4.1. Analysis of Second-Phase Routing
To solve the second-phase routing problem, we use two methods, as proposed in
Section 3.3. Mathematical models for CV routing and EV routing are implemented first, and a heuristic approach is further implemented for EV routing. Additionally, directly using the mathematical formulation for EV routing is equivalent to using the heuristic when the dispatch interval is the entire time span.
When using the heuristic, the EV routing problem is solved with different dispatch intervals
. If the total time span is
, then the number of intervals
can range from 1 to
. When
, for the EV routing subproblem, EVs leave their ID at time 0 and can only serve 1 route. The charging technology used at an ID can either be a slow charger over the night or a fast charger in less than 0.5 h. Moreover, each ID is transformed into one Big-customer for the CD, with new demands and time windows determined by all customers assigned to this ID, as described in
Section 3.2. In the case when
is greater than 1, customers assigned to an ID are split into
subsets. A total number of
Big-customers are generated for this ID, one for each subset of assigned customers. EVs are dispatched at
time points from their ID. After an EV returns to its ID after visiting customers on a route, it may be recharged at the depot depending on its current battery level. If the same EV is going to serve the next route, and it has sufficient battery to cover the next route, it is not necessary to be charged; otherwise, this EV needs to be charged to the least battery level required to serve the next route. Fast charge technology is used to enable this feature. Examples of two different cases are presented in
Figure 6. In
Figure 6a, there are two routes of the same ID. Either of them is associated with a time window of the depot and a travel distance. The maximum travel distance of an EV is 79.7. If an EV leaves at time 0 and returns at time 552.6 after completing the route
with a travel distance
, the same EV can continue to serve
without being charged at the depot. This is because the total distance of
and
, i.e., 58, is less than the maximum travel distance of the EV. However, in
Figure 6b, after the EV completes the route
, it needs to be charged before continuing to serve the route
because the sum of
and
, 82.8, is greater than 79.7.
Take instance c103_15 as an example of a small-sized instance; 1 customer is identified to be served by CV from CD while the rest are served by EVs from IDs. A total of 4 IDs exist, and the number of customers assigned to
is 3, 5, 3, and 3, respectively. Results for EV routing in second-phase routing are presented in
Table 3. With the same customer clustering from the first phase, both the E-VRP model and heuristic method are applied to the EV routing subproblem, while the C-VRP model is implemented for the CV routing subproblem. The second and third columns provide the number of EVs required and the routing distance at each ID, as determined by the optimal solution of the E-VRP model. For the heuristic method, dispatch interval
is selected in such a way that the number of intervals
takes values of 1, 2, 4, 8, and 16, as shown in the second row in
Table 3. With the same customer clustering, the EV routing problem is solved with a different number of intervals
for each ID, and results are also presented in
Table 3. Heuristic gives the best solution found in up to 20 iterations. For the same instance, using the E-VRP is equivalent to using the heuristic when the dispatch interval is the entire time span, i.e., only one interval that
, and this is observed in
Table 3. When there is only 1 interval, i.e.,
, all EVs leave the depot at time 0. However, when
is greater than 1, it means that EVs leave ID at
different time points, and the same EV can be used to serve multiple routes for the same ID.
It is observed from
Table 3 that the minimum number of EVs remains the same and the total travel distance of EV routing increases as m increases for this specific instance. The increase in total traveled distance can be explained by the number of customers assigned to each ID being too small and that having multiple dispatch times would force EVs to visit customers in separate routes. Meanwhile, the same minimum number of EVs indicates that an EV is used to serve multiple routes from different dispatch times for an ID.
An analysis is then conducted for the CV routing subproblem using heuristic, since the number of Big-customers is changing along with
value and the number of intervals. For the same instance c101_15, the results of CV routing with different
value are presented in
Table 4. The number of Big-customers generated for each ID increases as
increases. This is because customers assigned to an ID are split up into
subsets and one Big-customer is generated for each valid subset. While the number of CVs used remains the same, the total traveled distance of CV routing is reduced when
increases and has the smallest value when
4 or 8. Therefore, although
Table 3 shows that EV routing may not be improved with larger
, the CV routing could be improved instead.
Similarly, take instance c101_21 as an example of large-size instances. A total of 9 out of 100 customers are determined to be visited by CVs in first-phase clustering and results for EV routing and CV routing with the same customer assignments are obtained with
1, 2, 4, 8, and 16 as presented in
Table 5. The results presented in the table are the best found in 100 iterations. It is observed from
Table 5 that the total traveled distance of EVs becomes smaller as
becomes larger, which indicates that having additional dispatch times helps to give better routes. Additionally, the total traveled distance of EVs can be improved by at most 16.3%. The minimum number of EVs is always greater than or equal to 20 since a total of 20 IDs are available for the instance and each ID is assigned to at least 1 customer. It is shown in
Table 5 that the minimum number of EVs decreases as the number of intervals
increases. This is because EVs can be used for multiple routes if
is greater than 1. Additionally, the number of EVs needed for instance c101_21 can be reduced by up to 7 if the proposed heuristic is combined with the intelligent dispatch scheme. Additionally, since EVs can only be charged at IDs and the possible service time of EVs can spread through the entire time span, the number of chargers at an ID could be potentially reduced.
As for CV routing, it is shown in
Table 5 that at most, 4 additional CVs are required as
increases, while the total traveled distance of CVs is reduced by up to 7.3%. This can be explained by the fact that customers are split into more subsets with more dispatch times; thus, more Big-customers are visited by CVs. The overall routing distance including both EV routing distance and CV routing distance is displayed in the last column in
Table 5 and it can be improved by 11.7% as
increases to 16. The last observation from the experiments is that the larger
it takes, the longer the computational time to complete an iteration. This is mainly caused by the longer time to solve CV routing as the number of Big-customers increases when
is larger. Therefore, a larger
is not entirely beneficial to solving the problem, and a compromise between the value of
and satisfactory results is indeed necessary. It is noticeable from
Table 5 that the result from
and
are quite close, whereby both results are slightly worse than that obtained from
. However, the time it takes to compute for
4 and 8 is observed to be much less compared to that for
in the experiments. Thus, while considering computational efficiency,
is selected to obtain results and perform further analysis.