(a) It is planned to raise a research team to a strength of 50 chemists, which is to be maintained. The wastage of recruits…

(a) Let F_y be the cumulative percentage who left by the end of year y. The percentage leaving during year y is w_y = F_y - F_y-1, with F_0 = 0. Thus w₁=5, w₂=31, w₃=19, w₄=8, w₅=5, w₆=5, w₇=6, w₈=8, w₉=10, w₁₀=3. These sum to 100%. The expected length of service is μ = Σ y w_y /100 = (1×5 + 2×31 + 3×19 + 4×8 + 5×5 + 6×5 + 7×6 + 8×8 + 9×10 + 10×3)/100 = 437/100 = 4.37 years. Equivalently, μ = (100+95+64+45+37+32+27+21+13+3)/100 = 4.37 years.

For a maintained strength N = 50, the steady-state annual recruitment is R = N/μ = 50/4.37 = 5000/437 ≈ 11.44 chemists per year. So the average requirement is about 11.44 recruits per year, i.e. in practice 11 or 12 recruits per year, or 12 in some years and 11 in others.

For the 8 senior posts, promotion is by length of service. In steady state, a new entrant expects promotion when the expected number of employees with longer service equals 8. Thus R ∫_t^∞ S(u) du = 8, where S(u) is the survival function. Here R = 50/4.37, so ∫_t^∞ S(u) du = 8/R = 8×4.37/50 = 0.6992 years. Using the survival values, for t between 5 and 6 years, 0.32(6-t) + (0.27+0.21+0.13+0.03) = 0.6992. So 0.32(6-t) + 0.64 = 0.6992, giving 6-t = 0.0592/0.32 = 0.185, hence t = 5.815 years. Thus the average length of service after which the next entrant expects promotion is about 5.82 years.

(b) A queuing system has:

• input process: how customers arrive, e.g. Poisson;
• service mechanism: number of servers and service-time distribution;
• queue discipline: order of service, e.g. FCFS;
• system capacity: finite or infinite waiting room;
• calling population: finite or infinite.

For an M/M/1 queue, arrivals are Poisson with rate λ, service times are exponential with rate μ, there is one server, infinite capacity, FCFS discipline. Let N be the number in the system and pₙ = P(N=n) in steady state. The birth-death balance equations are λ p₀ = μ p₁, (λ+μ)pₙ = λ pₙ₋₁ + μ pₙ₊₁, n ≥ 1. Let ρ = λ/μ. From the first equation, p₁ = ρ p₀. Substituting pₙ = ρⁿ p₀ in the second equation satisfies all balance equations. Normalising, Σₙ₌₀^∞ pₙ = p₀/(1-ρ) = 1, so p₀ = 1-ρ. Hence pₙ = (1-ρ)ρⁿ, n = 0,1,2,…, provided ρ < 1. Then L = ρ/(1-ρ) = λ/(μ-λ), L_q = ρ²/(1-ρ) = λ²/[μ(μ-λ)], W = 1/(μ-λ), W_q = λ/[μ(μ-λ)].

For the busy period B, start with one customer. Let b(s) = E(e^(-sB)). During the first service time X ~ Exp(μ), K arrivals occur with K|X ~ Poisson(λX). After X, the remaining busy period is the sum of K independent busy periods. Hence b(s) = ∫₀^∞ μ e^(-(μ+s)x) e^(-λx(1-b(s))) dx = μ/[μ+s+λ(1-b(s))]. Solving, b(s) = [μ+s+λ - √((μ+s+λ)² - 4λμ)]/(2λ). Inverting this Laplace transform gives the busy-period density f_B(t) = (1/(t√ρ)) e^(-(λ+μ)t) I₁(2t√(λμ)), t > 0, where ρ = λ/μ < 1 and I₁ is the modified Bessel function of order 1. Its mean is E(B) = 1/(μ-λ). If ρ ≥ 1, the busy period may be infinite with positive probability.

(c) Annual demand D = 8000 × 50 = 400000 metres per year. Ordering cost per replenishment K = 1050 + 1650 = ₹2700. Holding cost h = 25% of ₹240 = 0.25 × 240 = ₹60 per metre per year. Using the EOQ formula, Q* = √(2DK/h) = √(2 × 400000 × 2700 / 60) = √(36,000,000) = 6000 metres. So the optimal policy is to order 6000 metres per order. Number of orders per year = D/Q* = 400000/6000 = 66.6667 orders/year. Cycle time = Q*/D = 6000/400000 = 0.015 year = 0.75 week. Thus the company should replenish every 0.75 week. Annual ordering cost = (D/Q*)K = 66.6667 × 2700 = ₹180000. Annual holding cost = (Q*/2)h = 3000 × 60 = ₹180000. Total inventory cost = ₹360000 per year. Including purchase cost, total cost = 400000 × 240 + 360000 = ₹96,360,000 per year.

For profit maximisation, write annual profit as P(Q) = (selling price - cost price)D - KD/Q - hQ/2. Since the first term does not depend on Q, differentiating with respect to Q gives the same condition -KD/Q² + h/2 = 0, so Q* = √(2DK/h). Thus, under fixed demand and selling price, the optimal lot size is unchanged. The analysis would differ only if price affects demand, quantity discounts exist, or shortages/backorders are allowed.

Selling price = ₹360 per metre, so gross profit per metre = 360 - 240 = ₹120. Annual gross profit = 120 × 400000 = ₹48,000,000 per year. Net profit after ordering and holding costs = 48,000,000 - 360,000 = ₹47,640,000 per year.