(a) Solve the game whose payoff matrix is -1 & -2 & 8 7 & 5 & -1 6 & 0 & 12 (15 marks) (b) Use the penalty (Big M) method…

(a)

Let the row player be the maximizer. Row minima are −2, −1, 0, so maximin = 0. Column maxima are 7, 5, 12, so minimax = 5. No saddle point.

Column 2 dominates column 1 because −2 < −1, 5 < 7, 0 < 6. Row 3 dominates row 1 because 6 > −1, 0 > −2, 12 > 8. The reduced game is

R2: 5, −1 R3: 0, 12

Let row player use R2 with probability p and R3 with probability 1−p. Equating payoffs against column 2 and column 3:

5p = −p + 12(1−p) 5p = 12 − 13p 18p = 12 p = 2/3.

Thus row strategy = (0, 2/3, 1/3), and value v = 5(2/3) = 10/3.

Let column player use column 2 with probability q and column 3 with probability 1−q. Equating payoffs against rows 2 and 3:

5q − 1(1−q) = 0q + 12(1−q) 6q − 1 = 12 − 12q 18q = 13 q = 13/18.

Thus column strategy = (0, 13/18, 5/18). Value v = 10/3.

Final: optimal strategies (0, 2/3, 1/3) and (0, 13/18, 5/18); value = 10/3.

(b)

Use Big M method. For minimization, penalise artificial variables by +M. Add slack s₁, surplus s₂ and artificials a₁, a₂:

2x₁ + 4x₂ + s₁ = 12 2x₁ + 2x₂ + a₁ = 10 5x₁ + 2x₂ − s₂ + a₂ = 10

Minimize Z = 5x₁ + 3x₂ + M a₁ + M a₂, M very large.

Initial basis: s₁ = 12, a₁ = 10, a₂ = 10. With M large, x₁ enters; minimum ratio min(12/2, 10/2, 10/5) = 2, so a₂ leaves. After pivot:

(16/5)x₂ + s₁ + (2/5)s₂ − (2/5)a₂ = 8 (6/5)x₂ + (2/5)s₂ + a₁ − (2/5)a₂ = 6 x₁ + (2/5)x₂ − (1/5)s₂ + (1/5)a₂ = 2

Next x₂ enters; minimum ratio is 8/(16/5) = 5/2, so s₁ leaves. After pivot:

x₂ + (5/16)s₁ + (1/8)s₂ − (1/8)a₂ = 5/2 −(3/8)s₁ + (1/4)s₂ + a₁ − (1/4)a₂ = 3 x₁ − (1/8)s₁ − (1/4)s₂ + (1/4)a₂ = 1

Next s₂ enters; minimum ratio is 3/(1/4) = 12, so a₁ leaves. Final tableau:

x₂ + (1/2)s₁ − (1/2)a₁ = 1 s₂ − (3/2)s₁ + 4a₁ − a₂ = 12 x₁ − (1/2)s₁ + a₁ = 4

Optimality check: expressing Z in the non-basic variables gives Z = 23 + s₁ + (M − 7/2)a₁ + M a₂, so no non-basic variable can lower Z. At optimum, artificials and non-basic variables are zero: x₁ = 4, x₂ = 1, s₂ = 12. Check constraints:

2(4) + 4(1) = 12, 2(4) + 2(1) = 10, 5(4) + 2(1) = 22 ≥ 10.

Objective: Z = 5(4) + 3(1) = 23.

Final: x₁ = 4, x₂ = 1, minimum Z = 23.

(c)

(i) A nonconforming unit is an entire unit that fails to satisfy one or more specified requirements. A nonconformity is a single departure from a specified requirement; one nonconforming unit may contain several nonconformities.

For a control chart for nonconformities, assume the number of nonconformities in an inspection unit follows a Poisson distribution, nonconformities occur independently, the opportunity for nonconformities is constant, and the inspection unit is fixed or its sample size is known. If c is the number of nonconformities in a sample of n inspection units, then c ~ Poisson(nλ). The average number per unit is u = c/n, with E(u) = λ and Var(u) = λ/n. Estimating λ by ū, the three-sigma limits are:

UCL = ū + 3√(ū/n), CL = ū, LCL = ū − 3√(ū/n), taking LCL = 0 if negative.

If n = 1, this reduces to the c-chart: UCL = c̄ + 3√c̄, CL = c̄, LCL = c̄ − 3√c̄, with LCL = 0 if negative.

(ii) In a unit-by-unit sequential sampling plan by attributes, inspect units one by one and classify each as conforming or nonconforming. Let p₁ be AQL, p₂ be LTPD, with p₂ > p₁; let α and β be producer’s and consumer’s risks. After m units, let D_m be the cumulative number of nonconforming units. Define

den = ln[p₂(1−p₁)/(p₁(1−p₂))], s = ln[(1−p₁)/(1−p₂)] / den, h₁ = ln[(1−α)/β] / den, h₂ = ln[(1−β)/α] / den.

Accept if D_m ≤ s m − h₁. Reject if D_m ≥ s m + h₂. Continue sampling if s m − h₁ < D_m < s m + h₂.

Its unique feature is that the sample size is not fixed in advance; it is a random variable determined by the observations. It often requires fewer units on average than a comparable fixed-sample plan for the same α and β.

(iii) For a Weibull distribution with shape β and scale θ,

Mean = θ Γ(1 + 1/β).

Given β = 1/2, θ = 5000 hours:

Mean = 5000 Γ(1 + 2) = 5000 Γ(3) = 5000 × 2 = 10000 hours.

Survival function:

S(t) = exp[−(t/θ)^β].

For t = 20000 hours:

S(20000) = exp[−(20000/5000)^(1/2)] = exp[−4^(1/2)] = exp(−2) = e⁻² ≈ 0.1353.

Final: mean time to failure = 10000 hours; fraction surviving beyond 20000 hours = e⁻² ≈ 0.1353, i.e. 13.53%.