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

(a) Statistical Quality Control (SQC) is the use of statistical methods to monitor, control and improve the quality of products and processes. It treats quality as a measurable characteristic affected by random and assignable causes, and uses sampling inspection, acceptance sampling, control charts, process capability studies and design of experiments to decide whether a process is stable, capable and improving. SQC is not only final inspection; it is process control. It converts quality decisions from subjective judgment into evidence-based decisions, and it is especially useful when production is continuous, costly, or customer-sensitive. Control charts plot sample statistics such as mean, range or standard deviation against time and use upper and lower control limits. If points stay within limits and show no non-random pattern, the process is in statistical control. If a point exceeds a limit or a trend, cycle or run appears, a special cause is investigated. Its need in industry is to reduce defects, scrap, rework and warranty costs; to detect abnormal shifts before large batches are spoiled; to give objective evidence of process stability; to ensure consistent customer satisfaction; and to support continuous improvement and certification. Its utility is that it separates normal variation from abnormal variation, focuses management on process causes rather than only final inspection, and gives a common quantitative language for quality decisions. Causes of variation are of two types. Common-cause variation is inherent and random, arising from the system itself: small differences in material, machine wear, temperature, humidity, operator technique, measurement error and method. Special-cause or assignable-cause variation is due to identifiable disturbances: tool breakage, wrong material, power fluctuation, operator mistake, maintenance fault or a change in procedure. The 6M framework, Man, Machine, Material, Method, Measurement and Mother Nature, is often used to classify these sources. SQC aims to remove special causes and reduce common causes.

(b) Let T be the lifetime and Z(t)=t/(t+1) be the failure rate for t≥0. The formula is valid for t≥0 and assumes Z(t) is the hazard rate of a non-negative lifetime. By definition, Z(t) = -d/dt ln R(t), where R(t)=P(T>t). Hence ln R(t) = -∫ from 0 to t Z(u) du. Now ∫ from 0 to t u/(u+1) du = ∫ from 0 to t (1 - 1/(u+1)) du = [u - ln(u+1)] from 0 to t = t - ln(t+1). Therefore R(t) = exp[-t + ln(t+1)] = (t+1)exp(-t), t≥0. This satisfies R(0)=1 and R(t)→0. The density is f(t)=Z(t)R(t)=t exp(-t). MTTF = E[T] = ∫ from 0 to ∞ R(t) dt = ∫ from 0 to ∞ (t+1)exp(-t) dt = ∫ from 0 to ∞ t exp(-t) dt + ∫ from 0 to ∞ exp(-t) dt = 1 + 1 = 2 time units. For the conditional survivor function, let s≥0 be current age and t≥0 be additional time. R(t | s) = P(T > s+t | T > s) = R(s+t)/R(s) = [(t+s+1)exp(-(t+s))]/[(s+1)exp(-s)] = ((t+s+1)/(s+1))exp(-t). MRL at age s is m(s)=E[T-s | T>s] = ∫ from 0 to ∞ R(u | s) du = ∫ from 0 to ∞ [(u+s+1)/(s+1)]exp(-u) du = (1/(s+1))[∫ from 0 to ∞ u exp(-u) du + (s+1)∫ from 0 to ∞ exp(-u) du] = (1/(s+1))[1 + (s+1)] = (s+2)/(s+1) time units. At s=0, m(0)=2, agreeing with MTTF.

(c) The equality x₃ - x₁ - x₂ = 0 gives x₃ = x₁ + x₂. Since x₁, x₂ ≥ 0, x₃ ≥ 0 automatically. Substitute in the objective: 4x₁ + 5x₂ + 6x₃ = 4x₁ + 5x₂ + 6(x₁+x₂) = 10x₁ + 11x₂. The problem becomes minimize C = 10x₁ + 11x₂ subject to x₁ + x₂ ≥ 11, x₁ - x₂ ≤ 5, x₁ ≥ 0, x₂ ≥ 0. The graphical method is applicable because the equality constraint removes one variable, so the original three-dimensional feasible set is represented in the x₁-x₂ plane. The constraint 7x₁ + 12x₂ ≥ 35 is redundant because x₁ + x₂ ≥ 11 and x₂ ≥ 0 imply 7x₁ + 12x₂ = 7(x₁+x₂) + 5x₂ ≥ 77 > 35. Graphically, draw L₁: x₁ + x₂ = 11 with intercepts (11,0) and (0,11); the feasible side is above this line. Draw L₂: x₁ - x₂ = 5, or x₂ = x₁ - 5, with intercepts (5,0) and (0,-5); the feasible side is x₂ ≥ x₁ - 5, i.e. above this line in the x₁-x₂ plane. The intersection of L₁ and L₂ is found from x₁ + x₂ = 11 and x₁ - x₂ = 5, giving 2x₁ = 16, so x₁ = 8 and x₂ = 3. The feasible region is an unbounded convex polygonal region, with boundary segment from (0,11) to (8,3), a vertical ray x₁ = 0, x₂ ≥ 11, and a ray x₂ = x₁ - 5, x₁ ≥ 8. Because the objective is linear, if a finite minimum exists it must occur at an extreme point or along an edge; checking the extreme points and the directions of the unbounded edges is sufficient. Objective contours are 10x₁ + 11x₂ = C, straight lines of slope -10/11. Decreasing C moves the contour toward the origin; the first point of contact with the feasible region is the corner (8,3). Check corner values: at (0,11), C = 121; at (8,3), C = 113. Along the ray x₁ = 0, C = 11x₂, increasing for x₂ ≥ 11. Along the ray x₂ = x₁ - 5, C = 10x₁ + 11(x₁-5) = 21x₁ - 55, increasing for x₁ ≥ 8. Hence the minimum occurs at x₁ = 8, x₂ = 3, and x₃ = 8 + 3 = 11. The minimum value is 4(8) + 5(3) + 6(11) = 32 + 15 + 66 = 113 cost units.

(d) In a two-person zero-sum game, let player A have m pure strategies and player B have n pure strategies. The payoff matrix to A is written in general notation as A = (a(i,j)), i = 1,...,m, j = 1,...,n, where a(i,j) is the amount received by A when A chooses row i and B chooses column j. A maximizes payoff and B minimizes it. The payoff matrix is written from A's perspective; B's payoff would be the negative of this matrix. For the coin game, take A's rows as H and T, and B's columns as H and T. If HH or TT occurs, B pays ₹7 to A, so A's payoff is +7. If HT or TH occurs, A pays ₹3 to B, so A's payoff is -3. Thus A's payoff matrix is A = [7, -3; -3, 7]. The Max Min criterion for A is: for each row, find the minimum payoff, the worst outcome if B chooses optimally against that row, and then choose the row with the largest of these row minima. Row H has minimum min(7,-3) = -3. Row T has minimum min(-3,7) = -3. Therefore A's maximin value is max(-3,-3) = -3 ₹. For B, using the same A-payoff matrix, the minimax value is min(max(7,-3), max(-3,7)) = min(7,7) = 7 ₹. A saddle point is an entry a(i,j) that is simultaneously the minimum of its row and the maximum of its column; equivalently, the maximin value equals the minimax value, and that common value is the value of the game. If a saddle point existed, both players would have optimal pure strategies and no randomization would be needed. Here maximin -3 is not equal to minimax 7, so there is no saddle point in pure strategies.

(e) Let P(i,j) be the probability of moving from state i to state j in one step, with rows as current state and columns as next state. From the transition diagram, the one-step transition matrix is P = [3/5, 1/5, 1/5; 2/5, 0, 3/5; 0, 4/5, 1/5], i.e. [0.6, 0.2, 0.2; 0.4, 0.0, 0.6; 0.0, 0.8, 0.2]. The transition matrix is row-stochastic: each row sums to 1, so it is a valid Markov transition matrix. The calculation uses the Markov property and time-homogeneous transition probabilities, so the two-step probability is the (1,3) entry of P squared. The required probability is Pᵣ[X₂ = 3 | X₀ = 1] = (P²)(1,3). Compute the (1,3) entry of P²: (P²)(1,3) = (3/5)(1/5) + (1/5)(3/5) + (1/5)(1/5) = 3/25 + 3/25 + 1/25 = 7/25. This is also the sum of the two-step paths 1→1→3, 1→2→3, and 1→3→3.