(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) Recruitment needed and promotion prospects

Assumptions: R recruits join at the start of every year, and strength is counted at the start of each year of service. The proportion of a recruiting cohort still present in its n-th year of service is 1 − (cumulative proportion that left by the end of year n − 1): 1, 0.95, 0.64, 0.45, 0.37, 0.32, 0.27, 0.21, 0.13, 0.03 for years 1 to 10, with sum 4.37.

Recruitment: strength = R × 4.37 = 50, so R = 50/4.37 = 11.44, that is, about 11 to 12 recruits a year (12 give a strength of 52.4, 11 give 48.1).

Chemists by year of service = 11.44 × proportion: 11.44, 10.87, 7.32, 5.15, 4.23, 3.66, 3.09, 2.40, 1.49, 0.34 (total 50).

Promotion: senior posts go to the longest-serving men. Those in their 7th to 10th year number 3.09 + 2.40 + 1.49 + 0.34 = 7.32, which is less than 8, while those in their 6th to 10th year number 10.98, which is more than 8. So the 8 senior posts are held by everyone with at least 6 completed years of service and by the most senior 0.68 of the 3.66 men in their sixth year. By interpolation the cut-off is 6 − 0.68/3.66 ≈ 5.8 years, so a new entrant who stays can expect promotion after about 6 years of service (roughly 5.8 years).

(b) Queuing system, M/M/1 and busy period

Structure: a queuing system has (1) an input source, the calling population, whose customers arrive according to an inter-arrival distribution; (2) the queue, with its capacity and discipline (FCFS, LIFO, priority); (3) the service mechanism, with the number of servers (in parallel or in series) and the service-time distribution; (4) departure of served customers. Kendall's notation A/B/c/K/N/D describes it.

M/M/1: Poisson arrivals at rate λ, exponential service at rate μ, one server, FCFS, infinite capacity; N(t) is a birth–death process. In steady state (ρ = λ/μ < 1) the balance equations are λp₀ = μp₁ and (λ + μ)pₙ = λpₙ₋₁ + μpₙ₊₁, giving pₙ = ρⁿp₀; Σpₙ = 1 then gives pₙ = (1 − ρ)ρⁿ, n = 0, 1, 2, … Hence L = ρ/(1 − ρ), Lq = ρ²/(1 − ρ), W = 1/(μ − λ), Wq = ρ/(μ − λ).

Busy period B (from an arrival at an empty system until it is empty again): B = S + B₁ + … + B_N, where S is the first service time and N ~ Poisson(λS) is the number of arrivals during S, each starting an independent sub-busy period with the law of B. Hence B*(s) = μ/(s + μ + λ − λB*(s)), that is, λB*² − (λ + μ + s)B* + μ = 0, so B*(s) = [λ + μ + s − √((λ + μ + s)² − 4λμ)]/(2λ). Inversion gives the density f(t) = e^(−(λ+μ)t) I₁(2t√(λμ))/(t√ρ), t > 0 (I₁ = modified Bessel function), with E(B) = 1/(μ − λ) and Var(B) = (λ + μ)/(μ − λ)³.

(c) Optimal inventory policy and gross profit

D = 8000 × 50 = 400,000 m a year; unit cost c = ₹240; cost per replenishment A = 1,050 + 1,650 = ₹2,700; holding cost h = 0.25 × 240 = ₹60 per metre a year; no shortages.

Q* = √(2AD/h) = √(2 × 2700 × 400,000/60) = √36,000,000 = 6,000 m. Orders a year = D/Q* = 66.7; cycle time = Q*/8000 = 0.75 week. Ordering cost = (D/Q*)A = ₹180,000; holding cost = (Q*/2)h = ₹180,000; together ₹360,000. Purchase cost = 400,000 × 240 = ₹96,000,000, so total annual cost = ₹96,360,000. Optimal policy: order 6,000 m whenever stock reaches zero, that is, every 0.75 week.

Profit maximisation: revenue = price × D does not depend on Q, since demand is fixed and no shortages are allowed. Maximising profit = revenue − total cost is therefore the same as minimising cost, and the same Q* = 6,000 m results. The analysis would differ only if demand depended on price or shortages were allowed, because revenue would then vary with the decision.

Gross profit at ₹360 a metre: revenue = 400,000 × 360 = ₹144,000,000; profit = 144,000,000 − 96,360,000 = ₹47,640,000 (about ₹4.76 crore). Before ordering and holding costs the margin is 400,000 × 120 = ₹48,000,000.