(a) Use penalty method to solve the following linear programming problem : Maximize Z = x₁ + 2x₂ + 3x₃ - x₄ subject to the…

(a) Penalty (Big-M) method

Constraint 3 already has x₄ as a unit column; constraints 1 and 2 need artificial variables A₁ and A₂ with penalty −M (M very large): Maximise Z = x₁ + 2x₂ + 3x₃ − x₄ − MA₁ − MA₂ subject to x₁ + 2x₂ + 3x₃ + A₁ = 15 2x₁ + x₂ + 5x₃ + A₂ = 20 x₁ + 2x₂ + x₃ + x₄ = 10 The initial basis (A₁, A₂, x₄) = (15, 20, 10) gives Z = −35M − 10, and the z_j − c_j row for (x₁, x₂, x₃) is (−3M − 2, −3M − 4, −8M − 4).

1. x₃ enters (most negative, −8M − 4). The ratios 15/3, 20/5, 10/1 are 5, 4, 10, so A₂ leaves and is dropped. New basis (A₁, x₃, x₄) = (3, 4, 6), Z = 6 − 3M.
2. z_j − c_j for (x₁, x₂) is (M/5 − 2/5, −7M/5 − 16/5), so x₂ enters. The ratios 3/(7/5), 4/(1/5), 6/(9/5) are 15/7, 20, 10/3, so A₁ leaves. New basis (x₂, x₃, x₄) = (15/7, 25/7, 15/7), Z = 90/7.
3. z_j − c_j for x₁ is −6/7, so x₁ enters. The ratios are (25/7)/(3/7) = 25/3 and (15/7)/(6/7) = 5/2 (the x₂ row has −1/7, so no ratio), so x₄ leaves. New basis (x₂, x₃, x₁) = (5/2, 5/2, 5/2).
4. Now z_j − c_j for x₄ is 1 ≥ 0 and no artificial variable is in the basis, so the solution is optimal.

Optimal solution: x₁ = x₂ = x₃ = 5/2, x₄ = 0, maximum Z = 5/2 + 5 + 15/2 = 15. (Check: constraint 1 gives Z = 15 − x₄ ≤ 15.)

(b) Crew pairing with minimum layover

Rule: pair Delhi–Jaipur flight i with Jaipur–Delhi flight j. A Delhi-based crew flies i and waits at Jaipur for j: layover = dep(j) − arr(i). A Jaipur-based crew flies j and waits at Delhi for i: layover = dep(i) − arr(j). A value under 5 hours is increased by 24 hours (daily service). The cost of the pair is the smaller layover, and the base is the city that gives it.

Layovers in hours with base (D = Delhi, J = Jaipur), for flights 101, 102, 103, 104:

• Flight 1: 21.75 J, 21.25 J, 17.75 J, 9.5 D
• Flight 2: 22.75 J, 22.25 J, 18.75 J, 8.5 D
• Flight 3: 17.5 D, 18 D, 21.5 D, 18.75 J
• Flight 4: 9.25 J, 8.75 J, 5.25 J, 22 D

Hungarian method: subtracting the row minima (9.5, 8.5, 17.5, 5.25) and then 0.5 from column 102 leaves zeros at (1,104), (2,104), (3,101), (3,102), (4,103). Three lines (column 104, rows 3 and 4) cover them, so subtract the uncovered minimum 8.25 from the uncovered cells and add it at the line intersections; three lines again suffice, and the uncovered minimum 3 is used likewise. The final matrix has four independent zeros.

Optimal pairings (total layover 52.5 hours):

1. Flight 1 with 103: Jaipur-based, 17.75 h
2. Flight 2 with 104: Delhi-based, 8.5 h
3. Flight 3 with 101: Delhi-based, 17.5 h
4. Flight 4 with 102: Jaipur-based, 8.75 h An alternative with the same total uses 1–102 (Jaipur, 21.25 h) and 4–103 (Jaipur, 5.25 h).

(c) Sequential sampling plans and the plan for the given risks

In a sequential sampling plan the sample size is not fixed in advance: items are inspected one at a time (or in small groups) and after each inspection the cumulative number of defectives dₙ among n items is compared with an acceptance line and a rejection line; the lot is accepted, rejected, or sampling continues. It is Wald's sequential probability ratio test of p₁ against p₂ with producer's risk α and consumer's risk β, and for the same risks it needs fewer items on average than single or double sampling.

Plan: k = log[p₂(1 − p₁)/(p₁(1 − p₂))] = log 6 + log(0.99/0.94) = 0.7782 + 0.0225 = 0.8007 h₁ = log[(1 − α)/β]/k = log 9.5/0.8007 = 0.9777/0.8007 = 1.221 h₂ = log[(1 − β)/α]/k = log 18/0.8007 = 1.2553/0.8007 = 1.568 s = log[(1 − p₁)/(1 − p₂)]/k = 0.0225/0.8007 = 0.0281 Acceptance line aₙ = −1.221 + 0.0281n; rejection line rₙ = 1.568 + 0.0281n. Rule: after n items accept if dₙ ≤ aₙ, reject if dₙ ≥ rₙ, otherwise inspect one more item. In practice, accept after 44 items with no defective or after 80 with at most one; reject as soon as 2 defectives appear within 15 items, 3 within 50, or 4 within 86.