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Step 1 — Convert to canonical form. The primal is a Max problem, so every constraint must become ≤.

Three printed constraints, but the equality splits into two — so the dual has four variables. Counting them before you start is the single best guard against losing this mark.

Step 2 — Set up the dual. Max primal → Min dual. The primal's RHS values (8, −10, 6, −6) become the dual's objective coefficients:

Step 3 — One dual constraint per primal variable, reading down each column of the canonical form. Max primal ⟹ dual constraints are ≥, with the primal objective coefficients as RHS.

Final answer — write exactly this:

✓ Matches the synoptic key exactly Key p. 1.

Step 1 — Canonical form. This one is a Min problem, so every constraint must become ≥ — the opposite direction to the Final's version.

Step 2 — Dual objective. Min primal → Max dual, RHS values become the coefficients:

Step 3 — Dual constraints, one per primal variable, direction ≤:

Identify the type first. There is no supply or demand row — every worker takes exactly one task. This is an assignment problem, m = 3 workers, n = 3 tasks.

✓ Matches the key: "Yes Problem is balanced as Number of rows = number of column; m+n = 6; mn = 9" Key p. 1.

Identify the type. A supply column and a demand row are given ⟹ transportation problem.

Balance test — always compute both totals explicitly:

✓ Matches the key Key p. 1, which also counts m + n and m × n on the original 3 × 3 matrix rather than the balanced 4 × 3 one.

Step 0 — build the resource table. Do this before writing anything else:

The question says "Formulate" — so stop here. No graph, no simplex.

✓ Matches the key exactly Key p. 3, including the same 1 / 1 / 2 / 1 mark split.

(1) Which constraints are binding? Give the evidence.

All three are binding. The test has two halves and you should quote both:

Zero slack means the resource is fully consumed; a non-zero shadow price means more of it would still be worth something. Both conditions hold for all three, so every resource is a genuine bottleneck.

(2) Interpret the shadow price 14.69 for Constraint 1 in economic terms.

(3) If the RHS of Constraint 2 increases by 10 units, what is the change in objective value? Is it valid?

Because the change is within range, the shadow price is still the right multiplier and the same variables remain basic.

(4) All variables have reduced cost = 0. What does that imply?

All three variables are basic — they are in the optimal solution at positive levels (X₁ = 36.25, X₂ = 35, X₃ = 21.25). A reduced cost of zero means no improvement to that variable's objective coefficient is needed to justify producing it; it is already worth producing.

(5) If the coefficient of X₂ rises from ₹45 to ₹55, does the basis stay optimal?

But the objective value does change. Add this line to separate a good answer from a full one:

(1) Which constraints are binding?

Material and Labour are binding — slack = 0 and shadow price > 0. Machine Hours is not: it has 40 units of unused capacity and a shadow price of 0, so it is not limiting the solution at all.

(2) Interpret the shadow price of 22 for Material.

(3) If Machine Hours RHS increases by 20 units, will profit change?

No — profit is unchanged. Two reasons, both worth stating: the shadow price is 0, so ΔZ = 0 × 20 = 0; and there are already 40 unused machine hours. Adding more of a resource you are not fully using cannot help. Machine Hours is a non-binding constraint.

(4) The reduced cost of X₃ is 4 — interpret it.

X₃ has value 0: it is not produced in the optimal solution. The reduced cost of 4 can be stated two equivalent ways, and either earns the mark:

• Forcing one unit of X₃ into the solution would reduce total profit by ₹4.
• Equivalently, X₃'s profit coefficient would have to rise by ₹4 — from ₹30 to ₹34 — before it becomes worth producing.

Notice the internal check: X₃'s upper bound in the table is 34 = 30 + 4. The report is consistent with itself, and quoting that agreement is a strong touch.

(5) If the profit coefficient of X₁ increases by ₹20, does the solution stay optimal?

Quantities stay X₁ = 30, X₂ = 40, X₃ = 0, but profit rises:

Step 1 — spot the variable type. "Which projects should be undertaken" ⟹ each project is a yes/no decision ⟹ binary variables, not quantities.

Step 2 — objective function. Maximise total expected profit:

Step 3 — resource constraints. One per quarter, read straight down the table columns:

Step 4 — the logical constraint. This is the sentence buried at the end of the stem: "Projects C and E cannot be undertaken together."

Step 5 — the binary condition. Without this line the model is an LP, not an ILP:

The synoptic key prints its Q4A answer as an image with no extractable text Key p. 3, but its published mark split is identical to the Final's: variables 1, objective 1, constraints 2, binary condition 1.

Step 1 — work out when each outbound flight lands. Flight time is 2 hours 15 minutes.

Step 2 — build the idle-time (cost) matrix. Idle time = return departure − arrival, rolling to the next day whenever the return leaves before the aircraft lands.

Sample working: flight 102 lands at 15:15; return 203 leaves at 19:45; idle = 19:45 − 15:15 = 4 h 30 m = 4.5. Flight 104 lands 23:15; return 202 leaves 14:30 the next day; idle = 15 h 15 m = 15.25.

Step 3 — balance the problem. There are 5 outbound flights but only 4 returns — the stem tells you flight 105 has no direct return option. So:

Step 4 — write the assignment LPP.

Step 1 — arrival times. Flight time 2.5 hours.

Step 2 — idle time, net of the 30-minute turnaround. This is what makes this version harder than the Final's. A pairing is only feasible if the gap is more than the 30-minute turnaround; otherwise the aircraft must wait until the same flight the next day. The cost recorded is the idle time beyond the required turnaround, i.e. gap − 0.5.

Sample working: 202 lands 14:00, return 303 departs 18:00 ⟹ gap 4 h, minus the 0.5 h turnaround = 3.5. Where the gap is only 0.5 h — 201→301 and 203→303 and 204→304 — the turnaround exactly consumes it, so the key rolls the aircraft to the same service next day, giving 24. ✓ Every cell above matches the key Key p. 5.

Step 3 — the formulation. Here it is 4 × 4, already balanced — no dummy needed.

(1) The special case is DEGENERACY.

(2) Justification — three observations, each pointing at a specific part of the tableau:

(3) One important implication for the simplex method:

✓ Matches the key's three-part structure — identification, justification, implication — almost word for word Key p. 7.

Step 1 — spot the twist. This is not a plain cost-minimisation. You are given plant costs, transportation costs and selling prices — so the cell value must be a profit, and the objective becomes Max.

Step 2 — build the profit matrix. Compute cell by cell:

Step 3 — lay it out in transportation format.

✓ This profit matrix matches the synoptic key cell for cell Key p. 8.

Step 4 — balance check and the counts the question asks for.

Step 5 — the LPP.

Seven constraints — exactly the m + n you computed. Because supply and demand are perfectly balanced, the supply constraints may be written as = rather than ≤.