Explain Single Sampling Plan with the help of an example. Also, write the importance of an Operating Characteristics Curve in a sampling plan. (10 marks)

Solve the above assignment problem.

Depot
I II III IV V
A 160 130 175

Question

Explain Single Sampling Plan with the help of an example. Also, write the importance of an Operating Characteristics Curve in a sampling plan. (10 marks)

Solve the above assignment problem.

Depot
        I    II   III  IV   V
A      160  130  175  190  200
Town B 135  120  130  160  175
C      140  110  145  170  185
D       50   50   80   80  110
E       55   35   80   80  105
(10 marks)

Use algebraic method to solve the above game.

Player B
        B₁     B₂     B₃     B₄
A₁     0·25   0·20   0·14   0·30
Player A A₂     0·27   0·16   0·12   0·14
A₃     0·35   0·08   0·15   0·19
A₄    −0·02   0·08   0·13   0·00
(10 marks)

Consider the Markov Chain with transition probability matrix:

0    1    2
0    (0    1    0)
1    (½    0    ½)
2    (0    1    0)

Show that the states are periodic and persistent non-null. (10 marks)

State the importance of the hazard function. If the hazard rate of a component is given by:

h(t) = { 0.015,  t ≤ 200
       { 0.025,  t > 200

then find an expression for the reliability function of the component. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Single Sampling Plan
A single sampling plan is defined by the lot size N, sample size n, and acceptance number c. Draw a random sample of n items from the lot and count defectives d. If d ≤ c, accept the lot; if d > c, reject it. Example: N = 1000, n = 50, c = 1. Accept if 0 or 1 defective is found; reject if 2 or more defectives are found. If the lot fraction defective is p, then
Pa(p) = Σ₍d=0₎¹ C(50,d) p^d (1−p)^(50−d) = (1−p)^50 + 50p(1−p)^49.
At p = 0.01, Pa ≈ 0.91; at p = 0.05, Pa ≈ 0.28. The Operating Characteristics curve plots Pa(p) against p. Its importance is that it measures the discriminating power of the plan, shows producer’s risk α and consumer’s risk β, helps fix AQL and LTPD, compares alternative plans, and guides choice of n and c balancing inspection cost and quality protection.

(b) Assignment Problem
Assuming costs are to be minimised. The cost matrix is:

A: 160, 130, 175, 190, 200
B: 135, 120, 130, 160, 175
C: 140, 110, 145, 170, 185
D: 50, 50, 80, 80, 110
E: 55, 35, 80, 80, 105

Row minima are 130, 120, 110, 50, 35. Subtract them. Column minima after row reduction are 0, 0, 10, 30, 55. Subtract them. Cover all zeros by rows B, D and column II; the smallest uncovered entry is 15. Subtract 15 from uncovered entries and add 15 at intersections BII and DII. A zero matching is then:
A→V, B→III, C→II, D→I, E→IV.
Total minimum cost = 200 + 130 + 110 + 50 + 80 = 570 cost units.

(c) Game
The payoff matrix is to Player A. A₃ dominates A₄, so delete A₄. Then B₂ dominates B₁ and B₃ dominates B₄; delete B₁ and B₄. In the remaining matrix, A₁ dominates A₂. The reduced 2×2 game is:
A₁: 0.20, 0.14
A₃: 0.08, 0.15

Let A play A₁ with probability p and A₃ with 1−p. For B₂ and B₃ payoffs to be equal:
0.20p + 0.08(1−p) = 0.14p + 0.15(1−p)
⇒ 0.08 + 0.12p = 0.15 − 0.01p
⇒ 0.13p = 0.07
⇒ p = 7/13, 1−p = 6/13.
Let B play B₂ with probability q and B₃ with 1−q. Equating A₁ and A₃ payoffs:
0.20q + 0.14(1−q) = 0.08q + 0.15(1−q)
⇒ 0.14 + 0.06q = 0.15 − 0.07q
⇒ 0.13q = 0.01
⇒ q = 1/13, 1−q = 12/13.
Value v = 0.14 + 0.06(1/13) = 47/325 ≈ 0.1446.
Optimal strategies: A: (7/13, 0, 6/13, 0); B: (0, 1/13, 12/13, 0). Value of game to A = 47/325.

(d) Markov Chain
The transition matrix is
0: (0, 1, 0)
1: (½, 0, ½)
2: (0, 1, 0)
Every transition goes from {0,2} to {1} or from {1} to {0,2}. Hence a return to any state is possible only in an even number of steps. Also two-step returns are positive, e.g. 0→1→0, 1→0→1 or 1→2→1, 2→1→2. Thus every state has period 2. The chain is irreducible because 0↔1↔2. A finite irreducible chain is persistent. For non-null, solve πP = π:
π₀ = ½π₁, π₁ = π₀ + π₂, π₂ = ½π₁, π₀ + π₁ + π₂ = 1.
This gives π = (¼, ½, ¼). Since the stationary distribution is positive, all states are persistent non-null.

(e)(i) Importance of Hazard Function
The hazard function is h(t) = f(t)/R(t). It gives the instantaneous failure rate at time t given survival up to t. It is important because it describes ageing or wear-out behaviour, distinguishes increasing/decreasing/constant failure rates, helps derive R(t) = exp(−∫₀^t h(u)du), and is used in reliability, survival analysis and maintenance decisions.

(e)(ii) Reliability Function
R(t) = exp(−∫₀^t h(u)du).
For 0 ≤ t ≤ 200:
R(t) = exp(−∫₀^t 0.015 du) = e^(−0.015t).
For t > 200:
∫₀^t h(u)du = ∫₀^200 0.015 du + ∫₂₀₀^t 0.025 du
= 3 + 0.025(t−200) = 0.025t − 2.
Hence
R(t) = e^(−(0.025t−2)) = e^(2−0.025t).
Therefore,
R(t) = e^(−0.015t) for 0 ≤ t ≤ 200, and R(t) = e^(2−0.025t) for t > 200, with t in the same time unit as the hazard rate.

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