Bus Fleet Arithmetic Under a Workshop Constraint
A city transport undertaking runs one trunk route, 14 km from end to end. Buses average 28 km/h including all stops, and at each terminus a bus must stand for 5 minutes before starting the return trip. The sanctioned peak-hour plan is a 10-minute headway in both directions; off-peak the undertaking runs a 20-minute headway. For the next month the workshop has grounded 2 buses of the peak-hour requirement for body repairs, and no hired vehicles are available. The traffic cell's counted peak-hour peak-direction demand is 480 passengers, a standard bus seats and stands 90, and the fare cell has asked whether running the surviving buses harder - cutting terminus stand time to 2 minutes - would be enough to see the month through.
1. Compute the round-trip cycle time and the minimum number of buses needed for the 10-minute peak headway and the 20-minute off-peak headway.
Read the worked answer
Round trip = 14 × 2 = 28 km. At 28 km/h that is 1 hour = 60 minutes of running.
Standing time = 5 min at each terminus = 10 minutes. Cycle time = 60 + 10 = 70 minutes.
Buses needed = cycle ÷ headway:
- Peak: 70/10 = 7 buses.
- Off-peak: 70/20 = 3.5 → 4 buses (3 buses would give 70/3 = 23.3 min, not 20).
The fleet is set by the peak, so the sanctioned fleet is 7, with 3 units surplus in the off-peak period and available for maintenance - which is precisely why grounding 2 leaves the peak, not the off-peak, in trouble.
Answer: 70-minute cycle; 7 buses at peak, 4 off-peak.
2. With 5 buses available, what headway can actually be held, and can the route still carry 480 passengers per hour at 90 per bus?
Read the worked answer
Headway = cycle ÷ buses = 70/5 = 14 minutes.
Departures per hour in one direction = 60/14 = 4.29. Capacity = 4.29 × 90 = 386 passengers per hour, against a demand of 480.
Shortfall = 480 − 386 = 94 passengers per hour, about 20 per cent. Even rounding generously to 5 departures, 5 × 90 = 450, still below 480.
Answer: a 14-minute headway can be held, and it fails. The residual passengers do not vanish - they force loads above 90 per bus, which lengthens dwell time at stops, stretches the actual cycle beyond 70 minutes, and pushes the achieved headway past 14 minutes. The failure compounds during the peak rather than settling.
3. Does cutting terminus stand time to 2 minutes solve the month? State what is actually required.
Read the worked answer
New cycle = 60 + (2 × 2) = 64 minutes. With 5 buses: headway = 64/5 = 12.8 min; departures/hour = 60/12.8 = 4.69; capacity = 4.69 × 90 = 422 < 480.
So no. The saving buys roughly 36 seats per hour, about a third of the gap.
What is required: departures ≥ 480/90 = 5.33 → at least 6 departures per hour → headway ≤ 10 minutes → buses ≥ 64/10 = 6.4 → 7 buses, even with the shortened stand time. With 7 buses on a 64-minute cycle the headway becomes 9.1 minutes, giving 6.6 departures and 594 seats - a load factor near 81 per cent, which is a defensible design point rather than a crush load.
Answer: the stand-time cut is a useful capacity margin, not a substitute for vehicles. The undertaking must restore the fleet to 7; if that is impossible, the honest options are hired cover or an explicit temporary service reduction, not a 12.8-minute headway advertised as the peak plan.