Heuristic Optimization for Multi-Depot Vehicle Routing Problem in ATM Network Model

Abstract

Encashment process in the ATM network is an essential part of the banks’ activities. Timely encashment process plays an important role in building a good reputation of the bank that helps attract new clients. Uninterrupted work of money collector teams and stable functioning of ATM network requires solving the following major problems: one of them is the optimal location of the ATMs, the bank and its encashment centers, while another is, minimization of servicing costs of ATM network. In a large city such as St.Petersburg, encashment process of ATM network is associated with a problem of long distances between ATMs and encashment centers. As a possible solution to this problem, Multi-Depot Location Routing Problem with Multi-Depot Vehicle Routing Problem has been combined to design optimal or almost optimal routes for money collector teams and define the optimal location of encashment centers. To design optimal or almost optimal routes for money collector teams and define the optimal location of encashment centers, Multi-Depot Location Routing Problem with Multi-Depot Vehicle Routing Problem have been combined as a possible solution. In this research, we apply two heuristic methods for designing routes of the money collector teams located in the several depots (encashment centers). These approaches have been tested using the existing geo-location data for depots and ATMs of a bank in St.Petersburg, Russia.

Appendices

Appendix 1

A list of ATMs included in the claim of servicing: Vitebskiy av., 53/4; Zvezdnaya st., 6; Leninskiy av., 129; Novatorov blvd., 11/2; Nevskiy av, 49; Gagarina av., 27; Kosmonavtov av., 28; Krasnoputilovskaya st., 121; Leninskiy av, 151; Koli Tomchaka st., 27; Dumskaya st., 4; Bukharestskaya st, 89; Basseynaya st., 17; Moskovskiy av., 200; Novosmolenskaya emb., 1/3; Veteranov av., 43; Veteranov av., 89; Leni Golikova st., 3; Izmaylovskiy av., 4; Moskovskiy av., 133 (Figs. 14, 15).

Fig. 14

Matrix of distances C for 30 ATM and 4 encashment center

Fig. 15

Matrix of distances C for 30 ATM and 4 encashment center

Appendix 2

In the current case study we apply both modification of genetic algorithms GA1 and GA 2 respectively for the problem of $109\times 109$, where we have 10 depots and 99 ATMs. By using GA1 we obtain

• $A - 11 - 35 - 64 - A$, $A - 16 - 31 - 4 - A$, length is 78.467 km;

• $B - 42 - 70 - 80 - B$, $B - 46 - 12 - 14 - 2 - B$, $B - 56 - 62 - 21 - 37 - B$, $B - 60 - 18 - 36 - 71 - B$, length is 86.149 km;

• $C - 42 - 85 - 13 - 61 - C$, $C - 81 - 44 - C$, length is 39.988 km;

• $D - 22 - 24 - 47 - D$, $D - 39 - 38 - 6 - D$, $D - 75 - 72 - 17 - 10 - D$, $D - 82 - 5 - 89 - 66 - D$, length is 87.649 km;

• $E - 8 - 15 - 25 - 73 - E$, $E - 83 - 84 - 74 - 79 - E$, $E - 86 - 59 {-} 87 {-} E$, length is 65.516 km;

• $F - 3 - 41 - 48 - F$, $F - 54 - 69 - 58 - 88 - F$, length is 42.953 km;

• $G - 7 - 77 - 20 - G$, $G - 63 - 26 - G$, length is 39.801 km;

• $H - 52 - 76 - 32 - 68 - H$, length is 27.390 km;

• $I - 27 - 57 - 55 - 65 - I$, $I - 49 - 40 - 19 - 9 - I$, $I - 51 - 23 - 45 - 34 - I$, $I - 78 - I$, length is 112.126 km;

• $J - 28 - 30 - 50 - 1 - J$, $J - 53 - 33 - 29 - 67 - J$, length is 47.477 km.

Total distance is 627.516 km and costs are 56883 rubles.

By GA2 approach we receive the next solution:

• $A - 72 - 1 - 38 - 7 - A$, $A - 19 - 56 - A$; $A - 77 - 20 - 84 - 29 - A$;

• $B - 59 - 10 - 40 - 4 - B$

• $C - 85 - 36 - 17 - 41 - C$;

• $G - 8 - 21 - 14 - 46 - G$;

• $E - 74 - 26 - 71 - 35 - E$;

• $D - 18 - 30 - D$; $D - 87 - 78 - 88 - 6 - D$;

• $D - 58 - 89 - 49 - 13 - D$;

• $D - 44 - 76 - 5 - 50 - D$, $D - 23 - 63 - 82 - 73 - D$;

• $E - 42 - 34 - 65 - E$;

• $E - 9 - 53 - 61 - 54 - E$; $E - 64 - 69 - 37 - 51 - E$,

• $G - 60 - 63 - 57 - 75 - G$;

• $H - 43 - 45 - 67 - 33 - H$;

• $H - 16 - H$, $H - 66 - 81 - 28 - 3 - H$, $H - 22 - 12 - 15 - 52 - H$;

• $H - 27 - 90 - 31 - 68 - H$;

• $H - 32 - 24 - H$, $H - 39 - 2 - 80 - 83 - H$;

• $H - 48 - 25 - 47 - 11 - H$.

Total distance is 452.119 km and costs are 39857 rubles.

From the computation we can see that the total distance on the routes for money collector teams received by GA2 are shorter that total distance received by GA1 for $18\%$.

Appendix 3

The savings algorithm is a heuristic algorithm, and therefore it does not provide an optimal solution to the problem with certainty. However it often gives a relatively good solution. The basic savings concept depicts the cost savings obtained by joining small routes into more large route. Consider the depot D and n demand points. Suppose that initially the solution to the VRP consists of using n vehicles and dispatching one vehicle to each one of the n demand points. The total tour length of this solution is, $2 \sum \limits _{i=1}^n d(D, i)$. If now we use a single vehicle to serve two points, say i and j, on a single trip, the total distance traveled is reduced by the amount

$$ S_{ij}=c_{i0}+c_{0j}-c_{ij}. $$

 

Stage 1.:

Calculate the savings $S_{ij} = d(D, i) + d(D, j) - d(i, j)$ for every pair (i, j) of demand points.

Stage 2.:

Rank the savings $S_{ij}$ and list them in descending order of magnitude. This creates the “savings list.” Process the savings list beginning with the topmost entry in the list (the largest $S_{ij})$.

Stage 3.:

For the savings $S_{ij}$ under consideration, include link (i, j) in a route if no route constraints will be violated through the inclusion of (i, j) in a route, and if:  

a.:

Either, neither i nor j have already been assigned to a route, in which case a new route is initiated including both i and j.

b.:

Or, exactly one of the two points (i or j) has already been included in an existing route and that point is not interior to that route (a point is interior to a route if it is not adjacent to the depot D in the order of traversal of points), in which case the link (i, j) is added to that same route.

c.:

Or, both i and j have already been included in two different existing routes and neither point is interior to its route, in which case the two routes are merged.

 

Stage 4.:

If the savings list $S_{ij}$ has not been exhausted, return to Stage 3, processing the next entry in the list; otherwise, stop: the solution to the VRP consists of the routes created during Stage 3. (Any points that have not been assigned to a route during Stage 3 must each be served by a vehicle route that begins at the depot D visits the unassigned point and returns to D.)

 

Appendix 4

See Figs. 16, 17, 18, 19, 20.

Fig. 16

Routes for depot B. $B - 4 - 16 -17 - 18 - B$. Length of route is 31.4 km

Fig. 17

Routes for depot B. $B - 10 - 20 - 13 - 6 - B$. Length of route is 16 km

Fig. 18

Routes for depot B. $B - 14 - 3 - 9 - 8 - B$. Length of route is 19.9 km

Fig. 19

Routes for depot C. $C - 12 - 1 - 7 - 2 - C$. Length of route is 19 km

Fig. 20

Routes for depot D. $D - 5 - 11 - 15 - 19 - D$. Route length is 27.2 km