Simple Monocentric City: No Housing, Land in Utility Function

In this R Markdown notebook I draw some figures illustrating the basic ideas in the monocentric city model (Alonso-Muth-Mills model). To keep things simple, I use a model without a housing construction industry and where consumers just consume land directly. The city is a line and the amount of land at every point is equal to 1.

Assumptions and Functional Forms

Consumers have Cobb-Douglas utility over land \(l\) and a numeraire good \(z\).

\[U(l,z)=l^{\alpha}z^{1-\alpha}\tag{1}\]

For a consumer living at distance \(x\) from the center, the budget constraint is: \[y=z+r(x)l(x)-\tau*x\tag{2}\] The city is closed, with population \(N\), and agricultural land rent \(r_A\).

Solving the Model

In equilibrium all consumers have the same (endogenous) utility, \(\bar{u}\), and the land rent at the fringe of the city is equal to the agricultural rent, \(r(\bar{x})=r_A\). Further, there is one unit of land at every location, \(L(x)=1\) for \(x=0\) to \(x=\bar{x}\), and everyone is housed between the center of the city and the fringe.

From Cobb-Douglas demand we know that \(z(x)=(1-\alpha)(y-\tau*x)\) and \(l(x)=\frac{\alpha(y-\tau*x)}{r(x)}\). Substituting these into the spatial equilibrium condition (equal utility) gives:

\[U(l,z)=\left[\frac{\alpha(y-\tau*x)}{r(x)}\right]^{\alpha}\left[(1-\alpha)(y-\tau*x)\right]^{1-\alpha}=\bar{u}\tag{3}\] We can rewrite this equation to express land rent at \(x\) as a function of \(y\), \(\tau\), \(\bar{u}\), and \(\alpha\). \[r(x)=\left(\frac{y-\tau*x}{\bar{u}}\right)^{\frac{1}{\alpha}}\alpha(1-\alpha)^{\frac{1-\alpha}{\alpha}}\tag{4}\] The Alonso-Muth condition tells us the gradient of the land rent, which we could also derive by taking the derivative of the above equation: \[\frac{dr(x)}{dx}=-\tau((1-\alpha)(y-\tau*x))^{\frac{1-\alpha}{\alpha}}\bar{u}^{\frac{-1}{\alpha}}=\frac{-\tau}{l(x)}\tag{5}\] There are \(1/l(x)\) people living at every location \(x\). Combining the Alonso-Muth condition with the requirement that everyone is housed in the city implies that: \[\displaystyle\int_{x=0}^{\bar{x}}\frac{1}{l(x)}dx=\frac{1}{-\tau}\displaystyle\int_{x=0}^{\bar{x}}\frac{dr(x)}{dx}dx=\frac{r(\bar{x})-r(0)}{-\tau}=N\tag{6}\] Using \(r(\bar{x})=r_A\) gives \(r(0)=\tau*N+r(\bar{x})\). We can then find the utility of a consumer at the center by inserting this expression for \(r(0)\) into the indirect utility function (eq 3), thus giving us \(\bar{u}\). Finally, with \(\bar{u}\) we can solve for \(\bar{x}\) by the land rent equation (eq 4) equal to \(r_A\).

Solving for equilibrium

First, let’s set the exogenous parameters: \(\alpha\), \(y\), \(\tau\), \(r_A\), and \(N\).

alpha<-0.5
y<-100
tau<-1
r.A<-50
N<-1000

Next, let’s write functions for the indirect utility and land rent, so that we can solve for the equilibrium values of \(\bar{u}\) and \(\bar{x}\).

V<-function(x,r.x,alpha,y,tau){
  l.x<-alpha*(y-tau*x)/r.x #demand for land
  z.x<-(1-alpha)*(y-tau*x) #demand for numeraire
  l.x^alpha*z.x^(1-alpha) #returns the utility at x
}
R<-function(x,alpha,y,tau,u.bar) {
  inner<-(y-tau*x)/u.bar
  const<-alpha*(1-alpha)^((1-alpha)/alpha)
  inner^(1/alpha)*const #returns the rent at x
}

Now let’s solve for the equilibrium values. To find the fringe we will look for the value \(\bar{x}\) such that \(R(\bar{x})-r_A=0\).

r.0<-r.A+tau*N
u.bar<-V(0,r.0,alpha,y,tau)
#Define a function so that we can find x.bar
find.xbar<-function(x,alpha,y,tau,r.A,u.bar) {
  R(x,alpha,y,tau,u.bar)-r.A
}
x.bar<-as.numeric(uniroot(find.xbar,lower=0,upper=y/tau,alpha=alpha,y=y,tau=tau,r.A=r.A,u.bar=u.bar)[1])

The equilibrium values are \(\bar{u}\)=1.5430335 and \(\bar{x}\)=78.178211. Is this correct? Let’s check if integrating the population at each location is equal to the total, \(N\).

l.demand<-function(x,alpha,y,tau,r.x) { #demand for land
  alpha*(y-tau*x)/r.x
}
pop<-function(x,alpha,y,tau,u.bar) {
  r.x<-R(x,alpha,y,tau,u.bar)
  1/l.demand(x,alpha,y,tau,r.x)
}
N.check<-integrate(pop,lower=0,upper=x.bar,alpha=alpha,y=y,tau=tau,u.bar=u.bar) #this should be equal to N

Integrating the population over the city yields a total population of N.check=1000

Graphing

First, let’s graph the equilibrium rent curve

curve(R(x,alpha,y,tau,u.bar),from=0,to=x.bar,xlab="Distance from center (x)",ylab="Land rent, r(x)",main="Land rent over the city")

Next we can look at land consumption:

curve(l.demand(x,alpha,y,tau,R(x,alpha,y,tau,u.bar)),from=0,to=x.bar,xlab="Distance from center (x)",ylab="Land consumption, l(x)",main="Land consumption over the city")

Comparative Statics

Now let’s look at some comparative statics.

y.1<-500
tau.1<-0.7
r.A.1<-100
N.1<-5000

How does a change in the transporation cost \(\tau\) from \(\tau_0\)=1 to \(\tau_1\)=0.7 affect land consumption at every point?

r.0.1<-r.A+tau.1*N
u.bar.1<-V(0,r.0.1,alpha,y,tau.1)
x.bar.1<-as.numeric(uniroot(find.xbar,lower=0,upper=y/tau.1,alpha=alpha,y=y,tau=tau.1,r.A=r.A,u.bar=u.bar.1)[1])

curve(R(x,alpha,y,tau.1,u.bar.1),from=0,to=x.bar.1,xlab="Distance from center (x)",ylab="Land rent, r(x)",main="Land rent over the city",col="red",ylim=c(0,max(r.0,r.0.1)),xlim=c(0,max(x.bar,x.bar.1)))
curve(R(x,alpha,y,tau,u.bar),from=0,to=x.bar,add=TRUE,col="black")

text(x.bar,l.demand(x.bar,alpha,y,tau,r.A),"tau.0")
text(x.bar.1,l.demand(x.bar.1,alpha,y,tau.1,r.A),"tau.1",col="red")

curve(l.demand(x,alpha,y,tau.1,R(x,alpha,y,tau.1,u.bar.1)),from=0,to=x.bar.1,xlab="Distance from center (x)",ylab="Land consumption, l(x)",main="Land consumption over the city",col="red",ylim=c(0,0.3),xlim=c(0,max(x.bar,x.bar.1)))
curve(l.demand(x,alpha,y,tau,R(x,alpha,y,tau,u.bar)),from=0,to=x.bar,add=TRUE,col="black")

text(x.bar,l.demand(x.bar,alpha,y,tau,r.A),"tau.0")
text(x.bar.1,l.demand(x.bar.1,alpha,y,tau.1,r.A),"tau.1",col="red")

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