Congestion Control for a System with Parallel Stations and Homogeneous Customers Using Priority Passes

Abstract

We consider a game theoretic congestion model with parallel nodes and homogeneous customers. The purpose of this paper is to examine how the priority passes improve social welfare for such a system. To this end, we prove the existence of an equilibrium. The system with no priority pass has a unique equilibrium. With the introduction of priority passes, the uniqueness of the equilibrium may be destroyed. We provide a sufficient condition under which the system with priority passes outperforms the system with no priority passes. The problem is explored numerically as well.

Appendix

Proof of Lemma 1

The dual linear program to (1) is

$$\begin{array} [c]{rl} & \min\limits_{a,\boldsymbol{z}}Ha+\sum\limits_{j = 1}^{J}z_{j}\\ {\text{s.t.}} & t_{j}a+z_{j}\geq b_{j}-ct_{j},\quad j = 1,2,\ldots,J,\\ & a\geq0,\quad\boldsymbol{z}\geq\boldsymbol{0}. \end{array} $$

Note that x is optimal if and only if x is feasible and there exists a dual feasible solution (a, z) satisfying the complementarity condition, i.e., $a\cdot \left ({\sum }_{j = 1}^{J}t_{j}x_{j}-H\right ) = 0$, z_j (1 − x_j) = 0 and x_j (t_ja + z_j − (b_j − ct_j)) = 0. This is equivalent to the condition of the lemma. □

Proof of Theorem 1

Let x ∈ B (T(x)). Then, from Lemma 1, x is a feasible solution of (1), and there exists a ≥ 0 such that the complementarity condition and (3) hold true. To show that x = x(a) holds, we will prove that (a) u_j (0) ≤ a ⇒ x_j = 0; (b)$u_{j}(1)<a<u_{j}(0)\Rightarrow x_{j}=u_{j}^{-1}(a)$; and (c) u_j (1) ≥ a ⇒ x_j = 1. Suppose that (a) does not hold, so that u_j (0) ≤ a and x_j > 0. Lemma 1 and x_j > 0 imply u_j (x_j) ≥ a. Since u_j is strictly deceasing, u_j (0) > a, contradiction. Thus, (a) holds. To prove (b), we note from Lemma 1 that u_j (1) < a < u_j (0) implies 0 < x_j < 1(otherwise,x_j = 0 or 1, which implies u_j (0) ≤ a or u_j (1) ≥ a, contradiction). This in turn implies from Lemma 1 that u_j (x_j) = a. To prove (c), suppose otherwise, i.e., u_j (1) ≥ a and x_j < 1. Lemma 1 and x_j < 1 implies u_j (x_j) ≤ a, so that u_j (1) < a, contradiction. To prove the converse of Theorem 1, we will show that x = x(a) satisfies (a) x_j = 0 ⇒ u_j (x_j) ≤ a; (b) 0 < x_j < 1 ⇒ u_j (x_j) = a; and (c) x_j = 1 ⇒ u_j (x_j) ≥ a. To prove (a), suppose x_j = 0 and u_j (x_j) > a. Since u_j (0) > a, (5) implies x_j (a) > 0, contradiction. To prove (b), suppose 0 < x_j < 1 and u_j (x_j) ≠ a. Note that u_j (x_j) ≠ a and (5) imply x_j = 0 or 1, contradiction. To prove (c), suppose x_j = 1 and u_j (x_j) < a. Thus, u_j (1) < a, which implies x_j (a) < 1 (see (5)), contradiction. □

Proof of Theorem 2

Let $h(\boldsymbol {x})\equiv {\sum }_{j = 1}^{J}x_{j}T_{j}(x_{j})$. As h(x) and x(a) are increasing and nonincreasing, respectively, h(x(a)) is nonincreasing. Furthermore, for a large enough, x(a) = 0 so that h(x(a)) = 0. Hence, either (a)s there exists $\bar {a}>0$ such that $h(\boldsymbol {x}(\bar {a}))=H$, or (b)h(x(a)) < H holds for all a > 0, but not both. We first examine Case (a). Let

$$S\equiv\{\bar{a}:\bar{a}>0,h(\boldsymbol{x}(\bar{a}))=H\}. $$

Suppose that S is a single ton with $S=\{\bar {a}\}$. Then, $\boldsymbol {x}=\boldsymbol {x}(\bar {a})$ satisfies (7), and no other x with x = x(a), a > 0, does. Note that x(0) is not an equilibrium because it does not satisfy the time constraint of (1). This completes the subcase of$S=\{\bar {a}\}$. Suppose that S is not asingleton. As $h(\boldsymbol {x}(\bar {a}))$ is constant over $\bar {a}\in S$, x(a)is non increasing and h(x) is increasing, we conclude that $\boldsymbol {x}(\bar {a})$ is constant over $\bar {a}\in S$. Thus, $\boldsymbol {x}(\bar {a})$, $\bar {a}\in S$, is the only x with x = x(a), a > 0, satisfying (7).To show that x(0) is not another equilibrium, suppose contrarily that x(0),$\boldsymbol {x}(0)\neq \boldsymbol {x}(\bar {a})$, $\bar {a}\in S$, is an equilibrium. As h(x) and x(a) are increasing and non increasing, respectively, $\boldsymbol {x}(0)\neq \boldsymbol {x}(\bar {a})$, $\bar {a}\in S$, implies that $H\geq h(\boldsymbol {x}(0))>h(\boldsymbol {x} (\bar {a}))=H$, which is a contradiction. Thus, x(0) is not an equilibrium. We next examine Case (b). In this case, a = 0 is the only way that (7) is satisfied. Clearly, (b) implies that h(x(0)) ≤ H. Thus, x(0) satisfies the constraint of (1). □

To prove Theorem 3, we require some mathematical tools.

Definition 1

A correspondence ϕ : X → 2^Y is upper semicontinuous if for any sequence (xⁿ), (yⁿ) the conditions $\lim _{n\rightarrow \infty }\boldsymbol {x}^{n}=\boldsymbol {x},\lim _{n\rightarrow \infty }\boldsymbol {y}^{n}=\boldsymbol {y},\boldsymbol {y}^{n} \in \phi (\boldsymbol {x}^{n})$ for all n imply that y ∈ ϕ(x). ϕ is lower semicontinuous if $\lim _{n\rightarrow \infty }\boldsymbol {x}^{n}=\boldsymbol {x}\,$ and y ∈ ϕ(x) implies that there exists y_n ∈ ϕ(x_n) such that $\lim _{n\rightarrow \infty }\boldsymbol {y}_{n}=\boldsymbol {y}$. ϕ is continuous if it is both upper and lower semi continuous (see, e.g., Wets 1985).

Let Q be defined by the optimal value of a linear program as follows:

$$ Q(\boldsymbol{e})=\sup\{\boldsymbol{\gamma}\cdot\boldsymbol{x}:A\boldsymbol{x} \leq\boldsymbol{\beta},\boldsymbol{x}\geq0\} $$

(29)

where $\boldsymbol {e}=(\boldsymbol {\gamma },A,\boldsymbol {\beta })\in {\mathbb {R}}^{n}\times {\mathbb {R}}^{m\times n}\times {\mathbb {R}}^{m}$ represents the parameters of the linear program. Let E = {e : −∞ < Q(e) < ∞}. Define the feasible regions of the primal and the dual programs as

$$\begin{array}{@{}rcl@{}} K(\boldsymbol{e}) & =&\{\boldsymbol{x}:A\boldsymbol{x}\leq\boldsymbol{\beta },\text{ }\boldsymbol{x}\geq0\}, \end{array} $$

(30)

$$\begin{array}{@{}rcl@{}} D(\boldsymbol{e}) & =&\{\boldsymbol{y}:\boldsymbol{y}A\geq\boldsymbol{\gamma },\text{ }\boldsymbol{y}\geq0\}. \end{array} $$

(31)

Theorem 6

• 1. Lete ∈ E. If correspondencesK andD are continuous ate, thenQ is continuous ate (Wets 1985).

• 2. IfintK(e) ≠ ∅for alle ∈ Eand no row of (A, β) is identically0for alle ∈ E, thenK is continuous onE. Similarly, ifintD(e) ≠ ∅ for alle ∈ Eand no column of$\binom {A}{\boldsymbol {\gamma }}$isidentically0for alle ∈ E, thenD is continuous on E (Wets 1985).

• 3. Ifϕ₁ : X → 2^Yandϕ₂ : Y → 2^Zare upper semicontinuous andY is compact, then the compositecorrespondence$\phi (\boldsymbol {x})=\bigcup \{\phi _{2}(\boldsymbol {y}):{\boldsymbol {y}\in \phi _{1}(\boldsymbol {x})}\}$isupper semicontinuous (Hogan 1973).

• 4. Let$X\subset {\mathbb {R}}^{n}$bea compact and convex set, and letϕ : X → 2^Xbe an upper semicontinuous correspondence such that the setϕ(x) ⊂ Xis convex and nonempty for allx ∈ X. Thenϕhas a fixed point, i.e.,x ∈ ϕ (x) (see, e.g., p.423 of Arrow and Hahn (1971)).

Let $\hat {Q}(\boldsymbol {t})$ be the optimal value of problem P(t;F), t = (t⁺, t⁻), in (16). Define

$$\begin{array}{@{}rcl@{}} A & =&\left( \begin{array} [c]{cc} \begin{array} [c]{cccc} t_{1}^{+} & t_{2}^{+} & {\cdots} & t_{J}^{+}\\ 1 & 1 & {\cdots} & 1 \end{array} &\begin{array} [c]{cccc} t_{1}^{-} & t_{2}^{-} & {\cdots} & t_{J}^{-}\\ 0 & 0 & {\cdots} & 0 \end{array} \\ \boldsymbol{I} & \boldsymbol{I} \end{array} \right) , \end{array} $$

(32)

$$\begin{array}{@{}rcl@{}} \boldsymbol{\beta} & =&(H,F,1,1,\ldots,1)^{T}\in{\mathbb{R}}^{2+J} , \end{array} $$

(33)

$$\begin{array}{@{}rcl@{}} \boldsymbol{\gamma} & =&(b_{1}-ct_{1}^{+},\ldots,b_{J}-ct_{J}^{+} ,b_{1}-ct_{1}^{-},\ldots,b_{J}-ct_{J}^{-}) \end{array} $$

(34)

where I is an identity matrix of order J. Then, the feasible region $\hat {K}(\boldsymbol {t})$ of (16) and the feasible region $\hat {D}(\boldsymbol {t})$ of its dual are given by

$$\begin{array}{@{}rcl@{}} \hat{K}(\boldsymbol{t}) & =&\{\boldsymbol{x}\in{\mathbb{R}}^{2J} :A\boldsymbol{x}\leq\boldsymbol{\beta},\text{ }\boldsymbol{x}\geq0\},\\ \hat{D}(\boldsymbol{t}) & =&\{\boldsymbol{y}\in{\mathbb{R}}^{2+J} :\boldsymbol{y}A\geq\boldsymbol{\gamma},\text{ }\boldsymbol{y}\geq0\}. \end{array} $$

As $\hat {K}(\boldsymbol {t})$ is compact and nonempty, linear program (16) has an optimal solution for all $\boldsymbol {t}\in {\mathbb {R}}_{+}^{2J}$ where ${\mathbb {R}}_{+}=[0,\infty )$. Thus, P(t;F)≠∅ for all $\boldsymbol {t}\in {\mathbb {R}}_{+}^{2J}$.

Lemma 2

Suppose thatF > 0 . Then, (a) correspondences$\hat {K}$and$\hat {D}$arecontinuous, (b) function$\hat {Q}$iscontinuous, and (c) correspondenceP(t;F)is upper semicontinuous with respect tot = (t⁺, t⁻).

Proof

For a sufficiently small ε > 0, $\boldsymbol {x}^{\ast } =(\varepsilon ,\varepsilon ,\ldots ,\varepsilon )\in {\mathbb {R}}_{+}^{2J}$ is an interior point $\hat {K}(\boldsymbol {t})$. Equations (32) and (33) show that no row of (A, β) is 0. Thus, the conditions of Theorem 6 (2) are met for $\hat {K}(\boldsymbol {t})$. Similarly, $\hat {D}(\boldsymbol {t})$ too satisfies the conditions of Theorem 6 (2). Thus, $\hat {K}$ and $\hat {D}$ are continuous on ${\mathbb {R}}_{+}^{2J}$, proving (a). Part (b) follows from part (a) and Theorem 6 (1). Clearly, P(t; F) can be written as

$$P(\boldsymbol{t};F)=\{\boldsymbol{x}:\hat{Q}(\boldsymbol{t})=\pi (\boldsymbol{x},\boldsymbol{t}),\boldsymbol{x}\in\hat{K}(\boldsymbol{t})\} $$

where

$$\pi(\boldsymbol{x},\boldsymbol{t})=\sum\limits_{j = 1}^{J}(b_{j}-ct_{j}^{+})x_{j} ^{+}+(b_{j}-ct_{j}^{-})x_{j}^{-}. $$

Let tⁿ → t, xⁿ → x as n →∞ and xⁿ ∈ P(tⁿ; F). Then, $\hat {Q}(\boldsymbol {t}^{n} )=\pi (\boldsymbol {x}^{n},\boldsymbol {t}^{n})$ and $\boldsymbol {x}^{n}\in \hat {K}(\boldsymbol {t}^{n})$. Since $\hat {K}(\boldsymbol {t})$ and $\hat {Q}(\boldsymbol {t})$ are continuous from Parts (a) and (b), $\hat {Q}(\boldsymbol {t})=\pi (\boldsymbol {x},\boldsymbol {t}),$$\boldsymbol {x} \in \hat {K}(\boldsymbol {t})$, so that x ∈ P(t; F) holds true, proving that P(t; F) is upper semi continuous in t. □

Proof of Theorem 3

If F = 0, then Theorem 2 can be invoked. So, suppose F > 0. Define

$$\bar{T}_{j}^{k}(x_{j}^{+},x_{j}^{-})=\min\{{T_{j}^{k}}(x_{j}^{+},x_{j} ^{-}),H+\varepsilon\},\quad k=+,-,j = 1,2,\ldots,J $$

where ε > 0 is arbitrarily fixed. Clearly, the equilibrium set $\mathcal {P}(F)$ remains the same (regardless of whether $\mathcal {P}(F)$ is empty or not) after replacing ${T_{j}^{k}}(x_{j}^{+},x_{j}^{-})$ with $\bar {T}_{j}^{k}(x_{j}^{+},x_{j}^{-})$. Define a function $\boldsymbol {\bar {T}}_{p}:[0,1]^{2J}\rightarrow \lbrack 0,H+\varepsilon ]^{2J}$ by $\boldsymbol {\bar {T}}_{p}(\boldsymbol {x}^{+},\boldsymbol {x}^{-})=((\bar {T}_{j}^{+}(x_{j}^{+},x_{j}^{-})),(\bar {T}_{j}^{-}(x_{j}^{+},x_{j}^{-}))$. We note that the range of $\boldsymbol {\bar {T}}_{p}$ is compact. Hence, from Theorem 6 (3) and Lemma 2, the composite correspondence $P(\boldsymbol {\bar {T}}_{p}(\cdot );F):[0,1]^{J}\rightarrow 2^{[0,1]^{J}}$ is upper semi continuous. Thus, using Theorem 6 (4) we see that correspondence$P(\boldsymbol {\bar {T} }_{p}(\cdot );F)$ hasa fixed point. □

We need a lemma to prove Theorem 5, which is a direct consequence of the duality condition of (16).

Lemma 3

Let$\boldsymbol {x}=(\boldsymbol {x}^{+} ,\boldsymbol {x}^{-})\in \mathcal {P}(F)$andleta be the dual variable associated with the time constraint of linear programP(T_p (x⁺, x⁻); F). Then,

$$\begin{array}{@{}rcl@{}} x_{j}^{+}<1,x_{j}^{-}= 0\quad\Rightarrow\quad u_{j}^{-}(x_{j}^{+},0)\leq a \end{array} $$

(35)

$$\begin{array}{@{}rcl@{}} x_{j}^{+}+x_{j}^{-}<1,0<x_{j}^{-}\quad\Rightarrow\quad u_{j}^{-}(x_{j}^{+},x_{j}^{-})=a \end{array} $$

(36)

$$\begin{array}{@{}rcl@{}} x_{j}^{+}+x_{j}^{-}= 1,0<x_{j}^{-}\quad\Rightarrow\quad u_{j}^{-}(x_{j}^{+},x_{j}^{-})\geq a \end{array} $$

(37)

$$\begin{array}{@{}rcl@{}} 0<x_{j}^{+}\quad\Rightarrow\quad u_{j}^{+}(x_{j}^{+},x_{j}^{-})\geq a \end{array} $$

(38)

Proof of Theorem 5

• (Part 1) Let $\boldsymbol {\bar {x}}$ and $(\boldsymbol {\bar {x}}^{+},\boldsymbol {\bar {x}}^{-})$ be the equilibria of $\mathcal {B}$ and $\mathcal {P}(F)$, respectively. We prove that

  $$ (b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{+} +(b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{-}\geq (b_{j}-cT_{j}(\bar{x}_{j}))\bar{x}_{j} $$

  (39)

  for each j. Since the time constraint is not binding,$\boldsymbol {\bar {x}}$ is an equilibrium of $\mathcal {B}$ if and only if (18) holds true for every j. Similarly, the equilibrium $(\boldsymbol {\bar {x} }^{+},\boldsymbol {\bar {x}}^{-})$ satisfies the conditions in Lemma 3 with a = 0. Based on $\boldsymbol {\bar {x}}^{+}$, we consider twocases of attraction j: (a) $\bar {x}_{j}^{+}= 0$; (b) $\bar {x}_{j}^{+}>0$. Suppose that (a) is the case. Consider the following subcases: (a-1)$\bar {x}_{j}^{-}= 0$; (a-2)$\bar {x}_{j}^{-}\in (0,1)$; (a-3)$\bar {x}_{j}^{-}= 1$. In subcase(a-1) $[\bar {x}_{j}^{+}= 0,\bar {x}_{j}^{-}= 0]$, (13), (35) and the continuity of the congestion function simply

  $$0\geq b_{j}-cT_{j}^{-}(0,0)=b_{j}-cT_{j}(0). $$

  Thus, if x_j > 0, then b_j − cT_j (x_j) <  0 sinceT_j is increasing. Hence, we see from (18) that $\bar {x}_{j}= 0$. We now see that both sides of (39) are zero and that (39) holds. Consider subcase (a-2) $[\bar {x}_{j}^{+}= 0,\bar {x}_{j}^{-}\in (0,1)]$. Equations (36) and (13) imply that

  $$0=b_{j}-cT_{j}^{-}(0,\bar{x}_{j}^{-})=b_{j}-cT_{j}(\bar{x}_{j}^{-}). $$

  Hence, from (18) and the monotonicity ofT_jwe see that$\bar {x}_{j}=\bar {x}_{j}^{-}$. (To see why, assume that $\bar {x}_{j} >\bar {x}_{j}^{-}$ holds. Then, $b_{j}-cT_{j}(\bar {x}_{j})<0$, which implies $\bar {x}_{j}= 0$ from (18). This contradicts $\bar {x}_{j}>\bar {x}_{j}^{-}$. A similar contradiction is obtained if we assume that $\bar {x}_{j}<\bar {x}_{j}^{-}$.) Thus, (39) holds.Consider subcase (a-3) $[\bar {x}_{j}^{+}= 0,\bar {x}_{j}^{-}= 1]$. In a similar manner, one has from (37) and (13) that

  $$0\leq b_{j}-cT_{j}^{-}(0,1)=b_{j}-cT_{j}(1). $$

  Thus, using (18) again, we have $\bar {x}_{j}= 1$. Hence (39) holds. Next suppose that (b) is the case. We examine the following subcases inturn: (b-1) $\bar {x}_{j}^{+}= 1$; (b-2) $\bar {x}_{j}^{+}<1,\bar {x}_{j}^{+}+\bar {x}_{j}^{-}= 1$; (b-3)$\bar {x}_{j}^{+}<1$ and $\bar {x}_{j}^{-}= 0$; and (b-4) $\bar {x}_{j}^{+}+\bar {x}_{j}^{-}<1$ and $\bar {x}_{j}^{-}>0$. For subcase(b-1) $[\bar {x}_{j}^{+}= 1]$, it follows from (38) and (13) that

  $$0\leq b_{j}-cT_{j}^{+}(1,0)=b_{j}-cT_{j}(1). $$

  Thus from (18) we see that $\bar {x}_{j}= 1$ (otherwise it contradicts the monotonicity of T_j).Hence, (39) holds with equality. For subcase (b-2) $[\bar {x}_{j}^{+}\in (0,1),\bar {x}_{j}^{+}+\bar {x}_{j}^{-}= 1]$,

  $$\begin{array}{@{}rcl@{}} T_{j}(1) & =&\bar{x}_{j}^{+}T_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}^{-} )+\bar{x}_{j}^{-}T_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-})\\ & <&\bar{x}_{j}^{+}T_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-})+\bar{x}_{j} ^{-}T_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-})\\ & =&T_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}) \end{array} $$

  where the first equality follows from (13) and the inequality follows from $x_{j}^{+}>0$ and (14). Thus, from (37)

  $$0\leq b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-})<b_{j}-cT_{j}(1). $$

  Using the same argumentas above, $\bar {x}_{j}= 1$ and(39) follows. Consider subcase (b-3)$[\bar {x}_{j}^{+}\in (0,1)$ and $\bar {x}_{j}^{-}= 0]$. Suppose that $b_{j}-cT_{j}^{+}(\bar {x}_{j}^{+},0)<0$ holds. Then, (38) implies $\bar {x}_{j}^{+}= 0$. Contradiction, i.e., $b_{j}-cT_{j}^{+}(\bar {x}_{j}^{+},0)<0$ does not arise. Next suppose that $b_{j}-cT_{j}^{+}(\bar {x}_{j}^{+},0)= 0$ holds. Then, $b_{j}-cT_{j}(\bar {x}_{j}^{+})= 0$ follows, which together with (18) implies that $\bar {x}_{j}\in (0,1)$ and $\bar {x}_{j}=\bar {x}_{j}^{+}$. This implies (39). Next suppose that $b_{j}-cT_{j}^{+}(\bar {x}_{j}^{+},0)>0$ holds. Then, one has $b_{j}-cT_{j}(\bar {x}_{j}^{+})>0$, and thus from (18), $\bar {x}_{j}>\bar {x}_{j}^{+}$ follows (otherwise, $\bar {x}_{j}^{+}\geq \bar {x}_{j}$, which implies $b_{j}-cT_{j}(\bar {x}_{j})>0$, thus $\bar {x}_{j}= 1$. This contradicts $\bar {x}_{j}^{+}<1$). Since $\bar {x}_{j}^{+}\in (0,1)$ and $\bar {x}_{j}-\bar {x}_{j}^{+}>0$, it follows from (15) and (35) that

  $$ b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+})<b_{j}-cT_{j} ^{-}(\bar{x}_{j}^{+},0)\leq0. $$

  (40)

  Thus, using $\bar {x}_{j}^{-}= 0$, (15) and (40), we see that

  $$\begin{array}{@{}rcl@{}} && (b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{+} +(b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{-}\\ & =&(b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j} ^{+}\\ & \geq&(b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+}))\bar {x}_{j}^{+}\\ & >&(b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+}))\bar{x}_{j}^{+}+(b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+} ))(\bar{x}_{j}-\bar{x}_{j}^{+})\\ & =&(b_{j}-cT_{j}(\bar{x}_{j}))\bar{x}_{j} \end{array} $$

  (41)

  Hence, (39) holds true with strict inequality. Consider subcase (b-4) $[0<\bar {x}_{j}^{+},0<\bar {x}_{j}^{-},\bar {x}_{j}^{+}+\bar {x}_{j}^{-}<1]$. From (36) and (14), one seesthat

  $$0=b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-})<b_{j}-cT_{j}^{+}(\bar {x}_{j}^{+},\bar{x}_{j}^{-}). $$

  Thus, $b_{j}-c(\bar {x}_{j}^{+}+\bar {x}_{j}^{-})T_{j}(\bar {x}_{j}^{+}+\bar {x}_{j}^{-})>0$. Hence, from (18),$\bar {x}_{j}>\bar {x}_{j}^{+}+\bar {x}_{j}^{-}$ (otherwise, $b_{j}-c\bar {x}_{j}T_{j}(\bar {x}_{j})>0$ holds true, so that $\bar {x}_{j}= 1$, which contradicts $\bar {x} _{j}\leq \bar {x}_{j}^{+}+\bar {x}_{j}^{-}<1$). We note from (15) and (36) that $\bar {x}_{j}-\bar {x}_{j}^{+}>\bar {x}_{j}^{-}$ and $(b_{j}-cT_{j}^{-}(\bar {x}_{j}^{+},\bar {x}_{j}-\bar {x}_{j}^{+}))<b_{j} -cT_{j}^{-}(\bar {x}_{j}^{+},\bar {x}_{j}^{-})= 0$. Thus,

  $$\begin{array}{@{}rcl@{}} & &(b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{+} +(b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}^{-}))\bar{x}_{j}^{-}\\ & >&(b_{j}-cT_{j}^{+}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+}))\bar{x}_{j}^{+}+(b_{j}-cT_{j}^{-}(\bar{x}_{j}^{+},\bar{x}_{j}-\bar{x}_{j}^{+} ))(\bar{x}_{j}-\bar{x}_{j}^{+})\\ & =&(b_{j}-cT_{j}(\bar{x}_{j}))\bar{x}_{j}. \end{array} $$

  This implies that (39) is satisfied with strict inequality.

• (Part 2) If $\bar {x}_{j}^{+}>0$ and $0<x_{j}^{+}+\bar {x}_{j}^{-}<1$, then station j satisfies either subcase (b-3) or (b-4) in Part 1. Hence,(39) holds true with strict inequality for station j. Other stations in $\mathcal {P}(F)$ are at least as good as those in $\mathcal {B}$, see Part 1. Thus, the statement follows.

• (Part 3) We note that $(\boldsymbol {0},\boldsymbol {\bar {x}})\in \mathcal {P} (0)$. Since we assumed that cT_j (0) < b_j for all j and there is no time constraint, we have $\boldsymbol {\bar {x}}>\boldsymbol {0}$. Note that we assumed that $\bar {x}_{j}<1$ for all j. For a sufficiently small F > 0, $(\boldsymbol {0},\boldsymbol {\bar {x}})\in \mathcal {P}(0)$ and $(\boldsymbol {\bar {x}}^{+},\boldsymbol {\bar {x}}^{-})\in \mathcal {P}(F)$ are close, so that $0<\bar {x}_{j}^{-}$ and $\bar {x}_{j}^{+}+\bar {x}_{j}^{-}<1$ for all j. Let $\hat {\jmath }$ be the station that maximizes the attractiveness $u_{j}^{+}(0,\bar {x}_{j})$. Then, the pare use dat $\hat {\jmath }$ so that $0<\bar {x}_{\hat {\jmath }}^{+}$. Thus, station $\hat {\jmath }$ satisfiessubcase (b-4) in Part 1. The rest of the argument is same as Part 2.

□

Remark 1 (Numerical Evaluation of $\mathcal {P}(F)$)

Let A, β and γ be as defined in (32) to (34). Then, linear program P(t⁺, t⁻;F) is written as

$$\max_{\boldsymbol{x}}\boldsymbol{\gamma}\cdot\boldsymbol{x}\text{\quad s.t.\quad}A\boldsymbol{x}\leq\boldsymbol{\beta}\text{ and }\boldsymbol{x} \geq\boldsymbol{0} $$

where $\boldsymbol {x}=\binom {\boldsymbol {x}^{+}}{\boldsymbol {x}^{-}} \in {\mathbb {R}}^{2J}$. Thus, from the complementary slackness theorem, x and $\boldsymbol {y}\in {\mathbb {R}}^{2+J}$ are the optimal solutions of (16) and its dual program, respectively, if and only if they satisfy the complementarity condition and the feasibility condition (see, e.g., Chvatal 1983), i.e.,

$$\begin{array}{@{}rcl@{}} & \left\{ \begin{array} [c]{c} \left( \boldsymbol{\beta}-A\boldsymbol{x}\right)^{T}\boldsymbol{y}= 0,\\ \left( \boldsymbol{\gamma}-\boldsymbol{y}A\right) \boldsymbol{x}= 0, \end{array} \right. \end{array} $$

(42)

$$\begin{array}{@{}rcl@{}} & \left\{ \begin{array} [c]{c} A\boldsymbol{x}\leq\boldsymbol{\beta},\quad\\ \boldsymbol{y}A\geq\boldsymbol{\gamma},\quad \end{array} \right. \end{array} $$

(43)

$$\begin{array}{@{}rcl@{}} & \left. \boldsymbol{x}\geq\boldsymbol{0},\boldsymbol{y}\geq\boldsymbol{0} .\right. \end{array} $$

(44)

Hence, the equilibria $\mathcal {P}(F)$ can be evaluated by solving (42) to (44) with (t⁺, t⁻) = T_p (x⁺, x⁻) with respect to x, y, and (t⁺, t⁻). For numerical computation, we examine all possible cases of which constraints are binding and non-binding in (43). To be more specific, we follow the following procedure. Denote the set of inequalities (43) by T.

• 1. Generate the power set $\mathcal {T}$ of the set T of the inequalities. For each $S\in \mathcal {T}$, repeat the following three steps.

• 2. Replace each inequality in S by an equality. Denote the resulting equality constraints by S_E. If the dual variable y_k is associated with an inequality constraint in T − S, then set y_k = 0. Similarly, if the primal variable $x_{j}^{+}$ ($x_{j}^{-}$) is associated with an inequality constraint in T − S, then set $x_{j}^{+}= 0$ ($x_{j}^{-}= 0$).

• 3. Solve the system of equations consisting of S_E and (t⁺, t⁻) = T_p (x⁺, x⁻) with respect to (t⁺, t⁻), x and y (excluding those $x_{j}^{\pm }$ and y_k set to zero in Step 2 where $\boldsymbol {x}=\binom {\boldsymbol {x}^{+} }{\boldsymbol {x}^{-}}$.

• 4. Check if the solution satisfies the inequality constraints T − S and the nonnegativity condition. If it does, then the solution is an equilibrium.

The numerical experiments in Section 4 are implemented using Mathematica 10. The system of equations in Step 3 is solved by NSolve of Mathematica. When the number of stations is large, this procedure suffers a combinatorial explosion, so that a more sophisticated method would be necessary.