(a) A Company ships truckloads of grain from three silos to four mills. The supply (in truckloads) and the demand (also in…

(a) Minimum-cost shipping schedule

Total supply = 250 + 300 + 400 = 950 and total demand = 200 + 225 + 275 + 250 = 950, so the problem is balanced. Step 1: Starting basic feasible solution by the north-west corner rule: A to D 200, A to E 50, B to E 175, B to F 125, C to F 150, C to G 250. These are 6 = 3 + 4 − 1 allocations. Cost = 200 × 11 + 50 × 13 + 175 × 18 + 125 × 14 + 150 × 13 + 250 × 10 = 12200. Step 2: Optimality test (MODI). With u_A = 0 and cᵢⱼ = uᵢ + vⱼ on occupied cells: v_D = 11, v_E = 13, u_B = 5, v_F = 9, u_C = 4, v_G = 6. Step 3: Opportunity costs dᵢⱼ = cᵢⱼ − uᵢ − vⱼ on empty cells: AF = 8, AG = 8, BD = 0, BG = −1, CD = 6, CE = 7. The negative value at BG means the solution can be improved. Step 4: Loop BG(+), BF(−), CF(+), CG(−). The smallest allocation on the minus cells is 125, so shift 125: B to G 125, B to F 0, C to F 275, C to G 125. Cost falls by 125 to 12075. Step 5: Re-test with u_A = 0, v_D = 11, v_E = 13, u_B = 5, v_G = 5, u_C = 5, v_F = 8: opportunity costs are AF = 9, AG = 9, BD = 0, BF = 1, CD = 5, CE = 6, all non-negative, so the solution is optimal. Optimal schedule: A to D 200 and A to E 50; B to E 175 and B to G 125; C to F 275 and C to G 125, with minimum cost 12075. (BD = 0 shows an alternative optimum with the same cost.)

(b)(i) Maximum allowable σ for the gadget life

Step 1: P(120 < X < 200) = P(−40/σ < Z < 40/σ) = 2Φ(40/σ) − 1 = 0·80. Step 2: Hence Φ(40/σ) = 0·90, and from the normal table z = 1·2816. Step 3: 40/σ = 1·2816, so σ = 31·2 hours. A larger σ would make the probability fall below 0·80, so the maximum allowable σ is about 31·2 hours.

(b)(ii) Probability for the log-normal concrete strength

Step 1: Y = ln X ~ N(3, 0·2²), so P(X ≤ 10) = P(Y ≤ ln 10) = P(Y ≤ 2·3026). Step 2: Z = (2·3026 − 3)/0·2 = −3·487. Step 3: P(X ≤ 10) = Φ(−3·487) ≈ 0·00024. The probability is about 0·00024, or 0·024%.

(c) Steady-state probabilities of the checkout queue

Arrival rate λ = 10 per hour; each counter serves μ = 60/12 = 5 per hour. Step 1: The service rate depends on the number of counters open: μₙ = 5 for n = 1 to 3, 10 for n = 4 to 6 and 15 for n ≥ 7. Step 2: Balance equations for the birth-death process give pₙ = (λ/μₙ)pₙ₋₁. Step 3: The ratios λ/μₙ are 2 for n = 1 to 3, 1 for n = 4 to 6 and 2/3 for n ≥ 7. Step 4: Therefore p₁ = 2p₀, p₂ = 4p₀, p₃ = 8p₀, p₄ = p₅ = p₆ = 8p₀, and pₙ = 8(2/3)^(n−6) p₀ for n ≥ 7. Step 5: Normalising, 1 = p₀[1 + 2 + 4 + 8 + 8 + 8 + 8 + 8 × (2/3)/(1 − 2/3)] = p₀(39 + 16) = 55p₀, so p₀ = 1/55. Result: p₀ = 1/55, p₁ = 2/55, p₂ = 4/55, p₃ = p₄ = p₅ = p₆ = 8/55 and pₙ = (8/55)(2/3)^(n−6) for n ≥ 7. A steady state exists because λ/(3μ) = 2/3 < 1.