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Consider a mono-centric city, where all workers earn a wage $W$ in the centre of the circular city and rent out land $L$.

Rent $r(d)$ and transport costs $t(d)$ vary with distance from the centre, $d$.

Utility is constant by the spatial equilibrium assumption:

$U(C, L) = U(W - t(d) - r(d)L, L) = \underline{U}$

Differentiating with respect to $d$ gives:

$r'(d) = \frac{-t'(d)}{L}$

Two transport technologies are available:

$t(d) = \bar{t}d$ (no fixed cost)

$t(d) = \underline{t}d + K$ (fixed cost)

Workers will choose the no-fixed-cost technology at $d < K/(\bar{t}-\underline{t})$ and the fixed cost technology further out.

The rent is $\underline{r}$ at the edge of the city, $\bar{d}$.

What is $r(d)$? (For $d < K/(\bar{t}-\underline{t})$ and $d > K/(\bar{t}-\underline{t})$)

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  • $\begingroup$ Welcome to Econ.SE. Have you tried solving for $r(d)$ on your own? $\endgroup$
    – Herr K.
    Commented Mar 25, 2017 at 7:22
  • $\begingroup$ @herr-k. Hi. Thanks for answering. Yes I have tried myself. I use the rent gradient $r′(d)=−t′(d)/L$ in each case. Integrating in the inner and outer circles I get: $r(d)=−\bar{t}d/L + A$ in the inner circle and $r(d)=−\underline{t}d/L + B$ in the outer circle. To solve for the constants A and B, I use the condition $\underline{r} = r(\bar{d})$ at the city's edge to find B. I'm not sure if there is a discontinuity in rent at the boundary between the inner and outer circles. The book I'm looking at gives the answers but I am not convinced. $\endgroup$
    – StevenRJClarke1985
    Commented Mar 25, 2017 at 14:44
  • $\begingroup$ (i) Based on the information given, $t'(d)$ is a step function: $$t'(d)=\begin{cases}\overline t & \text{if }d\in[0,K/(\overline t-\underline t)]\\\underline t &\text{if }d\in[K/(\overline t-\underline t),\overline d]\end{cases}$$ Note that step functions are Riemann-integrable (despite the discontinuity). (ii) You can use the indifference condition at the inner-outer boundary (i.e. $d=K/(\overline t-\underline t)$) to get $K=(A-B)L$. $\endgroup$
    – Herr K.
    Commented Mar 26, 2017 at 1:13
  • $\begingroup$ @Herr. Thanks, that's what I tried. I get: $$r(d)=\begin{cases}\underline{r} + \frac{K}{L} + \frac{1}{L}(\underline{t}\bar{d} - \bar{t}d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$ $\endgroup$
    – StevenRJClarke1985
    Commented Mar 27, 2017 at 0:34
  • $\begingroup$ @Herr. This differs from the book I'm looking at ('Cities, Agglomeration and Spatial Equilibrium' by Edward Glaeser) which gives: $$r(d)=\begin{cases}\underline{r} + \frac{\bar{t}}{L}(\bar{d} - d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{K}{L} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$ $\endgroup$
    – StevenRJClarke1985
    Commented Mar 27, 2017 at 0:41

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