<p>Consider a mono-centric city, where all workers earn a wage <span class="math-container">$W$</span> in the centre of the circular city and rent out land <span class="math-container">$L$</span>.</p>
<p>Rent <span class="math-container">$r(d)$</span> and transport costs <span class="math-container">$t(d)$</span> vary with distance from the centre, <span class="math-container">$d$</span>.</p>
<p>Utility is constant by the spatial equilibrium assumption:</p>
<blockquote>
<p><span class="math-container">$U(C, L) = U(W - t(d) - r(d)L, L) = \underline{U}$</span></p>
</blockquote>
<p>Differentiating with respect to <span class="math-container">$d$</span> gives:</p>
<blockquote>
<p><span class="math-container">$r'(d) = \frac{-t'(d)}{L}$</span></p>
</blockquote>
<p>Two transport technologies are available:</p>
<blockquote>
<p><span class="math-container">$t(d) = \bar{t}d$</span> (no fixed cost)</p>
<p><span class="math-container">$t(d) = \underline{t}d + K$</span> (fixed cost)</p>
</blockquote>
<p>Workers will choose the no-fixed-cost technology at <span class="math-container">$d &lt; K/(\bar{t}-\underline{t})$</span> and the fixed cost technology further out.</p>
<p>The rent is <span class="math-container">$\underline{r}$</span> at the edge of the city, <span class="math-container">$\bar{d}$</span>.</p>
<p>What is <span class="math-container">$r(d)$</span>? (For <span class="math-container">$d &lt; K/(\bar{t}-\underline{t})$</span> and <span class="math-container">$d &gt; K/(\bar{t}-\underline{t})$</span>)</p>

CC BY-SA 3.0
2017-03-25T07:22:27.277Z
Welcome to Econ.SE. Have you tried solving for $r(d)$ on your own?
Herr K.
CC BY-SA 3.0
2017-03-25T14:44:30.087Z
@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.
StevenRJClarke1985
CC BY-SA 3.0
2017-03-26T01:13:36.427Z
(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 $K=(A-B)L$.
Herr K.
CC BY-SA 3.0
2017-03-27T00:34:05.143Z
@Herr. Thanks, that's what I tried. I get:  $$r(d)=\begin{cases}\underline{r} + \frac{K}{L} + \frac{1}{L}(\underline{t}\bar{d} - \bar{t}d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$
StevenRJClarke1985
CC BY-SA 3.0
2017-03-27T00:41:11.443Z
@Herr. This differs from the book I'm looking at ('Cities, Agglomeration and Spatial Equilibrium' by Edward Glaeser) which gives: $$r(d)=\begin{cases}\underline{r} + \frac{\bar{t}}{L}(\bar{d} - d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{K}{L} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$
StevenRJClarke1985
CC BY-SA 3.0
2017-03-27T03:03:58.613Z
Hmm, interesting... I got the same answer as yours. Since I'm not very familiar with the model presented herein, I can't say if there's any additional assumptions/restrictions that may result in the book's answer. Just by inspecting the book's answer, however, there are at least two things that don't make sense to me. (1) The book's solution is inconsistent with the boundary condition $r(\overline d)=\underline r$; it equals $\underline r+\frac{K}{L}$. (2) People at the inner-outer boundary are not indifferent between the two methods of transportation.
Herr K.
CC BY-SA 3.0
2017-03-27T08:16:31.347Z
@Herr. That's what didn't make sense to me. It's worth noting the indifference condition will apply at all points. Since $U(C(d),L)$ where $C(d) = W - t(d) - r(d)L$,  we have $W -\underline{t}d - r_{i}(d)L = W - \bar{t}d - K - r_{o}(d)L$ (subscripts denote inner and outer). Our answer satisfies this, the book's doesn't. It wouldn't be the first algebra mistake I've spotted. Anyway, thank you for responding. I feel I can move on in the book now.
StevenRJClarke1985
<mathematical-economics>
Rent gradient in Alonso-Muth-Mills monocentric city with two transport technologies
StevenRJClarke1985
# Rent gradient in Alonso-Muth-Mills monocentric city with two transport technologies

(Asked by: username_0 on 2017-03-23)

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

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: Welcome to Econ.SE. Have you tried solving for $r(d)$ on your own?  
> **username_0**: @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.  
> **username_1**: (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 $K=(A-B)L$.  
> **username_0**: @Herr. Thanks, that's what I tried. I get:  $$r(d)=\begin{cases}\underline{r} + \frac{K}{L} + \frac{1}{L}(\underline{t}\bar{d} - \bar{t}d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$  
> **username_0**: @Herr. This differs from the book I'm looking at ('Cities, Agglomeration and Spatial Equilibrium' by Edward Glaeser) which gives: $$r(d)=\begin{cases}\underline{r} + \frac{\bar{t}}{L}(\bar{d} - d) &\text{if } d < K/(\bar{t}-\underline{t})\\\underline{r} + \frac{K}{L} + \frac{\underline{t}}{L}(\bar{d} - d) &\text{if } d > K/(\bar{t}-\underline{t})\end{cases}$$  
> **username_1**: Hmm, interesting... I got the same answer as yours. Since I'm not very familiar with the model presented herein, I can't say if there's any additional assumptions/restrictions that may result in the book's answer. Just by inspecting the book's answer, however, there are at least two things that don't make sense to me. (1) The book's solution is inconsistent with the boundary condition $r(\overline d)=\underline r$; it equals $\underline r+\frac{K}{L}$. (2) People at the inner-outer boundary are not indifferent between the two methods of transportation.  
> **username_0**: @Herr. That's what didn't make sense to me. It's worth noting the indifference condition will apply at all points. Since $U(C(d),L)$ where $C(d) = W - t(d) - r(d)L$,  we have $W -\underline{t}d - r_{i}(d)L = W - \bar{t}d - K - r_{o}(d)L$ (subscripts denote inner and outer). Our answer satisfies this, the book's doesn't. It wouldn't be the first algebra mistake I've spotted. Anyway, thank you for responding. I feel I can move on in the book now.  

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