Algorithms for non-linear and stochastic resource constrained shortest path

Abstract

Resource constrained shortest path problems are usually solved thanks to a smart enumeration of all the non-dominated paths. Recent improvements of these enumeration algorithms rely on the use of bounds on path resources to discard partial solutions. The quality of the bounds determines the performance of the algorithm. The main contribution of this paper is to introduce a standard procedure to generate bounds on paths resources in a general setting which covers most resource constrained shortest path problems, among which stochastic versions. In that purpose, we introduce a generalization of the resource constrained shortest path problem where the resources are taken in a monoid. The resource of a path is the monoid sum of the resources of its arcs. The problem consists in finding a path whose resource minimizes a non-decreasing cost function of the path resource among the paths that respect a given constraint. Enumeration algorithms are generalized to this framework. We use lattice theory to provide polynomial procedures to find good quality bounds. These procedures solve a generalization of the algebraic path problem, where arc resources belong to a lattice ordered monoid. The practical efficiency of the approach is proved through an extensive numerical study on some deterministic and stochastic resource constrained shortest path problems.

Appendix

1.1 Proof of Proposition 1

The proof relies on the $\mathcal {NP}$-hardness of the following Resource Constrained Shortest Path Problem (Handler and Zang 1980).

Resource Constrained Shortest Path Problem

Input. A digraph $D=(V,A)$, two vertices $o,d \in V$, collections $(c_{a})$ and $(w_{a})$ in $\mathbb {Z}_+^{A}$, and a threshold $\tau \in \mathbb {Z}_+$.

Output. An o–d path P such that $\displaystyle \sum _{a \in P}w_{a} \le \tau$ and with minimum $\displaystyle \sum _{a \in P}c_{a}$.

Proof

(Proposition 1) Consider the set $\mathcal {R}= [0,1]^{2} \cup \{+\infty \}$. We endow it with the partial order

$$\begin{aligned} x\leqslant + \infty , \forall x\quad \text {and} \quad (x^{1},x^{2}) \leqslant ({{\tilde{x}}}^{1},{{\tilde{x}}}^{2}) \text { if } x^{i} \le {\tilde{x}}^{i} , \quad \text {for } i=1,2, \end{aligned}$$

(13)

and the sum operator

$$\begin{aligned} \left| \begin{array}{l} x\oplus + \infty = +\infty \oplus x= +\infty , \forall x\quad \text {and} \\ (x^{1},x^{2}) \oplus ({{\tilde{x}}}^{1},{{\tilde{x}}}^{2}) = \left\{ \begin{array}{ll} (x^{1} + {{\tilde{x}}}^{1}, x^{2} + {{\tilde{x}}}^{2}),&{} \text {if} x^{i} + {{\tilde{x}}}^{i} \le 1, \forall i, \\ +\infty &{} \text {otherwise.} \end{array}\right. \end{array}\right. \end{aligned}$$

(14)

Operator $\oplus$ is associative and commutative because

$$\begin{aligned} x_{a} \oplus x_{b} \oplus x_{c} = \left\{ \begin{array}{ll} (x_{a}^{i} + x_{b}^{i} + x_{c}^{i})_{i} &{} \text {if } x_{a}^{i} + x_{b}^{i} + x_{c}^{i} \le 1 ,\text {for } i \in \{1,2\}\\ + \infty &{} \text {otherwise.} \end{array}\right. \end{aligned}$$

The order $\leqslant$ is compatible with $\oplus$ because, if $x_{a} \le x_{b}$ and $x_{a} \oplus x= +\infty$, then $x_{b} \oplus x= +\infty$. Hence, $(\mathcal {R},\oplus ,\leqslant )$ is a lattice ordered monoid.

We now prove that we can reduce the $\mathcal {NP}$-hard Resource Constrained Shortest Path Problem to the problem of computing $b_{v}^{\mathrm {opt}}$ on a digraph with resources in $(\mathcal {R},\oplus ,\leqslant )$. Let D, o, d, $w_{a}^{i}$ and $R^{i}$ be an instance of the Resource Constrained Shortest Path Problem. Let $x_{a}=(\frac{w_{a}^{i}}{R^{i}})_{i=1,2}$ if $\frac{w_{a}^{i}}{R^{i}} \le 1$ for $i=1,2$ and $+\infty$ otherwise. The $x_{P} = \sum _{a \in P} x_{a} \ne +\infty$ if and only if $\sum _{a \in P}x_{a}^{i} \le R_{i}$ for all i. As a consequence, $\bigwedge _{P\in \mathcal {P}_{od}} x_{P} < +\infty$ if and only if there exists and o–d path P such that $\sum _{a \in P}x_{a}^{i} \le R_{i}$ for all i. The value of $\bigwedge _{P\in \mathcal {P}_{od}} x_{P}$ gives the solution of the Resource Constrained Shortest Path Problem. $\square$

1.2 Proof of Proposition 2

The proof relies on the following technical lemma.

Lemma 7

The quantity $\min _{v}{\tilde{n}}_{v}$ does not decrease along the algorithm.

Proof

Let ${\hat{n}}_{v}$ be the value of ${\tilde{n}}_{v}$ right before being extended if v has been extended, and ${\hat{n}}_{v} = 0$ otherwise. We prove the lemma by showing the following results: at any time during the algorithm, for each pair of vertices (u, v), we have ${\hat{n}}_{u} < {\tilde{n}}_{u}$ and ${\hat{n}}_{u} \le {\tilde{n}}_{v}$. This result implies the lemma because it implies that for each vertex u, we have ${\hat{n}}_{u} < {\tilde{n}}_{u}$, and hence ${\tilde{n}}_{u}$ is non-decreasing.

The proof is by iteration on the steps of the algorithm. The result is true at the beginning of the algorithm. Let v be the vertex currently extended, and suppose that the result is true before the extension of v. For each vertex u, let $n_{u}^{\alpha }$ be the value of ${\tilde{n}}_{u}$ before the extension of v and $n_{u}^{\beta }$ the value of ${\tilde{n}}_{u}$ after the extension. After the extension of v, we have ${\hat{n}}_{v} = n_{v}^{\alpha } < \infty = n_{v}^{\beta }$. Let u be a vertex distinct from v. The index ${\hat{n}}_{u}$ is not modified during the extension of v. If (u, v) is not an arc, or if ${\tilde{b}}_{v} \leqslant x_{(u,v)} \oplus {\tilde{b}}_{u}$ before the extension, then ${\tilde{n}}_{u}$ is not updated, and $n_{u}^{\beta } = n_{u}^{\alpha } > {\hat{n}}_{u}$ and ${\hat{n}}_{u} \le n_{v}^{\alpha } \le n_{u}^{\alpha } = n_{u}^{\beta }$. If on the contrary u is updated, $n_{u}^{\beta } = \min (n_{u}^{\alpha },n_{v}^{\alpha }+1)$. As $n_{v}^{\alpha } \le n_{u}^{\alpha }$, there are two possibilities. In the first case, $n_{u}^{\alpha } = n_{v}^{\alpha }$, which implies $n_{u}^{\beta } = n_{u}^{\alpha }$, and by induction hypothesis ${\hat{n}}_{u} < n_{u}^{\alpha } = n_{v}^{\alpha } \le n_{u}^{\beta }$. In the second case $n_{u}^{\alpha } > n_{v}^{\alpha }$, and we have $n_{u}^{\beta } = n_{v}^{\alpha } +1 > n_{v}^{\alpha } \ge {\hat{n}}_{u}$. Finally, in both cases ${\hat{n}}_{u} \le n_{v}^{\alpha }$. We have thus proved that ${\hat{n}}_{u} < n_{u}^{\beta }$ and ${\hat{n}}_{u} \le n_{v}^{\alpha } \le n_{u}^{\beta }$ for each vertex u, which gives the result and the lemma. $\square$

The proof of Proposition 2 is now relatively straightforward, as Lemma 7 enables to link the values taken by $b_{v}'$ along the algorithm to the sequence $(b_{v}^{i})$ defined in Eq. 6.

Proof

(Proposition 2) Lemma 7 ensures that $\min _{v}{\tilde{n}}_{v}$ does not decrease. Based on this results, the update rule ensures that for each vertex u in L, if ${\tilde{n}}_{u} \ne + \infty$, we have either ${\tilde{n}}_{u} = \min _{v}{\tilde{n}}_{v}$ or ${\tilde{n}}_{u} = 1+\min _{v}{\tilde{n}}_{v}$. As a consequence, $\min _{v}{\tilde{n}}_{v}$ increases by at most one between two iterations. For each vertex u and index i, define $b_{u}^{i}$ and $n_{v}^{i}$ to be equal to the values of ${\tilde{b}}_{u}$ and ${\tilde{n}}_{u}$ when $\min _{v}{\tilde{n}}_{v} = i$ for the first time. Due to the update rule, we obtain by induction on i that $b_{u}^{i} = b_{u}^{i-1} \wedge \bigwedge _{(u,v)\in \delta ^{+}(u)} \left( x_{(u,v)} \oplus b_{v}^{i-1}\right)$. Indeed, suppose that the result is true up to $i-1$, and consider a vertex u, and an arc $(u,v) \in \delta ^{+}(u)$. The update rule then implies $b_{u}^{i} = b_{u}^{i-1} \wedge \bigwedge _{v \in U_{u}^{i}} \left( x_{(u,v)} \oplus b_{v}^{i-1}\right)$ and $n_{v}^{i-1} = +\infty$, where $U_{u}^{i}$ is the set of all v such that $(u,v)\in \delta ^{+}(u)$ and $n_{v}^{i-1} = i-1$. Besides, if $n_{v}^{i-1} \ne i-1$, the $b_{v}^{i-1} = b_{v}^{i-2}$ and the induction hypothesis gives the result. As a consequence, the $b_{v}^{i}$ correspond to those defined by Eq. (6), and we obtain the first part of the proposition and the dioid case. Besides the update rule ensures that if there is no vertex u in L such that $(v,u)\in \delta ^{+}(v)$, then ${\tilde{b}}_{v} = \bigwedge _{(v,u)\in \delta ^{+}(v)} b_{v}$. Thus, if L is empty at the end of the algorithm, then $\left( b_{v}\right) _{v}$ defines a solution of Eq. (7), which gives the second part of the proposition. $\square$

1.3 Proof of Proposition 3

Proof

(Proposition 3) Let $\xi$ and ${\tilde{\xi }}$ be such that $\xi \le _{\mathrm {st}}{\tilde{\xi }}$. Let F and G be their respective cumulative distribution functions, and $F^{-1}$ and $G^{-1}$ be their right continuous inverses. The definition of $\le _{\mathrm {st}}$ implies $F^{-1}(t) \le G^{-1}(t)$ for all t in [0, 1]. Let U be a uniform [0, 1] random variable, ${\widehat{\xi }} = F^{-1}(U)$, and $\widehat{{\tilde{\xi }}} = G^{-1}(U)$. For any atom $\omega$, we have ${\widehat{\xi }}(\omega ) = F^{-1}(U(\omega )) \le G^{-{1}}(U(\omega )) = \widehat{{\tilde{\xi }}}(\omega )$. By monotonicity of probability functional $\rho$, this implies $\rho ({\widehat{\xi }}) \le \rho (\widehat{{\tilde{\xi }}})$.

As $\xi$ and ${\widehat{\xi }}$ (resp. ${\tilde{\xi }}$ and $\widehat{{\tilde{\xi }}}$) have the same cumulative distribution, the hypothesis that $\rho$ is version independent implies that $\rho (\xi ) = \rho ({\widehat{\xi }})$ (resp. $\rho ({\tilde{\xi }}) = \rho (\widehat{{\tilde{\xi }}})$). The inequality $\rho ({\widehat{\xi }}) \le \rho (\widehat{{\tilde{\xi }}})$ then gives the proposition. $\square$