Equilibrium commuting

Abstract

We consider the role of a nonlinear commuting cost function in determination of the equilibrium commuting pattern where all agents are mobile. Previous literature has considered only linear commuting cost, where in equilibrium, all workers are indifferent about their workplace location. We show that this no longer holds for nonlinear commuting cost. The equilibrium commuting pattern is completely determined by the concavity or convexity of commuting cost as a function of distance. We show that a monocentric equilibrium exists when the ratio of the firm agglomeration externality to commuting cost is sufficiently high. We also show that equilibrium residential land rent is decreasing and convex in distance to the business district under concave commuting cost given that the total land used for offices is smaller than that used for housing. Finally, we find empirical evidence of both long and short commutes in equilibrium, implying that the commuting cost function is likely concave.

Appendix

Proof of Proposition 3

We focus on locations in $\mathbb {R} _{+}$. The technique of proof that equilibrium exists is guess and verify. For $b<x\le B$, define

$$\begin{aligned} f(x)=\frac{Ms_\mathrm{f}}{Ns_\mathrm{w}}\left( B-x\right) \end{aligned}$$

(9)

The function f is arbitrary otherwise. Hence, for $0\le y\le b$, $f^{-1}(y)=B-\frac{Ns_\mathrm{w}}{Ms_\mathrm{f}}y$. For $x>B$, define $r(x)=0$.

Integrating (7), we obtain:

$$\begin{aligned} w(y)=-\int T^{\prime }(f^{-1}(y)-y)\hbox {d}y=\frac{b}{B}T\left( f^{-1}(y)-y\right) +C_{a} \end{aligned}$$

(10)

where $C_{a}$ is a constant of integration.

From (6) and (10),

$$\begin{aligned} r(x)= & {} \frac{1}{s_\mathrm{w}}\left[ w(y)-T(x-y)-\overline{z}\right] =\frac{1}{s_\mathrm{w}} \left[ -\frac{B-b}{B}T\left( x-f(x)\right) +C_{a}-\overline{z}\right] \nonumber \\= & {} -\frac{ N}{2B}T\left( x-f(x)\right) +C_{b} \end{aligned}$$

(11)

where $C_{b}=\frac{1}{s_\mathrm{w}}\left( C_{a}-\overline{z}\right)$.

Observe that the residential land rent is decreasing and convex in x.

Since $r(B)=0$ in (11), we get $C_{b}=\frac{N}{2B}T\left( B\right)$. Plugging $C_{b}$ into (11) yields our final expression for rent (12):

$$\begin{aligned} \text {For }b<x\le B,\text { define }r(x)=\frac{N}{2B}\left[ T\left( B\right) -T\left( x-f(x)\right) \right] \end{aligned}$$

(12)

Using the profit function and the fact that the rent for consumers and producers must be equal at b,

$$\begin{aligned} \text {For }0\le & {} y\le b,\text { define} \nonumber \\ r(y)= & {} \frac{1}{s_\mathrm{f}}\left\{ A(y)-A\left( b\right) +\frac{N}{M}\left[ w\left( b\right) -w(y)\right] \right\} +\frac{N}{2B}T\left( B\right) \end{aligned}$$

(13)

If we substitute (10) into (13), we obtain

$$\begin{aligned} r(y)=\frac{1}{s_\mathrm{f}}\left\{ A(y)-A\left( b\right) +\frac{N}{M}\left[ w\left( b\right) -\frac{b}{B}T\left( f^{-1}(y)-y\right) -C_{a}\right] \right\} + \frac{N}{2B}T\left( B\right) \end{aligned}$$

Using $r(y)=r(x)$ evaluated at $y=x=b$, we obtain

$$\begin{aligned} r(y)=\frac{M}{2b}\left[ A(y)-A\left( b\right) \right] -\frac{N}{2B}\left[ T\left( f^{-1}(y)-y\right) -T\left( B\right) \right] \end{aligned}$$

(14)

Then

$$\begin{aligned} R = 2\int _{0}^{b}r(y)\hbox {d}y+2\int _{b}^{B}r(x)\hbox {d}x, \end{aligned}$$

a function of only exogenous parameters. So $\overline{z}$ can be found as only a function of exogenous parameters by plugging R into (5).

$$\begin{aligned} \text {For }0\le y\le b,\text { define }w(y)=\overline{z} +r(f^{-1}(y))s_\mathrm{w}+T(f^{-1}(y)-y) \end{aligned}$$

Thus, w(y) ($0\le y\le b$) is a function of only exogenous variables.

Notice that (4) and (14) imply that profits are constant on $0\le y\le b$, namely

$$\begin{aligned} \varPi (y)=\overline{\varPi }=A\left( b\right) -\frac{N}{M}w\left( b\right) -\frac{ Nb}{MB}T\left( B\right) \end{aligned}$$

Hence,

$$\begin{aligned} \text {For }0\le y\le b,\text { define }w(y)=\frac{M}{N}\left[ A(y)- \overline{\varPi }-r(y)s_\mathrm{f}\right] \end{aligned}$$

To show that this represents an equilibrium, we must verify that $\overline{ \varPi }\ge 0$, $\overline{z}\ge 0$, that no consumer wishes to move to [0, b] , and that no firm wants to move to $(b,\infty )$.

We verify that what we have constructed is an equilibrium. In the case of $T^{\prime \prime }<0$, the total land rent is

$$\begin{aligned} R=\int _{0}^{b}\frac{M}{b}\left[ A(y)-A\left( b\right) \right] \hbox {d}y+\frac{N}{ B-b}\int _{b}^{B}\left[ T\left( B\right) -T\left( x-f(x)\right) \right] \hbox {d}x \end{aligned}$$

So (5) can be rewritten as

$$\begin{aligned} \overline{z}=\frac{1}{N}\left[ \frac{M}{b}\int _{0}^{b}A(y)\hbox {d}y-\frac{N}{B-b} \int _{b}^{B}T(x-f(x))\hbox {d}x-R\right] =\frac{M}{N}A\left( b\right) -T\left( B\right) \end{aligned}$$

The condition for spatial equilibrium is then given by

$$\begin{aligned} MA\left( b\right) \ge NT\left( B\right) \end{aligned}$$

(15)

(i) First, we seek no-deviation condition of a firm. Suppose a firm deviates to from $y\in [0,b]$ to $x\in (b,B]$. We compute the wage w(x) that makes a worker indifferent if she residing at $x_{1}\in [b,B]$ shifts her workplace from $y\in [0,b]$ to $x\in [x_{1},B]$. We can focus on the interval of $x_{1}\in [x,B]$.

If she works at $y=f(x_{1})$, her consumption of composite good is

$$\begin{aligned} \overline{z}_{b}=w(f(x_{1}))-r(x_{1})s_\mathrm{w}-T\left( x_{1}-f(x_{1})\right) =w(0)-T\left( B\right) +\frac{1}{N}\int _{-\infty }^{\infty }\varPi (y^{\prime })m(y^{\prime })\hbox {d}y^{\prime } \end{aligned}$$

On the other hand, if she works at x, her consumption of composite good is

$$\begin{aligned} \overline{z}_{a}=w(x)-r(x_{1})s_\mathrm{w}-T\left( x_{1}-x\right) +\frac{1}{N} \int _{-\infty }^{\infty }\varPi (y^{\prime })m(y^{\prime })\hbox {d}y^{\prime } \end{aligned}$$

Because $\overline{z}_{b}=\overline{z}_{a}$ for equal utility, the wage offered by a firm at location x for a worker at location $x_{1}$ should satisfy

$$\begin{aligned} w(x,x_{1})=w(0)-T\left( B\right) +r(x_{1})s_\mathrm{w}+T\left( x_{1}-x\right) \end{aligned}$$

The profit of a firm relocating from y to x is

$$\begin{aligned} \varPi (x,x_{1})=A(x)-\frac{N}{M}\left[ w(0)-T\left( B\right) +r(x_{1})s_\mathrm{w}+T\left( x_{1}-x\right) \right] -r(x)s_\mathrm{f} \end{aligned}$$

The no-deviation condition is given by $\varPi (x,x_{1})\le \overline{\varPi },\ \forall b\le x,x_{1}\le B$. We have

$$\begin{aligned} \frac{\partial \varPi (x,x_{1})}{\partial x_{1}}=\frac{N}{M}\left[ T^{\prime }\left( x_{1}-f(x_{1})\right) -T^{\prime }\left( x_{1}-x\right) \right] >0 \end{aligned}$$

because $x_{1}-f(x_{1})>x_{1}-x$ and $T^{\prime \prime }\left( x\right) >0$. This implies $x_{1}=B$ is the minimizer of $\varPi (x,x_{1})$. Hence, the no-deviation condition is:

$$\begin{aligned} \varPi (x,B)\le \overline{\varPi },\qquad \forall x\in (b,B] \end{aligned}$$

(16)

We have

$$\begin{aligned} \varPi (x,B)= & {} A(x)-A(0)+\frac{Nb}{MB}\left[ T\left( x-f(x)\right) -T\left( B\right) \right] +\overline{\varPi }\\&+\frac{N}{M}\left[ T\left( B\right) -T\left( B-x\right) \right] \end{aligned}$$

Then, (16) can be rewritten as

$$\begin{aligned} \frac{M}{N}\left[ A(0)-A\left( x\right) \right] +\left[ T(B-x)-T\left( B\right) \right] +\frac{b}{B}\left[ T(B)-T\left( x-f(x)\right) \right] \ge 0\quad \end{aligned}$$

(17)

for all $x\in (b,B]$. This is rewritten as

$$\begin{aligned} \frac{\gamma }{t}\ge & {} \max _{y, y_{1}}G_{1}(x,s_\mathrm{f},s_\mathrm{w},M,N) \quad \text {where }G_{1}(x,s_\mathrm{f},s_\mathrm{w},M,N) \\\equiv & {} \frac{Ns_\mathrm{f}}{M}\frac{\frac{Ns_\mathrm{w}}{Ns_\mathrm{w}+Ms_\mathrm{f}}\delta \left( \frac{Ns_\mathrm{w}}{2}+\frac{Ms_\mathrm{f}}{2}\right) +\frac{Ms_\mathrm{f}}{Ns_\mathrm{w}+Ms_\mathrm{f}}\delta \left( x-f(x)\right) -\delta \left( \frac{Ns_\mathrm{w}}{2}+\frac{Ms_\mathrm{f}}{2} -x\right) }{\int _{-Ms_\mathrm{f}/2}^{Ms_\mathrm{f}/2}-h\left( \left| y^{\prime }\right| \right) +h\left( x-y^{\prime }\right) \hbox {d}y^{\prime }} \end{aligned}$$

and $G_{1}$ is finite and differentiable with respect to $x\in (b,B]$.

If we plug $x=B$ into (17), we get $M\left[ A(0)-A\left( B\right) \right] -NT\left( B\right) \ge 0$, which is stricter than the previous condition (15).

(ii) Second, we consider no-deviation condition of a worker. Suppose a worker deviates to from $x\in [b,B]$ to $y_{1}\in [0,b)$. The consumption of composite good before deviation was $\overline{z} _{b}(y)=w(y)-r(f^{-1}(y))s_\mathrm{w}-T\left( f^{-1}(y)-y\right)$. On the other hand, the consumption of composite good after deviation is $\overline{z} _{a}(y,y_{1})=w(y)-r(y_{1})s_\mathrm{w}-T\left( \left| y-y_{1}\right| \right)$. Let $y_{1}=y_{1}^{*}$ be the maximizer of $\overline{z} _{a}(y,y_{1})$. For a symmetric monocentric spatial equilibrium, the consumption of composite good before deviation is not smaller than that after deviation. That is, $\overline{z}_{b}(y)\ge \overline{z} _{a}(y,y_{1}^{*}),\ \forall y\in [0,b)$, or

$$\begin{aligned} \min _{y_{1}}\left[ r(y_{1})s_\mathrm{w}+T\left( \left| y-y_{1}\right| \right) \right] -r(f^{-1}(y))s_\mathrm{w}-T\left( f^{-1}(y)-y\right) \ge 0 \end{aligned}$$

(18)

for all $y\in [0,b)$. This is rewritten as

$$\begin{aligned} \frac{\gamma }{t}\ge & {} \max _{y,y_{1}}G_{2}(y,y_{1},s_\mathrm{f},s_\mathrm{w},M,N)\quad \text {where }G_{2}(y,y_{1},s_\mathrm{f},s_\mathrm{w},M,N) \\\equiv & {} \frac{s_\mathrm{f}^{2}}{s_\mathrm{w}}\frac{\frac{Ms_\mathrm{f}}{Ns_\mathrm{w}+Ms_\mathrm{f}}\delta \left( f^{-1}(y)-y\right) +\frac{Ns_\mathrm{w}}{Ns_\mathrm{w}+Ms_\mathrm{f}}\delta \left( f^{-1}(y_{1})-y_{1}\right) -\delta \left( \left| y-y_{1}\right| \right) }{\int _{-Ms_\mathrm{f}/2}^{Ms_\mathrm{f}/2}-h\left( \left| y^{\prime }-y_{1}\right| \right) +h\left( Ms_\mathrm{f}/2-y^{\prime }\right) \hbox {d}y^{\prime }} \end{aligned}$$

and $G_{2}$ is finite and differentiable with respect to y, $y_{1}\in [0,b)$.

Hence, the conditions for a symmetric monocentric spatial equilibrium are given by (17) and (18).

Next, we show that there is no cross commuting in any symmetric, monocentric equilibrium. Thus, the equilibrium specified above is the only one.

In equilibrium, for a worker residing at x to prefer commuting to y instead of $\widetilde{y}$, the following should hold:

$$\begin{aligned} w\left( y\right) -T\left( \left| x-y\right| \right) \ge w\left( \widetilde{y}\right) -T\left( \left| x-\widetilde{y}\right| \right) \end{aligned}$$

In equilibrium, for a worker residing at $\widetilde{x}$ to prefer commuting to $\widetilde{y}$ instead of y, the following should hold:

$$\begin{aligned} w\left( y\right) -T\left( \left| \widetilde{x}-y\right| \right) \le w\left( \widetilde{y}\right) -T\left( \left| \widetilde{x}-\widetilde{y} \right| \right) \end{aligned}$$

Hence,

$$\begin{aligned} \varphi (x)=T\left( \left| \widetilde{x}-y\right| \right) +T\left( \left| x-\widetilde{y}\right| \right) -T\left( \left| x-y\right| \right) -T\left( \left| \widetilde{x}-\widetilde{y} \right| \right) \ge 0 \end{aligned}$$

(19)

should hold for all x.

Suppose that there is parallel commuting. Then for some $y<\widetilde{y} \le x<\widetilde{x}$, we have

$$\begin{aligned} \varphi ^{\prime }(x)=T^{\prime }\left( x-\widetilde{y}\right) -T^{\prime }\left( x-y\right) >0 \end{aligned}$$

because $x-\widetilde{y}<x-y$ and $T^{\prime \prime }(x)<0$. We also have $\varphi (\widetilde{x})=0$, and thus $\varphi (x)<0$ for all $x<\widetilde{x}$, which contradicts $\varphi (x)\ge 0$.

Hence, there is only cross commuting at a symmetric, monocentric equilibrium. Thus, we conclude that the only such equilibrium is the one we have specified. $\square$