Consider a village (A) that has two round bus services a day to the next town (B).
The timetable might look like this:
Depart A 7:00am
Arrive B 7:30am
Depart B 7:30am
Arrive A 8:00am
Depart A 7:00pm
Arrive B 7:30pm
Depart B 7:30pm
Arrive A 8:00pm
Assuming that no overnight stays are desired, the number of trip choices from either place to the other is one. Thus the service is inflexible and would only cater for the minority of travellers for whom the timetable was suitable, or those who had to make the trip and had no other alternative.
Now supposing an extra trip was added so it departed A at 1:00pm, called at B at 1:30pm and returned to A at 2:00pm.
The level of service has increased by 50% (3 services instead of 2) but what has happened to the travel choices possible?
The answer is that instead of one choice only, there are now three combinations available for visitors to B from A (leave 7:00am, return 2:00pm, leave 7:00am, return 8:00pm, or leave 1:00pm, return 8:00pm).
Adding a fourth trip (at say 9:00am) increases choice dramatically. Double in fact, so that instead of three possible combinations there are now six.
Similarly five trips allows ten choices and six trips gives fifteen choices.
Maths boffins will see a pattern in this. Each extra service increases the number of combinations added by one. Hence the pattern (1), (1+2 = 3), (1+2+3 = 6), (1+2+3+4 = 10) and so on. As a point of trivia note that all answers are 'triangular numbers' due to the pattern formed (below).
O Total = 1
OO Total = 3
OOO Total = 6
OOOO Total = 10
OOOOO Total = 15
At this level a relatively small service increase (say two extra services) dramatically increases travel choices and the practicality of using public transport for many trips. This is something to bear in mind when planning services, particularly in areas that may currently get only one or two a day.
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