<p>Since you commented you only needed the agents’ demands as a function of the relative prices, I’m giving my answer.</p>
<p>I take as numeraire <span class="math-container">$p_x = 1$</span>.</p>
<ul>
<li>Agent <span class="math-container">$A$</span></li>
</ul>
<p>The optimization program is</p>
<p><span class="math-container">$\max x + 100 (1-e^{-\frac{y}{10}})$</span></p>
<p>s.t. <span class="math-container">$x + y p_y = 40$</span></p>
<p>The optimality condition is <span class="math-container">$MRS_A = \frac{p_x}{p_y}$</span></p>
<p><span class="math-container">$\frac{1}{10 e^{-\frac{y}{10}}} = \frac{1}{p_y} \implies \frac{p_y}{10} = e^{-\frac{y}{10}} \implies -10 \ln(\frac{p_y}{10}) = y \implies {y_A}^\star = 10 \ln(\frac{10}{p_y})$</span></p>
<p>Isolating <span class="math-container">$x$</span> from the budget constraint we get</p>
<p><span class="math-container">$x = 40 - p_y y$</span></p>
<p>Substituting Agent <span class="math-container">$A$</span>’s <span class="math-container">$y$</span>-demand we get</p>
<p><span class="math-container">${x_A}^\star = 40 - 10 p_y \ln(\frac{10}{p_y})$</span></p>
<p>Notice our expression for <span class="math-container">${y_A}^\star$</span> is negative exactly when <span class="math-container">$p_y &gt;10$</span>.</p>
<p>So when <span class="math-container">$p_y \leq 10$</span>, Agent <span class="math-container">$A$</span>’s demands are the expressions we got.</p>
<p>On the other hand, when <span class="math-container">$p_y &gt; 10$</span>, the actual demands are</p>
<p><span class="math-container">${y_A}^\star = 0$</span></p>
<p><span class="math-container">${x_A}^\star = 40$</span></p>
<p>We can show with calculus that the expression for <span class="math-container">${x_A}^\star$</span> attains its minimum value at <span class="math-container">$p_y = \frac{10}{e}$</span>.</p>
<p>Evaluating <span class="math-container">${x_A}^\star$</span> at this argument, we get its minimum value is <span class="math-container">${x_A}^\star = 4 - \frac{10}{e} &gt; 4 - \frac{10}{2.5} = 0$</span>.</p>
<p>Since the minimum value for the expression for <span class="math-container">${x_A}^\star$</span> is positive, there is no need to worry about the non-negativity constraint on this one.</p>
<ul>
<li>Agent <span class="math-container">$B$</span></li>
</ul>
<p>The optimization program is</p>
<p><span class="math-container">$\max y + 110 (1-e^{-\frac{x}{10}})$</span></p>
<p>s.t. <span class="math-container">$x + p_y y = 50 p_y$</span></p>
<p>The optimality condition is <span class="math-container">$MRS_B = \frac{p_x}{p_y}$</span></p>
<p><span class="math-container">$11 e^{-\frac{x}{10}} = \frac{1}{p_y} \implies x = -10 \ln(\frac{1}{11 p_y}) \implies {x_B}^\star = 10 \ln(11 p_y)$</span></p>
<p>Isolating <span class="math-container">$y$</span> from the budget constraint we get</p>
<p><span class="math-container">$y = 50 - \frac{x}{p_y}$</span></p>
<p>Substituting Agent <span class="math-container">$B$</span>’s <span class="math-container">$x$</span>-demand we get</p>
<p><span class="math-container">${y_B}^\star = 50 - \frac{10}{p_y} \ln(11 p_y)$</span></p>
<p>Notice our expression for <span class="math-container">${x_B}^\star$</span> is negative exactly when <span class="math-container">$p_y &lt; 1$</span>.</p>
<p>So when <span class="math-container">$p_y \leq 1$</span>, Agent <span class="math-container">$B$</span>’s demands are the expressions we got.</p>
<p>On the other hand, when <span class="math-container">$p_y &lt;1$</span>, the actual demands are</p>
<p><span class="math-container">${x_B}^\star = 0$</span></p>
<p><span class="math-container">${y_B}^\star = 50$</span></p>
<p>We can show with calculus that the expression for <span class="math-container">${y_B}^\star$</span> attains its minimum value at <span class="math-container">$p_y = \frac{e}{11}$</span>.</p>
<p>Evaluating <span class="math-container">${y_B}^\star$</span> at this argument, we get its minimum value is <span class="math-container">${y_B}^\star = 5 - \frac{11}{e} &gt; 5 - \frac{11}{2.2} = 0$</span>.</p>
<p>Since the minimum value for the expression for <span class="math-container">${y_B}^\star$</span> is positive, there is no need to worry about the non-negativity constraint on this one.</p>

CC BY-SA 4.0
2023-05-11T00:56:26.037Z
2023-05-11T02:33:51.850Z
2023-05-11T02:33:51.850Z
Nicolas Torres
<p>I'm having difficulties solving for multiple equilibria for competitive exchange economies.</p>
<p>Considering a quasi linear preference as such:</p>
<p><span class="math-container">$U_{A}(x,y)=x+100(1-e^{-y/10})$</span></p>
<p><span class="math-container">$U_{B}(x,y)=y+110(1-e^{-x/10})$</span></p>
<p>with initial endowments:
<span class="math-container">$e_{A}=(40,0),
e_{B}=(0,50)$</span></p>
<p>when I attempted to solve this I found:</p>
<p><span class="math-container">$MRS_{A}= {\frac{1}{10e^{-y/10}}} = {\frac{p_1}{p_2}}$</span></p>
<p><span class="math-container">$MRS_{B}= 11e^{-x/10} = {\frac{p_1}{p_2}}$</span></p>
<p>and then I plugged this into the endowment constraint but I'm stuck. Is my approach to this problem wrong?</p>

CC BY-SA 4.0
2023-05-10T05:58:56.740Z
You are ignoring nonnegativity cobditions. Also, only relative prices matter; you can normalize them.
Michael Greinecker
CC BY-SA 4.0
2023-05-10T14:12:02.237Z
No matter what I try, I always get transcendental equations, such as a product of a linear term and a log, or in terms of the function $x^x$
Nicolas Torres
CC BY-SA 4.0
2023-05-10T14:30:18.823Z
I used the non negativity constraints as Michael Greinecker pointed out, it only gave me that $p_y \in [1,10]$. Here I took the numeraire as $p_x = 1$.
Nicolas Torres
CC BY-SA 4.0
2023-05-10T14:33:54.080Z
I get the equation $\ln(11 p_y) = p_y \ln(\frac{10}{p_y})$, which can’t be solved algebraically.
Nicolas Torres
CC BY-SA 4.0
2023-05-10T14:35:14.337Z
Using GeoGebra, I get that there are $2$ valid solutions: one near $p_y = 1.33$ and another one near $p_y = 3.597$
Nicolas Torres
CC BY-SA 4.0
2023-05-10T14:37:21.633Z
Do you actually have to solve for everything explicitly?
Nicolas Torres
CC BY-SA 4.0
2023-05-10T17:40:22.703Z
the question asks for each agent's demand as a function of P
Lily B
<microeconomics><competitive-equilibrium>
Multiple Equilibria
Lily B
# Multiple Equilibria

(Asked by: username_0 on 2023-05-09)

I'm having difficulties solving for multiple equilibria for competitive exchange economies.

Considering a quasi linear preference as such:

$U\_{A}(x,y)=x+100(1-e^{-y/10})$

$U\_{B}(x,y)=y+110(1-e^{-x/10})$

with initial endowments:
$e\_{A}=(40,0),
e\_{B}=(0,50)$

when I attempted to solve this I found:

$MRS\_{A}= {\frac{1}{10e^{-y/10}}} = {\frac{p\_1}{p\_2}}$

$MRS\_{B}= 11e^{-x/10} = {\frac{p\_1}{p\_2}}$

and then I plugged this into the endowment constraint but I'm stuck. Is my approach to this problem wrong?

**Tags:** microeconomics, competitive-equilibrium
> **Comments:**
> **username_1**: You are ignoring nonnegativity cobditions. Also, only relative prices matter; you can normalize them.  
> **username_2**: No matter what I try, I always get transcendental equations, such as a product of a linear term and a log, or in terms of the function $x^x$  
> **username_2**: I used the non negativity constraints as Michael Greinecker pointed out, it only gave me that $p_y \in [1,10]$. Here I took the numeraire as $p_x = 1$.  
> **username_2**: I get the equation $\ln(11 p_y) = p_y \ln(\frac{10}{p_y})$, which can’t be solved algebraically.  
> **username_2**: Using GeoGebra, I get that there are $2$ valid solutions: one near $p_y = 1.33$ and another one near $p_y = 3.597$  
> **username_2**: Do you actually have to solve for everything explicitly?  
> **username_0**: the question asks for each agent's demand as a function of P  

---

## 1 answers

### Answer by username_2 on 2023-05-11. Score: 1

Since you commented you only needed the agents’ demands as a function of the relative prices, I’m giving my answer.

I take as numeraire $p\_x = 1$.

* Agent $A$

The optimization program is

$\max x + 100 (1-e^{-\frac{y}{10}})$

s.t. $x + y p\_y = 40$

The optimality condition is $MRS\_A = \frac{p\_x}{p\_y}$

$\frac{1}{10 e^{-\frac{y}{10}}} = \frac{1}{p\_y} \implies \frac{p\_y}{10} = e^{-\frac{y}{10}} \implies -10 \ln(\frac{p\_y}{10}) = y \implies {y\_A}^\star = 10 \ln(\frac{10}{p\_y})$

Isolating $x$ from the budget constraint we get

$x = 40 - p\_y y$

Substituting Agent $A$’s $y$-demand we get

${x\_A}^\star = 40 - 10 p\_y \ln(\frac{10}{p\_y})$

Notice our expression for ${y\_A}^\star$ is negative exactly when $p\_y >10$.

So when $p\_y \leq 10$, Agent $A$’s demands are the expressions we got.

On the other hand, when $p\_y > 10$, the actual demands are

${y\_A}^\star = 0$

${x\_A}^\star = 40$

We can show with calculus that the expression for ${x\_A}^\star$ attains its minimum value at $p\_y = \frac{10}{e}$.

Evaluating ${x\_A}^\star$ at this argument, we get its minimum value is ${x\_A}^\star = 4 - \frac{10}{e} > 4 - \frac{10}{2.5} = 0$.

Since the minimum value for the expression for ${x\_A}^\star$ is positive, there is no need to worry about the non-negativity constraint on this one.

* Agent $B$

The optimization program is

$\max y + 110 (1-e^{-\frac{x}{10}})$

s.t. $x + p\_y y = 50 p\_y$

The optimality condition is $MRS\_B = \frac{p\_x}{p\_y}$

$11 e^{-\frac{x}{10}} = \frac{1}{p\_y} \implies x = -10 \ln(\frac{1}{11 p\_y}) \implies {x\_B}^\star = 10 \ln(11 p\_y)$

Isolating $y$ from the budget constraint we get

$y = 50 - \frac{x}{p\_y}$

Substituting Agent $B$’s $x$-demand we get

${y\_B}^\star = 50 - \frac{10}{p\_y} \ln(11 p\_y)$

Notice our expression for ${x\_B}^\star$ is negative exactly when $p\_y < 1$.

So when $p\_y \leq 1$, Agent $B$’s demands are the expressions we got.

On the other hand, when $p\_y <1$, the actual demands are

${x\_B}^\star = 0$

${y\_B}^\star = 50$

We can show with calculus that the expression for ${y\_B}^\star$ attains its minimum value at $p\_y = \frac{e}{11}$.

Evaluating ${y\_B}^\star$ at this argument, we get its minimum value is ${y\_B}^\star = 5 - \frac{11}{e} > 5 - \frac{11}{2.2} = 0$.

Since the minimum value for the expression for ${y\_B}^\star$ is positive, there is no need to worry about the non-negativity constraint on this one.


---

