Evolutionary algorithm based approach for solving transportation problems in normal and pandemic scenario

1.1. Motivation

From the literature survey, it is evident that most countries imposed restrictions which greatly affect the transportation system of items (both essential and non-essential). Thus, the existing models of transportation problem (TP) are based on the assumption that there is no such restrictions in the movement of vehicles and suitable for normal scenario only. Moreover, in most of the works on TP (in particular fixed-charge transportation problem (FCTP)), it is considered that a vehicle can avail at most one trip to a destination. However, in real-world scenario, the amount of an item available at an origin may exceed the total capacity of all the vehicles, and hence, one or more vehicles may need more than one trip to satisfy the demand at a destination. In the existing works, none of the researchers has considered that the number of trips of a vehicle to a destination can be more than one, except Balaji et al. [16]. However, the number of trips of a vehicle to a destination can be more than one, and must be considered into the formulation, since for a FCTP, the number of trips contribute to the fixed cost, total time and total profit (in case of shipping of perishable items).

In pandemic scenario, regions are categorized in different groups depending upon the level of restriction in a region persistent over a certain period of time. The level of restriction in a region is dependent on various factors such as number of active cases (i.e., number of people infected), number of deaths, population density, number of COVID-19 hospitals and number of migrant workers returned or might return to a region. Thus, in pandemic scenario, for origins and destinations that are situated in regions with higher restrictions, the number of trips of vehicles need to be reduced in such a way that, a balance between the supply and demand of item(s) is maintained. However, reducing the number of trips of vehicles may increase the transportation cost to an unrealistic bound. Thus, a transportation company needs to find a proper balance between the transportation cost and the reduction in number of trips of vehicles considering the levels of restriction imposed in different regions. Hence, planning of transportation scheme in a pandemic scenario is a challenging task for transportation companies, and consideration of a new type of transportation plan becomes necessary. However, there is no such work available in the literature, which motivated us to formulate a transportation model for pandemic scenario and to solve it. This model is also applicable in emergency scenarios such as major earthquakes, floods and other natural calamities in which only a limited number of trips of some particular types of vehicles can be availed.

1.2. Our proposed contribution

In this paper, we first formulate a fixed cost transportation model for a homogeneous item in COVID-19 pandemic scenario, in which regions are categorized in groups depending upon the level of restrictions on the mobility of freight vehicles. It is also considered that more than one type of vehicles are available at each origin, and each vehicle may take more than one trip to the same or different destinations, where the capacity of each vehicle is not the same. For an origin and a destination, the variable cost of a unit item and the fixed-charge varies for each vehicle, which also varies for different pairs of an origin and a destination. The aim of the problem is to obtain a minimum cost transportation plan with minimum number of trips of vehicles moving from origins to destinations that are situated in regions with higher levels of restrictions. For this, the problem is posed as a single-objective optimization problem (SOOP), in which minimization of transportation cost is considered as the objective function. Moreover, to minimize the number of trips of vehicles from origins to destinations situated in regions with higher levels of restrictions, a penalty is imposed in the objective function for each trip of a vehicle that depends upon the level of restrictions of the two regions. To keep the transportation cost within realistic bound, a constraint is imposed with an upper bound on transportation cost. Then, the problem is solved using a Genetic Algorithm (GA) based approach, in which newly designed genetic operators (crossover and mutation) are incorporated to handle multiple trips and capacity constraints of vehicles. Some numerical examples of the proposed model are generated artificially, in which three levels of restrictions are considered for the regions associated with the origins and destinations. To prove that the imposed constraint plays a crucial role, the same examples are solved without considering the constraint. Thereafter, the same examples are solved in normal scenario, i.e., ignoring any categorization of regions and the results are analyzed. The performance of our algorithm is also compared with three existing works, considering a particular instance of our proposed FCTP model. Finally, some future research directions are discussed.

The organization of the rest of the paper is as follows. In Section 2, the notations and abbreviations are presented. The mathematical model of the FCTP in pandemic scenario is given in Section 3. The solution methodology is discussed in Section 4. Section 5 contains the experimental results with discussion. Finally, in Section 6, conclusions are drawn with the lines of further research directions are discussed.

3.1. Fixed-charge transportation problem in pandemic scenario

Consider a transportation network consisting of $m$ origins, say, $O_1,O_2,\dots ,O_m$ and $n$ destinations, say, $D_1,D_2,\dots ,D_n$. Let in COVID-19 pandemic, the regions associated with the origins and destinations be divided in $K$ categories, say, $G_1,G_2,\dots ,G_K$, arranged in increasing order of levels of restrictions. Let there be $l$ types of vehicles available at each origin, where each vehicle may take one or more trips to same or different destinations and the capacity of each vehicle being different. The unit variable cost $c_{\lambda \mu \eta }$ of the item and the fixed cost $h_{\lambda \mu \eta }$ corresponding to the vehicle $V_{\eta }$ to transport from origin $O_{\lambda }$ to destination $D_{\mu }$ vary for different pairs of origins and destinations. Let for a trip of a vehicle from an origin $O_{\lambda }$ to a destination $D_{\mu }$ situated in regions $G_r$ and $G_s$, respectively, let $P_{rs}$ be the penalty to be imposed on the objective function. The penalty $P_{rs}$ depends only on the level of restrictions in the two regions $G_r$ and $G_s$, i.e., the penalty is large if the level of restrictions is high and vice-versa. A higher value of penalty will restrict the vehicles to take less number of trips between an origin and a destination.

subject to

Here, $f\left(x_{\lambda \mu \eta u}\right)$ represents the total transportation cost (the sum of the total variable cost and the total fixed-charge) associated with the transportation of $x_{\lambda \mu \eta u}$ units of the item from origin $O_{\lambda }$ to destination $D_{\mu }$ in trip $u$ of the vehicle $V_{\eta }$, and is given by $f\left(x_{\lambda \mu \eta u}\right)=c_{\lambda \mu \eta }.x_{\lambda \mu \eta u}+h_{\lambda \mu \eta }.g_{\lambda \mu \eta u}$ for the linear form of FCTP, and $f\left(x_{\lambda \mu \eta u}\right)=c_{\lambda \mu \eta }.x_{\lambda \mu \eta u}^2+h_{\lambda \mu \eta }.g_{\lambda \mu \eta u}$ for the quadratic form of FCTP (non-linear). From here on, we shall call the quadratic form of FCTP as the non-linear FCTP.

The objective function (1) represents the minimization of the total transportation cost (i.e., the sum of total variable cost and total fixed-charge) associated with the transportation of different units of the item from all the origins to all the destinations using one or more trips of the vehicles. Eqs. (2), (3) represent, respectively the supply and demand constraints of the item at the origins and destinations. Eq. (4) represents the capacity constraint of the vehicles, Eq. (5) shows that the FCTP is balanced, whereas, the non-negativity restrictions of the decision variables $x_{\lambda \mu \eta u}$ are given in (6).

A special case:

If in the proposed model of FCTP, we consider the restrictions of each region to be in zero level (i.e., the LSR value of each region is considered as zero), then the problem gets reduced to a FCTP in normal scenario given as follows.

subject to the same constraints and non-negativity restrictions considered in the FCTP without any upper limit on transportation cost. In this case, the penalty for each pair of origin and destination becomes zero.

4.1. Generation of chromosome

Many researchers have used different encoding procedures to represent individual chromosomes, such as matrix representation [17], [28], [44], spanning tree [9], [10], [11], [13] and priority-based encoding [14] to solve FCTP and its variants. Among these, the encoding procedures, namely, spanning tree and priority-based encoding are suitable for FCTPs, in which only one type of vehicle with no capacity constraint are available for each pair of an origin and a destination, and a vehicle can ship items to a destination in one trip at most. Thus, it is very difficult to incorporate these encoding procedures into our proposed FCTP. Moreover, these representations need encoding and decoding procedure to understand the transportation scheme corresponding to a solution. So, we use the matrix representation to represent an individual chromosome. As the decision variable $x_{\lambda \mu \eta u}$ has four indices, a four-dimensional matrix is used to represent a chromosome. The process of generation of a chromosome is given in Algorithm 1.

To constitute an initial population of size $X_0$, Algorithm 1 is repeatedly used. We now illustrate the process of generation of a chromosome for the proposed model of FCTP with two origins, three destinations and two vehicles using Algorithm 1.

Example 1

Let us consider a transportation network consisting of two origins $O_1,O_2$, two destinations $D_1,D_2$ and $V_1$, $V_2$ be two vehicles capable of carrying 10 and 20 units of the item, respectively, are available at each origin. Let the availability of the item at the origins $O_1,O_2$ be 30, 50 units and the demand for the items at the destinations be $D_1,D_2$ and 45, 35 units, respectively.

The process of generation of a chromosome for Example 1 is described below.

Iteration 1: Initially, Set $a_1\leftarrow 30$, $a_2\leftarrow 50$, $b_1\leftarrow 45$, $b_2\leftarrow 35$. Then $sum=80$. Also, set $N_{\eta }^{\lambda }\leftarrow 0(\lambda =1,2;\eta =1,2)$; $x_{\lambda \mu \eta u}\leftarrow 0\forall \lambda ,\mu ,\eta ,u$; ${marko}_{\lambda }\leftarrow 0(\lambda =1,2)$ and ${markd}_{\mu }\leftarrow 0(\mu =1,2)$ (Step 1). Let the origin $O_1$, the destination $D_2$ and the vehicle $V_2$ be selected (Step 2). Since $N_2^1=0$, the value of $N_2^1$ is changed to 1 (Step 3). Now, $Q=20(=minimum\{30,35,20-0\})$ and the updated values are $x_{1221}=20,a_1=10,a_2=50,b_1=45,b_2=15,sum=60$ and $N_2^1=2$ (Step 4). Since the value of $sum=60>0$, we go to Step 2.

Iteration 2: Let the origin $O_1$, the destination $D_2$ and the vehicle $V_1$ be selected (Step 2). Since $N_1^1=0$, the value of $N_1^1$ is changed to 1 (Step 3). Now, $Q=10\left(=minimum\left\{10,15,10-0\right\}\right)$ and the updated values are $x_{1211}=10,a_1=0,a_2=50,b_1=45,b_2=5,sum=50$ and $N_1^1=2$ (Step 4). Since $a_1$ becomes 0, the value of ${marko}_1$ is changed to 1 and the updated values are ${marko}_1=1,{marko}_2=0,{markd}_1=0,{markd}_2=0$. Since the value of $sum=50>0$, we go to Step 2.

Iteration 3: Let the origin $O_2$, the destination $D_1$ and the vehicle $V_1$ be selected (Step 2). Since $N_1^2=0$, the value of $N_1^2$ is changed to 1 (Step 3). Now, $Q=10\left(=minimum\left\{50,45,10-0\right\}\right)$ and the updated values are $x_{2111}=10,a_1=0,a_2=40,b_1=35,b_2=5,sum=40$ and $N_1^2=2$ (Step 4). Since the value of $sum=40>0$, we again go to Step 2.

Iteration 4: Let the origin $O_2$, the destination $D_2$ and the vehicle $V_1$ be selected (Step 2). Now, $N_1^2=2$. Thus, $Q=5(=minimum\left\{40,5,10-0\right\})$ and the updated values become $x_{2212}=5,a_1=0,a_2=35,b_1=35,b_2=0$ and $sum=35$ (Step 4). Since $b_1$ becomes 0, the value of ${markd}_1$ is changed to 1 and the updated values are ${marko}_1=1,{marko}_2=0,{markd}_1=1,{markd}_2=0$. Since the value of $sum=53>0$, we go to Step 2.

Iteration 5: Let the origin $O_2$, the destination $D_1$ and the vehicle $V_2$ be selected (Step 2). Since $N_2^2=0$, the value of $N_2^2$ is changed to 1 (Step 3). Now, $Q=20\left(=minimum\left\{35,35,20-0\right\}\right)$ and the updated values become $x_{2121}=20,a_1=0,a_2=15,b_1=15,b_2=0$, $sum=15$ and $N_2^2=2$ (Step 4). Since the value of $sum=15>0$, we go to Step 2.

Iteration 6: Let the origin $O_2$, the destination $D_1$ and the vehicle $V_2$ be selected (Step 2). Now, $N_2^2=2$. Thus, $Q=15(=minimum\left\{15,15,20-0\right\})$ and the updated values are $x_{2122}=15,a_1=0,a_2=0,b_1=0,b_2=0$ and $sum=0$ (Step 4). Since the value of $sum=0$, the process of generation of chromosome is completed.

The generated chromosome is given in Table 1 and the transportation scheme is represented diagrammatically in Fig. 1.

After the initial population is constituted, the fitness value of each chromosome is evaluated. In this paper, the binary tournament selection is used.

4.2. Crossover

In this paper, we develop a new crossover for the proposed model of FCTP. In this crossover, two child chromosome are obtained from two parent chromosomes, the selection of parent chromosomes being random from the mating pool. The process of generation of a child chromosome say, $ch\leftarrow \left(z_{\lambda \mu \eta u}\right)$ from two parent chromosomes $c{h}_1\leftarrow \left(x_{\lambda \mu \eta u}\right)$ and $c{h}_2\leftarrow \left(y_{\lambda \mu \eta u}\right)$ using the proposed crossover is provided in Algorithm 2.

After both the child chromosomes are obtained, the best two chromosomes among the parent and child chromosomes are selected to constitute the population of next generation.

Let us now illustrate the procedure of the proposed crossover two particular chromosomes $P_1\left(x_{\lambda \mu \eta u}\right)$ and $P_2\left(y_{\lambda \mu \eta u}\right)$ of the transportation network considered in Example 1. The chromosomes $P_1$ and $P_2$ are given in Table 2, the transportation network for which are represented diagrammatically in Fig. 2. Here, we illustrate the process of generation of a child chromosome $Q_1\left(z_{\lambda \mu \eta u}\right)$ only, the process of generation of the other child $Q_2\left(z_{\lambda \mu \eta u}^{\prime }\right)$ being similar.

Generation of a child chromosome from the parent chromosomes $P_1$ and $P_2$ :

At first, assign $a_1^{\prime }\leftarrow 30$, $a_2^{\prime }\leftarrow 50$, $b_1^{\prime }\leftarrow 45$, $b_2^{\prime }\leftarrow 35$, $N_{\eta }^{\lambda }\left(z\right)\leftarrow 0(\lambda =1,2;\eta =1,2)$, $sum\leftarrow 80(={\sum }_{\lambda =1}^m{a}_{\lambda }^{\prime })$ and $z_{\lambda \mu \eta u}\leftarrow 0\forall \lambda ,\mu ,\eta ,u$(Step 1). We have, $count=4$ and $LIST\left[4\right]=\left\{\left(O_1,D_1\right),\left(O_1,D_2\right),\left(O_2,D_1\right)\text{ and}\left(O_2,D_2\right)\right\}$ (Step 2). Assign $sig{n}_{\lambda }=0(\lambda =1,2,3,4)$ (Step 3).

Let us choose $id=1$. Thus, $\alpha =1$ and $\beta =1$. Also select $\xi =2$ (Step 4). Since $N_2^1\left(z\right)=0$, we change the value of $N_2^1\left(z\right)$ to 1 (Step 5). Then $u=1$ and $Q\leftarrow 20\left(minimum\{30,45,20\}\right)$, i.e., an amount of 20 units of the items is transported from the origin $O_1$ to the destination $D_1$ in first trip of vehicle $V_2$ originating from $O_1$. Then $z_{1121=}20$, $a_1^{\prime }=10$, $a_2^{\prime }=50$, $b_1^{\prime }=25$, $b_2^{\prime }\leftarrow 35$, $sum=60$ and $N_2^1\left(z\right)=2$(Step 6). Since $sum=60>0$, we go to Step 4.

Since $sig{n}_{\lambda }=0\forall \lambda =1,2,3,4$, we can choose $id\in \{1,2,3,4\}$. Let us choose $id=4$. Thus, $\alpha =2$ and $\beta =2$. Also select $\xi =2$ (Step 4). Since $N_2^2\left(z\right)=0$, we change the value of $N_2^2\left(z\right)$ to 1 (Step 5). Then $u=1$ and $Q\leftarrow 20\left(minimum\{50,35,20\}\right)$, i.e., an amount of 20 units of the items is transported from the origin O₂ to the destination $D_2$ in first trip of vehicle $V_2$ originating from $O_2$. Then $z_{2221=}20$, $a_1^{\prime }=10$, $a_2^{\prime }=30$, $b_1^{\prime }=25$, $b_2^{\prime }\leftarrow 15$, $sum=40$ and $N_2^2\left(z\right)=2$(Step 6). Since $sum=40>0$, we go to Step 4.

Since $sig{n}_{\lambda }=0\forall \lambda =1,2,3,4$, we can choose $id\in \{1,2,3,4\}$. Let us choose $id=2$. Thus, $\alpha =1$ and $\beta =2$. Also select $\xi =1$ (Step 4). Since $N_1^1\left(z\right)=0$, we change the value of $N_1^1\left(z\right)$ to 1 (Step 5). Then $u=1$ and $Q\leftarrow 10\left(minimum\{10,15,10\}\right)$, i.e., an amount of 10 units of the items is transported from the origin $O_1$ to the destination $D_2$ in first trip of vehicle $V_1$ originating from $O_1$. Then $z_{1211=}10$, $a_1^{\prime }=0$, $a_2^{\prime }=30$, $b_1^{\prime }=25$, $b_2^{\prime }\leftarrow 5$, $sum=30$ and $N_2^2\left(z\right)=2$(Step 6). Since $a_1^{\prime }$ becomes 0, $sig{n}_1=1$ and $sig{n}_2=1$ (Step 7). Again, $sum=30>0$, we go to Step 4.

Since $sig{n}_1=1$, $sig{n}_2=1$, $sig{n}_3=0$ and $sig{n}_4=0$, we can choose $id\in \{3,4\}$. Let us choose $id=4$. Thus, $\alpha =2$ and $\beta =2$. Also select $\xi =1$ (Step 4). Since $N_1^2\left(z\right)=0$, we change the value of $N_1^2\left(z\right)$ to 1 (Step 5).

Then $u=1$ and $Q\leftarrow 5\left(minimum\{30,5,10\}\right)$, i.e., an amount of 5 units of the item is transported from the origin O₂ to the destination $D_2$ in first trip of vehicle $V_1$ originating from $O_2$. Then, $z_{2211=}5$, $a_1^{\prime }=0$, $a_2^{\prime }=25$, $b_1^{\prime }=25$, $b_2^{\prime }=0$, $sum=25$ and $N_1^2\left(z\right)=2$(Step 6). Since $b_2^{\prime }$ becomes 0, $sig{n}_1=1$, $sig{n}_2=1$, $sig{n}_3=0$ and $sig{n}_4=1$ (Step 7). Again, since $sum=25>0$, we go to Step 4.

Since $sig{n}_1=1$, $sig{n}_2=1$, $sig{n}_3=0$ and $sig{n}_4=1$, the only value that can be chosen is $id=3$. Thus, $\alpha =2$ and $\beta =1$. Also, select $\xi =1$ (Step 4). We have $N_1^2\left(z\right)=2$. Thus, $u=2$ and $Q\leftarrow 10\left(minimum\{25,25,10\}\right)$, i.e., an amount of 10 units of the items is transported from the origin O₂ to the destination $D_1$ in second trip of vehicle $V_1$ originating from $O_2$. Then $z_{2112=}10$, $a_1^{\prime }=0$, $a_2^{\prime }=15$, $b_1^{\prime }=15$, $b_2^{\prime }=0$, $sum=15$ and $N_1^2\left(z\right)=3$(Step 6). Since $sum=15>0$, we go to Step 4.

Since $sig{n}_1=1$, $sig{n}_2=1$, $sig{n}_3=0$ and $sig{n}_4=1$, the only value that can be chosen is $id=3$. Thus, $\alpha =2$ and $\beta =1$. Let us choose $\xi =2$ (Step 4). We have $N_2^2\left(z\right)=2$. Thus, $u=2$ and $Q\leftarrow 15\left(minimum\{15,15,20\}\right)$, i.e., an amount of 15 units of the items is transported from the origin O₂ to the destination $D_1$ in the second trip of vehicle $V_2$ originating from $O_2$. Then $z_{2122=}15$, $a_1^{\prime }=0$, $a_2^{\prime }=0$, $b_1^{\prime }=0$, $b_2^{\prime }=0$, $sum=0$ and $N_1^2\left(z\right)=3$(Step 6). Since $sum=0$, the generation of the child chromosome $Q_1\left(z_{\lambda \mu \eta u}\right)$ is completed.

The child chromosomes $Q_1$ and $Q_2$ obtained from the parent chromosomes $P_1$ and $P_2$ are given in Table 3. The diagrammatic representation of $Q_1$ and $Q_2$ are given in Fig. 3.

4.3. Mutation

In this paper, a new mutation suitable for the proposed problem is developed. The process of the proposed mutation operation is described in Algorithm 3.

Let us illustrate the process of the proposed mutation for a particular chromosome $ch\leftarrow \left(x_{\lambda \mu \eta u}\right)$, as given in Table 4, for which the transportation scheme is represented diagrammatically in Fig. 4(a). Let the chromosome to be obtained after the mutation be $c{h}^{\prime }\leftarrow \left(x_{\lambda \mu \eta u}^{\prime }\right)$.

Assign $N_{\eta }^{\lambda }\left(x^{\prime }\right)\leftarrow N_{\eta }^{\lambda }\left(x\right)(\lambda =1,2;\eta =1,2)$ and $x_{\lambda \mu \eta u}^{\prime }\leftarrow x_{\lambda \mu \eta u}\forall \lambda ,\mu ,\eta ,u$ (Step 1). Let us select ${\beta }_1=1$ and ${\alpha }_1=1$, ${\xi }_1=2$. We get $u_1=2$ (Step 2). Again, let us select ${\beta }_2=2$ and ${\alpha }_2=2$, ${\xi }_2=2$ and $u_2=2$ (Step 3-5). Then $Q=10\left(minimum\{x_{1122}^{\prime }=10,x_{2222}^{\prime }=15\}\right)$ and the updated values are obtained as $x_{1122}^{\prime }=0$, $x_{2222}^{\prime }=5$. Also, since $u_1=N_2^1\left(x^{\prime }\right)$ and $x_{1122}^{\prime }=0$, the value of $N_2^1\left(x^{\prime }\right)$ is decreased by 1 i.e., $N_2^1\left(x^{\prime }\right)=1$ (Step 6). Next, we have $Q_1=10$ and select the vehicle say, $V_1$(Step 7). Since $N_1^1\left(x^{\prime }\right)=0$, the value of $N_1^1\left(x^{\prime }\right)$ is increased by 1 i.e., $N_1^1\left(x^{\prime }\right)=1$ (Step 8). Then $q_1=10\left(minimum\{10,10-0\}\right)$ and $x_{1211}^{\prime }=10,Q_1=0$ (Step 9). Since $Q_1=0$, we go to Step 11 ( Step 10 ).

We have $Q_2=10$ and select the vehicle say, $V_1$(Step 11). Now, $N_1^2\left(x^{\prime }\right)=2$ and we obtain $q_2=5\left(minimum\{10,10-5\}\right)$ and hence $x_{2112}^{\prime }=10$, $Q_2=5$. Also, since $e_1=x_{1212}$, the value of $N_1^2\left(x^{\prime }\right)$ is increased by 1 i.e., $N_1^2\left(x^{\prime }\right)=3$(Step 13). Since $Q_2=5>0$, we again go to Step 11 and select a vehicle say, $V_1$. Now, $N_1^2\left(x^{\prime }\right)=3$ and we obtain $q_2=5\left(minimum\{5,10-0\}\right)$ and hence $x_{2113}^{\prime }=5$, $Q_2=0$ (Step 13). Since $Q_2=0$, the process is completed and is given in Table 4. A diagrammatic representation of the chromosome after mutation is given in Fig. 4(b).

5.1. Dataset

The proposed model is different from the existing models of FCTP, and so, we generate new datasets according to our model. For experimental purpose, we consider five numerical examples of the same size given in Lofti & Tavakkoli-Moghaddam [14] (i.e., $4\times 5,5\times 10,10\times 10,10\times 20$ and 20 × 30). Thus, for these numerical examples, we take the availability and demands for the item as given in Lofti & Tavakkoli-Moghaddam [14]. We consider two vehicles, say, $V_1$ and $V_2$ with capacities 10 and 20 units, respectively, corresponding to each numerical example. The variable and fixed costs corresponding to the vehicles $V_1$ and $V_2$ for the numerical example with 20 origins and 30 destinations are generated randomly within the ranges $[4,12]$ and $[50,135]$, and are presented in Appendix. The variable and fixed cost matrices for a numerical example of smaller size, say, $m^{\prime }\times n^{\prime }$ (where $m^{\prime }\le 20,n^{\prime }\le 30$) is taken as the sub-matrix of order $m^{\prime }\times n^{\prime }$, starting from the north-west corner of the corresponding matrix of size 20 × 30. For each numerical example, we categorize the regions in three groups, in which the level of restrictions are ‘high’, ‘medium’ and ‘low’, and are marked in ‘Red’, ‘Orange’ and ‘Green’, respectively. The list of origins and destinations belonging to each group are presented in Table 5.

5.2. Parameter settings

To obtain the best possible solution using the algorithm, the control parameters of the algorithm such as $X_0,I{t}_{max},p_{cros}$ and $p_{mut}$ are set to values that produce promising results in preliminary testing. The parameter values of $X_0,I{t}_{max}$ used to solve the numerical examples of different size are shown in Table 6. The values of the parameters $p_{cros}$ and $p_{mut}$ are taken as 0.8 and 0.15 for each numerical example.

5.3. Results and discussion

In this section, we describe the method for computation of penalty for the proposed FCTP, which depends upon the level of restriction of the regions in which the origins and the destinations are located. The purpose of imposing penalty in the objective function is to lower the number of trips of vehicles if the level of restriction in the regions are ‘high’. For this purpose, we associated a numerical value corresponding to each category of regions, and term as Level of Severity of Restriction (LSR) value. The process of computation of penalty is given as follows.

In a pandemic scenario, if the regions be categorized in $K$ different groups, say, $G_1,G_2,\dots ,G_K$, then the region $G_{\lambda }$ is assigned a LSR value $v_{\lambda }$ that lies between 1 and $K$ in the relative ranking of the regions when arranged in increasing order of level of restrictions. The LSR value of a region is assigned zero, when no restrictions are imposed in a region. For a trip of any vehicle from an origin $O_{\lambda }$ to a destination $D_{\mu }$ located in regions $G_r$ and $G_s$ respectively, the penalty is denoted by $P_{rs}$, and computed as $\left[max\left\{v_r,v_s\right\}+|v_r-v_s|\right]\ast M$, where $M$ is a large positive number and $v_r,v_s$ are the LSR values of the regions $G_r$ and $G_s$ respectively. For solving the numerical examples, we have chosen the value of $M$ as 100.0.

Explanation:

The reason of including the terms $max\left\{v_r,v_s\right\}$ and $|v_r-v_s|$ in the penalty function is to consider higher penalty when either of the situation occurs, (i) at least one of the regions in which an origin or a destination situated takes large LSR value, i.e., the restriction is high (ii) the difference in LSR values of the two regions associated with a tour of any vehicle is large, i.e., the level of restriction in one of the two regions is low, whereas, the level restriction in the other region is high.

Let us illustrate the process of computation of penalty for a particular example. For this, let us consider the transportation network given in Example 1 (Ref. Section 4.1.). Let us consider that the regions be categorized in three groups, and are marked in ‘Red’, ‘Orange’ and ‘Green’. Then, the penalty for a single trip of a vehicle for different possible combinations of LSR values corresponding to an origin and a destination is given in Table 7.

In this section, we discuss the results obtained for the five numerical examples solved for each of the problems, namely, the proposed FCTP (given in (7)), the corresponding problem without any constraint on transportation cost and the problem in normal scenario (given in (8)). Each of the problems are solved taking three different forms of the cost function, viz., the linear fixed-charge form, quadratic fixed-charge form (non-linear) and the reduced classical form (a special case of the fixed charge forms in which the fixed costs are taken to be zero). Consequently, a total of 15 instances are solved for each problem, and a total of 45 $\left(=15\times 3\right)$ instances are solved in this paper. The best found objective function value, penalty value and the total number of trips for each example corresponding to the linear and quadratic form of cost function are presented in Table 8 and Table 9, respectively. The corresponding results for the reduced CTP are presented in Table 10. Due to the randomness nature of GA, 20 independent runs are taken for each instance of a numerical example.