Algorithms for a special class of state-dependent shortest path problems with an application to the train routing problem

Abstract

We study the state-dependent shortest path problem and focus on its application to the Single Train Routing Problem consisting of a rail network with only double-track segments, where the objective is to route one train through an empty network as fast as possible. We show that the Single Train Routing Problem is NP-hard. We investigate the solution properties and present sufficient conditions for optimality. Different conditions on the parameters are given to guarantee that certain local route selection is optimal. Then, a dynamic programming heuristic is introduced and conditions when the proposed heuristic can obtain the optimal solution in polynomial time are also discussed. Experimental results show the efficiency of the proposed heuristics for general problem settings.

Appendix

Proof of Lemma 1:

When $t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) {<}+\infty$, $t( v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} )$ has two forms as follows:

• 1. If $v_{\mathrm{exit}} \le \bar{v} ,v_{\mathrm{enter}} \le \bar{v} ,v_{\mathrm{exit}}^2 -v_{\mathrm{enter}}^2 \le 2r_\mathrm{a} l,v_{\mathrm{enter}}^2 -v_{\mathrm{exit}}^2 \le 2r_\mathrm{d}l,\sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a}r_\mathrm{d} l}{r_\mathrm{a} +r_\mathrm{d} }}\le \bar{v}$

  $$\begin{aligned}&t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) =\\&\quad -\frac{v_{\mathrm{enter}} }{r_\mathrm{a}}-\frac{v_{\mathrm{exit}}}{r_\mathrm{d}}+\left( {\frac{1}{r_\mathrm{a} } +\frac{1}{ r_\mathrm{d}}} \right) \\&\quad \sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d}}}\\&\frac{\partial t}{\partial v_{\mathrm{enter}}}=\frac{1}{r_\mathrm{a} }\left( {\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d}}} \right) ^{-1/2}\\&\quad \left( {v_{\mathrm{enter}} -\sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 +r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} +r_\mathrm{d} }}} \right) \\&\frac{\partial t}{\partial v_{\mathrm{exit}} }=\frac{1}{r_\mathrm{d} } \left( {\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d}}} \right) ^{-\frac{1}{2}}\\&\quad \left( {v_{\mathrm{exit}} -\sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} +r_\mathrm{d}}}} \right) \\&\frac{\partial t}{\partial l}=\left( {\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d}l}{r_\mathrm{a} +r_\mathrm{d}}} \right) ^{-1/2}\\&\frac{\partial t}{\partial \bar{v} }=0 \end{aligned}$$

  Since $v_{\mathrm{enter}}^2 -v_{\mathrm{exit}}^2 \le 2r_\mathrm{d}l$, we have

  $$\begin{aligned} v_{\mathrm{enter}}^2 \le \frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a}r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d} }, \end{aligned}$$

  i.e.,

  $$\begin{aligned} v_{\mathrm{enter}} \le \sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 +r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d}}} \end{aligned}$$

  which implies $\frac{\partial {t}}{\partial {v}_{\mathrm{enter}}}\le 0$. And $\frac{\partial {t}}{\partial {v}_{\mathrm{enter}}}=0$ only if $v_{\mathrm{enter}}^2 -v_{\mathrm{exit}}^2 =2r_\mathrm{d}l$.

  Similarly from $v_{\mathrm{exit}}^2 -v_{\mathrm{enter}}^2 \le 2r_\mathrm{a}l$, we have

  $$\begin{aligned} v_{\mathrm{exit}} \le \sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 +r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} + r_\mathrm{d} }} \end{aligned}$$

  which implies $\frac{\partial \mathrm{t}}{\partial {v}_{\mathrm{exit}}}\le 0$. And $\frac{\partial t}{\partial v_{\mathrm{exit}} }=0$ only if $v_{\mathrm{exit}}^2 -v_{\mathrm{enter}}^2 =2r_\mathrm{a} l$. And obviously in this case we have $\frac{\partial t}{\partial l}>0$ and $\frac{\partial t}{\partial \bar{v} }\le 0$.

• 2. If $v_{\mathrm{exit}} \le \bar{v} ,v_{\mathrm{enter}} \le \bar{v} ,v_{\mathrm{exit}}^2 -v_{\mathrm{enter}}^2 \le 2r_\mathrm{a} l,v_{\mathrm{enter}}^2 -v_{\mathrm{exit}}^2 \le 2r_\mathrm{d} l,\sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 +r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a}r_\mathrm{d}l}{r_\mathrm{a} + r_\mathrm{d}}}>\bar{v}$

  $$\begin{aligned}&t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) =\frac{\bar{v} -v_{\mathrm{enter}} }{r_\mathrm{a} }+\frac{\bar{v} -v_{\mathrm{exit}} }{ r_\mathrm{d}}\\&\quad +\frac{1}{\bar{v} }\left( {l-\frac{\bar{v}^{2}-v_{\mathrm{enter}}^2 }{2r_\mathrm{a} }-\frac{\bar{v}^{2}-v_{\mathrm{exit}}^2 }{2r_\mathrm{d}}} \right) \\&\frac{\partial t}{\partial v_{\mathrm{enter}} }=\frac{v_{\mathrm{enter}} -\bar{v} }{r_\mathrm{a} *\bar{v} }\\&\frac{\partial t}{\partial v_{\mathrm{exit}} }=\frac{v_{\mathrm{exit}} -\bar{v} }{ r_\mathrm{d}*\bar{v} }\\&\frac{\partial t}{\partial l}=\frac{1}{\bar{v} }\\&\frac{\partial t}{\partial \bar{v} }=\left( {\frac{1}{2r_\mathrm{a} }+\frac{1}{2 r_\mathrm{d}}} \right) -\frac{1}{\bar{v}^{2}}\left( {l+\frac{v_{\mathrm{enter}}^2 }{2r_\mathrm{a} }+\frac{v_{\mathrm{exit}}^2 }{2 r_\mathrm{d}}} \right) \end{aligned}$$

  We have $\frac{\partial t}{\partial v_{\mathrm{enter}} }\le 0$. And $\frac{\partial t}{\partial v_{\mathrm{enter}} }=0$ only if $v_{\mathrm{enter}} =\bar{v}$. Similarly we have$\frac{\partial t}{\partial v_{\mathrm{exit}} }\le 0$. And $\frac{\partial {t}}{\partial {v}_{\mathrm{exit}} }=0$ only if $v_{\mathrm{exit}} =\bar{v}$. And obviously we have $\frac{\partial {t}}{\partial {l}}>0$.

From $\sqrt{\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 +r_\mathrm{d} v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d} l}{r_\mathrm{a} +r_\mathrm{d} }}>\bar{v}$, we have

$$\begin{aligned} \frac{\partial {t}}{\partial \bar{v} }=\left( {\frac{1}{2r_\mathrm{a} }+\frac{1}{2 r_\mathrm{d}}} \right) \frac{1}{\bar{v}^{2}}\left( {\bar{v}^{2}-\frac{r_\mathrm{a} v_{\mathrm{exit}}^2 + r_\mathrm{d}v_{\mathrm{enter}}^2 +2r_\mathrm{a} r_\mathrm{d}l}{r_\mathrm{a} + r_\mathrm{d}}} \right) <0 \end{aligned}$$

Since $t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right)$ is a continuous function when it is finite, and from both cases we have $\frac{\partial {t}}{\partial {v}_{\mathrm{enter}} }\le 0\,(\frac{\partial {t}}{\partial {v}_{\mathrm{enter}} }=0$ only at finite points), $\frac{\partial {t}}{\partial {v}_{\mathrm{exit}} }\le 0\,(\frac{\partial {t}}{\partial {v}_{\mathrm{exit}} }=0$ only at finite points), $\frac{\partial {t}}{\partial {l}}>0$ and $\frac{\partial {t}}{\partial \bar{v} }\le 0$. Therefore, $t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right)$ is a strictly decreasing function of $v_{\mathrm{enter}}$ and $v_{\mathrm{exit}}$, a strictly increasing function of l and a non-increasing function of $\bar{v}$. $\square$

Proof of Corollary 1

From Property 1, we know $v_{\mathrm{exit}}^2 -v_{\mathrm{enter}}^2 \le 2r_\mathrm{a} l$ and $v_{\mathrm{enter}}^2 -v_{\mathrm{exit}}^2 \le 2 r_\mathrm{d}l$.

If $v_{\mathrm{enter}} \ge v_{\mathrm{exit}}$, let $l^\prime =\frac{v_{{\mathrm{enter}}}^2-v_{{\mathrm{exit}}}^2}{2r_\mathrm{d}}\le l$. From Lemma 1 we have

$$\begin{aligned} t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v}} \right) \ge t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,{l}',\bar{v}} \right) =\frac{v_{{\mathrm{enter}}}-v_{{\mathrm{exit}}}}{r_\mathrm{d}} \end{aligned}$$

We also have $t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) >0\ge \frac{v_{\mathrm{exit}} -v_{\mathrm{enter}} }{r_\mathrm{a} }$. So the claim is true.

If $v_{\mathrm{exit}} \ge v_{\mathrm{enter}}$, let $l'=\frac{v_{{\mathrm{exit}}}^2-v_{\mathrm{enter}}^2}{2r_\mathrm{a}}\le l$. From Lemma 1 we have

$$\begin{aligned} t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) \ge t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,{l}',\bar{v}} \right) =\frac{v_{\mathrm{exit}}-v_{\mathrm{enter}}}{r_\mathrm{a}} \end{aligned}$$

We also have $t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) >0\ge \frac{v_{\mathrm{enter}} -v_{\mathrm{exit}} }{ r_\mathrm{d}}$. So the claim is true. $\square$

Proof of Lemma 2

We first prove (20) and (21) by induction.

First we consider the scenario when $m=1$. Since $T\left( {0,v_\mathrm{e} ,p_1 } \right) <+\infty$, from Property 1 we have $v_\mathrm{e}^2 \le 2r_\mathrm{a} l_1$. So $v^{2}\le v_\mathrm{e}^2 \le 2r_\mathrm{a} l_1$, which implies $T\left( {0,v,p_1 } \right) <+\infty$. And $v_\mathrm{e}^2 \le 2r_\mathrm{a} l_1 \le 2r_\mathrm{a} l_1^{\prime }$, which implies $T\left( {0,v_\mathrm{e} ,p_1^{\prime } } \right) <+\infty$. From Lemma 1 we know $T\left( {0,v_\mathrm{e} ,p_1} \right) <T\left( {0,v,p_1 } \right)$ and $T\left( {0,v_\mathrm{e} ,p_1} \right) =t\left( {0,v_\mathrm{e} ,l_1 ,\bar{v}_1 } \right) <t\left( {0,v_\mathrm{e} ,l_1^{\prime } ,\bar{v}_1 } \right) =T\left( {0,v_\mathrm{e} ,p_1^{\prime }} \right)$. Now consider a special scenario when $l_1^{\prime } =l_1 +\frac{v_\mathrm{e}^2 -v^{2}}{2*r_\mathrm{d} }$. Let the train first travel through $l_1$ distance from velocity 0 to velocity $v_\mathrm{e}$, which takes $T\left( {0,v_\mathrm{e} ,p_1 } \right)$, and for the rest $\frac{v_\mathrm{e}^2 -v^{2}}{2*r_\mathrm{d}}$ decelerate from velocity $v_\mathrm{e}$ to v, which takes $\frac{v_\mathrm{e} -v}{r_\mathrm{d}}$. $T\left( {0,v,p_1^{\prime } } \right)$ is the optimal travel time starting from velocity 0 to v with distance $l_1^{\prime }$, so we have

$$\begin{aligned} T\left( {0,v,p_1^{\prime } } \right) \le T\left( {0,v_\mathrm{e} ,p_1 } \right) +\frac{v_\mathrm{e} -v}{ r_\mathrm{d}} \end{aligned}$$

Left-hand side is the optimal travel time while the right-hand side is the travel time for one candidate solution. So $T\left( {0,v_\mathrm{e} ,p_1 } \right) <T\left( {0,v_\mathrm{e} ,p_1^{\prime } } \right) \le T\left( {0,v_\mathrm{e} ,p_1 } \right) +\frac{v_\mathrm{e} -v}{r_\mathrm{d} }$. Therefore, (20) and (21) are valid when m=1.

Assume the claim is true for m = k. We consider the scenario when $m=k+1$. Since $T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) <+\infty$, the corresponding problem (1)–(4) for this instance is feasible. Let the optimal solution for the corresponding problem (1)–(4) be $v_i^*\left( {i=0,\ldots ,k+1} \right)$. Since $v_\mathrm{e}^2 -v_k^{*2} \le 2r_\mathrm{a} l_{k+1} <2r_\mathrm{a} l_{k+1}^{\prime }$ and $v_k^{*2} -v_\mathrm{e}^2 \le 2 r_\mathrm{d}l_{k+1} <2 r_{d} l'_{k+1}$, we have $t\left( {v_k^*,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) <+\infty$. So

$$\begin{aligned}&T\left( {0,v_\mathrm{e} ,p_{k+1}^{\prime } } \right) \le T\left( {0,v_k^*,p_k } \right) \\&\quad +\,t\left( v_k^*,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} \right) <+\infty . \end{aligned}$$

After changing the length of the $\left( {k+1} \right)$th segment to $l_{k+1}^{\prime } (l_{k+1}^{\prime } >l_{k+1} )$, let the optimal solution for the corresponding problem (1)–(4) be $v_i^{{\prime }*} \left( {i=0,\ldots ,k+1} \right)$. Then

$$\begin{aligned}&T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) =T\left( {0,v_k^*,p_k } \right) +t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) \\&T\left( {0,v_\mathrm{e} ,p_{k+1}^{\prime } } \right) =T\left( {0,v_k^{{\prime }*} ,p_k } \right) +t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) \end{aligned}$$

If $v_k^*\ge v_k^{{\prime }*}$, $T\left( {0,v_k^*,p_k } \right) \le T\left( {0,v_k^{{\prime }*} ,p_k } \right)$ holds due to the assumption that (20) is true for m = k. Also

$$\begin{aligned}&t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) <t\left( {v_k^*,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) \\&\quad \le t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) \end{aligned}$$

holds due to Lemma 1. So $T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) <T\left( {0,v_\mathrm{e} ,p_{k+1} } \right)$.

If $v_k^*<v_k^{{\prime }*}$, $T\left( {0,v_k^{{\prime }*} ,p_k } \right)<T\left( {0,v_k^*,p_k } \right) <T\big ( 0,v_k^{{\prime }*} , p_k \big )+\frac{v_k^{{\prime }*} -v_k^*}{r_\mathrm{d} }$ holds due to the assumption that (20) is true for m = k.

Also since $T\left( {0,v_k^*,p_k } \right) <+\infty$, $\forall v\in \left( {v_k^*,v_k^{{\prime }*} } \right)$, we have $T\left( {0,v,p_k } \right)<T\left( {0,v_k^*,p_k} \right) <+\infty$. From Proposition 1 we know $v_k^*$ is the largest velocity among all the feasible velocities at the end of the kth segment, which implies v is not a feasible velocity, i.e., $t\left( {v,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) =+\infty$. From $t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) <+\infty$ and $\left( {v,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) =+\infty \forall v\in \left( {v_k^*,v_k^{{\prime }*} } \right)$, from Property 1 we know

$$\begin{aligned}&v_k^{*2} -v_\mathrm{e}^2 =2 r_\mathrm{d} {l}_{k+1}\\&\quad t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) =\frac{v_k^*-v_\mathrm{e} }{ r_\mathrm{d}} \end{aligned}$$

Also from Corollary 1, we have

$$\begin{aligned}&t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) \ge \frac{v_k^{{\prime }*} -v_\mathrm{e} }{ r_\mathrm{d}}\\&\quad =t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) +\frac{v_k^{{\prime }*} -v_k^*}{ r_\mathrm{d}}, \end{aligned}$$

i.e., $t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) \le t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) -\frac{v_k^{{\prime }*} -v_k^*}{r_\mathrm{d} }$. So

$$\begin{aligned}&T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) =T\left( {0,v_k^*,p_k } \right) +t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) \\&\quad <T\left( {0,v_k^{{\prime }*} ,p_k } \right) +\frac{v_k^{{\prime }*} -v_k^*}{r_\mathrm{d} }+t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1} } \right) \\&\quad -\frac{v_k^{{\prime }*} -v_k^*}{ r_\mathrm{d}}=T\left( {0,v_k^{{\prime }*} ,p_k } \right) +t\left( {v_k^{{\prime }*} ,v_\mathrm{e} ,l_{k+1}^{\prime } ,\bar{v}_{k+1}} \right) \\&\quad =T\left( 0,v_\mathrm{e},p^\prime _{k+1}\right) \end{aligned}$$

$\forall v\in \left[ {0,v_\mathrm{e} } \right] ,$ take $l_{k+1}^{\prime } =l_{k+1} +\frac{v_\mathrm{e}^2 -v^{2}}{2r_\mathrm{d} }$. We can break the $\left( {k+1} \right)$th segment into two pieces such that the length of the first piece is $l_{k+1}$ and that the length for the second piece is $\frac{v_\mathrm{e}^2 -v^{2}}{2r_\mathrm{d} }$. So the path from the beginning of the first segment to the break point at the $\left( {k+1} \right)$th segment is exactly the same as the original path. Due to the optimality of $T\left( {0,v,p_{k+1}^{\prime } } \right)$, we have

$$\begin{aligned} T\left( {0,v,p_{k+1}^{\prime } } \right) \le T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) +\frac{v_\mathrm{e} -v}{ r_\mathrm{d}} \end{aligned}$$

which implies

$$\begin{aligned} T\left( {0,v,p_{k+1} } \right) <T\left( {0,v,p_{k+1}^{\prime } } \right) \!\le \!T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) \!+\!\frac{v_\mathrm{e} -v}{ r_\mathrm{d}} \end{aligned}$$

Let $v_k^{\prime }$ satisfy $T\left( {0,v,p_{k+1} } \right) =T\big ( {0,v_k^{\prime } ,p_k } \big )+t\left( v_k^{\prime } ,\right. \left. v,l_{k+1} ,\bar{v}_{k+1} \right)$, i.e., $v_k^{\prime }$ is the optimal velocity at the end of the kth segment when the train travels from the beginning of the first segment at velocity 0 to the end of the $\left( {k+1} \right)$th segment at velocity v.

If $v_k^{\prime } >v_k^*$, then $v_\mathrm{e}^2 -v_k^{{\prime }2} \le v_\mathrm{e}^2 -v_k^{*2} \le 2r_\mathrm{a} l_{k+1}$ and $v_k^{{\prime }2} -v_\mathrm{e}^2 \le v_k^{{\prime }2} -v^{2}\le 2r_\mathrm{d} l_{k+1}$, which implies $t\left( {v_k^{\prime } ,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1}} \right) <+\infty$. So $v_k^{\prime }$ is a feasible velocity at the end of the kth segment. However, from Proposition 1 we know $v_k^*\ge v_k^{\prime }$, which is a contradiction. So $v_k^{\prime } \le v_k^*$, and we have

$$\begin{aligned}&T\left( {0,v_\mathrm{e} ,p_{k+1} } \right) =T\left( {0,v_k^*,p_k } \right) +t\left( {v_k^*,v_\mathrm{e} ,l_{k+1} ,\bar{v}_{k+1} } \right) \\&\quad <T\left( {0,v_k^{\prime } ,p_k } \right) +t\left( {v_k^{\prime } ,v,l_{k+1} ,\bar{v}_{k+1} } \right) \\&\quad =T_{k+1} \left( {0,v} \right) \end{aligned}$$

Therefore, (20) and (21) are valid when m = k+1.

In summary, $\forall m\in N^{+}$

$$\begin{aligned}&T\left( {0,v_\mathrm{e} ,p_m } \right)<T\left( {0,v,p_m } \right)<T\left( {0,v_\mathrm{e} ,p_m } \right) +\frac{v_\mathrm{e} -v}{r_\mathrm{d} }\\&T\left( {0,v_\mathrm{e} ,p_m } \right)<T\left( {0,v_\mathrm{e} ,p_m^{\prime } } \right) <+\infty \end{aligned}$$

Using the same argument we can have that if $T\left( {v_\mathrm{s} ,0,q_1 } \right) <+\infty$, then $\forall v\in \left[ {0,v_\mathrm{s} } \right)$ we have

$$\begin{aligned}&T\left( {v_\mathrm{s} ,0,q_1 } \right)<T\left( {v,0,q_1 } \right)<T\left( {v_\mathrm{s} ,0,q_1 } \right) +\frac{v_\mathrm{s} -v}{r_\mathrm{a} }\\&T\left( {v_\mathrm{s} ,0,q_1 } \right)<T\left( {v_\mathrm{s} ,0,q_1^{\prime } } \right) <+\infty \end{aligned}$$

$\square$

Proof of Lemma 3

Given $t\left( {v_{i-1} ,v_i ,l_{i2} ,\bar{v}_{i2} } \right) <+\infty$ from Property 1, we have $t\left( {v_{i-1} ,v_i ,l_{i1} ,\bar{v}_{i1} } \right) <+\infty$ (this statement is shown to be valid in the proof of Proposition 5). So the definition of $\Delta \left( {v_{i-1} ,v_i } \right)$ is valid.

We first consider the scenario when $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d} }}{>}\bar{v}_{i2}$, i.e., the following holds.

$$\begin{aligned}&\left( {\frac{1}{2r_\mathrm{a} }+\frac{1}{2r_\mathrm{d} }} \right) \bar{v}_{i2}^2 <\frac{v_{i-1}^2 }{2r_\mathrm{a} }+\frac{v_i^2 }{2r_\mathrm{d} }+l_{i2}\nonumber \\&\quad t\left( {v_{i-1} ,v_i ,l_{i2} ,\bar{v}_{i2} } \right) =\frac{\bar{v} -v_{i-1} }{r_\mathrm{a} }\nonumber \\&\quad +\frac{\bar{v} -v_i }{r_\mathrm{d}}+\frac{1}{\bar{v} }\left( {l-\frac{\bar{v}^{2}-v_{i-1}^2 }{2r_\mathrm{a} }-\frac{\bar{v}^{2}-v_i^2 }{2 r_\mathrm{d}}} \right) \nonumber \\&\quad \frac{\hbox {d}t\left( {v_{i-1} ,v_i ,l_{i2} ,\bar{v}_{i2} } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} \bar{v}_{i2} }-\frac{1}{r_\mathrm{a}} \end{aligned}$$

(25)

Also if $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }{r_\mathrm{a} +r_\mathrm{d} }}>\bar{v}_{i1}$,

$$\begin{aligned}&\frac{\hbox {d}t\left( {v_{i-1} ,v_i ,l_{i1} ,\bar{v}_{i1} } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1}}{r_\mathrm{a} \bar{v}_{i1} }-\frac{1}{r_\mathrm{a}}\\&\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} \bar{v}_{i1} \bar{v}_{i2} }\left( {\bar{v}_{i2} -\bar{v}_{i1} } \right) <0 \end{aligned}$$

where the last inequality is from Condition 2.

If $\sqrt{\frac{r_\mathrm{a}r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }{r_\mathrm{a} +r_\mathrm{d}}}\le \bar{v}_{i1}$,

$$\begin{aligned}&\frac{\hbox {d}t\left( {v_{i-1} ,v_i ,l_{i1} ,\bar{v}_{i1} } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} }\sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 + r_\mathrm{d}v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }}-\frac{1}{r_a }\\&\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} }\!=\!\frac{v_{i-1} }{r_\mathrm{a} }\left( {\sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 \!+\! r_\mathrm{d}v_{i-1}^2\! +\!2r_\mathrm{a} r_\mathrm{d} l_{i1} }}-\frac{1}{\bar{v}_{i2} }} \right) \!<\!0 \end{aligned}$$

since

$$\begin{aligned}&\frac{r_\mathrm{a}+r_\mathrm{d}}{r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i1}} -\frac{1}{\bar{v}_{i2}^{2}}\\&\quad =\frac{2r_\mathrm{a}r_\mathrm{d}}{\bar{v}_{i2}^{2}\left( r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i1} \right) }\\&\quad \left( \left( \frac{1}{2r_\mathrm{a}}+\frac{1}{2r_\mathrm{d}} \right) \bar{v}_{i2}^{2} -\left( \frac{v_{i-1}^{2}}{2r_\mathrm{a}}+\frac{v_{i}^{2}}{2r_\mathrm{d}}+l_{i1} \right) \right) \\&\quad<\frac{2r_\mathrm{a}r_\mathrm{d}}{\bar{v}_{i2}^{2}\left( r_\mathrm{a}v_{i}^{2} +r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i1} \right) }\\&\quad \left( \left( \frac{1}{2r_\mathrm{a}}+\frac{1}{2r_\mathrm{d}} \right) \bar{v}_{i2}^{2}-\left( \frac{v_{i-1}^{2}}{2r_\mathrm{a}}+\frac{v_{i}^{2}}{2r_\mathrm{d}}+l_{i2} \right) \right) <0 \end{aligned}$$

where the last inequality is from (25).

So we have $\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} }<0$ when $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d} }}{>}\bar{v}_{i2}$ holds.

Then we consider the scenario when $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d} }}\le \bar{v}_{i2}$, i.e., where the following holds.

$$\begin{aligned}&\left( {\frac{1}{2r_\mathrm{a} }+\frac{1}{2 r_\mathrm{d}}} \right) \bar{v}_{i2}^2 \ge \frac{v_{i-1}^2 }{2r_\mathrm{a} }+\frac{v_i^2 }{2 r_\mathrm{d}}+l_{i2}\nonumber \\&\quad t\left( {v_{i-1} ,v_i ,l_{i2} ,\bar{v}_{i2} } \right) \nonumber \\&\quad =-\frac{v_{i-1} }{r_\mathrm{a} }-\frac{v_i }{r_\mathrm{d} }+\left( {\frac{1}{r_\mathrm{a} }+\frac{1}{r_\mathrm{d} }} \right) \sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d} }}\nonumber \\&\quad \frac{\hbox {d}t\left( {v_{i-1} ,v_i ,l_{i2} ,\bar{v}_{i2} } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} }\sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }}-\frac{1}{r_\mathrm{a} }\nonumber \\ \end{aligned}$$

(26)

Therefore, if $\sqrt{\frac{r_\mathrm{a} v_i^2 + r_\mathrm{d}v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }{r_\mathrm{a} +r_\mathrm{d} }}\le \bar{v}_{i1}$,

$$\begin{aligned}&\frac{\hbox {d}t(v_{{i-1}}, v_{i},l_{{i1}}, \bar{v}_{i1})}{\hbox {d}v_{i-1}}=\frac{v_{i-1}}{r_\mathrm{a}}\nonumber \\&\quad \times \sqrt{ \frac{r_\mathrm{a}+r_\mathrm{d}}{r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i1}}} -\frac{1}{r_\mathrm{a}}\\&\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} }\left( \sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }}\right. \\&\quad \left. -\sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }} \right) <0 \end{aligned}$$

since $l_{i1}{>}l_{i2}$.

If $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i1} }{r_\mathrm{a} +r_\mathrm{d}}}{>}{\bar{v}}_{i1}$,

$$\begin{aligned}&\frac{\hbox {d}t\left( {v_{i-1} ,v_i ,l_{i1} ,\bar{v}_{i1} } \right) }{\hbox {d}v_{i-1} }=\frac{v_{i-1} }{r_\mathrm{a} \bar{v}_{i1} }-\frac{1}{r_\mathrm{a}} \frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} } \\&\quad =\frac{v_{i-1} }{r_\mathrm{a} }\left( {\frac{1}{\bar{v}_{i1} }-\sqrt{\frac{r_\mathrm{a} +r_\mathrm{d} }{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }}} \right) <0 \end{aligned}$$

since

$$\begin{aligned}&\frac{1}{\bar{v}_{i1}^{2}}-\frac{r_\mathrm{a}+r_\mathrm{d}}{r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i2}} \\&\quad =\frac{2r_\mathrm{a}r_\mathrm{d}}{\bar{v}_{i1}^{2}\left( r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i2} \right) }\\&\quad \times \left( \left( \frac{v_{i-1}^{2}}{2r_\mathrm{a}} +\frac{v_{i}^{2}}{2r_\mathrm{d}}+l_{i2} \right) -\left( \frac{1}{2r_\mathrm{a}}+\frac{1}{2r_\mathrm{d}} \right) \bar{v}_{i1}^{2} \right) \\&\quad <\frac{2r_\mathrm{a}r_\mathrm{d}}{\bar{v}_{i1}^{2}\left( r_\mathrm{a}v_{i}^{2}+r_\mathrm{d}v_{i-1}^{2}+2r_\mathrm{a}r_\mathrm{d}l_{i2} \right) }\\&\quad \left( \left( \frac{v_{i-1}^{2}}{2r_\mathrm{a}}+\frac{v_{i}^{2}}{2r_\mathrm{d}}+l_{i2} \right) -\left( \frac{1}{2r_\mathrm{a}}+\frac{1}{2r_\mathrm{d}} \right) \bar{v}_{i2}^{2} \right) \le 0 \end{aligned}$$

where the last inequality is due to (26).

So we have $\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_{i-1} }<0$ when $\sqrt{\frac{r_\mathrm{a} v_i^2 +r_\mathrm{d} v_{i-1}^2 +2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} + r_\mathrm{d}}}\le \bar{v}_{i2}$ holds. Therefore, ${\Delta }\left( {v_{i-1} ,v_i } \right)$ is a strictly decreasing function of $v_{i-1}$. Similarly we can also show $\frac{\hbox {d}{\Delta }\left( {v_{i-1} ,v_i } \right) }{\hbox {d}v_i }<0$ and that ${\Delta }\left( {v_{i-1} ,v_i } \right)$ is a strictly decreasing function of $v_i$. $\square$

Proof of Proposition 8

Suppose it is not true, i.e., $\sqrt{\frac{2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d}}}\le \bar{v}_{i2}$, then

$$\begin{aligned} t\left( {0,0,l_{i2} ,\bar{v}_{i2} } \right) =\left( {\frac{1}{r_\mathrm{a} }+\frac{1}{r_\mathrm{d} }} \right) \sqrt{\frac{2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} +r_\mathrm{d} }} \end{aligned}$$

which is not dependent on $\bar{v}_{i2}$. So $t\left( {0,0,l_{i2} ,\bar{v}_{i2} } \right) =t\left( {0,0,l_{i2} ,\bar{v}_{i1} } \right)$ since $\sqrt{\frac{2r_\mathrm{a} r_\mathrm{d} l_{i2} }{r_\mathrm{a} + r_\mathrm{d}}}\le \bar{v}_{i2} \le \bar{v}_{i1}$ holds. From Lemma 1 and Condition 2

$$\begin{aligned} t\left( {0,0,l_{i2} ,\bar{v}_{i2} } \right) =t\left( {0,0,l_{i2} ,\bar{v}_{i1} } \right) <t\left( {0,0,l_{i1} ,\bar{v}_{i1} } \right) \end{aligned}$$

which contradicts Condition 5. So we have $l_{i2} >\left( {\frac{1}{2r_\mathrm{a} }+\frac{1}{2 r_\mathrm{d}}} \right) \bar{v}_{i2}^2$. $\square$

Proof of Lemma 4

Take $l^{'}=\frac{v_{{\mathrm{enter}}}^{\prime 2} -v_{{\mathrm{enter}}}^{2}}{{2r}_\mathrm{a}}+l+ \frac{v_{{\mathrm{exit}}}^{\prime 2}-v_{{\mathrm{exit}}}^{2}}{2^{*}r_\mathrm{d}}$. Then

$$\begin{aligned}&t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l,\bar{v} } \right) <t\left( {v_{\mathrm{enter}} ,v_{\mathrm{exit}} ,l^{\prime },\bar{v}} \right) \\&\quad \le t\left( {v_{\mathrm{enter}}^{\prime } ,v_{\mathrm{exit}}^{\prime } ,l,\bar{v} } \right) +\frac{v_{\mathrm{enter}}^{\prime } -v_{\mathrm{enter}} }{r_\mathrm{a} }+\frac{v_{\mathrm{exit}}^{\prime } -v_{\mathrm{exit}} }{r_\mathrm{d}} \end{aligned}$$

The first inequality is due to Lemma 1, and the second inequality is based on the optimality of $t\left( v_{\mathrm{enter}} ,v_{\mathrm{exit} } ,l^{\prime },\bar{v} \right)$. $\square$