Shuttle Headway, Wait Time, and Induced Demand
A campus shuttle runs one bus on a closed loop. Driving time is 12 minutes per loop regardless of load. Boarding and alighting take 0.4 minutes per passenger, and the loop currently carries an average of 30 passengers per trip. The bus holds 40 passengers. Students arrive at stops randomly through the day, so the average wait is modelled as half the headway, where headway is the time between consecutive bus arrivals at a stop. The transport office proposes adding a second identical bus and claims this will halve the average wait. A student volunteer points out that shorter waits may attract more riders; the office's demand study estimates a 15% increase in passengers per loop when the average wait is halved. The office has budget for one additional bus and one driver for 8 hours, and it wants a defensible estimate before committing.
1. Calculate the current loop time and average wait, and the average wait with a second bus if demand does not change.
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Loop time (T=12+0.4(30)=12+12=24) min. With one bus, headway (h=T=24) min, so average wait (=h/2=12) min. With two identical buses evenly spaced, headway (=T/2=12) min, average wait 6 min, ignoring any change in demand. So the office's claim is correct only under the fixed-demand assumption.
2. Incorporate the 15% ridership increase. What is the new average wait, and does the second bus still halve the wait?
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With 15% more passengers, (30(1.15)=34.5) per loop. New loop time (T=12+0.4(34.5)=12+13.8=25.8) min. Two buses give headway (25.8/2=12.9) min and average wait (6.45) min. Compared with 12 min, the reduction is ((12-6.45)/12=0.4625), or 46.25%, not 50%. The claim “halves the wait” is therefore slightly optimistic. Capacity is still adequate on average (34.5<40), but the margin is only 5.5 passengers, so peak loads could exceed capacity.
3. Identify the main limitations of this model and suggest a sensitivity check the office could present alongside the 6.45-minute estimate.
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The model assumes random passenger arrivals (wait = half headway), constant 0.4 min per passenger, no traffic or dwell variability, evenly spaced buses, and a fixed 15% demand response. These are strong. Class schedules create bursts, not random arrivals; boarding time may rise with crowding; two buses can bunch, making actual headway uneven; and ridership response may be larger or smaller. A sensitivity check: a 30% ridership increase gives 39 passengers, (T=27.6) min, headway 13.8, wait 6.9 min-still below 12 but close to capacity. The office should present the 6.45 min estimate as a central case with a range, and monitor peak loads before treating the second bus as sufficient.