We present a 2E-IRP whose distribution network is made up of a supplier, several DCs and several retailers. Figure 1 gives a graphical representation of the distribution network as well as the different types of routes. In this problem, the transport of a single type of product is ensured by a homogeneous fleet of vehicles parked in the DCs. Retailer procurement is carried out only from the DCs at the second level (DCs-retailer). At the first level (supplier-DCs), the supply of the DCs is carried out by the same fleet of vehicles parked in the DCs. Three types of routes are considered in the route of a vehicle associated with a given DC. We have the supply round, which consists of a vehicle associated with a DC leaving this DC to collect products from the supplier and return to the DC. The second type of round is a delivery round. This tour consists of a vehicle associated with a DC delivering orders from a set of retailers and returning to the DC. The third type of round is a delivery and collection round. In this round, a vehicle first delivers orders from retailers and then goes to the supplier for product collection to supply the DC with which it is associated. The objective of this work is to minimize the overall cost of vehicle routing and inventories in the DCs and at retailers over a finite and discreet planning horizon.

G(N, A) is a complete graph in which “N” represents the set of vertices materialized by the supplier, the DCs and the retailers. “A” represents the set of arcs of the graph $A\left(N\right)=\left\{\left(i,j\right):i,j\in N,i\ne j\right\}$. The supplier is identified by the index “0”. Retailers and DCs are represented by the sets $\left\{1,\cdots ,n\right\}$ and $\left\{n+1,\cdots ,N\right\}$ respectively where n is the number of retailers. Figure 1 represents a distribution network with a supplier (0), two DCs (16 and 17) and 15 retailers (from 1 to 15).

The sets:

$N=\left\{0,\cdots ,n,n+1,\cdots ,\left|N\right|\right\}$: The set of vertices representing the supplier, the retailers and the DCs.

$N_R=\left\{1,\cdots ,n\right\}$: All retailers.

$N_D=\left\{n+1,\cdots ,\left|N\right|\right\}$: All DCs.

$N_{0R}=\left\{0,\cdots ,n\right\}$: The whole constituted by the supplier and the retailers.

$N_{RD}=\left\{1,\cdots ,\left|N\right|\right\}$: The set made up of retailers and DCs.

$T=\left\{1,\cdots ,l\right\}$: All periods of the planning horizon (in days).

$K_y=\left\{1,\cdots ,m_y\right\}$: All the vehicles associated with the DC y.

$K={\displaystyle \cup K_y}$.

The indexes:

$i$ and $j$ are the indices of all the vertices of graph G.

$y$ is an index associated with the DC.

$t$ represents the index of the periods of the planning horizon.

$k$ is the index of vehicles.

The settings:

$h_i$ is the unit cost of inventories at vertex i.

$d_{it}$ represents the demand of retailer i at period t.

$L_i$ is the maximum storage capacity at vertex i.

$I_{i0}$ is the initial stock level at vertex i.

$c_{ij}$ is the cost of transportation when a vehicle travels directly from vertex i to vertex j.

$Q$ is the maximum load of a vehicle.

$M{C}_{it}=\min \left\{L_i,Q,{\displaystyle {\sum }_{\tau =t}^l{d}_{i\tau }}\right\}$ is the maximum quantity of products that can be delivered to a retailer i during a period t.

Decision variables:

$I_{it}$: level of stocks at vertex i during period t.

$q_{ikt}^y$: quantity delivered to node i by vehicle k associated with DC y during period t.

$Z_{ikt}^y$: binary variable, equal to 1 if vertex i is visited by vehicle k associated with DC y during the period t or 0 otherwise.

$x_{ijkt}^y$: binary variable, equal to 1 if vehicle k associated with DC y travels directly from the vertex i to vertex j or 0 otherwise.

Mathematics formulation:

$Z=\min {\displaystyle {\sum }_{i\in N_{RD}}{\displaystyle {\sum }_{t\in T}{h}_i{I}_{i,t}}}+{\displaystyle {\sum }_{\left(i,j\right)\in A}{\displaystyle {\sum }_{k\in K_y}{\displaystyle {\sum }_{y\in N_D}{\displaystyle {\sum }_t{x}_{ijkt}^y}}}}$ (1)

S.T

${\displaystyle {\sum }_{k\in K}{q}_{ykt}^y}+I_{yt-1}={\displaystyle {\sum }_{i\in N_R}{\displaystyle {\sum }_{k\in K}{q}_{ikt}^y}}+I_{yt},\forall y\in N_D,\forall t\in T$ (2)

${\displaystyle {\sum }_{k\in K}{\displaystyle {\sum }_{i\in N_R}{q}_{ikt}^y}}\le I_{y,t-1},\forall y\in N_D,\forall t\in T$ (3)

$q_{ykt}^y\le Q{z}_{0kt}^y,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (4)

${\displaystyle {\sum }_{y\in N_D}{\displaystyle {\sum }_{k\in K_y}{q}_{ikt}^y}}+I_{it-1}=d_{it}+I_{it},\forall i\in N_R,\forall t\in T$ (5)

${\displaystyle {\sum }_{i\in N_R}{q}_{ikt}^y}\le Q{z}_{ykt}^y,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (6)

$q_{ikt}^y\le M{C}_{it}{z}_{ikt}^y,\forall i\in N_R,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (7)

$I_{it}\le L_i,\forall i\in N_{RD}$ (8)

$q_{0kt}^y=0,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (9)

${\displaystyle {\sum }_{i\in N_{0R}}{x}_{iykt}^y}+{\displaystyle {\sum }_{j\in N_{0R}}{x}_{yjkt}^y}=2Z_{ykt}^y,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (10)

${\displaystyle {\sum }_{i\in N_{RD}}{x}_{iakt}^y}+{\displaystyle {\sum }_{j\in N}{x}_{ajkt}^y}=2Z_{akt}^y,\forall a\in N_R,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (11)

${\displaystyle {\sum }_{i\in N_{RD}}{x}_{i0kt}^y}+x_{0ykt}^y=2Z_{0kt}^y,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (12)

${\displaystyle {\sum }_{y\in N_D}{\displaystyle {\sum }_{k\in K_y}{z}_{ikt}^y}}\le 1,\forall i\in N_R,\forall t\in T$ (13)

${\displaystyle {\sum }_{k\in K}{\displaystyle {\sum }_{y\in N_D}{\displaystyle {\sum }_{t\in T}{Z}_{ikt}^y}}}\ge 1,\forall i\in N_R$ (14)

${\displaystyle {\sum }_{j\in N}{x}_{ijkt}^y}=z_{ikt}^y,\forall i\in N,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (15)

${\displaystyle {\sum }_{i\in S}{\displaystyle {\sum }_{j\in S}{x}_{ijkt}^y}}\le \left|S\right|-1,\forall S\subseteq N_{0R},\left|S\right|\ge 2,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (16)

${\displaystyle {\sum }_{y\in N_D}{\displaystyle {\sum }_{k\in K}{Z}_{ykt}^y}}\le m,\forall t\in T$ (17)

$I_{it},q_{ikt}^y\ge 0,\forall i\in N,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (18)

$x_{ijkt}^y,z_{ikt}^y\in \left\{0,1\right\},\forall \left(i,j\right)\in A\left(N\right),\forall i\in N,\forall y\in N_D,\forall k\in K_y,\forall t\in T$ (19)

Equation (1) is the Objective function. It allows us to simultaneously minimize the overall cost of inventories (DCs and retailers) and the cost of the different vehicle routes associated with each DC. Constraints (2) reflect the conservation of the flow of products in each DC. They state that in each DC, the quantity of products received per period and the stock of products from the previous period allow deliveries to be made to retailers and to build up new stock for the current period. Constraints (3) stipulate that in a DC, deliveries for a given period are made with the previous period’s stock. Constraints (4) indicate that any vehicle that makes a delivery in a DC must first visit the supplier before making its delivery in the DC. Constraints (5) represent the maintenance of product flows at retailers. They express the fact that the quantity of products received during a period t and the quantity of products stored during period t − 1 make it possible to satisfy demand and to constitute the stock of products in period t. Inequalities (6) limit the maximum load of a vehicle leaving a DC to a quantity Q. Constraints (7) are used to limit the quantity of products that can be delivered to a retailer for a given period. Inequalities (8) impose a limit on the level of stocks both in retailers and in DCs. Constraints (9) indicate that the load of a vehicle intended for the collection of products from the supplier is zero. Figure 2 shows the predecessor and successor of each vertex for the formalization of vehicle routing constraints. The constraints associated with vehicle routing are as follows: a vehicle that visits a DC must be directly from the supplier or a retailer, and any vehicle that leaves a DC immediately visits a retailer or the supplier. So, for ease of understanding, we will say that the predecessor of a DC is the supplier or a retailer and the successor of a DC is the supplier or a retailer. This situation is reflected in Constraints (10). Similarly, the predecessor of a retailer is either a DC or a retailer and the successor of a retailer is either a retailer, a supplier or a DC (see Constraint (11)). As far as the supplier is concerned, he accepts a retailer or a DC as his predecessor and is succeeded by a DC (Constraint (12)). Constraints (13) indicate that each retailer must be visited no more than once per period. Constraints (14) require that each retailer be visited at least once over the planning horizon. Constraints (15) indicate the activation of a path to the successor of an i-vertex if that vertex is visited by a vehicle during a given period. Inequalities (16) are the subtours elimination constraints (SECs) on the vehicle paths. Constraints (17) limit the total number of vehicles assigned to distribute products in the supply chain. Constraints (18) indicate that the variables representing stock levels in the DCs and at retailers as well as the quantities of products transported by the vehicles must have positive values. And Constraints (19) represent the constraints for defining the binary variables of the mathematical model. In the rest of this work, we present a Branch and Cut (B&C) algorithm to solve the MILP defined in this session.

These inequalities prevent a vehicle from returning empty to the DC after visiting the supplier.

Under these constraints, any DC that has been supplied at least once over the planning horizon is bound to make product deliveries to retailers.

These constraints state that the path of a vehicle between two vertices i and j is an arc. Thus, if a vehicle travels directly from vertex i to vertex j then it can no longer travel directly from vertex j to vertex i.

These constraints make it possible to ensure that the use of vehicle k + 1 is conditional on the use of vehicle k, so that vehicle k + 1 can be used only after the use of vehicle k.

These constraints prevent a vehicle associated with a DC from visiting another DC.

These constraints make it possible to ensure that a vehicle visiting a retailer delivers a quantity of products at least equal to its demand.

To implement our B&C algorithm, we add the valid inequalities expressed by Constraints (20)-(25) in the initial model expressed by the objective function (1) and the constraints ranging from (2) to (19) and we subtract the SECs expressed by Constraints (16). Next, the new linear program (LP) is solved. If the LP solution contains a subtour in the path of a vehicle k associated with a DC for a given period t, then all the subtours in the path of this vehicle are identified and each SEC relative to the vertices concerned by each subtour is generated and added to the PL for a new reoptimization. Table 1 gives the details of our B&C algorithm. In this algorithm, the star symbol (*) represents a component of the PL solution at a node in the B&C search tree. Our B&C algorithm is an improvement of the algorithm developed by [48] for a 2E-PRP with a supplier, a depot and several customers. In this work, the authors attempt to separate all the vertices visited by a vehicle k at a time t given in two partitions. The first partition consists of the subset of vertices forming a Hamiltonian circuit with the depot, and the second partition is the complement of this first partition in the set of vertices visited by the vehicle k at the same time t. Thus, when the second partition is empty then the vehicle k does not contain any subtour in its period path t. However, if the second partition is non-empty then a subtour is detected in the path of vehicle k at the period t. In the latter case, SECs are generated and added to the model for reoptimization. In the algorithm developed in our work, we exhaustively determine all the subsets of vertices forming a Hamiltonian circuit in the path of vehicle k associated with the depot y during the period t. If the number of subsets of vertices forming a Hamiltonian circuit is greater than 1, then a SEC is generated and added to the model for each vertex subset.

We use the Concert Technology library from IBM ILOG CPLEX Optimization Studio and the C++ programming language to implement the B&C algorithm exposed in Session 4. The tests are conducted on an AMD Athlon Silver 3050U LAPTOP with Radeon Graphics 2.30 GHz and 8.00 GB of installed RAM. The dataset used for the tests is drawn from the work of [40] . From this work, we have retained the dataset for which the instances are composed of a single supplier. Table 2 gives a description of the dataset used to perform the tests with our B&C algorithm. In this table, the first column indicates the typology of inventory costs. In this typology, we have instances with high unit inventory costs over a 3-period planning horizon (H3) and instances with low unit inventory costs over a 3-period planning horizon (L3). The second column shows the different instance classes from abs1 to abs5. Similarly, going from the third column to the last column, we have respectively the supplier, the DCs, the customers and the homogeneous vehicles used to make the deliveries of the products on the planning horizon. The charge of each Q vehicle in our tests is equal to the average of the charge of a vehicle in the first echelon ( $Q_1$) and a vehicle in the

second echelon ( $Q_2$). So, $Q=\left(Q_1+Q_2\right)/2$. By multiplying the cardinal of each instance parameter, we obtain a total of 2 * 5 * 1 * 1 * 5 * 4 = 200 instances (see Table 2 ).

Considering that a solution is optimal when its integrality gap is less than 0.05% [40] , our algorithm resolved 168 instances to the optimal (83%), 29 instances to non-optimal (14.5%) and 3 unresolved instances (1.5%). Table 3 illustrates the distribution of the solutions obtained in relation to their integrality gap. To show the effectiveness of the results of our B&C algorithm on the 2E-IRP with ETR policy, we present the details of the result obtained on the H3_abs1_1S_2M_5C_2V-M instance. The instance designates a set of data with a high inventory cost over a planning horizon of three days (H3) in the abs1 type instance register. The instance considers one supplier (1S), two DCs (2M) and two medium-sized vehicles (2V-M). According to the details of the results contained in Table A1 , the cost of inventories is equal to 222.32 and the cost of vehicle routes is equal to 2187. This gives a total cost of 2411.32. To verify these costs, we start by presenting the vehicle route file for the H3_abs1_1S_2M_5C_2V-M instance represented in Figure 3 . The Readers will find the file for Table A1 and the rest of the files for Figure 3 in link A of Appendix. In this figure, the vehicle k = 1 associated with the DC of index 7 (second DC) leaves the DC in the first period and successively visits retailers (customers) 5 and 2, giving them respectively the quantities of products 22 and 35. After the visit to the last retailer (2), the vehicle goes to the supplier (0) and collects a quantity of products equal to 194 for the supply of the DC. In period two, the vehicle k = 1 leaves the second DC of index 7 and then successively visits retailers 3, 4, 1 delivering them respectively quantities 116, 24, 65 before returning empty to the DC. In the third period, there is no delivery of products. Let’s start with the determination of the cost of inventory over the three periods of the planning horizon. Tables 4-6 summarize the calculations of inventory costs over these three periods.

By postponing vehicle routes and the quantities received in Figure 3 represented respectively by line (i) and line (q_i) of Table 4 , we determine the level of inventories at the end of the period (I_i) as follows: at vertex i, we have I_i = level of inventories of the previous period + quantity of products received during the current period − quantity of products consumed for the same period. And we determine the costs of inventories by multiplying the inventory level of the period by the unit cost of storage.

This is translated by the equation IC_i = I_i * h_i. Let’s calculate these values for DC 7, visited customer 5, and unvisited customer 1 for period t = 1. We have I₇ = 68 + 194 − (22 + 35) = 205 and IC₇ = 205 * 0.3 = 61.5. For retailer 5, we have I₅ = 11 + 22 – 11 = 22 and IC₅ = 22 * 0.18 = 3.96. for the unvisited retailer (1), we have I₁ = 130 + 0 – 65 = 65 and IC₁ = 65 * 0.23 = 14.95. Similarly, we calculate the inventory costs for each DC and each retailer over the three periods of the planning horizon. The total cost of inventory is therefore equal to the sum of the periodic inventory costs, i.e. IC = 128.73 + 73.19 + 20.4 = 222.32. For determination of the overall cost of vehicle routes, we organize the calculations in Table 7 . In this table, 7 → 5 indicates that the vehicle travels directly from vertex 7 to vertex 5. Vertex 7 has coordinates (1, 360) and vertex 5 has coordinates (38, 152). This allows us to calculate the distance from vertex 7 to vertex 5 (d(7, 5)). Thus, $d\left(7,5\right)=\mathrm{int}\left(\sqrt{{\left(1-38\right)}^2+{\left(360-152\right)}^2}+0.5\right)=\sqrt{{\left(1-38\right)}^2+{\left(360-152\right)}^2}+0.5$, so d(7, 5) = 211. We can therefore determine the overall cost of vehicle routes by summing the costs of routes per period. Hence the overall cost of vehicle rounds Trans_Cost = 1431 + 758 = 2189. Finally, the overall cost is equal to the sum of the cost of transportation and the cost of inventory. Thus, we have Total_Cost = Inv_Cost + Trans_Cost = 222.32 + 2189 = 2411.32 monetary unit. In the continuation of this work, we compare the results of our B&C algorithm with the results of the branch and price (B&P) developed by [40] . In both jobs, it is forbidden to deliver to a customer (or retailer) from the supplier. However, the two models differ in the strategy adopted in the procurement of DCs (or Satellites). In the procurement strategy of these authors, the vehicle dedicated to the supply of DCs starts its run at the supplier and visits one or more DCs before ending its run at the supplier. Whereas in our model, the same fleet of vehicles based at the DC is used to make the delivery to the customers and if necessary, one or more vehicles from this fleet go directly to the supplier or after the last retailer’s visit to collect products that will be used to supply the DC (ETR policy). In Tables 8-12 , the rows represent the average of the results obtained with the two, three, four and five vehicle instances. The time (B&P) and time (B&C) columns represent the time taken by the B&P algorithm and the B&C algorithm to find the respective upper bounds (UB) respectively. The Time Gap % column determines the rate of increase (positive value) or rate of reduction (negative value) of the time associated with the execution of the B&C algorithm relative to the B&P algorithm. The values in this column are calculated as follows: Time GAP % = (time (B&C) − time (B&P))/time (B&P). Similarly, UB GAP % = (UB (B&C) − UB (B&P))/UB(B&P). In Table 8 representing instances of 5 retailers, we see that the average B&C GAP % for each line is equal to 0. This means that each different fleet instance associated with each instance line of five retailers is resolved to the optimum. This is also reflected in the average of the average GAPs of the B&C at the AVERAGE line. In addition, the values of the numbers in the UB GAP % column are all negative. This means that cost reductions are made in favor of the B&C algorithm. For example, we have a discount of 23.93% on average on type H3_abs1_1S_2M_5C instances (4 instances) and overall, we get an average discount of 16.69% on instances of 5 retailers (AVERAGE line). As far as the calculation time is concerned, we also get an overall average decrease of 15.55% (Time GAP%) in favor of the B&C algorithm. Similarly, Table 9 shows that all instances of ten retailers were resolved to the optimal (GAP B&C % column) with an overall average cost (UB) decrease of 23.45% (AVERAGE UB GAP%) and an overall decrease in computation time of 90.06%. Table 10 summarizes the results for the instances of fifteen retailers. This table also shows the same trends as Table 8 and Table 9 . It also shows that all instances of fifteen retailers are resolved to the optimal with a decrease in average overall cost by 20.66%. We also see an overall average decrease in computation time of 67.90%. Table 11 shows the results obtained on the instances of 20 retailers. Unlike the previous tables, this table has a non-zero overall average of the GAP B&C % (2.76%). This implies that some instances of twenty retailers have not been resolved to the optimal level. Table A1 of the details in Appendix does show that the instances for twenty retailers H3_abs1_1S_2M_20C_3V, H3_abs1_1S_2M_20C_5V, H3_abs2_1S_2M_20C_5V, and instance H3_abs3_1S_2M_20C _5V were not resolved to the optimal level. However, we achieve a 19.87% reduction in average overall cost in favor of the B&C algorithm compared to the B&P algorithm. We also have a decrease in computation time of 85.59%. Table 12 highlights the results obtained on the instances of twenty-five retailers with an average overall B&C GAP % of 15.46%, we can say that several instances have not been resolved to the optimal. The UB GAP % column shows that the B&C algorithm allows for an average overall cost reduction of 9.13% across instances of 25 retailers. It should be noted that for instances with twenty-five retailers, our B&C algorithm resulted in a cost increase of 12.31, 4.86 and 4.47 respectively over the average of the H3_abs2_1S_2M_25C, H3_abs3_1S_2M_25C and H3_abs5_1S_2M_25C instances. This represents a total of 9 instances of twenty-five retailers. In addition, we also see an overall average decrease in calculation time of 63.96%. Overall, our ETR policy and B&C algorithm resulted in 185 (93.9%) best solutions out of the 197 instances for which we were able to obtain a solution.