(a) What do you understand by Statistical Quality Control (SQC)? Discuss briefly its need and utility in Industry. Discuss the…

(a) Statistical quality control: meaning, need, utility and causes of variation

Statistical Quality Control (SQC) is the use of statistical methods, mainly control charts and acceptance sampling, to monitor and improve the quality of products and processes. Shewhart introduced control charts in the 1920s. Need: it detects abnormal shifts before large batches are spoiled, reduces defects, scrap, rework and warranty cost, and gives objective evidence that a process is stable and meets specifications. Utility: it separates normal variation from abnormal variation, so management acts on the process rather than only inspecting the final product; it also supports supplier acceptance decisions and continuous improvement. Causes of variation: Chance (common) causes are many small, unavoidable influences inherent in the process, such as slight differences in material, machine vibration, temperature and operator technique. The pattern is stable and predictable. Assignable (special) causes are identifiable disturbances such as tool breakage, wrong material, faulty setting or operator error. They produce points outside control limits or non-random patterns and should be found and removed. The aim of SQC is to remove assignable causes and keep the process within chance-cause variation.

(b) Survivor function, MTTF, conditional survival and mean residual life for Z(t) = t/(t + 1)

Step 1: Since Z(t) = −d ln R(t)/dt, ln R(t) = −∫₀ᵗ u/(u + 1) du = −[u − ln(u + 1)]₀ᵗ = −t + ln(t + 1). Step 2: Hence R(t) = (t + 1)e^(−t), t ≥ 0. Step 3: MTTF = ∫₀^∞ R(t) dt = ∫₀^∞ t e^(−t) dt + ∫₀^∞ e^(−t) dt = 1 + 1 = 2. Step 4: Conditional survival at age s: R(t | s) = R(s + t)/R(s) = [(s + t + 1)/(s + 1)] e^(−t). Step 5: Mean residual life: m(s) = ∫₀^∞ R(t | s) dt = [1/(s + 1)][1 + (s + 1)] = (s + 2)/(s + 1). At s = 0, m(0) = 2, which equals the MTTF.

(c) Graphical solution of the linear programme

Step 1: The equality gives x₃ = x₁ + x₂, which is automatically non-negative. The objective becomes 4x₁ + 5x₂ + 6(x₁ + x₂) = 10x₁ + 11x₂. Step 2: The problem is to minimise C = 10x₁ + 11x₂ subject to x₁ + x₂ ≥ 11, x₁ − x₂ ≤ 5, x₁, x₂ ≥ 0. Step 3: The constraint 7x₁ + 12x₂ ≥ 35 is redundant, since x₁ + x₂ ≥ 11 gives 7x₁ + 12x₂ = 7(x₁ + x₂) + 5x₂ ≥ 77. Step 4: Plot x₁ + x₂ = 11 (through (11, 0) and (0, 11), feasible side above it) and x₁ − x₂ = 5 (through (5, 0), feasible side x₂ ≥ x₁ − 5). They meet at (8, 3). The feasible region is unbounded with corners (0, 11) and (8, 3). Step 5: C = 121 at (0, 11) and C = 113 at (8, 3). Along both unbounded edges C increases (11x₂ along x₁ = 0, and 21x₁ − 55 along x₂ = x₁ − 5). The minimum is at x₁ = 8, x₂ = 3, x₃ = 11, with minimum value 4(8) + 5(3) + 6(11) = 113.

(d) Payoff matrix, maximin criterion and saddle point

General notation: if A has m strategies and B has n strategies, the payoff matrix is (aᵢⱼ), i = 1..m, j = 1..n, where aᵢⱼ is what A receives (and B pays) when A plays i and B plays j. Coin game: A's payoff is +7 for HH or TT and −3 for HT or TH, so with rows H, T and columns H, T the matrix is [[7, −3], [−3, 7]]. Maximin criterion: A finds the minimum of each row (the worst outcome of that strategy) and chooses the row whose minimum is largest. Here both row minima are −3, so A's maximin value is −3. B's minimax value is the least of the column maxima, which is 7. Saddle point: an entry that is the minimum of its row and the maximum of its column, which exists exactly when maximin = minimax; that common value is the value of the game. Here −3 ≠ 7, so there is no saddle point and the players must mix strategies; with probabilities 1/2 each, the value of the game is (49 − 9)/(7 + 7 + 3 + 3) = 2.

(e) Transition matrix and two-step probability

Step 1: Read the arrows from the figure: from state 1, stay with 0·6, go to 2 with 0·2 and to 3 with 0·2; from state 2, go to 1 with 0·4 and to 3 with 0·6; from state 3, go to 2 with 0·8 and stay with 0·2. Step 2: P = [[0·6, 0·2, 0·2], [0·4, 0, 0·6], [0, 0·8, 0·2]]; each row sums to 1. Step 3: P(X₂ = 3 | X₀ = 1) = (P²)₁₃ = p₁₁p₁₃ + p₁₂p₂₃ + p₁₃p₃₃ = 0·6 × 0·2 + 0·2 × 0·6 + 0·2 × 0·2 = 0·12 + 0·12 + 0·04 = 0·28 = 7/25.