Fest Ticket Mix: Where the Relaxed Linear Program Goes Wrong
A student council is planning a two-day tech fest and must choose a ticket mix for two paid events. Workshop passes sell at ₹150 and admit holders to Hall A, which seats 400. Concert passes sell at ₹400 and admit holders to Hall B, which seats 250. Ushers and volunteers can handle at most 600 tickets in total across both events. Marketing costs ₹40 per workshop pass and ₹25 per concert pass, and the council has exactly ₹20,000 to spend on it. The faculty adviser asks for the mix that maximises revenue and suggests starting with the two 'obvious' constraints only - total tickets and marketing budget - leaving the halls out of the model. That relaxed problem returns a solution the council finds absurd. The student must explain why, redo the optimisation with every real constraint in place, give an integer plan, and advise on a late offer: a 350-seat hall for the concert at an extra ₹18,000.
1. Solve the relaxed linear program (total tickets and budget only) and explain why its optimum is not credible.
Read the worked answer
Let x = workshop passes, y = concert passes. Maximise R = 150x + 400y subject to x + y ≤ 600 and 40x + 25y ≤ 20,000, x, y ≥ 0.
Vertices:
- (0, 0): R = 0
- (500, 0) from the budget line: R = ₹75,000
- (0, 600) from the ticket line, budget 25 × 600 = ₹15,000 ≤ ₹20,000: R = ₹2,40,000
- Budget ∩ ticket line: 40x + 25(600 − x) = 20,000 → x = 333.3, y = 266.7: R = ₹1,56,667
The maximum sits at (0, 600): sell no workshop passes at all and 600 concert passes. That is the tell. A model whose optimum is to abandon an entire event has omitted something structural, and here it is capacity - Hall B seats 250, so y ≤ 250, and Hall A caps x ≤ 400. The relaxed solution also quietly assumes the volunteer bottleneck, not the halls, is what limits sales, which is backwards.
2. Re-solve with the hall capacities included and give a practical integer ticket plan.
Read the worked answer
Add x ≤ 400 and y ≤ 250. Feasible vertices:
- (0, 0): ₹0
- (400, 0): ₹60,000
- (0, 250): ₹1,00,000, marketing ₹6,250 ✓
- (400, 160) from budget: marketing 16,000 + 4,000 = ₹20,000 ✓, R = ₹1,24,000
- Budget ∩ y = 250: 40x = 20,000 − 6,250 = 13,750 → x = 343.75, R = 150(343.75) + 100,000 = ₹1,51,562.50
- Ticket line ∩ y = 250 gives x = 350, but marketing would be 14,000 + 6,250 = ₹20,250 > ₹20,000 - infeasible, which is why the budget line, not the volunteer cap, binds here.
Integer plan: 343 workshop passes and 250 concert passes, revenue ₹51,450 + ₹1,00,000 = ₹1,51,450, marketing ₹19,970. Rounding down by one pass keeps marketing inside the cap; rounding x up to 344 breaches it by ₹30.
3. Evaluate the 350-seat hall offer at ₹18,000 and state the attendance threshold below which it loses money.
Read the worked answer
With y ≤ 350, the binding pair becomes the volunteer cap and the new hall. Best vertex: x = 250, y = 350 (total 600, marketing 10,000 + 8,750 = ₹18,750 ✓), revenue = 37,500 + 1,40,000 = ₹1,77,500.
Incremental revenue over the 250-seat plan = 1,77,500 − 1,51,450 = ₹26,050. Net of the ₹18,000 hire = ₹8,050 - positive, but thin, roughly a 4.5% improvement on total revenue.
Break-even: 18,000/400 = 45 concert passes. If fewer than about 305 of the 350 seats sell, the hire destroys value. Two further cautions: the volunteer cap of 600 is now binding, so any extra concert seats must be paid for by giving up workshop passes, and those give up ₹150 each. Recommendation: accept the hire only against a presale or sponsor commitment covering at least 305 concert passes; otherwise stay in Hall B.