<p>If the prices $p_1$ and $p_2$ are positive than as you pointed out the equations
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}
hold. This is troublesome, because subtracting the first equation from the second yields
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; -10
\end{eqnarray*}
which has no solution for positive prices. However if a good has zero price, than the demand equations are different as no one minds buying superflous goods as long as they are free. For $p_2 = 0$ you would have
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; \leq &amp; 20.
\end{eqnarray*}
This is because $\frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2}$ is a minimum that the buyers need but they do not mind having more as good 2 is free. There is also a formulation of equilibrium, the 'value-equilibrium' which is that for good $i$
\begin{eqnarray*}
p_i \cdot D_i(p) = p_i \cdot S_i(p) \mbox{ and } D_i(p) \leq S_i(p)
\end{eqnarray*}
meaning that there can be oversupply of zero price goods. <br>
(Think about air. Would you buy air from someone?)
<br>
Unfortunately this equation system would still imply
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; \leq &amp; -10
\end{eqnarray*}
which has no solution in non-negative prices. On the other hand, for $p_1 = 0$ you would have
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; \leq &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}
which implies
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; \geq &amp; -10.
\end{eqnarray*}
As $p_1 = 0$ any positive $p_2$ is a solution for this.</p>

CC BY-SA 3.0
2016-03-19T20:04:59.997Z
So, does it mean that as good 1 becomes free, it wont contribute to the wealth of anyone, but still (a1, b1) will be, say, (30-t, t) given t is between 5 and 30? I'm still confused...
Ktp
CC BY-SA 3.0
2016-03-20T08:28:30.897Z
I suggest you draw the indifference curves and the budget set with $p_1 = 0$. Perhaps it will help you see the optimal consumptions better.
Giskard
CC BY-SA 3.0
2016-03-19T19:35:01.237Z
2016-03-19T19:35:01.237Z
Giskard
<p>Set of feasible allocations is</p>
<p><span class="math-container">$\mathcal{F}=\{((a_1,b_1),(a_2,b_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|a_1+a_2=30 \ \wedge b_1+b_2=20\}$</span></p>
<p>As can be seen from the picture below, set of competitive equilibrium allocations is given by</p>
<p><span class="math-container">$\mathcal{CE}=\{((a_1,b_1),(a_2,b_2))\in\mathcal{F}|0\leq a_1\leq 25 \ \wedge b_1=0\}$</span></p>
<p>and is supported by <span class="math-container">$p_A=0$</span>, <span class="math-container">$p_B&gt;0$</span>.</p>
<p><a href="https://i.sstatic.net/4wSoQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4wSoQ.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-10-30T10:42:34.320Z
For further details on the methodology, please visit https://youtu.be/WvHuzrFDxxY?feature=shared
Amit
CC BY-SA 4.0
2023-10-30T10:18:40.247Z
2023-10-30T10:18:40.247Z
Amit
<p>[![page 1st<a href="https://i.sstatic.net/TGpd0.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TGpd0.jpg" alt="page 2nd" /></a>
We solve this question with considering the marked for b
Consider Pb = 1 , and equate the dd and ss of the good b
M1/(Pa+Pb)  +  4M2/(Pa+4Pb) = 20 ( total no of good b )</p>

CC BY-SA 4.0
2022-06-30T12:36:34.170Z
Hi!

It's quite hard to tell what you are going for. Can you provide a bit more explanation? Also, typing your equations in MathJax would probably make them more readable.
Giskard
CC BY-SA 4.0
2022-06-28T20:13:46.277Z
2022-06-28T20:13:46.277Z
Aneja
<p>For a homework , I struggled to solve the following question but couldn't go further:</p>

<pre><code>endowment of person 1 = (30,0)

endowment of person 2 = (0,20)
</code></pre>

<p>utility functions are such that:
\begin{eqnarray*}
    U (a_1,b_1) &amp; = &amp; \min(a_1,b_1) \\
\\
    U (a_2,b_2) &amp; = &amp; \min(4a_2,b_2).
\end{eqnarray*}<br>
What I am doing is setting a1 equal to b1 and 4a2 equal to b2. After that I'm writing these;
$$
p_1 a_1 + p_2 b_1 = 30p_1 \mbox{ and } p_1a_2 + p_24a_2 = 20p_2
$$
Finally, by looking at feasibility condition, I'm writing down:
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}</p>

<p>Here, if I do some calculations, they result in $p_1 = p_2$ and accordingly $a_1=b_1=15, a_2=4$ and $b_2=16.$ But then the problem is that there is excess demand for good2 and there isn't a WE.</p>

<p>Alternatively, I am considering to have $p_2 = 0$ so that there exist a Walrasian Equilibrium.</p>

<p>BUT, I'm stuck at this point and lack the correct intuition for solving next steps. Or, I might be in a completely incorrect way. Please, explain me what will be done with this excess demand?</p>

<p>Thank you in advance.</p>

CC BY-SA 3.0
2016-03-19T18:21:45.090Z
Why are you considering $p_2 = 0$? Notice that everyone likes good 2 'more' than good 1. And the total supply of good 1 exceeds that of good 2. This means that good 2 will always become scarce before good 1 does. Hence you should consider $p_1 = 0$.
Giskard
CC BY-SA 3.0
2016-03-19T18:35:00.653Z
How do we see that both person 1 and person 2 likes good 2 more?
Ktp
CC BY-SA 3.0
2016-03-19T19:27:34.013Z
From the complement ratios.
Giskard
<perfect-complements><walrasian>
Perfect Complements - Walrasian Equilibrium
Ktp
# Perfect Complements - Walrasian Equilibrium

(Asked by: username_0 on 2016-03-19)

For a homework , I struggled to solve the following question but couldn't go further:
endowment of person 1 = (30,0)

endowment of person 2 = (0,20)

utility functions are such that:
\begin{eqnarray*}
    U (a_1,b_1) & = & \min(a_1,b_1) \\
\\
    U (a_2,b_2) & = & \min(4a_2,b_2).
\end{eqnarray*}
What I am doing is setting a1 equal to b1 and 4a2 equal to b2. After that I'm writing these;
$$
p_1 a_1 + p_2 b_1 = 30p_1 \mbox{ and } p_1a_2 + p_24a_2 = 20p_2
$$
Finally, by looking at feasibility condition, I'm writing down:
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} & = & 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} & = & 20.
\end{eqnarray*}
Here, if I do some calculations, they result in $p_1 = p_2$ and accordingly $a_1=b_1=15, a_2=4$ and $b_2=16.$ But then the problem is that there is excess demand for good2 and there isn't a WE.
Alternatively, I am considering to have $p_2 = 0$ so that there exist a Walrasian Equilibrium.
BUT, I'm stuck at this point and lack the correct intuition for solving next steps. Or, I might be in a completely incorrect way. Please, explain me what will be done with this excess demand?
Thank you in advance.

**Tags:** perfect-complements, walrasian

> **Comments:**
> **username_1**: Why are you considering $p_2 = 0$? Notice that everyone likes good 2 'more' than good 1. And the total supply of good 1 exceeds that of good 2. This means that good 2 will always become scarce before good 1 does. Hence you should consider $p_1 = 0$.  
> **username_0**: How do we see that both person 1 and person 2 likes good 2 more?  
> **username_1**: From the complement ratios.  

---

## 3 answers

### Answer by username_1 on 2016-03-19. Score: 2

If the prices $p\_1$ and $p\_2$ are positive than as you pointed out the equations
\begin{eqnarray\*}
\frac{30p\_1}{p\_1+p\_2} + \frac{20p\_2}{p\_1+4p\_2} & = & 30 \\
\\
\frac{30p\_1}{p\_1+p\_2} + \frac{4 \cdot 20p\_2}{p\_1+4p\_2} & = & 20.
\end{eqnarray\*}
hold. This is troublesome, because subtracting the first equation from the second yields
\begin{eqnarray\*}
\frac{3 \cdot 20p\_2}{p\_1+4p\_2} & = & -10
\end{eqnarray\*}
which has no solution for positive prices. However if a good has zero price, than the demand equations are different as no one minds buying superflous goods as long as they are free. For $p\_2 = 0$ you would have
\begin{eqnarray\*}
\frac{30p\_1}{p\_1+p\_2} + \frac{20p\_2}{p\_1+4p\_2} & = & 30 \\
\\
\frac{30p\_1}{p\_1+p\_2} + \frac{4 \cdot 20p\_2}{p\_1+4p\_2} & \leq & 20.
\end{eqnarray\*}
This is because $\frac{30p\_1}{p\_1+p\_2} + \frac{4 \cdot 20p\_2}{p\_1+4p\_2}$ is a minimum that the buyers need but they do not mind having more as good 2 is free. There is also a formulation of equilibrium, the 'value-equilibrium' which is that for good $i$
\begin{eqnarray\*}
p\_i \cdot D\_i(p) = p\_i \cdot S\_i(p) \mbox{ and } D\_i(p) \leq S\_i(p)
\end{eqnarray\*}
meaning that there can be oversupply of zero price goods.   
(Think about air. Would you buy air from someone?)
  
Unfortunately this equation system would still imply
\begin{eqnarray\*}
\frac{3 \cdot 20p\_2}{p\_1+4p\_2} & \leq & -10
\end{eqnarray\*}
which has no solution in non-negative prices. On the other hand, for $p\_1 = 0$ you would have
\begin{eqnarray\*}
\frac{30p\_1}{p\_1+p\_2} + \frac{20p\_2}{p\_1+4p\_2} & \leq & 30 \\
\\
\frac{30p\_1}{p\_1+p\_2} + \frac{4 \cdot 20p\_2}{p\_1+4p\_2} & = & 20.
\end{eqnarray\*}
which implies
\begin{eqnarray\*}
\frac{3 \cdot 20p\_2}{p\_1+4p\_2} & \geq & -10.
\end{eqnarray\*}
As $p\_1 = 0$ any positive $p\_2$ is a solution for this.

> **Comments:**
> **username_0**: So, does it mean that as good 1 becomes free, it wont contribute to the wealth of anyone, but still (a1, b1) will be, say, (30-t, t) given t is between 5 and 30? I'm still confused...  
> **username_1**: I suggest you draw the indifference curves and the budget set with $p_1 = 0$. Perhaps it will help you see the optimal consumptions better.  

---

### Answer by username_2 on 2023-10-30. Score: 1

Set of feasible allocations is

$\mathcal{F}=\{((a\_1,b\_1),(a\_2,b\_2))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|a\_1+a\_2=30 \ \wedge b\_1+b\_2=20\}$

As can be seen from the picture below, set of competitive equilibrium allocations is given by

$\mathcal{CE}=\{((a\_1,b\_1),(a\_2,b\_2))\in\mathcal{F}|0\leq a\_1\leq 25 \ \wedge b\_1=0\}$

and is supported by $p\_A=0$, $p\_B>0$.

[![enter image description here](https://i.sstatic.net/4wSoQ.png)](https://i.sstatic.net/4wSoQ.png)

> **Comments:**
> **username_2**: For further details on the methodology, please visit https://youtu.be/WvHuzrFDxxY?feature=shared  

---

### Answer by username_3 on 2022-06-28. Score: 0

[![page 1st[![page 2nd](https://i.sstatic.net/TGpd0.jpg)](https://i.sstatic.net/TGpd0.jpg)
We solve this question with considering the marked for b
Consider Pb = 1 , and equate the dd and ss of the good b
M1/(Pa+Pb) + 4M2/(Pa+4Pb) = 20 ( total no of good b )

> **Comments:**
> **username_1**: Hi!  It's quite hard to tell what you are going for. Can you provide a bit more explanation? Also, typing your equations in MathJax would probably make them more readable.  

---

