The Split Bus Fleet
A school hires buses for a three-day camp. Each large bus seats 54 students and costs $420 per day. Each small bus seats 22 students and costs $190 per day. Exactly 310 students attend, and every student must have a seat. The school wants at least 8 spare seats across the whole fleet in case of last-minute joiners, but no more than 20 spare seats, because unused capacity is seen as wasteful. Buses cannot be shared between days, and the same fleet must be used on all three days. The camp budget for transport is $2,400 in total for the three days. The organiser also insists that the number of large buses must be greater than the number of small buses.
1. Find a fleet mix that satisfies every constraint, and show why it is the only one.
Read the worked answer
Let L = large buses and S = small buses.
Seating: 54L + 22S must be between 310 + 8 = 318 and 310 + 20 = 330.
Cost per day: 420L + 190S, and over three days this must be ≤ 2400, so per day ≤ 800.
Ordering: L > S.
Test small values of L, since cost caps L at 800/420 ≈ 1.9, so L can only be 1.
- L = 1: seats 54 + 22S. For 318 ≤ 54 + 22S ≤ 330, we need 264 ≤ 22S ≤ 276, so 12 ≤ S ≤ 12.5. Only S = 12 works for the lower bound, but then seats = 54 + 264 = 318 (valid) and cost = 420 + 12×190 = 420 + 2280 = 2700 per day - far over 800.
So L = 1 fails cost. L must be at least 2 to be greater than S only if S ≤ 1, but L = 2 with S ≤ 1 gives at most 108 + 22 = 130 seats, below 318. No mix satisfies all constraints.
Conclusion: the constraints are mutually inconsistent. The cost cap of $800 per day cannot seat 318 students at these prices (cheapest per seat is the large bus at about $7.78, so 318 seats cost at least about $2,470 per day). The correct judgement is to report that no feasible fleet exists and that the budget or the spare-seat rule must be relaxed.
2. The budget is raised. What is the cheapest three-day fleet that meets the seating and ordering rules?
Read the worked answer
Ignore the old $800/day cap. Minimise 420L + 190S subject to 318 ≤ 54L + 22S ≤ 330 and L > S.
Cost per seat: large = 420/54 ≈ $7.78; small = 190/22 ≈ $8.64. Large buses are cheaper per seat, so favour large buses.
Try L = 6: 324 seats, need 318–330, so 324 is in range with S = 0. But L > S holds (6 > 0). Cost = 6 × 420 = $2,520 per day, $7,560 for three days.
Try L = 5: 270 seats, need 48–60 more from small buses: 22S in 48–60 gives S = 3 (66 seats, total 336 - too many). S = 2 gives 44, total 314 - too few. So L = 5 fails.
Try L = 4: 216 seats, need 102–114: 22S in that range gives S = 5 (110, total 326 - valid, but L > S fails since 4 < 5).
So the only valid cheap option is L = 6, S = 0: $2,520 per day, $7,560 total. Check L = 7: 378 seats exceeds 330. Therefore the minimum is 6 large buses, no small buses, at $7,560 for three days.