Section 3 : Inter-timing Services

Organising the timetable to achieve exact inter-timing

If the combination of minimum and excess layovers is the same on the two separate services, then inter-timing will happen as a matter of course.  It’s simply a matter of writing out the timetables so that the two separate services involved provide a ‘clock face’ headway over the common portion of route.  Layovers must be applied equally at the common terminus.

Where the combination of minimum and excess layovers is not the same, as in the above example, the following steps are taken (at the start apply only the minimum layover at the outer terminus).

	Service No. 8	Service No. 8a
Total layover	4 minimum + 4 excess	4 minimum + 6 excess
Layover at outer terminus	2 minimum	2 minimum
Layover at common terminus	2  minimum + 4 excess	2 minimum + 6 excess

The layover on service 8a at the common terminus exceeds the layover on service 8 at the common terminus by 2 minutes.  Equalise the layovers at the common terminus by removing 2 minutes excess layover from 8a.

Layover at the common terminus is now 6 minutes for the two services (2 minimum + 4 excess). The 2 minutes excess layover that were taken away from the layover of service 8a at the common terminus are now added or reassigned to the minimum layover at the outer terminus of service 8a as follows:

	Service No. 8	Service No. 8a
Total layover	4 minimum + 4 excess	4 minimum + 6 excess
Layover at outer terminus	2 minimum	2 minimum + 2 additional
Layover at common terminus	2  minimum + 4 excess	2 minimum + 4 excess

Increasing the layover at the outer terminus of service 8a  has the effect of holding back departures from Minton by an extra 2 minutes which means that buses on services 8 and 8a pass Fixton, the common mid-route timing point, at exact 6 minute intervals.  In effect the round journey time of service 8a is increased from 30 minutes to 32 minutes.

The calculations for the number of buses needed now looks like this:

Service No. 8

26 + 2 + 26 + 2 / 12 = 56 / 12 = 5 buses + 4 minutes

Service No. 8A

13 + (2+2) + 13 + 2 / 12 = 32 / 12 = 3 buses + 4 minutes

* The total layover at the outer terminus of service 8A is made up of the minimum layover (in this case 2 minutes) PLUS the difference between total layovers on the two services i.e. the reassigned excess layover (in this case 2 minutes).

Note that the time difference between the two round journey times is a multiple of the headway – this has to be the case for inter-timing to be achieved.

Service 8 Round Journey Time       =             56 minutes

Service 8a Round Journey Time     =             32 minutes

Difference                                              =         24 minutes

Working Timetable

Bus No.	1	6	2	7	3	8	4	6	5	7	1
Service No	8	8a	8	8a	8	8a	8	8a	8	8a	8
Newborne	0628	…….	0640	…….	0652	…….	0704	…….	0716	…….	0728
Minton	…….	0647	…….	0659	…….	0711	…….	0723	…….	0735	…….
Fixton	0644	0650	0656	0702	0708	0714	0720	0726	0732	0738	0744
Town Centre	0654	0700	0706	0712	0718	0724	0730	0736	0742	0748	0754
Service No	8	8a	8	8a	8	8a	8	8a	8	8a	8
Town Centre	0700	0706	0712	0718	0724	0730	0736	0742	0748	0754	0800
Fixton	0710	0716	0722	0728	0734	0740	0746	0752	0758	0804	0810
Minton	…….	0719	…….	0731	…….	0743	…….	0755	…….	0807	…….
Newborne	0726	…….	0738	…….	0750	…….	0802	…….	0814	…….	0826

Rules for exact inter-timing

• layovers at a common terminus must be the same on two services
• where the layover on one service is greater than the other and there is excess layover available,  layovers at the common terminus are equalised
• any remaining excess layover is added to the outer terminus of the service which had the greater total layover