Is it still possible to save a bus by interworking?
Let’s take another look at the calculation that was arrived at after having adjusted excess layovers to achieve inter-timing:
Service No. 10
28 + (5+5) + 27 + 5 / 15 = 70 / 15 = 5 buses + 5 minutes
Service No. 11
15 + 5 + 15 + 5 / 15 = 40 / 15 = 3 buses + 5 minutes
Now that 5 of the original 15 minutes excess layover have been added to the minimum layover at Long Green, only 10 excess minutes remain and so, if the revised calculation is applied, interworking will not save a bus because 15 or more excess minutes are needed.
Saving a bus will therefore mean sacrificing exact inter-timing. So, if a bus IS saved, what will the combined timetable for the two services be like?
If the two services are interworked, it must follow that a bus will operate a round journey on service 10 followed by a round journey on service 11.
If a bus leaves, say, at 0700 on service 10, applying the known round journey time of 65 minutes would mean that the next journey on service 11 operated by this bus is at 0805 (65 minutes later). It can be concluded therefore, that if headways of 15 minutes are operated on the two services, departures past each hour from the common terminus will be as follows:
Service 10 00, 15, 30, 45
Service 11 05, 20, 35, 50
So instead of a straight 7½ (8/7) headway over the common portion of route, the split will be 5/10. A very simple rule emerges.
The individual excess layovers are 10 on service 10, and 5 on service 11. Interworking takes out this excess layover, but these figures give the clue as to what the split will be other than a straight 7½ (8/7) headway over the common portion of route – it will be 10 and 5 which coincides with the two excess layovers. But for this to be true, the layover at the common terminus must be the same on the two services, and the total excess layover must be equal to the headway.
Working Timetable – interworking with no inter-timing
The best possible timetable to give interworking with 7 buses will look like this with gaps of 10 and 5 minutes over the common portion of route:
| Bus No. | 1 | 2 | 3 | 4 | 5 | 6 | 2 | 7 | 4 | 1 |
| Service No | 11 | 10 | 11 | 10 | 11 | 10 | 11 | 10 | 11 | 10 |
| Long Green | .. | 0633 | .. | 0648 | 0703 | .. | 0718 | .. | 0733 | |
| Hillbury | 0640 | .. | 0655 | .. | 0710 | .. | 0725 | .. | 0740 | .. |
| Gilmore | 0645 | 0650 | 0700 | 0705 | 0715 | 0720 | 0730 | 0735 | 0745 | 0750 |
| Town Centre | 0655 | 0700 | 0710 | 0715 | 0725 | 0730 | 0740 | 0745 | 0755 | 0800 |
| Service No | 10 | 11 | 10 | 11 | 10 | 11 | 10 | 11 | 10 | 11 |
| Town Centre | 0700 | 0705 | 0715 | 0720 | 0730 | 0735 | 0745 | 0750 | 0800 | 0805 |
| Gilmore | 0710 | 0715 | 0725 | 0730 | 0740 | 0745 | 0755 | 0800 | 0810 | 0815 |
| Hillbury | .. | 0720 | .. | 0735 | .. | 0750 | .. | 0805 | .. | 0820 |
| Long Green | 0728 | .. | 0743 | .. | 0758 | .. | 0813 | .. | 0828 |
Conclusions
There are two options:
- achieve exact inter-timing over the common portion of route with a 7½ (8/7) headway using 8 buses
- reduce the number of buses from 8 to 7 by interworking which means giving up the opportunity to achieve exact inter-timing over the common portion of route
On this occasion it isn’t possible to save a bus and at the same time achieve exact inter-timing.