Equilibrium analysis of observable express service with customer choice

Abstract

We study a stylized queueing model motivated by paid express lanes on highways. There are two parallel, observable first-come, first-served queues with finitely many servers: one queue has a faster service rate, but charges a fee to join, and the other is free but slow. Upon arrival, customers see the state of each queue and choose between them by comparing the respective disutility of time spent waiting, subject to random shocks. This framework encompasses both the multinomial logit and exponomial customer choice models. Using a fluid limit approximation, we give a detailed characterization of the equilibrium in this system. We show that social welfare is optimized when the express queue is exactly at (but not over) full capacity; however, in some cases, revenue is maximized by artificially creating congestion in the free queue. The latter behavior is caused by changes in the price elasticity of demand as the service capacity of the free queue fills up.

Proofs

Below, we give the complete proofs of all results that were stated in the main text.

1.1 Proof of Theorem 1

Again, we assume $Q_i\left( 0\right) = Q^n_i\left( 0\right) = q_i\left( 0\right) = 0$. We follow [29] in using the following result from [24]:

Lemma 1

A standard Poisson process $\varPi$ and a standard Brownian motion B can be constructed on the same probability space in such a way that the limit

$$\begin{aligned} Z = \sup _{t\ge 0} \frac{\left| \varPi \left( t\right) - t - B\left( t\right) \right| }{\log \left( \max \left\{ 2,t\right\} \right) } \end{aligned}$$

is a.s. finite and has finite mean and finite moment-generating function in the neighborhood of zero.

Lemma 1 allows us to rewrite $Q^n_i$ in terms of two standard Brownian motions $B^\mathrm{arr}_i,B^\mathrm{dep}_i$ via

$$\begin{aligned}&\frac{1}{n}\varPi ^\mathrm{arr}_i\left( n\int ^t_0 \lambda p_i(Q^n_e(s),Q^n_r(s))\hbox {d}s\right) \nonumber \\&\quad = \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\nonumber \\&\qquad +\,\frac{1}{n}B^\mathrm{arr}_i\left( n \int ^t_0 \lambda p_i(Q^n_e(s),Q^n_r(s)) \hbox {d}s\right) + O\left( \frac{\log n}{n}\right) \nonumber \\&\quad = \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) , Q^n_r\right) \hbox {d}s\nonumber \\&\qquad +\, \frac{1}{\sqrt{n}}B^\mathrm{arr}_i\left( \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) + O\left( \frac{\log n}{n}\right) \end{aligned}$$

(44)

and

$$\begin{aligned}&\frac{1}{n}\varPi ^\mathrm{dep}_i\left( n\int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) \nonumber \\&\quad = \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\nonumber \\&\qquad +\, \frac{1}{n}B^\mathrm{dep}_i\left( n\int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) + O\left( \frac{\log n}{n}\right) \nonumber \\&\quad = \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\nonumber \\&\qquad +\, \frac{1}{\sqrt{n}}B^\mathrm{dep}_i\left( \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) + O\left( \frac{\log n}{n}\right) . \end{aligned}$$

(45)

Now, we calculate the difference between the scaled length process for queue $i \in \left\{ e,r\right\}$ and its fluid limit, given by

$$\begin{aligned} Q^n_i\left( t\right) - q_i\left( t\right)= & {} \frac{1}{n}\varPi ^\mathrm{arr}_i\left( n\int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) - \int ^t_0 \lambda p_i\left( q_e\left( s\right) ,q_r\left( s\right) \right) \hbox {d}s\\&-\,\frac{1}{n}\varPi ^\mathrm{dep}_i\left( n \int ^t_0 \mu _i\min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) + \int ^t_0 \mu _i\min \left\{ q_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s, \end{aligned}$$

whence

$$\begin{aligned} \left| Q^n_i\left( t\right) - q_i\left( t\right) \right|\le & {} \left| \frac{1}{n}\varPi ^\mathrm{arr}_i\left( n\int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) - \int ^t_0 \lambda p_i\left( q_e\left( s\right) ,q_r\left( s\right) \right) \hbox {d}s\right| \nonumber \\&+\, \left| \frac{1}{n}\varPi ^\mathrm{dep}_i\left( n \int ^t_0 \mu _i\min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) - \int ^t_0 \mu _i\min \left\{ q_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right| .\nonumber \\ \end{aligned}$$

(46)

Substituting (44) into the first term of (46), we obtain

$$\begin{aligned}&\left| \frac{1}{n}\varPi ^\mathrm{arr}_i\left( n\int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) - \int ^t_0 \lambda p_i\left( q_e\left( s\right) ,q_r\left( s\right) \right) \hbox {d}s\right| \\&\quad \le \left| \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s - \int ^t_0 \lambda p_i\left( q_e\left( s\right) ,q_r\left( s\right) \right) \hbox {d}s\right| \\&\qquad + \left| \frac{1}{\sqrt{n}}B^\mathrm{arr}_i\left( \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) \right| + O\left( \frac{\log n}{n}\right) , \end{aligned}$$

with the Brownian term satisfying

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{t' \le t}\left| \frac{1}{\sqrt{n}}B^\mathrm{arr}_i\left( \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s\right) \right|\le & {} \lim _{n\rightarrow \infty }\left| \frac{1}{\sqrt{n}}B^\mathrm{arr}_i\left( \lambda t\right) \right| \\= & {} 0. \end{aligned}$$

Substituting (45) into the second term of (46) yields

$$\begin{aligned}&\left| \frac{1}{n}\varPi ^\mathrm{dep}_i\left( n \int ^t_0 \mu _i\min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) - \int ^t_0 \mu _i\min \left\{ q_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right| \\&\quad \le \left| \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s - \int ^t_0 \mu _i\min \left\{ q_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right| \\&\qquad + \left| \frac{1}{\sqrt{n}}B^\mathrm{dep}_i\left( \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) \right| + O\left( \frac{\log n}{n}\right) , \end{aligned}$$

with the Brownian term satisfying

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{t' \le t} \left| \frac{1}{\sqrt{n}}B^\mathrm{dep}_i\left( \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right) \right|\le & {} \lim _{n\rightarrow \infty }\left| \frac{1}{\sqrt{n}}B^\mathrm{dep}_i\left( \mu _i{\bar{q}}_i t\right) \right| \\= & {} 0. \end{aligned}$$

Thus, (46) has become

$$\begin{aligned}&\left| Q^n_i\left( t\right) - q_i\left( t\right) \right| \le \left| \int ^t_0 \lambda p_i\left( Q^n_e\left( s\right) ,Q^n_r\left( s\right) \right) \hbox {d}s - \int ^t_0 \lambda p_i\left( q_e\left( s\right) ,q_r\left( s\right) \right) \hbox {d}s\right| \\&+ \left| \int ^t_0 \mu _i \min \left\{ Q^n_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s - \int ^t_0 \mu _i\min \left\{ q_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right| + o\left( n\right) . \end{aligned}$$

The choice probability $p_i$ and the departure function both have uniformly bounded derivatives by assumption P1, so there exist constants C and $\varepsilon$ such that, for large enough n, we have

$$\begin{aligned} \left| Q^n_i\left( t\right) - q_i\left( t\right) \right| \le C \int ^t_0 \sup _{0\le s' \le s} \left| Q^n_i\left( s'\right) - q_i\left( s'\right) \right| \hbox {d}s + \varepsilon . \end{aligned}$$

Applying Gronwall’s lemma [18], we obtain

$$\begin{aligned} \sup _{0\le s \le t} \left| Q^n_i\left( s\right) - q_i\left( s\right) \right| \le \varepsilon \cdot \hbox {e}^{Ct}. \end{aligned}$$

We can then take $\varepsilon \rightarrow 0$ to obtain the desired result.

The solution of (6) and (7) is non-negative because, for $i\in \left\{ r,e\right\}$, $q_i=0$ implies $q'_i>0$ (the choice probabilities are always strictly positive). Uniqueness of the solution is shown by letting $q^{(1)}_i,q^{(2)}_i$ be two sets of solutions with the same initial values, and writing

$$\begin{aligned}&\left| q^{(1)}_i\left( t\right) - q^{(2)}_i\left( t\right) \right| \le \left| \int ^t_0 \lambda p_i\left( q^{(1)}_e\left( s\right) ,q^{(1)}_r\left( s\right) \right) \hbox {d}s - \int ^t_0 \lambda p_i\left( q^{(2)}_e\left( s\right) ,q^{(2)}_r\left( s\right) \right) \hbox {d}s\right| \\&+ \,\left| \int ^t_0 \mu \min \left\{ q^{(1)}_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s - \int ^t_0 \mu \min \left\{ q^{(2)}_i\left( s\right) ,{\bar{q}}_i\right\} \hbox {d}s\right| . \end{aligned}$$

Then, we again invoke the properties of $p_i$ and Gronwall’s lemma, as above, to conclude that $q^{(1)}_i=q^{(2)}_i$.

1.2 Proof of Theorem 2

Existence of the equilibrium We first show the existence of the equilibrium using Brouwer’s fixed point theorem, which states that, if f is a continuous function mapping a compact convex set to itself, there exists a point $x_0$ satisfying $f\left( x_0\right) = x_0$.

We rewrite the equilibrium conditions (8) and (9) as

$$\begin{aligned} \lambda p_e - \mu _e \min \left\{ q_e,{\bar{q}}_e\right\} + q_e= & {} q_e, \end{aligned}$$

(47)

$$\begin{aligned} \lambda p_r - \mu _r \min \left\{ q_r,{\bar{q}}_r\right\} + q_r= & {} q_r. \end{aligned}$$

(48)

We can then express the system (47) and (48) as $f\left( q\right) = q$, where $q = \left( q_e,q_r\right)$. Because we have assumed continuity of $p_e,p_r$ (assumption P1), it straightforwardly follows that f is continuous.

To show that $f = \left( f_e,f_r\right)$ maps a compact convex set to itself, let us consider the first component $f_e$ and suppose that $q_e < {\bar{q}}_e$. In this case, we have the bound $f_e\left( q_e\right) \le \lambda + {\bar{q}}_e$.

When $q_e \ge {\bar{q}}_e$, we have $f_e\left( q_e\right) = \lambda p_e\left( u\left( \frac{q_e}{\mu _e{\bar{q}}_e}\right) ,u_r,{\bar{u}}\right) - \mu _e {\bar{q}}_e + q_e$. Note that, if ${\bar{q}}_e \ge \frac{\lambda }{\mu _e}$, then $f_e\left( q_e\right) < q_e$ and the codomain of $f_e$ is automatically contained in the domain.

If ${\bar{q}}_e < \frac{\lambda }{\mu _e}$, let ${\tilde{q}}_e$ be a value satisfying

$$\begin{aligned} \lambda p_e\left( u\left( \frac{{\tilde{q}}_e}{\mu _e{\bar{q}}_e}\right) ,u\left( \infty \right) ,{\bar{u}}\right) = \mu _e {\bar{q}}_e. \end{aligned}$$

Then, for $q_e \ge {\tilde{q}}_e$, we have

$$\begin{aligned} \lambda p_e\left( u\left( \frac{q_e}{\mu _e{\bar{q}}_e}\right) ,u_r,{\bar{u}}\right)\le & {} \lambda p_e\left( u\left( \frac{{\tilde{q}}_e}{\mu _e{\bar{q}}_e}\right) ,u_r,{\bar{u}}\right) \\\le & {} \lambda p_e\left( u\left( \frac{{\tilde{q}}_e}{\mu _e{\bar{q}}_e}\right) ,u\left( \infty \right) ,{\bar{u}}\right) \\= & {} \mu _e{\bar{q}}_e, \end{aligned}$$

implying $f\left( q_e\right) \le q_e$. Finally, for ${\bar{q}}_e< q_e < {\tilde{q}}_e$, we have

$$\begin{aligned} \lambda p_e\left( u\left( \frac{q_e}{\mu _e{\bar{q}}_e}\right) ,u_r,{\bar{u}}\right) -\mu _e{\bar{q}}_e +q_e \le \lambda p_e\left( u\left( \frac{1}{\mu _e}\right) ,u_r\left( \infty \right) ,{\bar{u}}\right) -\mu _e{\bar{q}}_e +{\tilde{q}}_e. \end{aligned}$$

(49)

Denote by ${\hat{q}}_e$ the right-hand side of (49). Then, for any $0 \le q_e \le \max \left\{ {\bar{q}}_e,{\tilde{q}}_e,{\hat{q}}_e,\frac{\lambda }{\mu _e}\right\}$, we have $f\left( q_e\right)$ in the same interval, regardless of $q_e$. Thus, the conditions for Brouwer’s fixed point theorem hold and the equilibrium exists.

Uniqueness of the equilibrium Let $\lambda ,\mu _e,\mu _r$ and the disutility function u be given. Suppose that there are two non-identical equilibrium solutions $\left( q^{(1)}_e,q^{(1)}_r\right)$ and $\left( q^{(2)}_e,q^{(2)}_r\right)$. Let us focus on the case where $q^{(1)}_e < q^{(2)}_e$ (the other case where we start with $q^{(1)}_r < q^{(2)}_r$ is handled symmetrically).

We first show that, if $q^{(1)}_e < q^{(2)}_e$, then $q^{(1)}_r < q^{(2)}_r$ as well. To see this, let us assume the contrary, i.e., that $q^{(1)}_r \ge q^{(2)}_r$. We derive

$$\begin{aligned} 0= & {} \lambda p_e\left( u_e\left( q^{(1)}_e\right) +c,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) - \mu _e\min \left\{ q^{(1)}_e,{\bar{q}}_e\right\} \nonumber \\\ge & {} \lambda p_e\left( u_e\left( q^{(2)}_e\right) +c,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) - \mu _e\min \left\{ q^{(1)}_e,{\bar{q}}_e\right\} \end{aligned}$$

(50)

$$\begin{aligned}\ge & {} \lambda p_e\left( u_e\left( q^{(2)}_e\right) +c,u_r\left( q^{(2)}_r\right) ,{\bar{u}}\right) - \mu _e\min \left\{ q^{(1)}_e,{\bar{q}}_e\right\} \end{aligned}$$

(51)

$$\begin{aligned}\ge & {} \lambda p_e\left( u_e\left( q^{(2)}_e\right) +c,u_r\left( q^{(2)}_r\right) ,{\bar{u}}\right) - \mu _e\min \left\{ q^{(2)}_e,{\bar{q}}_e\right\} \nonumber \\= & {} 0, \end{aligned}$$

(52)

where (50) is obtained from $q^{(1)}_e < q^{(2)}_e$ and the fact that $u' > 0$ (assumption U3) while $p_e$ is decreasing in $u_e$; equation (51) follows from the assumption that $q^{(1)}_r \ge q^{(2)}_r$ and the fact that $u' > 0$ while $p_e$ is increasing in $u_r$; and (52) follows from $q^{(1)}_e < q^{(2)}_e$. However, since the first and last line both equal zero due to the equilibrium conditions, (50)–(52) must all hold with strict equality. Consequently, (51) and (52) imply that

$$\begin{aligned} \min \left\{ q^{(1)}_e,{\bar{q}}_e\right\} = \min \left\{ q^{(2)}_e,{\bar{q}}_e\right\} , \end{aligned}$$

(53)

whence we conclude ${\bar{q}}_e \le q^{(1)}_e < q^{(2)}_e$. From that, however, (50) yields

$$\begin{aligned} p_e\left( u\left( \frac{q^{(1)}_e}{\mu _e{\bar{q}}_e}\right) +c,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) = p_e\left( u\left( \frac{q^{(2)}_e}{\mu _e{\bar{q}}_e}\right) +c,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) , \end{aligned}$$

and this is impossible since $u' > 0$ with strict inequality. Therefore, $q^{(1)}_e < q^{(2)}_e$ implies $q^{(1)}_r < q^{(2)}_r$.

Next, we claim that $q^{(2)}_r > {\bar{q}}_r$. To see this, let us assume the opposite, i.e., that $q^{(2)}_r \le {\bar{q}}_r$, whence $u_r\left( q^{(1)}_r\right) = u_r\left( q^{(2)}_r\right) = u\left( \frac{1}{\mu _r}\right)$. We then have

$$\begin{aligned} \mu _e \min \left\{ q^{(2)}_e,{\bar{q}}_e\right\}= & {} \lambda p_e\left( u_e\left( q^{(2)}_e\right) +c,u\left( \frac{1}{\mu _r}\right) ,{\bar{u}}\right) \nonumber \\\le & {} \lambda p_e\left( u_e\left( q^{(1)}_e\right) +c,u\left( \frac{1}{\mu _r}\right) ,{\bar{u}}\right) \nonumber \\= & {} \mu _e \min \left\{ q^{(1)}_e,{\bar{q}}_e\right\} , \end{aligned}$$

(54)

and $q^{(1)}_e < q^{(2)}_e$ implies that (54) holds with strict equality. This again implies (53) and the same reasoning as before can be repeated to obtain a contradiction. Therefore, $q^{(2)}_r > {\bar{q}}_r$. A symmetric argument can be used to show $q^{(2)}_e > {\bar{q}}_e$.

Combining the previous facts, (10) yields

$$\begin{aligned} p_o\left( u_e\left( q^{(1)}_e\right) ,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) = p_o\left( u\left( \frac{q^{(2)}_e}{\mu _e{\bar{q}}_e}\right) ,u\left( \frac{q^{(2)}_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) . \end{aligned}$$

However, from $q^{(1)}_e < q^{(2)}_e$ and $q^{(1)}_r < q^{(2)}_r$ we obtain

$$\begin{aligned} p_o\left( u_e\left( q^{(1)}_e\right) ,u_r\left( q^{(1)}_r\right) ,{\bar{u}}\right) < p_o\left( u\left( \frac{q^{(2)}_e}{\mu _e{\bar{q}}_e}\right) ,u\left( \frac{q^{(2)}_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) , \end{aligned}$$

regardless of whether $q^{(1)}_e$ and $q^{(1)}_r$ are under or over capacity, because $p_o$ satisfies assumption P2. We conclude that it is impossible to have $q^{(1)}_e < q^{(2)}_e$ and still satisfy the equilibrium conditions for both solutions.

1.3 Proof of Theorem 3

We examine each of regimes R1–R4 separately. In each regime, we write (6) and (7) as

$$\begin{aligned} \left( q'_e,q'_r\right) = \left( f_e\left( q_e,q_r\right) ,f_r\left( q_e,q_r\right) \right) , \end{aligned}$$

obtain all of the first-order partial derivatives $\frac{\partial f_i}{\partial q_i}$ for $i \in \left\{ e,r\right\}$, put them into matrix form (the Jacobian) and evaluate this matrix at the equilibrium $\left( q^*_e,q^*_r\right)$, which we know exists and is unique from the preceding. The equilibrium is locally stable if both eigenvalues of the Jacobian are negative [33].

Regime R1 We have $q^*_e \ge {\bar{q}}_e$, $q^*_r \ge {\bar{q}}_r$ and the Jacobian is given by

$$\begin{aligned} J^{R1} = \lambda \left[ \begin{array}{c c} \frac{\partial p_e}{\partial u_e}\frac{\partial u_e}{\partial q_e} &{}\quad \frac{\partial p_e}{\partial u_r}\frac{\partial u_r}{\partial q_r}\\ \frac{\partial p_r}{\partial u_e}\frac{\partial u_e}{\partial q_e} &{}\quad \frac{\partial p_r}{\partial u_r}\frac{\partial u_r}{\partial q_r} \end{array} \right] . \end{aligned}$$

Letting $e_1,e_2$ be the eigenvalues, the characteristic equation is given by

$$\begin{aligned} \left( \lambda \frac{\partial p_e}{\partial u_e}\frac{\partial u_e}{\partial q_e} -e_1\right) \cdot \left( \lambda \frac{\partial p_r}{\partial u_r}\frac{\partial u_r}{\partial q_r}-e_2\right) - \lambda ^2 \frac{\partial p_e}{\partial u_r}\frac{\partial u_r}{\partial q_r} \cdot \frac{\partial p_r}{\partial u_e}\frac{\partial u_e}{\partial q_e} = 0, \end{aligned}$$

which can be rewritten as

$$\begin{aligned} \lambda ^2 \frac{\partial u_e}{\partial q_e}\frac{\partial u_r}{\partial q_r}\left( \frac{\partial p_e}{\partial u_e}\frac{\partial p_r}{\partial u_r} - \frac{\partial p_e}{\partial u_r}\frac{\partial p_r}{\partial u_e}\right) +e_1e_2= \lambda e_1 \frac{\partial p_r}{\partial u_r}\frac{\partial u_r}{\partial q_r} + \lambda e_2 \frac{\partial p_e}{\partial u_e}\frac{\partial u_e}{\partial q_e}. \end{aligned}$$

(55)

We argue that

$$\begin{aligned} \det \left( J^{R1}\right) = \lambda ^2 \frac{\partial u_e}{\partial q_e}\frac{\partial u_r}{\partial q_r}\left( \frac{\partial p_e}{\partial u_e}\frac{\partial p_r}{\partial u_r} - \frac{\partial p_e}{\partial u_r}\frac{\partial p_r}{\partial u_e}\right) \end{aligned}$$

is positive, which would imply that the product $e_1 e_2$ in (55) is also positive. We first observe that the product $\frac{\partial u_e}{\partial q_e}\frac{\partial u_r}{\partial q_r}$ is positive since, for example,

$$\begin{aligned} \frac{\partial u_e}{\partial q_e} = u'\left( \frac{q^*_e}{\mu _e{\bar{q}}_e}\right) \frac{1}{\mu _e{\bar{q}}_e} > 0 \end{aligned}$$

by assumption U3. The same is true of $\frac{\partial u_r}{\partial q_r}$.

Assumption P3 implies

$$\begin{aligned} \frac{\partial p_e}{\partial u_e} + \frac{\partial p_e}{\partial u_r} + \frac{\partial p_e}{\partial {\bar{u}}} = 0 \end{aligned}$$

since changing all the disutilities by the same amount does not change the probability of any choice. Since $\frac{\partial p_e}{\partial {\bar{u}}} > 0$ by assumption P2, it follows that $\frac{\partial p_e}{\partial u_e} + \frac{\partial p_e}{\partial u_r} < 0$, whence

$$\begin{aligned} \frac{\partial p_e}{\partial u_e} < - \frac{\partial p_e}{\partial u_r} \end{aligned}$$

(56)

and, symmetrically,

$$\begin{aligned} \frac{\partial p_r}{\partial u_r} < - \frac{\partial p_r}{\partial u_e}. \end{aligned}$$

(57)

From this we obtain

$$\begin{aligned} \frac{\partial p_e}{\partial u_e}\frac{\partial p_r}{\partial u_r}> - \frac{\partial p_e}{\partial u_r}\frac{\partial p_r}{\partial u_r} > \frac{\partial p_e}{\partial u_r}\frac{\partial p_r}{\partial u_e}, \end{aligned}$$

where the first inequality is obtained from (56) and the fact that $\frac{\partial p_e}{\partial u_e} < 0$, while the second inequality is obtained from (57) and the fact that $\frac{\partial p_e}{\partial u_r} > 0$. Thus, we conclude that $\det \left( J^{R1}\right) > 0$ and so both $e_1,e_2$ have the same sign.

From the preceding, it follows that the left-hand side of (55) is positive. On the right-hand side of (55), suppose that $e_1,e_2$ are both positive; then we have

$$\begin{aligned} \lambda e_1 \frac{\partial p_r}{\partial u_r}\frac{\partial u_r}{\partial q_r}< 0, \qquad \lambda e_2 \frac{\partial p_e}{\partial u_e}\frac{\partial u_e}{\partial q_e} < 0 \end{aligned}$$

since $\frac{\partial p_r}{\partial u_r},\frac{\partial p_e}{\partial u_e} < 0$ while $\frac{\partial u_r}{\partial q_r},\frac{\partial u_e}{\partial q_e} > 0$. Therefore, both $e_1,e_2$ must be negative, as required.

Regime R2 We have $q^*_e < {\bar{q}}_e$, $q^*_r < {\bar{q}}_r$ and the Jacobian is given by

$$\begin{aligned} J^{R2} = \lambda \left[ \begin{array}{c c} -\mu _e &{}\quad 0\\ 0 &{}\quad -\mu _r \end{array} \right] , \end{aligned}$$

from which the conclusion directly follows.

Regime R3 We have $q^*_e \ge {\bar{q}}_e$, $q^*_r < {\bar{q}}_r$ and the Jacobian is given by

$$\begin{aligned} J^{R3} = \lambda \left[ \begin{array}{c c} \lambda \frac{\partial p_e}{\partial u_e}\frac{\partial u_e}{\partial q_e} &{}\quad 0\\ \lambda \frac{\partial p_r}{\partial u_e}\frac{\partial u_e}{\partial q_e} &{}\quad -\mu _r \end{array} \right] , \end{aligned}$$

which is a lower triangular matrix, meaning that its eigenvalues are on the diagonal. It is easy to see that both are negative.

Regime R4 The proof is very similar to the previous case and is omitted.

1.4 Proof of Theorem 4

The assumptions on v imply that, for any $s_1 < s_2$, we have

$$\begin{aligned} v\left( s_2\right) - v\left( s_1\right) > u\left( s_2\right) - u\left( s_1\right) . \end{aligned}$$

(58)

This fact will be used to show the desired result in each of the relevant regimes.

Regime R2 Since both queues are under capacity, we have $u_e\left( q^u_e\right) = u\left( \frac{1}{\mu _e}\right)$ and $u_r\left( q^u_r\right) = u\left( \frac{1}{\mu _r}\right)$. From (58), we obtain

$$\begin{aligned} v\left( \frac{1}{\mu _e}\right) - u\left( \frac{1}{\mu _e}\right) \le v\left( \frac{1}{\mu _r}\right) - u\left( \frac{1}{\mu _r}\right) < {\bar{v}} - {\bar{u}}. \end{aligned}$$

(59)

We will now show that $q^u_e < q^v_e$ by contradiction. Suppose that $q^u_e \ge q^v_e$. It follows that the express queue continues to be under capacity when we switch to v. We then derive

$$\begin{aligned} \lambda p_e\left( v_e\left( q^v_e\right) +c,v_r\left( q^u_r\right) ,{\bar{v}}\right)= & {} \lambda p_e\left( v_e\left( q^u_e\right) +c,v_r\left( q^u_r\right) ,{\bar{v}}\right) \end{aligned}$$

(60)

$$\begin{aligned}= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{1}{\mu _r}\right) ,{\bar{v}}\right) , \end{aligned}$$

(61)

where (60) and (61) follow because both $q^u_e,q^v_e < {\bar{q}}_e$. Next, we let $\delta = v\left( \frac{1}{\mu _r}\right) - u\left( \frac{1}{\mu _r}\right)$, note that $\delta > 0$, and observe that

$$\begin{aligned} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{1}{\mu _r}\right) ,{\bar{v}}\right)= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c-\delta ,v\left( \frac{1}{\mu _r}\right) -\delta ,{\bar{v}}-\delta \right) \end{aligned}$$

(62)

$$\begin{aligned}= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c-\delta ,u\left( \frac{1}{\mu _r}\right) ,{\bar{v}}-\delta \right) \nonumber \\> & {} \lambda p_e\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{1}{\mu _r}\right) ,{\bar{u}}\right) \end{aligned}$$

(63)

$$\begin{aligned}= & {} \mu _e q^u_e \end{aligned}$$

(64)

$$\begin{aligned}\ge & {} \mu _e q^v_e. \end{aligned}$$

(65)

Above, (62) is due to assumption P3, (63) follows from assumption P2 combined with (59), and (64) follows by (8).

To obtain the desired contradiction, we consider two cases, one where $q^v_r < {\bar{q}}_r$ and one where $q^v_r \ge {\bar{q}}_r$. Suppose that $q^v_r < {\bar{q}}_r$. Then, both queues are under capacity with v as the disutility function, so (8) implies

$$\begin{aligned} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{1}{\mu _r}\right) ,{\bar{v}}\right) = \mu _e q^v_e. \end{aligned}$$

On the other hand, if $q^v_r \ge {\bar{q}}_r$, we have $q^v_r > q^u_r$ and

$$\begin{aligned} \lambda p_e\left( v_e\left( q^v_e\right) +c,v_r\left( q^u_r\right) ,{\bar{v}}\right)< & {} \lambda p_e\left( v_e\left( q^v_e\right) +c,v_r\left( q^v_r\right) ,{\bar{v}}\right) \\= & {} \mu _e q^v_e, \end{aligned}$$

by assumption P2. Either case, when combined with (65), yields $q^v_e < q^v_e$, which is impossible; therefore, we must have $q^u_e < q^v_e$.

Regime R4 In this regime, only the express queue is under capacity. There are two possible permutations

$$\begin{aligned} u\left( \frac{1}{\mu _e}\right)< {\bar{u}} \le u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) , \qquad u\left( \frac{1}{\mu _e}\right)< u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) < {\bar{u}}. \end{aligned}$$

Applying (58) to both of these yields

$$\begin{aligned} v\left( \frac{1}{\mu _e}\right) - u\left( \frac{1}{\mu _e}\right)< & {} {\bar{v}} - {\bar{u}} \le v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) , \end{aligned}$$

(66)

$$\begin{aligned} v\left( \frac{1}{\mu _e}\right) - u\left( \frac{1}{\mu _e}\right)< & {} v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) < {\bar{v}} - {\bar{u}}. \end{aligned}$$

(67)

First, let us suppose that permutation (66) is correct. Again, we proceed by contradiction and assume that $q^u_e \ge q^v_e$. Since the express queue is under capacity with either disutility function, we have

$$\begin{aligned} \lambda p_e\left( v_e\left( q^v_e\right) +c,v_r\left( q^u_r\right) ,{\bar{v}}\right)= & {} \lambda p_e\left( v_e\left( q^u_e\right) +c,v_r\left( q^u_r\right) ,{\bar{v}}\right) \\= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{v}}\right) . \end{aligned}$$

Letting $\delta = {\bar{v}} - {\bar{u}}$, we further derive

$$\begin{aligned} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{v}}\right)= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c-\delta ,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -\delta ,{\bar{v}}-\delta \right) \end{aligned}$$

(68)

$$\begin{aligned}= & {} \lambda p_e\left( v\left( \frac{1}{\mu _e}\right) +c-\delta ,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -\delta ,{\bar{u}}\right) \nonumber \\> & {} \lambda p_e\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) \end{aligned}$$

(69)

$$\begin{aligned}= & {} \mu _e q^u_e\nonumber \\\ge & {} \mu _e q^v_e, \end{aligned}$$

(70)

where, as before, (68) is due to assumption P3, while (69) follows from (66) combined with assumption P2. From (70) and assumption P2, we conclude that $q^u_r > q^v_r$, otherwise there will be no way to satisfy (8).

Now, (10) yields

$$\begin{aligned} 0= & {} \lambda p_o\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) - \lambda + \mu _e q^u_e + \mu _r\min \left\{ q^u_r,{\bar{q}}_r\right\} \nonumber \\\ge & {} \lambda p_o\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) - \lambda + \mu _e q^v_e + \mu _r\min \left\{ q^v_r,{\bar{q}}_r\right\} \nonumber \\= & {} \lambda p_o\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) - \lambda p_o\left( v\left( \frac{1}{\mu _e}\right) +c,v_r\left( q^v_r\right) ,{\bar{v}}\right) ,\qquad \end{aligned}$$

(71)

whence

$$\begin{aligned} p_o\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right) \le p_o\left( v\left( \frac{1}{\mu _e}\right) +c,v_r\left( q^v_r\right) ,{\bar{v}}\right) . \end{aligned}$$

(72)

At the same time, letting $\delta = v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right)$, we obtain

$$\begin{aligned} p_o\left( u\left( \frac{1}{\mu _e}\right) +c,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}\right)= & {} p_o\left( u\left( \frac{1}{\mu _e}\right) +c+\delta ,u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) +\delta ,{\bar{u}}+\delta \right) \nonumber \\= & {} p_o\left( u\left( \frac{1}{\mu _e}\right) +c+\delta ,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{u}}+\delta \right) \nonumber \\> & {} p_o\left( v\left( \frac{1}{\mu _e}\right) +c,v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) ,{\bar{v}}\right) \end{aligned}$$

(73)

$$\begin{aligned}> & {} p_o\left( v\left( \frac{1}{\mu _e}\right) +c,v_r\left( q^v_r\right) ,{\bar{v}}\right) , \end{aligned}$$

(74)

where the first equality is due to assumption P3, while (73) follows from (66) and assumption P2, while (74) follows from assumption P2 and $q^u_r > q^v_r$. Clearly (72) and (74) contradict each other, whence we conclude that $q^u_e < q^v_e$.

Finally, we suppose that permutation (67) is correct. In this case, however, the proof is nearly identical. The only difference is that, in order to obtain (70), we use $\delta = v\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right) -u\left( \frac{q^u_r}{\mu _r{\bar{q}}_r}\right)$, while (73) is obtained by using $\delta = {\bar{v}} - {\bar{u}}$. The same contradiction then follows.