Vehicle routing problem considering equity of demand satisfaction

Abstract

Planning of distribution processes has been represented by vehicle routing problems (VRP), where objective functions are commonly related to the cost of routes and delivery times involved in those processes. However, other objectives, such as equity of demand satisfaction, are of particular interest in social or non-profit distribution systems. This study addresses a VRP focused on an egalitarian distribution among all demand centroids considering a heterogeneous fleet. In particular, our goal is to maximize the minimum fraction of fulfilled demand to foster the equity of demand satisfaction while minimizing the delivery times as a secondary objective. We formulate our optimization problem as a mixed-integer linear program that is intractable by using a commercial solver for large instances. Therefore, we design three variants of the biased random-key genetic algorithm, and one of them can obtain near-optimal solutions for instances up to 120 demand points and 15 routes in less than one minute of computational time. Finally, we present an analysis of the trade-off between egalitarian demand satisfaction and total delivery time for a scenario based on an economically deprived region in Mexico.

A Example of decoder algorithms

This section presents an example of each decoder algorithm using a small instance of our proposed VRPE. To define that instance, we consider four demand points and the depot (i.e., $I=\{0,1,2,3,4\}$), two vehicles ($K=\{1,2\}$), and a supply of $s=500$. Moreover, the demand parameters are set as $b_1=126$, $b_2=251$, $b_3=191$, and $b_4=206$. Finally, vehicle capacities are $c^{1}=350$ and $c^2=380$.

1.1 A.1 Example of Decoder1

Each decoder’s input is a random-key Y of a predefined size based on the decoder’s definition. In the case of Decoder1, the random-key has $2\left| I'\right| +|K|=2(4)+2=10$ elements, see an example in Fig. 8.

Fig. 8

Example of a random-key for Decoder1

At the beginning of Decoder1, parameters are initialized as $\overline{c}_1=350$, $\overline{c}_2=380$, and $\overline{s}=500$. Moreover, the set $I'$ is increasingly ordered in terms of the values $Y_i$, for $i\in I'$; thus, $\overline{I}=\left\{ 1,3,4,2\right\}$.

Analogously, the last two elements of Y define the order to construct the vehicles’ routes, i.e., $\overline{K}=\{2,1\}$.

Now, we describe the iterations of Decoder1. Notice that there is available supply ($\overline{s}=500>0$) and there are vehicles with positive capacity ($\left| \overline{K}\right| =2>0$). Then, we construct the route of the first vehicle $k=2$ in $\overline{K}$. First, demand point 1 is chosen (first element in $\overline{I}$), and we compute $w_1^2=\min \left( \lceil b_{1}Y_{|I'|+1}\rceil ,\overline{c}^{2},\overline{s}\right) =\min \left( \lceil (126)(0.78)\rceil ,380,500\right) =99$. Next, the demand point 1 is assigned to the route of vehicle 2, and parameters are updated as $\overline{c}_2=281$, $\overline{s}=401$, and $\overline{I}=\left\{ 3,4,2\right\}$. We continue adding demand points since we did not reach the stop criteria, i.e., $\overline{c}_2>0$ and $\overline{s}>0$. We take first element $i=3$ in $\overline{I}$ and we compute $w_3^2=\min \left( \lceil b_{3}Y_{|I'|+3}\rceil ,\overline{c}^{2},\overline{s}\right) =\min \left( 165,281,401\right) =165$. Next, we add the demand point 3 to the route of vehicle 2, and we update parameters as $\overline{c}_2=116$, $\overline{s}=236$, and $\overline{I}=\left\{ 4,2\right\}$. Similarly, demand point 4 is the last point in the route of vehicle 2 with a delivered demand of $w_4^2=\min \left( 186,116,236\right) =116$. Notice that the stop criterion of the vehicle’s capacity is reached ($\overline{c}^2 = 0$), and available supply is $\overline{s}=120$. Since there are demand points not visited by routes ($|\overline{I}| > 0$) and vehicles with positive capacities ($|\overline{K}|>0$), we construct the route of the next vehicle $k=1$, leading to the solution in Table 3.

Table 3 Obtained solution using Decoder1

1.2 A.2 Example of Decoder2

The main idea of Decoder2 is to compute the minimum fraction of fulfilled demand using the last element of a random-key of size $\left| I \right|$. Figure 9 shows an example of a random-key of five elements for our example instance.

Fig. 9

Example of a random-key for Decoder2

In the first step of Decoder2, we initialize as Decoder1, i.e., $\overline{c}_1=350$, $\overline{c}_2=380$, $\overline{s}=500$, and $\overline{I}=\left\{ 1,3,4,2 \right\}$. Then, we construct routes for each vehicle $k=1,\dots , |K|$. To define the route of vehicle $k=1$, we take the first demand point $i=1$ in $\overline{I}$, and we compute the number of products to delivery as $\lceil b_1 Y_5\rceil =\lceil \left( 126\right) \left( 0.78\right) \rceil =99$. Since we have enough capacity in vehicle 1 ($350\ge 99$) and supply ($500\ge 99$) to deliver that demand, the point $i=1$ is assigned to de route of vehicle $k=1$, with $w_1^1=99$. Next, we update $\overline{c}_1=251$, $\overline{s}=401$, and $\overline{I}=\{3,4,2\}$. Since $\overline{c}_1>0$, it is possible to add more demand points to the route of vehicle 1. We take $i=3$ and compute $\lceil b_3 Y_5\rceil =\lceil \left( 191\right) \left( 0.78 \right) \rceil =149$. Similarly to point 1, we add demand point 3 to the route of vehicle 1 and we update $\overline{c}_1=102$, $\overline{s}=252$, and $\overline{I}=\left\{ 4,2\right\}$. Notice that in the case of the next demand point $i=4$, we have $\lceil b_4Y_5\rceil = \lceil (206)(0.78)\rceil =161$, which is greater than the capacity; thus, we can not satisfy the demand represented by $Y_5$ and we end the route of vehicle $k=1$ by updating $\overline{c}^1=0$. In the case of vehicle 2, we first add point $i=4$ leading to $\overline{c}_2=219$, $\overline{s}=91$, and $\overline{I}=\left\{ 2 \right\}$. Notice that the last demand point $i=2$ can not be added to the route since $\lceil b_2 Y_5 \rceil =\lceil \left( 251\right) \left( 0.78 \right) \rceil =196$ is greater than the available supply $\overline{s}=91$. Then, demand point $i=4$ is not visited and $r=0$. Table 4 shows the solution obtained by Decoder2.

Table 4 Obtained solution using Decoder2

1.3 A.4 Example of Decoder3

Our Decoder3 considers three phases to create the vehicles’ routes satisfying that the fraction of fulfilled demand is greater than 0 for all demand points. The input of algorithm is a random-key with size $2\left| I\right| =2\left( 4 \right) =8$ and p as a parameter in $\left( 0,1\right)$, but we will use $p=0.2$ in our example. A random-key example is shown in Fig. 10.

Fig. 10

Example of a random-key for Decoder3

At the beginning of Decoder3, we initialize as Decoder1, i.e., $\overline{c}_1=350$, $\overline{c}_2=380$, $\overline{s}=500$, and $\overline{I}=\left\{ 1,3,4,2 \right\}$. Moreover, we define a set of potential delivered demand as follows.

$\overline{b}=\{\lfloor b_{1}Y_{5}\rfloor ,\ \lfloor b_{2}Y_{8}\rfloor ,\ \lfloor b_{3}Y_{6}\rfloor ,\ \lfloor b_{4}Y_{7}\rfloor \}=\{98, 225, 74, 177\}$

In Phase I, we verify if demands in $\overline{b}$ can be satisfied in terms of the total vehicles’ capacity and the supply. In this example, it is not possible since $\sum _{i\in I'}\overline{b}_i=574$, which is greater than $\min \left( 730, 500\right) =500$. Then, we iteratively take the fraction $(1-p)$ for each demand $\overline{b}_i$ until demands in $\overline{b}$ can be satisfied. In particular, we update $\overline{b}_1=\lfloor (1-0.2)98\rfloor =78$, $\overline{b}_2=\lfloor (1-0.2)225\rfloor =180$, $\overline{b}_3=\lfloor (1-0.2)74\rfloor =59$, and $\overline{b}_4=\lfloor (1-0.2)177\rfloor =141$. Notice that, we finish Phase I since $\sum _{i\in I'}\overline{b}_{i}=402$ and $402<500$.

In Phase II, the vehicles’ routes are constructed as follows. We take vehicle $k=1$ with capacity $\overline{c}^1=350$, and we choose the first demand point $i=1$ in $\overline{I}$. Since $\overline{b_1}=78$ does not exceed the capacity $\overline{c}^1=350$, we add point 1 to the route of vehicle 1 by fixing $w_1^1=78$ and updating $\overline{c}^1=272$, $\overline{s}=422$, and $\overline{I}=\{3,4,2\}$. Similarly, we add demand points 3 and 4 leading to $\overline{c}^1=72$, $\overline{s}=222$, and $\overline{I}=\left\{ 2\right\}$. Since demand $\overline{b}_2=180$ is greater than capacity $\overline{c}^{1}=72$, the vehicle 1 can not visit the demand point 2, and we explore the next vehicle. We have $\overline{c}^2=200$ for vehicle 2 and $\overline{s}=222$; thus, we add point 2 to the route by fixing $w_2^2=180$ and we update $\overline{c^2}=20$, $\overline{s}=42$, and $\overline{I}=\emptyset$ (we finish Phase II). At the end of this phase, vehicle 1 visits the points 1, 3 y 4, and vehicle 2 visits point 2 as shown in Table 5.

Table 5 Obtained solution using Decoder2

Phase III is not done since all demand points are visited.