Four Errands, One Long Gap or Two Outings

Declare the windows before choosing the route

Call the stops A, B, C and D. A is a small collection open from 09:00 to 09:30 and needs five minutes. B is another collection available from 10:00 to 10:20 and also needs five minutes. C can be done from 09:00 to 11:00 and needs ten minutes. D is available from 10:30 to 11:00 and needs ten minutes. In this model, the whole service must fit inside its window.

The one-loop travel assumptions are home to A ten minutes, A to B fifteen, B to C ten, C to D fifteen and D home ten. They total sixty minutes. Services total thirty minutes. There are no queues, delays, carrying constraints or additional stops. Those omissions are boundaries of the invented calculation, not promises that errands really work that way.

One feasible loop contains thirty minutes of waiting

Leave home at 09:00, arrive at A at 09:10 and finish at 09:15. Travel fifteen minutes to B, arriving at 09:30. Because B does not start until 10:00, wait thirty minutes. Its service finishes at 10:05. Reach C at 10:15, finish at 10:25, then reach D at 10:40 and finish at 10:50. Ten minutes home gives an 11:00 return.

The two-hour absence is sixty minutes traveling, thirty in service and thirty waiting. All services fit the declared windows. Calling the route feasible therefore says something narrower than calling it preferable. A household wanting one departure and no return in between could value it; another might find the half-hour before B inconvenient.

A later start does not automatically fix the wait. If departure moves to 09:25 with the same first leg, arrival at A becomes 09:35, after its window closes. That is a specific missed-window case, not a small variation that can be hidden by moving the later services. No alternate opening permission is assumed.

Two outings use two additional travel assumptions

For the split option, declare A to C as fifteen minutes and C home as ten. Leave home at 09:00, complete A from 09:10 to 09:15, reach C at 09:30, finish at 09:40 and return home at 09:50. This outing lasts fifty minutes: thirty-five traveling and fifteen in service.

For the second outing, declare home to B as ten minutes and B to D as fifteen. Depart at 10:05, complete B from 10:15 to 10:20, reach D at 10:35, finish at 10:45 and return home at 10:55. It also lasts fifty minutes, again thirty-five traveling and fifteen in service. The B service ends exactly at the assumed closing time, leaving no margin in this fictional plan.

The split option therefore contains seventy minutes of travel and thirty of service, with no modeled waiting at a stop. Its total time away is one hundred minutes. The calendar from the first departure to final return spans one hundred and fifteen minutes because the person spends fifteen minutes at home between 09:50 and 10:05. Calendar span and absence are different quantities.

Let one stated preference decide the comparison

Imagine a household that values a short return home more than avoiding a second departure and can use that fifteen-minute interval. Under the declared timings, I would investigate the split option first. It adds ten travel minutes but removes thirty waiting minutes, reducing time away by twenty. The choice follows from that preference and these assumptions, not a universal instruction to split errands.

TfL’s journey planner distinguishes leaving and arriving inputs, a useful reminder to state which time a travel question concerns. No journey was generated for this exercise. Before applying any plan to real errands, replace the invented windows and durations with the actual arrangements and leave whatever margin the circumstances require. Buying a planning app cannot supply missing opening information or turn the exact-close B handoff into a guaranteed one.