Scheduling irregular distribution patterns

Although irregular distribution patterns are not actually being considered in our computational experiments, we still discuss the irregular distribution pat-tern scheduling problem shortly. This scheduling problem may look completely dierent from regular scheduling, but there is in fact a lot of similarity. In the irregular case, holding costs need to be minimized within a given cycle time T . This can be obtained by spreading dierent deliveries to the same customer as good as possible. In the regular case, a perfect spreading of the dierent deliveries is imposed and the cycle time T_sched is a variable. Thus, the regular schedule that is found is the ideal schedule for the irregular case, but it is in-feasible if Tschedis larger than the (approximated) optimal cycle time T , which is the makespan of the irregular schedule.

In regular scheduling, we start with the schedule time T_sched equal to the minimal cycle time T_minand then increase it to avoid overlaps of time intervals that need to be equidistant. In irregular scheduling, we just work the other way around: the makespan of the schedule is xed, and to avoid overlapping intervals, the intervals themselves are shifted forward or backward.

Illustrative example: the full-truckload distribution pattern

In our illustrative example, the so-called `full-truckload' distribution pattern has an optimal cycle time of 93 hours. However, the minimal schedule time when deliveries have to be equidistant is 120 hours. Thus, a cycle time of 93 hours can only be achieved if an irregular distribution pattern is allowed. Here, we construct the irregular schedule when the cycle time is rounded to 96 hours.

The tour to customer 1 has the highest frequency and is therefore inserted rst.

It is started every 24 hours, at times 0, 24, 48 and 72.

- t(h)

1 1 1 1

0 24 48 72 96

The tour to customer 2 has frequency 3 and is inserted next. Due to symmetry, both possible gaps give the same result and thus the gap at t^ = 6 is chosen.

This results in the following starting times for the second tour: 6, 38 and 70.

The last iteration has an overlap of 2 hours with the last delivery to customer 1. To avoid this overlap, the deliveries to customer 2 are shifted in time a bit.

The last delivery to customer 2 is started 2 hours earlier, at 68 hours and the second delivery to customer 2 is started 1 hour earlier, at 37 hours.

3.6 Scheduling tours in a distribution pattern 71

- t(h)

1 1 1 1

2 2 2

- t(h)

1 2 1 2 1 2 1

0 24 48 72 96

0 24 48 72 96

Next, the tour to customer 4 is added. There are three possible gaps: at t^ = 10, t^= 30 and t^ = 42. Inserting into the rst gap gives starting times 10 and 58, which gives no overlap, so this is immediately selected.

- t(h)

1 2 4 1 2 1 4 2 1

0 24 48 72 96

Finally, the tour to customer 3 is inserted. Again, the rst gap, at t^ = 14, does not cause problems and the starting time of this tour is thus at 14 hours.

The irregular schedule is then completed, and only the deliveries to customer 2 had to be slightly shifted.

- t(h)

1 2 4 3 1 2 1 4 2 1

0 24 48 72 96

The 4-tour distribution pattern with adjusted frequencies

In an alternative solution for our illustrative example, the customers are still in separate tours, but the frequencies are adjusted to (3; 3; 1; 2). Although a regular schedule can be found for the optimal cycle time of 84 hours, an irregular schedule for a cycle time of 48 hours is constructed here to illustrate the irregular scheduling heuristic. The tour to customer 1 is inserted rst, at times 0, 16 and 32.

- t(h)

1 1 1

0 16 32 48

Then, the tour to customer 4 is inserted in the rst gap, at t^= 6. This gives starting times 6 and 30. However, this second iteration overlaps with the third iteration of the rst tour, and is therefore started 2 hours earlier, after 28 hours.

- t(h)

1 4 1 4 1

0 16 32 48

The next tour to be inserted is the tour to customer 2, with a frequency of 3. It can only be assigned to the rst gap, at t^ = 10. The resulting starting times are 10, 26 and 42. This second delivery overlaps with the second delivery to customer 4 and is therefore shifted 2 hours forward. The other iterations of this tour are not shifted forward or backward, because this only increases the deviation from having equidistant deliveries.

- t(h)

1 4 2 1 2 4 1 2

0 16 32 48

Finally, the tour to customer 3 is inserted. This tour takes 8 hours, while the largest gap in this schedule is only 4 hours, at t^ = 38. To make a feasible schedule, the deliveries before and after this largest gap are shifted: the second delivery to customer 2, the second delivery to customer 4 and the third delivery to customer 3 are started 2 hours earlier and the third delivery to customer 2 is started 2 hours later. To compensate the earlier start of the third iterations of the tour to customer 1, its second iteration is started 1 hour earlier. In this schedule, the only customer receiving equidistant deliveries is customer 3, but this customer has equidistant deliveries by denition since it is only served once per cycle. This schedule is highly irregular because (i) frequencies 2 and 3 are incompatible, and (ii) only 2 hours of idle time are available.

- t(h)

1 4 2 1 2 4 1 3 2

0 16 32 48

The full-truckload distribution pattern with driving time constraints When driving time restrictions apply, the `full-truckload' distribution pattern with a cycle time of 96 hours or 12 days is feasible. The construction of its schedule is illustrated here.

The tour to customer 1 has the highest frequency and is inserted rst. The time between deliveries should be 3 days and therefore this tour is assigned to days 1, 4, 7 and 10.

- t (days)

1 1 1 1

1 2 3 4 5 6 7 8 9 10 11 12

The next tour to be inserted is the one visiting customer 2. It has frequency 3 and is to be repeated every 4 days. Since it is impossible to achieve inter-delivery times of exactly 4 days due to the presence of the rst tour, the fol-lowing happens. When assigning the tour to day t^ = 2, the third and last iteration overlaps with the last iteration of the rst tour. This third and last