# Rent gradient in Alonso-Muth-Mills monocentric city with two transport technologies

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 $$d > K/(\bar{t}-\underline{t})$$)

• Welcome to Econ.SE. Have you tried solving for $r(d)$ on your own? Commented Mar 25, 2017 at 7:22
• @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. Commented Mar 25, 2017 at 14:44
• (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 $K=(A-B)L$. Commented Mar 26, 2017 at 1:13
• @Herr. Thanks, that's what I tried. I get: $$r(d)=\begin{cases}\underline{r} + \frac{K}{L} + \frac{1}{L}(\underline{t}\bar{d} - \bar{t}d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$ Commented Mar 27, 2017 at 0:34
• @Herr. This differs from the book I'm looking at ('Cities, Agglomeration and Spatial Equilibrium' by Edward Glaeser) which gives: $$r(d)=\begin{cases}\underline{r} + \frac{\bar{t}}{L}(\bar{d} - d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{K}{L} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$ Commented Mar 27, 2017 at 0:41