Capacitated multiple allocation hub location problems under the risk of interdiction: model formulations and solution approaches

Abstract

Hub-and-spoke networks play a critical role in reducing cost and enhancing service levels in various infrastructural sectors since hubs act as the consolidation and transshipment points of the flows. The failure of hubs in such a network can cause severe disruptions. While disruptions can be natural or man-made, a disruption by a rational individual or entity can be significantly detrimental to the network and is often studied as an interdiction problem. It is important to take interdiction effects at the design stage; therefore, we study the three-level capacitated hub-and-spoke network design problem from the perspective of a defender who considers the risk of interdiction by a rational attacker. Within the three levels, the upper level represents the network design level, and the lower two levels represent the bi-level hub interdiction problem. The introduction of capacity constraints within an interdiction model dramatically increases the complexity of the problem, as there can be some unfulfilled flows post-interdiction. Moreover, a flow may or may not be fulfilled through the least-cost route using the nearest hubs. This work makes two major contributions: the first contribution is on the efficient handling of the bi-level hub interdiction problem using the Dual-based approach and the Penalty-based approach, and the second contribution is on solving the overall three-level problem using a super valid inequality. These two contributions allow us to solve large-scale versions of the capacitated multiple allocation p-median hub location problem under the risk of interdiction, which is otherwise mathematically intractable and can be handled only using complete enumeration techniques.

Appendix A Proofs of Propositions

1.1 A.1 Proof of Proposition 1

There are two possible cases:

• Case 1: Total capacity of the surviving hubs < Total flow demand of the network

  There will be at least one source-destination pair (i, j) for which Constraint (12) will have a strict inequality. And so, there is going to be some unfulfilled flow demand. For this case, any positive value of $P_{ij}$ will work to incur a penalty for not fulfilling the demand.

• Case 2: Total capacity of the surviving hubs $\ge$ Total flow demand of the network

  In this case, all the flow demands have to be satisfied by taking any route having an unsaturated hub. Suppose for a given source-destination pair (i, j), only $\eta$ fraction of the flow is remaining to be allocated. In the worst-case scenario, this demand will be satisfied through the only possible path having the highest transportation cost. Then, from the objective function (11), we can see that:

  Worst case cost to fulfill this demand = $\eta W_{ij} \max _{k,m} \{d_{ijkm}\}$.

  On the other hand, the cost of not fulfilling this demand = $\eta P_{ij}$.

  Hence, to ensure that the remaining unfulfilled demand is satisfied, as we have enough capacity left to accommodate that, the following condition needs to be satisfied

  $$\begin{aligned}&\eta P_{ij}> \eta W_{ij} \max _{k,m} \{d_{ijkm}\} \\&\quad \implies P_{ij} > W_{ij} \max _{k,m} \{d_{ijkm}\} \end{aligned}$$

Hence, $\tilde{P}_{ij} = W_{ij} \max _{k,m} \{d_{ijkm}\} + \epsilon$ is a valid value for $P_{ij}$ for any positive $\epsilon$.

1.2 A.2 Proof of Proposition 2

Suppose $\bar{P}_{ij}$ is the tight value of $P_{ij}$ such that $\bar{P}_{ij} < \tilde{P}_{ij}$. Then, $\bar{P}_{ij} = W_{ij} \max _{k,m} \{d_{ijkm}\}$. Now, consider case 1 of the proof for Proposition 1. As the cost for fulfilling this demand through the only available path having the highest cost is equal to the penalty for not fulfilling it, there is a possibility that the mathematical program leaves this demand unfulfilled even when the network has sufficient capacity to fulfill it. So, $\bar{P}_{ij}$ is not a valid value for $P_{ij}$. Therefore, $\tilde{P}_{ij}$ is the best-known value of $P_{ij}$, and any value less than this may lead to under-utilization of the available capacity.

1.3 A.3 Proof of Proposition 3

From constraint sets (21),(22) and (25), we get

$$\begin{aligned}&P_{ij} - W_{ij}d_{ijkm} -\phi _{ij} - W_{ij}\theta _{k} \le \delta _{ijk}&\forall i,j \in N; \forall k,m \in H; m = k \\&P_{ij} - W_{ij}d_{ijkm} -\phi _{ij} - W_{ij}\theta _{k} - \delta _{ijm} \le \delta _{ijk}&\forall i,j \in N; \forall k,m \in H; m \ne k \\&0 \le \delta _{ijk}&\forall i,j \in N; \forall k \in H \end{aligned}$$

As $\delta _{ijk}$ appears with a negative sign in the maximization type objective function (16), it will take the minimum possible value which satisfies the constraints having $\delta _{ijk}$. Now, rewriting these equations as follows:

$$\begin{aligned}&\delta _{ijk} \ge P_{ij} - W_{ij}d_{ijkk} -\hat{\phi }_{ij} - W_{ij}\hat{\theta }_{k}\\&\delta _{ijk} \ge P_{ij} - W_{ij}\min \limits _{m \mid m \ne k} \{d_{ijkm}\} -\hat{\phi }_{ij} - W_{ij}\hat{\theta }_{k}- \min \limits _{m \mid m \ne k} \{\hat{\delta }_{ijm}\} \\&\delta _{ijk} \ge 0 \\&\text {where }\hat{\phi }_{ij}, \hat{\delta }_{ijm} \hbox { and } \hat{\theta }_{k}\hbox { are the minimum possible values of } \phi _{ij}, \delta _{ijm} \hbox { and } \theta _{k}, \hbox { respectively.} \end{aligned}$$

$\phi _{ij}$ and $\theta _{k}$ are the dual variables associated with the constraint sets (12) and (14) in $\hbox {HIMP}_{{2L}}$, respectively. And hence their values will be the shadow prices associated with those constraints. Then, we can clearly see that

• $\hat{\phi }_{ij} = 0$ as the minimum possible change in the objective function (11) will happen when the complete flow demand from the source i to destination j can not be met post-interdiction. For a given source-destination pair (i, j), if the right-hand side of constraint (12) is 0, then it will cost objective function a value of $P_{ij}$; while if the right side is increased to 1, then because of the limited capacity this complete demand will incur a penalty of $P_{ij}$ for not being met. So, there will be no change in the objective function.

• $\hat{\theta }_{k} = 0$ as the minimum possible change in the objective function (11) will happen when there is more than enough capacity with surviving hub k. So, even if the right-hand side of constraint (14) is increased by 1 unit, there will be no change in the objective function.

Therefore, to satisfy the above set of conditions on $\delta _{ijk}$, we can set the worst case (maximum possible) value of $\delta _{ijk}$ as follows:

$$\begin{aligned} \delta _{ijk} \le \max \{P_{ij} - W_{ij}d_{ijkk}, P_{ij} - W_{ij}\min \limits _{m \mid m \ne k}\{d_{ijkm}\},0 \} = P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\} \end{aligned}$$

Hence, $\hat{M}_{ijk} = P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\}$ is a valid value of $M_{ijk}$ for the $\hbox {HIMP}_{{DF}}$ formulation.

1.4 A.4 Proof of Proposition 4

As $\hat{M}_{ijk} = P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\}$, so $\bar{M}_{ijkm} = P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\} + \epsilon$

$M_{ijkm}$ should take such value so that there is no flow through an interdicted hub. There are three possible cases for a given source-destination pair (i, j), a positive fraction ($\eta$) of flow demand between i and j via hub k and hub m in that order:

• Case 1: Both of the hubs are surviving

  Any value of $M_{ijkm}$ will work as $q_{k} = q_{m} = 1$ in the objective function (29) making the associated terms 0.

• Case 2: One of the two hubs is interdicted

  Suppose only hub k is interdicted; while hub m is surviving,i.e., $q_{k} = 0$ and $q_{m} = 1$ Then,

  • The total cost for fulfilling such flow = $\eta W_{ij} d_{ijkm} + \eta \bar{M}_{ijkm}$

  • On the other hand, the total cost for not fulfilling such flow = $\eta P_{ij}$

  In order to restrict the flow from passing through the interdicted hub, the cost for fulfillment should be greater than that of unfulfilled flow.

  $$\begin{aligned}&\eta W_{ij} d_{ijkm} + \eta \bar{M}_{ijkm}> \eta P_{ij} \\&\quad \implies W_{ij} d_{ijkm} + \bar{M}_{ijkm}> P_{ij} \\&\quad \implies W_{ij} d_{ijkm} + P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\} + \epsilon> P_{ij} \\&\quad \implies W_{ij} \left( d_{ijkm} - \min _{m} \{d_{ijkm}\}\right) + \epsilon > 0 \end{aligned}$$

  which is true and hence a valid value.

• Case 3: Both the hubs are interdicted

  Here, $q_k = q_m = 0$. Then,

  • The total cost for fulfilling such flow = $\eta W_{ij} d_{ijkm} + 2 \eta \bar{M}_{ijkm}$

  • On the other hand, the total cost for not fulfilling such flow = $\eta P_{ij}$

  In order to restrict the flow from passing through the interdicted hub, the cost for fulfillment should be greater than that of unfulfilled flow.

  $$\begin{aligned}&\eta W_{ij} d_{ijkm} + 2 \eta \bar{M}_{ijkm}> \eta P_{ij} \\&\quad \implies W_{ij} d_{ijkm} + 2 \bar{M}_{ijkm}> P_{ij} \\&\quad \implies W_{ij} d_{ijkm} + 2 \left( P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\} + \epsilon \right)> P_{ij} \\&\quad \implies W_{ij} (d_{ijkm} - \min _{m} \{d_{ijkm}\}) + \left( P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\}\right) + 2 \epsilon > 0 \end{aligned}$$

  which is true and hence a valid value.

1.5 A.5 Proof of Proposition 5

There are three possible cases for a given source-destination pair (i, j), a positive fraction ($\eta$) of flow demand between i and j through hub k and hub m:

• Case 1: Both of the hubs are surviving

  Any value of $M_{ijkm}$ will work as $q_{k} = q_{m} = 1$ in the objective function (29) making the associated terms 0.

• Case 2: One of the two hubs is interdicted

  Suppose only hub k is interdicted while hub m is surviving,i.e., $q_{k} = 0$ and $q_{m} = 1$ Then,

  • The total cost for fulfilling such flow = $\eta W_{ij} d_{ijkm} + \eta M_{ijkm}$

  • On the other hand, the total cost for not fulfilling such flow = $\eta P_{ij}$

• Case 3: Both the hubs are interdicted

  Here, $q_k = q_m = 0$. Then,

  • The total cost for fulfilling such flow = $\eta W_{ij} d_{ijkm} + 2 \eta M_{ijkm}$

  • On the other hand, the total cost for not fulfilling such flow = $\eta P_{ij}$

Here, we are concerned about only Case 2 and Case 3, as in Case 1, any value of $M_{ijkm}$ will work. The value of $M_{ijkm}$ applicable for Case 2 will also be a valid value for Case 3. So, we are going to analyze only Case 2.

The total cost of flow using the path containing an interdicted hub should be greater than the cost of not fulfilling the same, in order to discourage any flow using an interdicted hub. This means,

$$\begin{aligned}&\eta W_{ij} d_{ijkm} + \eta M_{ijkm}> \eta P_{ij} \\&\quad \implies W_{ij} d_{ijkm} + M_{ijkm}> P_{ij} \\&\quad \implies M_{ijkm} > P_{ij} - W_{ij} d_{ijkm} \end{aligned}$$

Hence, $\tilde{M}_{ijkm} = P_{ij} - W_{ij}d_{ijkm} + \epsilon$ is a valid as well as the best-known value of $M_{ijkm}$ for a given source-destination pair (i, j) and routing of flow through hubs k and m in $\hbox {HIMP}_{{PDF}}$, where $\epsilon$ is a positive infinitesimal quantity.

1.6 A.6 Proof of Proposition 6

We know,

$$\begin{aligned}&\bar{M}_{ijkm} = \hat{M}_{ijk} + \epsilon \\&\quad \implies \bar{M}_{ijkm} = P_{ij} - W_{ij}\min _{m} \{d_{ijkm}\} + \epsilon \\&\quad \implies \bar{M}_{ijkm} \ge P_{ij} - W_{ij}d_{ijkm} + \epsilon \\&\quad \implies \bar{M}_{ijkm} \ge \tilde{M}_{ijkm} \end{aligned}$$

1.7 A.7 Proof of Proposition 7

We know $\tilde{P}_{ij}$ is the best-known value of $P_{ij}$ for $\hbox {HIMP}_{{2L}}$ such that any other value of $P_{ij}$ less than $\tilde{P}_{ij}$ will result into under-utilization of the capacity of hubs. As

$$\begin{aligned} \bar{P}_{ij}&= \max \limits _{H \in \mathcal {H}} \tilde{P}_{ij}(H)&\\ \implies \bar{P}_{ij}&\ge \tilde{P}_{ij}(H)&\forall H \in \mathcal {H} \end{aligned}$$

Therefore, $\bar{P}_{ij}$ is a valid value of $P_{ij}$ for SP, which is a $\hbox {HIMP}_{{2L}}$.