 2, x₂ ≤ 4, x₁, x₂ ≥ 0. Step 1: Standard form: 2x₁ + x₂ − s₁ + a₁ = 2, x₁ + 3x₂ + s₂ = 2, x₂ + s₃ = 4, with surplus s₁, slacks s₂, s₃ and artificial a₁. Step 2: Phase I minimises W = a₁. Initial basis a₁ = 2, s₂ = 2, s₃ = 4, and W = 2 − 2x₁ − x₂ + s₁. Variable x₁ enters; the ratio test (2/2 = 1 against 2/1 = 2) makes a₁ leave. Step 3: After the pivot, x₁ = 1 − x₂/2 + s₁/2 − a₁/2, s₂ = 1 − 5x₂/2 − s₁/2 + a₁/2, s₃ = 4 − x₂, and W = a₁. W = 0, so Phase I ends with a feasible solution x₁ = 1, x₂ = 0, s₂ = 1, s₃ = 4, and a₁ is dropped. Step 4: Phase II maximises Z = 3x₁ − x₂ = 3 − 5x₂/2 + 3s₁/2. Variable s₁ has a positive coefficient and enters; only s₂ limits it (ratio 1/(1/2) = 2), so s₂ leaves, giving s₁ = 2 − 5x₂ − 2s₂. Step 5: Now x₁ = 2 − 3x₂ − s₂ and Z = 6 − 10x₂ − 3s₂. All non-basic coefficients are non-positive, so the solution is optimal. Optimal solution: x₁ = 2, x₂ = 0, maximum Z = 6.

(b)(i) Assignment problem

Costs (rows A to D, columns I to IV): A: 10, 12, 19, 11; B: 5, 10, 7, 8; C: 12, 14, 13, 11; D: 8, 15, 11, 9. Step 1: Row minima are 10, 5, 11, 8. Subtracting gives A: 0, 2, 9, 1; B: 0, 5, 2, 3; C: 1, 3, 2, 0; D: 0, 7, 3, 1. Step 2: Column minima are 0, 2, 2, 0. Subtracting gives A: 0, 0, 7, 1; B: 0, 3, 0, 3; C: 1, 1, 0, 0; D: 0, 5, 1, 1. Step 3: Four independent zeros can be chosen at A-II, B-III, C-IV and D-I (covering all zeros needs four lines), so this assignment is optimal. Assignment: A to II, B to III, C to IV, D to I, with minimum cost 12 + 7 + 11 + 8 = 38.

(b)(ii) Dual of the given primal

Let y₁, y₂, y₃, y₄ correspond to the four constraints of the maximisation problem. Rules: a ≤ constraint gives y ≥ 0, a ≥ constraint gives y ≤ 0 and an equality gives an unrestricted y. An unrestricted primal variable gives an equality dual constraint; a non-negative variable in a maximisation gives a ≥ dual constraint. Dual: Minimise W = 4y₁ + 5y₂ + 6y₃ + 7y₄ subject to 2y₁ + y₂ + 2y₃ + y₄ = 6 (x₁ unrestricted), 4y₁ − y₂ + 2y₃ + 8y₄ ≥ −5 (x₂ ≥ 0), −y₁ + 6y₂ + 4y₃ + y₄ ≥ 7 (x₃ ≥ 0), y₁ + 7y₂ + 5y₃ = 1 (x₄ unrestricted), with y₁ ≥ 0, y₂ ≤ 0, y₃ and y₄ unrestricted.

(c) Basic EOQ model and its calculation

The basic EOQ model finds the order quantity that minimises the sum of annual ordering cost and annual holding cost for one item. Assumptions: demand is known and constant; replenishment is instantaneous in a fixed lead time; no shortages are allowed; ordering cost per order and holding cost per unit per year are constant; price is constant, with no quantity discounts. Step 1: D = 12000 units per year, S = ₹100 per order, and H = 20% of ₹50 = ₹10 per unit per year. Step 2: EOQ Q = √(2DS/H) = √(2 × 12000 × 100/10) = √240000 ≈ 490 units (489·9). Step 3: Number of orders per year = D/Q = 12000/489·9 ≈ 24·5. Step 4: Time between orders = Q/D = 489·9/12000 ≈ 0·0408 year, about 14·9 days (about 15 days on a 365-day year).