Pricing strategies for the site-dependent vehicle routing problem

Abstract

The vehicle pricing game (VPG) which addresses the vehicles’ viewpoints within a vehicle routing problem (VRP) is introduced. Each vehicle acts as a player who demands a price per kilometer. That is, vehicles represent decentralized actors in transport systems, e.g., carriers under subcontracts. Based on these prices, a VRP is solved and profits are generated. Which price should a vehicle choose to maximize its own profit, considering the competition among vehicles? To answer this question, site dependencies leading to inhomogeneous vehicles are included. More detailed, skill-levels, e.g., relating to the size of a vehicle, are used to indicate a vehicle’s ability to carry out particular services. Moreover, penalty options are added. The VPG serves as an element of a vertical collaboration in a transport scenario and thus provides decision support for cooperative models. Theoretical results for the VPG are provided for a particular case of a two-player ring network game, for which the full set of equilibria is described and their uniqueness is discussed. It is shown that the uniqueness of the higher-skilled vehicle’s payoff is guaranteed even for multiple equilibria. The competition ratio is defined; it restricts a vehicle’s price to keep its competitiveness. Moreover, the acceptance ratio gives a lower bound on prices such that a loss of market share is still accepted. Experimental results are provided for general networks including the analysis of penalty options. It is demonstrated that strict site dependencies by tendency lead to monopolistic structures. In addition, particular penalty types show a positive effect regarding load imbalances caused by universally skilled vehicles.

Appendix

Proposition 1(see Sect. 4) Consider a two-player VPG applied to a Skill VRP on a ring with $K\ge 2$, $0<B<K$, and $\delta >0$. Furthermore, we have $\Sigma _1, \Sigma _2 \subseteq \mathbb R^+$. A feasible solution $x \ne x^I, x^{\textit{II}}$ will not be chosen by the central dispatcher.

Proof

First, we have that solution $x^I$ is feasible for any $B$, such that $\mathcal U =\ell ^I(\sigma ) = \delta \sigma _2 (K+1)$ is an upper bound on the optimal solution value. Furthermore, any vehicle $t$ is able to visit all nodes on a tour of length $\mathcal U^\mathrm{tour}=\delta (K+1)$. As we assume positive prices $\sigma _1,\sigma _2 \in \mathbb R^+,$ we conclude that $\mathcal U^\mathrm{tour}$ is an upper bound on the tour length of vehicle $t$ in an optimal solution. For a feasible solution $x$, let $V_1(x)$,$V_2(x)$ be the sets of nodes assigned to vehicles 1 and 2, respectively. It can be observed that once a vehicle $t$ serves $|V_t(x)|\ge (K+1)/2$ nodes, the respective tour length of vehicle $t$ is given by $\mathcal U^\mathrm{tour}$.

Assume that there is a feasible solution $x \ne x^I, x^{\textit{II}}$ that is chosen by the central dispatcher for a given $\sigma$, i.e., $x$ is optimal. By $K \ge 2$ and $B < K$, for $\sigma _1\ge \sigma _2$ the only optimal solutions are $x^I$ and $\bar{x}^I$. By Assumption 1, the dispatcher then chooses $x^I$, a contradiction, i.e., $\sigma _1 < \sigma _2$ has to hold.

For $V_2(x)=V{\setminus } \{0\},$ each feasible solution $x$ has total routing costs of at least $\ell ^I(\sigma )$. That is, by Assumption 1 we have that $x =x^I$, a contradiction. Thus $V_1(x)\ne \emptyset$ has to hold. From $B<K,$ we obtain that $\{B+1,B+2,\ldots ,K-1,K\} \subseteq V_2(x)$ is satisfied. Assume $V_2(x)=\{B+1,B+2,\ldots ,K-1,K\}$. But then, the total routing costs of any feasible solution $x$ are greater than or equal to $\ell ^{\textit{II}}(\sigma )$, i.e., by Assumption 1 we have $x= x^{\textit{II}}$, a contradiction.

Consequently, there has to be a node $M \in V_2(x)$ with $M \le B$. By $V_1(x)\ne \emptyset$ and $\mathcal U = \delta \sigma _2 (K+1),$ it has to hold that the tour length of vehicle 2 is less than $\delta (K+1)=U^\mathrm{tour}$. That is, vehicle 2 does not traverse the complete ring. Or, in other words, the tour of vehicle 2 is contained in a ring segment of $G'$ with less than $(K+1)/2$ nodes. Thus, vehicle 2 serves at most $K/2$ nodes. Consequently, vehicle 1 serves at least $K/2$ nodes. We distinguish two cases:

• i) Vehicle 1 is carrying out a complete tour, i.e., has a tour length of $\delta (K+1)$. Then we shift the node set $\{Q \in V_2(x):Q\le B\}$ from $V_2(x)$ to $V_1(x)$. The tour length of vehicle 1 will not be changed by this reassignment of nodes. However, the tour length of vehicle 2 will decrease by $\delta 2 (B-M+1)$. As we have positive strategies $\sigma _2 \in \mathbb R^+$, a decrease of the total routing costs will be obtained, a contradiction to the optimality of $x$.

• ii) Vehicle 1 does not carry out a complete tour, i.e., it has a tour length of less than $\delta (K+1)$. But then this tour is carried out on a ring segment containing less than $(K+1)/2$ nodes. As vehicle 1 serves at least $K/2$ nodes, this ring segment contains exactly $K/2$ nodes. By definition, node $K$ has to be in $V_2(x)$. Thus, the ring segment of vehicle 1 has to be $1,2,\ldots , K/2$. As there is an $M \le B$ in $V_2(x)$, we have that $B$ is in the ring segment of vehicle 2, i.e., $B \ge K/2+1$. Now, we shift the node set $\{Q \in V_2(x):Q\le B\}$ from $V_2(x)$ to $V_1(x)$. For vehicle 1, the tour length will increase by $\delta (K+1)-2 \delta K/2=\delta$. The tour length of vehicle 2 will decrease by $2 \delta K/2 - 2 \delta (K-B)=\delta (2B-K) \ge \delta (2 (K/2+1)-K)=2 \delta$. By $\sigma _1 < \sigma _2,$ we obtain a reduction of the total routing costs, a contradiction to the optimality of $x$. $\square$

Proof of Theorem 2 (see Sect. 4) Consider a two-player VPG applied to a Skill VRP on a ring with $K\ge 2$ and $\delta >0$. Furthermore, let $\Sigma _1, \Sigma _2 \subseteq \mathbb R^+$ be finite strategy sets with $|\Sigma _1|, |\Sigma _2| \ge 2$. Assume that $(K-1)/2<B<K$ is satisfied and that $\alpha >0$ and $\beta >0$ holds. Then we have the set of equilibria given as

$$\begin{aligned} \Sigma ^*=\bigcup _{k=1}^4 \Sigma ^k \end{aligned}$$

(42)

and we say that strategy profiles of types 1–4 are equilibria of types 1–4.

Proof

First note that for $(K-1)/2<B<K$, positive $\alpha$ and $\beta$ exist as given in Lemmas 2 and 4. Furthermore, $\hat{\sigma }_1$ and $\bar{\sigma }_2$ exist by definition of $\Sigma _1$ and $\Sigma _2$. We first prove that $\Sigma ^k\subseteq \Sigma ^*$ holds for $k=1,\ldots ,4$, i.e., strategy profiles of types 1–4 are equilibria, if they exist. Afterward, we prove that there are no further equilibria in that game.

$k=1$, a) Let there be no $\sigma _2\in \Sigma _2$ that satisfies $\sigma _2\alpha > \hat{\sigma }_1$, i.e., $\Sigma ^1$ is defined according to (27). For all $\sigma _2\in \Sigma _2,$ it holds that $\sigma _2\alpha \le \hat{\sigma }_1$. By $\hat{\sigma }_1=\min \Sigma _1,$ it follows that $\sigma _2 \alpha \le \sigma _1$ for all $\sigma \in \Sigma$. That is, the dispatcher chooses $x^I$ and we have $p_1(\sigma )=0$ and $p_2(\sigma )=\delta \sigma _2(K+1)$ for all $\sigma \in \Sigma$. Consider a $\sigma ^1\in \Sigma ^1$. We have that $\sigma _2^{1}=\max \Sigma _2$ maximizes $p_2(\sigma )$ for all $\sigma _1 \in \Sigma _1$. As $p_1(\sigma )$ is constant, it follows that $\Sigma ^1\subseteq \Sigma ^*$.

b) For all other cases, we have $\Sigma ^1=\emptyset$, i.e., $\Sigma ^1\subseteq \Sigma ^*$.

$k=2$, a) Let $\hat{\sigma }_2=\min \{\sigma _2\in \Sigma _2:\sigma _2\alpha > \hat{\sigma }_1\}$ and $\bar{\sigma }_2^{2}=\max \{\sigma _2\in \Sigma _2:\sigma _2\alpha \le \hat{\sigma }_1\}$ exist, i.e., $\Sigma ^2_1$ is well defined. Let furthermore $\bar{\sigma }_2^2 \beta \ge \bar{\sigma }_2$ hold. Then, $\Sigma ^2$ is nonempty according to (28). By definition of $\bar{\sigma }_2^2$ and $\hat{\sigma }_2,$ we have

$$\begin{aligned} \not \exists \ \sigma _2\in \Sigma _2: \bar{\sigma }_2^2<\sigma _2<\hat{\sigma }_2. \end{aligned}$$

(43)

Thus, we have for all $\sigma ^2_1\in \Sigma _1^2$

$$\begin{aligned}&\min \left\{ \sigma _2\in \Sigma _2:\sigma _2\alpha >\sigma ^2_1\right\} =\hat{\sigma }_2\end{aligned}$$

(44)

$$\begin{aligned}&\max \left\{ \sigma _2\in \Sigma _2:\sigma _2\alpha \le \sigma ^2_1\right\} =\bar{\sigma }^2_2. \end{aligned}$$

(45)

Thus, for all strategies $\sigma _1^2 \in \Sigma ^{2}_1$ it holds that for $\sigma ^2=(\sigma ^2_1,\bar{\sigma }_2^{2})$ the dispatcher chooses $x^I$ and for $(\sigma ^2_1,\hat{\sigma }_2)$ solution $x^{\textit{II}}$ is chosen. That is, we have $p_2(\sigma ^{2})=\delta \sigma _2^{2} (K+1)$ for all $\sigma ^2\in \Sigma ^2$. Recall that we have $\alpha >0$ by assumption. If vehicle 2 chooses any $\sigma _2<\sigma _2^{2}$ while vehicle 1 keeps $\sigma _1^2$, then we obtain $\sigma _2\alpha <\sigma _2^2\alpha =\bar{\sigma }_2^2\alpha \le \sigma _1^2$, i.e., solution $x^I$ is chosen and thus $p_2(\sigma _1^2,\sigma _2)=\delta \sigma _2 (K+1)<\delta \sigma _2^{2} (K+1)= p_2(\sigma ^2)$. Alternatively, consider $\sigma _2>\sigma _2^{2}$ which exists as $\hat{\sigma }_2$ exists and we obtain by (43) that $\sigma _2\ge \hat{\sigma }_2$ holds. Furthermore, $\sigma _2\le \bar{\sigma }_2$. Consequently, $\bar{\sigma }_2\alpha \ge \sigma _2\alpha \ge \hat{\sigma }_2\alpha >\sigma ^2_1$. Thus, for strategy profiles $(\sigma _1^2,\sigma _2)$ and $(\sigma _1^2,\bar{\sigma }_2)$, solution $x^{\textit{II}}$ is chosen, i.e.,

$$\begin{aligned} p_2(\sigma _1^2,\sigma _2)=\delta \sigma _2 2(K-B)\le & {} \delta \bar{\sigma }_2 2(K-B) = p_2(\sigma _1^2,\bar{\sigma }_2)\\\le & {} p_2(\sigma ^{2}). \end{aligned}$$

The last inequality holds due to (23) and (24) and as $\bar{\sigma }_2^2 \beta \ge \bar{\sigma }_2$ is satisfied. Summarizing, vehicle 2 has no possibility to increase its own payoff.

Next, consider vehicle 1 for which we have $p_1(\sigma ^{2})=0$. Choosing $\sigma _1>\sigma _1^{2}$ while vehicle 2 keeps $\sigma _2^2$ leads to solution $x^I$ being chosen as $\sigma _1>\sigma _1^2\ge \bar{\sigma }_2^2\alpha =\sigma _2^2\alpha$ holds. Thus, we have $p_1(\sigma _1,\sigma _2^2)=0=p_1(\sigma ^2)$. Alternatively, consider $\sigma _1<\sigma _1^{2}$. By definition of $\Sigma ^2_1$ we have $\hat{\sigma }_1 \in \Sigma _1^{2}$, i.e., we have that $\sigma _1 \in \Sigma _1^{2}$ as $\bar{\sigma }_2^2\alpha \le \hat{\sigma }_1\le \sigma _1 <\sigma _1^2<\hat{\sigma }_2\alpha$ does hold. That is, for $(\sigma _1,\sigma ^{2}_2),$ solution $x^I$ is chosen by the dispatcher and we have $p_1(\sigma _1,\sigma _2^{2})=0=p_1(\sigma ^2)$. Summarizing, vehicle 1 is not able to increase the own payoff. Consequently, $\sigma ^{2}$ is an equilibrium.

b) For all other cases, we have $\Sigma ^2=\emptyset$, i.e., $\Sigma ^2\subseteq \Sigma ^*$.

$k=3$, a) Let $\bar{\sigma }_1=\max \{\sigma _1\in \Sigma _1:\sigma _1<\bar{\sigma }_2\alpha \}$ exist and, furthermore, let there be no $\sigma _2 \in \Sigma _2$ that satisfies $\sigma _2\alpha \le \bar{\sigma }_1$, i.e., $\Sigma ^3$ is nonempty according to (29). For all $\sigma ^3\in \Sigma ^3$ we have that $x^{\textit{II}}$ is chosen by the dispatcher. Furthermore, as we have $\sigma _2^3=\max \Sigma _2$, vehicle 2 can only deviate by choosing $\sigma _2<\sigma _2^3$. However, for all $\sigma _2 \in \Sigma _2$ we have that $\sigma _2\alpha > \bar{\sigma }_1=\sigma ^3_1$ holds, i.e., solution $x^{\textit{II}}$ is chosen for $(\sigma _1^3,\sigma _2)$. Thus we have $p_2(\sigma _1^3,\sigma _2)= \delta \sigma _2 2(K-B)<\delta \sigma ^3_2 2(K-B)=p_2(\sigma ^3)$. Consequently, vehicle 2 is not able to increase its own payoff.

Next, consider vehicle 1. We have $p_1(\sigma ^3)=\delta \sigma _1^3 2 B$ for $B\le (K+1)/2$ and $p_1(\sigma ^3)=\delta \sigma _1^3 (K+1)$ for $B\ge (K+1)/2$. Thus, independent from $B$, $p_1(\sigma ^3)>0$ does hold. By definition of $\bar{\sigma }_1$, choosing $\sigma _1>\sigma ^3_1=\bar{\sigma }_1$ (if such a $\sigma _1$ exists) leads to $\sigma _1>\bar{\sigma }_2\alpha =\sigma _2^3\alpha$, i.e., solution $x^I$ is chosen. Consequently, we obtain $p_1(\sigma _1,\sigma _2^3)=0<p_1(\sigma ^3)$. On the other hand, choosing $\sigma _1<\sigma _1^3$ (if such a $\sigma _1$ exists) will lead to a reduction of $p_1$ independent of parameter $B$. Summarizing, also vehicle 1 is not able to increase its own payoff and $\sigma ^3$ is an equilibrium, i.e., $\Sigma ^3\subseteq \Sigma ^*$.

b) For all other cases we have $\Sigma ^3=\emptyset$, i.e., $\Sigma ^3\subseteq \Sigma ^*$.

$k=4$, a) Let $\bar{\sigma }_1=\max \{\sigma _1\in \Sigma _1:\sigma _1<\bar{\sigma }_2\alpha \}$ and $\tilde{\sigma }_2=\max \{\sigma _2 \in \Sigma _2: \sigma _2\alpha \le \bar{\sigma }_1\}$ exist. Let furthermore $\sigma _2^4 \ge \tilde{\sigma }_2\beta$ be satisfied, i.e., $\Sigma ^4$ is nonempty according to (30). For all $\sigma ^4\in \Sigma ^4$ we have that $x^{\textit{II}}$ is chosen by the dispatcher, i.e., we have $p_1(\sigma ^4)=\delta \sigma _1^4 2 B>0$ for $B\le (K+1)/2$ and $p_1(\sigma ^4)=\delta \sigma _1^4 (K+1)>0$ for $B\ge (K+1)/2$. Choosing $\sigma _1<\sigma _1^4$ leads to $\sigma _1<\sigma _1^4=\bar{\sigma }_1<\bar{\sigma }_2\alpha =\sigma _2^4\alpha$, i.e., solution $x^{\textit{II}}$ remains chosen by the dispatcher. Consequently, independent of $B,$ one obtains $p_1(\sigma _1,\sigma _2^4)<p_1(\sigma _1^4,\sigma _2^4)$. Choosing $\sigma _1>\sigma _1^4=\bar{\sigma }_1$, however, will result in $x^I$ being chosen by the dispatcher with $p_1(\sigma )=0$. That is, vehicle 1 has no incentive to change the chosen strategy.

Next, consider vehicle $2$ with $p_2(\sigma ^4)=\delta \sigma _2^4 2(K-B)$. A strategy $\sigma _2>\sigma _2^4=\bar{\sigma }_2=\max \Sigma _2$ does not exist in $\Sigma _2$. Alternatively, let $\sigma _2<\sigma _2^4$. For $\sigma _2>\tilde{\sigma }_2$, we have that $x^{\textit{II}}$ remains chosen by the dispatcher for $(\sigma _1^4,\sigma _2)$, then obviously $p_2(\sigma _1^4,\sigma _2)<p_2(\sigma ^4)$ does hold. Alternatively, for $\sigma _2\le \tilde{\sigma }_2$ we have that $(\sigma _1^4,\sigma _2)$ leads to solution $x^I$ and

$$\begin{aligned} p_2(\sigma _1^4,\sigma _2)=\delta \sigma _2 (K+1)\le & {} \delta \tilde{\sigma }_2 (K+1)=p_2(\sigma _1^4,\tilde{\sigma }_2)\\\le & {} p_2(\sigma ^4). \end{aligned}$$

The last inequality holds by (24) and (25), as $\sigma ^4_2 \ge \tilde{\sigma }_2\beta$ is satisfied. Summarizing, vehicle 2 has no possibility to increase the payoff, i.e., $\sigma ^4$ is an equilibrium and we have $\Sigma ^4\subseteq \Sigma ^*$.

b) For all other cases, we have $\Sigma ^4=\emptyset$, i.e., $\Sigma ^4\subseteq \Sigma ^*$.

For the second part of the proof, it remains to show that

$$\begin{aligned} \sigma '\notin \bigcup _{k=1}^4\Sigma ^k \Rightarrow \sigma ' \notin \Sigma ^*. \end{aligned}$$

We prove by contradiction: assume there is a $\sigma ' \in \Sigma ^*$ and $\sigma '\notin \bigcup _{k=1}^4\Sigma ^k$.

i) Let $\sigma _1'<\sigma _2'\alpha$, i.e., $x^{\textit{II}}$ is chosen by the dispatcher. Assume $\sigma _2'<\bar{\sigma }_2=\max \Sigma _2$. But then $p_2(\sigma ')=\delta \sigma _2'2 (K-B)< \delta \bar{\sigma }_2 2 (K-B)=p_2(\sigma _1',\bar{\sigma }_2),$ as by $\sigma _1'<\sigma _2'\alpha <\bar{\sigma }_2\alpha$ solution $x^{\textit{II}}$ remains chosen for $(\sigma _1',\bar{\sigma }_2)$, a contradiction to $\sigma '\in \Sigma ^*$. Thus, we have

$$\begin{aligned} \sigma _2'=\bar{\sigma }_2. \end{aligned}$$

(46)

Moreover, $\bar{\sigma }_1=\max \{\sigma _1\in \Sigma _1:\sigma _1<\bar{\sigma }_2\alpha \}$ exists as $\sigma _1'<\sigma _2'\alpha =\bar{\sigma }_2\alpha$. Note that solution $x^{\textit{II}}$ is chosen for $(\bar{\sigma }_1,\sigma _2')$. Assume that $\sigma _1'<\bar{\sigma }_1$ holds. But then we have $p_1(\sigma ')=\delta \sigma _1' 2 B<\delta \bar{\sigma }_1 2 B=p_1(\bar{\sigma }_1,\sigma _2')$ for $B\le (K+1)/2$ and $p_1(\sigma ')=\delta \sigma _1' (K+1)<\delta \bar{\sigma }_1 (K+1)=p_1(\bar{\sigma }_1,\sigma _2')$ for $B\ge (K+1)/2$, a contradiction to $\sigma '\in \Sigma ^*$. Thus, we have

$$\begin{aligned} \sigma _1'=\bar{\sigma }_1. \end{aligned}$$

(47)

i.i) Assume there is no $\sigma _2\in \Sigma _2$ such that $\sigma _2\alpha \le \bar{\sigma }_1$. But then we have $\sigma '\in \Sigma ^3$, a contradiction.

i.ii) Let $\tilde{\sigma }_2=\max \{\sigma _2\in \Sigma _2:\sigma _2\alpha \le \bar{\sigma }_1\}$ exist. If $\bar{\sigma }_2\ge \tilde{\sigma }_2\beta ,$ then we have $\sigma '\in \Sigma ^4$, a contradiction. Alternatively, if $\bar{\sigma }_2 < \tilde{\sigma }_2\beta ,$ we obtain $p_2(\sigma ')=p_2(\bar{\sigma }_1,\bar{\sigma }_2)< p_2(\bar{\sigma }_1,\tilde{\sigma }_2)=p_2(\sigma _1',\tilde{\sigma }_2)$ by (23), a contradiction to $\sigma '\in \Sigma ^*$.

ii) From the previous cases, we obtain that $\sigma _1'\ge \sigma _2'\alpha$ has to hold, i.e., $x^{I}$ is chosen by the dispatcher and we have $p_1(\sigma ')=0$. Consequently, $\sigma _2^{\max }=\max \{\sigma _2\in \Sigma _2:\sigma _2\alpha \le \sigma _1'\}$ exist and $x^I$ remains chosen for $(\sigma _1',\sigma _2^{\max })$. Assume that $\sigma _2'<\sigma _2^{\max }$. But then we have $p_2(\sigma ')=\delta \sigma _2' (K+1)<\delta \sigma _2^{\max } (K+1)= p_2(\sigma _1',\sigma _2^{\max })$, a contradiction to $\sigma '\in \Sigma ^*$. Thus, we have

$$\begin{aligned} \sigma _2'=\sigma _2^{\max }. \end{aligned}$$

(48)

Assume there is no $\sigma _2\in \Sigma _2$ such that $\sigma _2\alpha >\hat{\sigma }_1=\min \Sigma _1$. It follows for all $\sigma _2\in \Sigma _2$ that $\sigma _2\alpha \le \hat{\sigma }_1\le \sigma _1'$ and we obtain $\sigma _2'=\sigma _2^{\max }=\max \Sigma _2=\bar{\sigma }_2$. That is, we have $\sigma '\in \Sigma ^1$, a contradiction. Thus, $\hat{\sigma }_2=\min \{\sigma _2\in \Sigma _2:\sigma _2\alpha >\hat{\sigma }_1\}$ exists.

ii.i) Assume that there is no $\sigma _2\in \Sigma _2$ such that $\sigma _2\alpha \le \hat{\sigma }_1$. That is, for all $\sigma _2\in \Sigma _2$ we have $\sigma _2\alpha > \hat{\sigma }_1$ and solution $x^{\textit{II}}$ is chosen for $(\hat{\sigma }_1,\sigma _2)$ for all $\sigma _2\in \Sigma _2$. In particular, this holds true for $\sigma _2'$ and we obtain $p_1(\hat{\sigma }_1,\sigma _2')=\delta \hat{\sigma }_1 2 B>p_1(\sigma ')=0$ for $B\le (K+1)/2$ and $p_1(\hat{\sigma }_1,\sigma _2')=\delta \hat{\sigma }_1 (K+1)>p_1(\sigma ')=0$ for $B\ge (K+1)/2$. This is a contradiction to $\sigma '\in \Sigma ^*$.

ii.ii) Alternatively, let $\bar{\sigma }_2^2=\max \{\sigma _2\in \Sigma _2:\sigma _2\alpha \le \hat{\sigma }_1\}$ exist. Thus, $\Sigma ^2_1$ is well-defined and non-empty. Assume that $\sigma _1'\notin \Sigma _1^2$ holds. We have for all $\sigma _1 \in \Sigma$ that $\sigma _2^{2} \alpha \le \hat{\sigma }_1\le \sigma _1$ holds. That is, from $\sigma '_1 \notin \Sigma _1^{2}$ we obtain that $\hat{\sigma }_2\alpha \le \sigma _1'$. Consequently, $\hat{\sigma }_2 \le \max \{\sigma _2 \in \Sigma _2:\sigma _2\alpha \le \sigma '_1\}=\sigma _2^{\max }=\sigma _2'$. Thus, for all $\sigma _1\in \Sigma _1^{2},$ we have $\sigma _1 < \hat{\sigma }_2 \alpha \le \sigma '_2\alpha$. That is, for $(\sigma _1,\sigma _2')$, the solution $x^{\textit{II}}$ is chosen by the dispatcher and we obtain $p_1(\sigma ')=0<p_1(\sigma _1,\sigma _2')=\delta \sigma _1 2 B$ for $B\le (K+1)/2$ and $p_1(\sigma ')=0<p_1(\sigma _1,\sigma _2')=\delta \sigma _1 (k+1)$ for $B\ge (K+1)/2$. This is a contradiction to $\sigma '\in \Sigma ^*$. Thus, we have

$$\begin{aligned} \sigma _1'\in \Sigma _1^2. \end{aligned}$$

(49)

We obtain by (45) that $\bar{\sigma }_2^2=\max \{\sigma _2\in \Sigma _2:\sigma _2\alpha \le \sigma _1'\}=\sigma _2^{\max }=\sigma _2'$ holds. Let moreover $\bar{\sigma }_2^2\beta \ge \bar{\sigma }_2$ be satisfied. But then $\sigma '\in \Sigma ^2$, a contradiction.

Alternatively, let $\bar{\sigma }_2^2\beta <\bar{\sigma }_2$. But then we have $p_2(\sigma ')=p_2(\sigma _1',\bar{\sigma }_2^2)<p_2(\sigma _1',\bar{\sigma }_2)$ by (25), a contradiction to $\sigma '\in \Sigma ^*$. Consequently, $\sigma _1'\ge \sigma _2'\alpha$ cannot hold true and it can be concluded that there exists no $\sigma ' \in \Sigma ^*$ such that $\sigma '\notin \bigcup _{k=1}^4\Sigma ^k$. $\square$