(a) (i) Discuss the theoretical basis of the sequential probability ratio test (SPRT) in acceptance sampling. Explain how acceptance and rejection numbers are chosen to satisfy α and β risks. (15 marks)

(ii) Ten units are tested for 200 hours. Failu

Question

(a) (i) Discuss the theoretical basis of the sequential probability ratio test (SPRT) in acceptance sampling. Explain how acceptance and rejection numbers are chosen to satisfy α and β risks. (15 marks)

(ii) Ten units are tested for 200 hours. Failures occur at 50, 80, 120, and 150 hours (4 failures). Assume that the lifetimes are exponentially distributed with type-I censoring. Find the MLE of λ̂, where λ denotes the failure rate. (5 marks)

(b) (i) What is a Transportation problem ? Give a general mathematical model of a transportation problem. (5 marks)

(ii) Use Vogel’s approximation method to solve the following transportation problem :
Distribution Centre
Plant	D₁	D₂	D₃	D₄	Capacity
P₁	2	3	11	7	6
P₂	1	0	6	1	1
P₃	5	8	15	9	10
Demand	7	5	3	2 (10 marks)

(c) (i) Define a Poisson process and state at least three applications of the same. (5 marks)

(ii) In a local post office, a particular service is operated by one person. The arrival pattern of customers on a particular day follows a Poisson distribution with a mean arrival rate of 10 people per hour. Customers are served on a First-Come, First-Served (FCFS) basis and the service time is estimated to be exponentially distributed with an average time of 4 minutes. Determine :
(I) Probability (there is no queue)
(II) The average size of the queue
(III) The expected waiting time in the queue
(IV) Probability (a customer will spend less than 12 minutes in the queue) (10 marks)

UPSC Answer Check · Accepted Answer

(a) (i) SPRT in acceptance sampling

Wald's sequential probability ratio test treats lot acceptance as a test of H₀: p = p₁ (AQL) against H₁: p = p₂ (LTPD), p₂ > p₁, with producer's risk α = P(reject | p₁) and consumer's risk β = P(accept | p₂). Items are inspected one at a time; after n items containing d defectives the likelihood ratio is λₙ = [p₂^d (1 − p₂)^(n−d)] / [p₁^d (1 − p₁)^(n−d)]. Rule: accept the lot if λₙ ≤ B, reject if λₙ ≥ A, otherwise inspect one more item, with Wald's approximations A = (1 − β)/α and B = β/(1 − α). (These ignore the overshoot of the boundaries; the achieved risks satisfy α′ + β′ ≤ α + β.)

Taking logarithms, with k = ln[p₂(1 − p₁)/p₁(1 − p₂)], the continuation region becomes the band between two parallel lines in the (n, d) plane: acceptance line d = sn − h₁, rejection line d = sn + h₂, where s = ln[(1 − p₁)/(1 − p₂)]/k, h₁ = ln[(1 − α)/β]/k, h₂ = ln[(1 − β)/α]/k. At stage n the acceptance number is aₙ = the largest integer ≤ sn − h₁ and the rejection number is rₙ = the smallest integer ≥ sn + h₂; accept if d ≤ aₙ, reject if d ≥ rₙ, continue otherwise. The slope s (between p₁ and p₂) depends only on the two quality levels, while α and β enter only through the intercepts h₁ and h₂, which is how the two risks are enforced; no acceptance is possible before n > h₁/s. The OC function is L(p) = (Aʰ − 1)/(Aʰ − Bʰ), giving L(p₁) = 1 − α and L(p₂) = β, and the ASN is E(n | p) ≈ [L(p) ln B + (1 − L(p)) ln A]/E(z). By the Wald-Wolfowitz theorem the SPRT minimises the expected sample number at both p₁ and p₂ among all plans with the same α and β (often about half the single-sampling n); in practice the plan is truncated at some n₀.

(a) (ii) MLE of λ under type-I censoring

n = 10 units, test stopped at T₀ = 200 h, r = 4 failures at 50, 80, 120, 150 h and 6 survivors. L(λ) = Πᵢ₌₁⁴ λe^(−λtᵢ) × [e^(−200λ)]⁶ = λ⁴ e^(−λT), T = Σtᵢ + (n − r)T₀. ln L = 4 ln λ − λT; d ln L/dλ = 4/λ − T = 0 gives λ̂ = r/T (second derivative −4/λ² < 0). Total time on test T = (50 + 80 + 120 + 150) + 6 × 200 = 400 + 1200 = 1600 unit-hours. λ̂ = 4/1600 = 0.0025 failures per hour (estimated MTTF = 1/λ̂ = 400 hours).

(b) (i) Transportation problem and its model

A transportation problem is a special linear programming problem of shipping a single commodity from m sources (supplies aᵢ) to n destinations (demands bⱼ) at minimum total cost, cᵢⱼ being the cost per unit from source i to destination j and xᵢⱼ the quantity shipped. Minimise Z = Σᵢ₌₁ᵐ Σⱼ₌₁ⁿ cᵢⱼ xᵢⱼ subject to Σⱼ xᵢⱼ = aᵢ (i = 1, …, m), Σᵢ xᵢⱼ = bⱼ (j = 1, …, n), xᵢⱼ ≥ 0. A feasible solution exists when the problem is balanced, Σaᵢ = Σbⱼ; otherwise a dummy source or destination with zero costs is added. Only m + n − 1 constraints are independent, so a basic feasible solution has m + n − 1 occupied cells.

(b) (ii) Vogel's approximation method and optimality

Supply 6 + 1 + 10 = 17 = demand 7 + 5 + 3 + 2, so the problem is balanced; m + n − 1 = 6. Step 1: row penalties (difference of the two lowest costs) 1, 1, 3; column penalties 1, 3, 5, 6. Largest is 6 (D₄); its least cost is 1 at P₂: x₂₄ = 1, P₂ exhausted. Step 2: row penalties P₁ 1, P₃ 3; column penalties 3, 5, 4, 2. Largest 5 (D₂): least cost 3 at P₁: x₁₂ = 5, D₂ satisfied, P₁ has 1 left. Step 3: row penalties P₁ 5, P₃ 4; column penalties 3, 4, 2. Largest 5 (P₁): least cost 2 at D₁: x₁₁ = 1, P₁ exhausted. Step 4: only P₃ remains: x₃₁ = 6, x₃₃ = 3, x₃₄ = 1. VAM solution (6 cells, non-degenerate): x₁₁ = 1, x₁₂ = 5, x₂₄ = 1, x₃₁ = 6, x₃₃ = 3, x₃₄ = 1; cost = 2 + 15 + 1 + 30 + 45 + 9 = ₹102.

MODI test (cᵢⱼ = uᵢ + vⱼ on occupied cells, u₁ = 0): v₁ = 2, v₂ = 3, u₃ = 3, v₃ = 12, v₄ = 6, u₂ = −5. Opportunity costs dᵢⱼ = cᵢⱼ − uᵢ − vⱼ: d₁₃ = −1, d₂₃ = −1, d₁₄ = 1, d₂₁ = 4, d₂₂ = 2, d₃₂ = 2. Negative values, so the VAM solution is not optimal. Enter (1,3): loop (1,3)+ (3,3)− (3,1)+ (1,1)−, θ = 1, cost ₹101. Enter (2,3): loop (2,3)+ (3,3)− (3,4)+ (2,4)−, θ = 1, cost ₹100. Re-test (u₁ = 0): v₂ = 3, v₃ = 11, u₂ = −5, u₃ = 4, v₁ = 1, v₄ = 5; all dᵢⱼ ≥ 0, so the solution is optimal. Optimal: x₁₂ = 5, x₁₃ = 1, x₂₃ = 1, x₃₁ = 7, x₃₃ = 1, x₃₄ = 2; minimum cost ₹100 (the VAM starting solution cost ₹102).

(c) (i) Poisson process and applications

A counting process N(t), t ≥ 0 is a Poisson process with rate λ if N(0) = 0, its increments over disjoint intervals are independent and stationary, and it is orderly: P[N(h) = 1] = λh + o(h), P[N(h) ≥ 2] = o(h). Then P[N(t) = n] = e^(−λt)(λt)ⁿ/n!, n = 0, 1, 2, …, with E[N(t)] = V[N(t)] = λt; equivalently the inter-arrival times are independent exponential with mean 1/λ. Applications: arrivals of customers or calls in queueing systems; emissions in radioactive decay; failures of components in reliability studies; insurance claims or road accidents over time; defects per unit length or area (basis of the c chart).

(c) (ii) M/M/1 queue at the post office

λ = 10 per hour, μ = 60/4 = 15 per hour, ρ = λ/μ = 2/3 < 1. (I) Server idle (nobody in the system): P₀ = 1 − ρ = 1/3 ≈ 0.333. Read strictly as "nobody waiting", P(N ≤ 1) = 1 − ρ² = 5/9 ≈ 0.556. (II) L_q = ρ²/(1 − ρ) = (4/9)/(1/3) = 4/3 ≈ 1.33 customers (Lₛ = ρ/(1 − ρ) = 2). (III) W_q = L_q/λ = (4/3)/10 = 2/15 hour = 8 minutes (Wₛ = 1/(μ − λ) = 12 minutes). (IV) The waiting time in the queue has P(W_q = 0) = 1 − ρ and P(W_q > t) = ρe^(−(μ−λ)t). With t = 12 min = 0.2 h and μ − λ = 5 per hour: P(W_q > 0.2) = (2/3)e^(−1) = 0.245, so P(W_q < 12 minutes) = 1 − (2/3)e^(−1) ≈ 0.755. (If the time spent in the whole system is meant, P(Wₛ < 0.2) = 1 − e^(−1) ≈ 0.632.)

(a) (i) Discuss the theoretical basis of the sequential probability ratio test (SPRT) in acceptance sampling. Explain how…

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(a) (i) SPRT in acceptance sampling

(b) (i) Transportation problem and its model

(c) (i) Poisson process and applications

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