Local search algorithms for multiple-depot vehicle routing and for multiple traveling salesman problems with proved performance guarantees

Abstract

We consider two related problems: the multiple-depot vehicle routing problem (MDVRP) and the Multiple traveling salesman problem (mTSP). In both of them, given is the complete graph on n vertices $G = (V,E)$ with nonnegative edge lengths that form a metric on V. Also given is a positive integer k. In typical applications, V represents locations of customers and k represents the number of available vehicles. In MDVPR, we are also given a set of k depots $\{O_1,\ldots ,O_k\} \subseteq V$, and the goal is to find a minimum-length cycle cover of G of size k, that is, a collection of k (possibly empty) cycles such that each $v \in V$ is in exactly one cycle, and each cycle in the cover contains exactly one depot. In mTSP, no depots are given, so the goal is to find (any) minimum-length cycle cover of G of size k. We present local search algorithms for both problems, and we prove that their approximation ratio is 2.

Justification of Assumption 4.1

• (i) Assume that we know how to obtain a good packing for instances that satisfy Assumption 4.1(i). Given any instance $\langle G,\mathcal{V } \rangle$ that does not satisfy it, for each pair $(C,i)$ such that $C \subseteq E$ is a cycle with $V[C] \subseteq V_i$, contract C and identify it with a new vertex $v_i^C$ in $V_i$. This way, we obtain a new instance $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$ that satisfies Assumption 4.1(i). Let $\{H^{\prime }_i : i \in [k]\}$ be a good packing for $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$. For each pair $(C,i)$ defined above, add to $H_i^{\prime }$ all the edges of C except for one (arbitrarily); the edges that account for the vertices of C are the newly added $|C|-1$ edges and the edge that accounts for $v_i^C$ in $H^{\prime }_i$. Thus, it is easily verified that the resulting subset family is a good packing for $\langle G,\mathcal{V } \rangle$.

• (ii) For convenience, call a block $V_i \in \mathcal{V }$ “mixed” if it contains both an isolated vertex and a non-isolated vertex. Assume that we know how to obtain a good packing for instances that satisfy Assumption 4.1(ii), namely, with no mixed blocks. Given any instance $\langle G,\mathcal{V } \rangle$ with mixed blocks, we will show how to eliminate a cycle that traverses a mixed block. The justification of the assumption then follows by induction on the number of such cycles. Thus, let $C$ be a cycle that traverses a mixed block $V_1$, say. Construct a new instance $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$ where $G^{\prime } := (V^{\prime },E^{\prime })$ and $\mathcal{V^{\prime } } := \{ V^{\prime }_i : i \in [k] \}$ are as follows. The edge set is $E^{\prime } := E \setminus C$. As for the vertices, define $V^{\prime }_1 := (V_1 \setminus V[C]) + v_C$ where $v_C$ is a new isolated vertex, and for all $i=2,\ldots ,k$ define $V^{\prime }_i := V_i \setminus V[C]$. Note that if $V_i \subseteq V[C]$ (i.e., all the vertices of $V_i$ are covered by $C$), then $V^{\prime }_i = \emptyset$; this is the reason for allowing empty blocks in the definition of a good packing.

By this construction, $G^{\prime }$ consists of k (disjoint) cycles, one of them is the empty cycle $(v_C)$ which ‘replaces’ C from $\langle G,\mathcal{V } \rangle$. The number of cycles that traverse mixed blocks in $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$ is one less than that in $\langle G,\mathcal{V } \rangle$, so let $\{ H^{\prime }_i : i \in [k] \}$ be a good packing for $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$. Note that $H^{\prime }_i = \emptyset$ if $V_i \subseteq V[C]$.

Let $cont^{\prime (i)}$ denote the operation $cont^{(i)}$ with respect to $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$. Note that since $E^{\prime } \subseteq E$ and $V^{\prime } \setminus V$ consists of the single new isolated vertex $v_C$, it follows that for every subset $H^{\prime } \subseteq E^{\prime }$ and $i \in [k], cont^{\prime (i)}(H^{\prime })$ is also a subset of $E^{(i)}$ – the edges of $C[G,\mathcal{V }]^{(i)}$. It follows that $cont^{(i)}(H^{\prime }) = cont^{\prime (i)}(H^{\prime })$ (we emphasize that $H^{\prime } \subseteq E^{\prime } \equiv E \setminus C$). Therefore, the packing $\{ H^{\prime }_i : i \in [k] \}$, which is good for $\langle G^{\prime },\mathcal{V^{\prime } } \rangle$, satisfies:

Observation 6.1

For each $i \in [k], H^{\prime }_i$ is a forest in G, and $cont^{(i)}(H^{\prime }_i)$ is a forest in $C[G,\mathcal{V }]^{(i)}$.

We will use $\{H^{\prime }_i : i \in [k]\}$ to construct a packing $\{H_i : i \in [k]\}$ which is good for $\langle G,\mathcal{V } \rangle$, by the following procedure:

We claim that $\{ H_i : i \in [k] \}$ is a good packing for $\langle G,\mathcal{V } \rangle$. Fix $i \in [k]$. We first show that $H_i$ is a forest in G and $cont^{(i)}(H_i)$ is a forest in $C[G,\mathcal{V }]^{(i)}$, and $|H_i| = |cont^{(i)}(H_i)|$. The key observation is the following property of $C_i$, which is an immediate consequence of its construction:

Observation 6.2

If $C_i \ne \emptyset$, then $C_i$ is a collection of (vertex-)disjoint paths; each such path can be written as $(u_1,\ldots ,u_l)$, where:

• $u_1 \notin V_i, u_2,\ldots ,u_l \in V_i$,

• The degree of $u_l$ both in the graph $(V,H_i)$ and in $C[G,\mathcal{V }]^{(i)}$ is 1.

Now, by Observation 6.1, $H^{\prime }_i$ is a forest in G, and $cont^{(i)}(H^{\prime }_i)$ is a forest in $C[G,\mathcal{V }]^{(i)}$. Therefore, Observation 6.2 implies that $H_i$, which is $H^{\prime }_i \cup C_i$, is a forest in G and $cont^{(i)}(H_i)$ is a forest in $C[G,\mathcal{V }]^{(i)}$; note that $V[H^{\prime }_i] \cap V[C_i] = \emptyset$, and that $V[cont^{(i)}(H^{\prime }_i)] \cap V[cont^{(i)}(C_i)]$ need not be empty: $x_j \in V[cont^{(i)}(H^{\prime }_i)] \cap V[cont^{(i)}(C_i)]$ if and only if $u_1 \in V_j$ where $u_1$ is an end-vertex of one of $C_i$’s paths in the representation of Observation 6.2. Observation 6.2 also implies that $H_i$ satisfies conditions (i)–(iii) of Observation 2.1, hence $|H_i| = |cont^{(i)}(H_i)|$.

It remains to show that $|H_i| = |V_i| - 1 = n_i - 1$. Recall that by assumption, $|H^{\prime }_i| = |V^{\prime }_i| - 1$. Also note that after step 2.a., and hence at termination, $|C_i| = |V_i \cap V[C]|$. We distinguish four cases.

• C does not traverse $V_i$: in this case, $V^{\prime }_i = V_i$, so at termination of the Construction procedure, $H_i = H^{\prime }_i$ and $|H_i| = |H^{\prime }_i| = |V^{\prime }_i|-1 = |n_i| - 1$.

• C traverses $V_i$ and $V_i \subseteq V[C]$: in this case, $V^{\prime }_i = \emptyset$ and thus $H^{\prime }_i = \emptyset$. Then after step 2.b., $|H_i| = |C_i| = |V_i \cap V[C]| = |V_i|$, so after the edge deletion at 2.c., $|H_i| = |V_i| - 1 = n_i - 1$.

• C traverses $V_i, i \ne 1$, and $V_i$ contains vertices outside $C$ (either isolated or not): in this case, step 2.c. is not executed, so $|H_i| = |H^{\prime }_i| + |C_i| = |H^{\prime }_i| + |V_i \cap V[C]|$. But $|H^{\prime }_i| = |V^{\prime }_i| - 1$ and $|V^{\prime }_i| = |V_i| - |V_i \cap V[C]| = n_i - |V_i \cap V[C]|$. It follows that $|H_i| = n_i - 1$.

• $i=1$: Since $V^{\prime }_1 = (V_1 \setminus V[C]) + v_C, |V^{\prime }_1| = |V_1| - |V_1 \cap V[C]| + 1$, so $|H^{\prime }_1| = |V^{\prime }_1|-1 = |V_1| - |V_1 \cap V[C]|$. Therefore, after step 2, $|H_1| = |H^{\prime }_1| + |V_1 \cap V[C]| = |V_1| = n_1$, so the edge deletion at step 3 guarantees that $|H_1| = n_1 - 1$ as required.

This completes the proof.