<p>Maximizing the sum of utilities or more generally weighted sum of utilities will give you efficient allocations as solutions.</p>
<p>To get efficient allocations, we can solve the following problem:</p>
<p><span class="math-container">$\max_{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+}\alpha_1\min(x_1,y_1)+\alpha_2\min(2x_2,y_2)+\alpha_3\min(3x_3,y_3)$</span></p>
<p>subject to <span class="math-container">$x_1+x_2+x_3=\omega_X, \ y_1+y_2+y_3=\omega_Y$</span></p>
<p>where <span class="math-container">$(\alpha_1,\alpha_2,\alpha_3)\in\mathbb{R}^3_+\setminus (0,0,0)$</span> and <span class="math-container">$\omega_X&gt;0$</span> and <span class="math-container">$\omega_Y&gt;0$</span> are given.</p>
<p>Solution to the above problem is necessarily Pareto efficient. As we vary <span class="math-container">$(\alpha_1,\alpha_2,\alpha_3)$</span> and consider the union of set of solutions generated in the process, we get the set of all Pareto efficient allocations.</p>
<p>For example: maximising the sum of utilities for the case where <span class="math-container">$\omega_X=\omega_Y=\omega&gt;0$</span>,</p>
<p><span class="math-container">$\max_{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+}\min(x_1,y_1)+\min(2x_2,y_2)+\min(3x_3,y_3)$</span></p>
<p>subject to <span class="math-container">$x_1+x_2+x_3=\omega, \ y_1+y_2+y_3=\omega$</span></p>
<p>will give the following set of feasible allocations as solutions to the above problem:
<span class="math-container">$\{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathcal{F}|y_1\leq x_1 \wedge y_2 \leq 2x_2 \wedge y_3 \leq 3x_3\}$</span></p>
<p>Here <span class="math-container">$\mathcal{F}$</span> is the set of feasible allocations i.e.
<span class="math-container">$\mathcal{F}=\{((x_1,y_1),(x_2,y_2),(x_3,y_3))\in\mathbb{R}^2_+\times\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1 +x_2+x_3=y_1+y_2+y_3=\omega\}$</span></p>
<p>These solutions are Pareto efficient, but not necessarily always the only efficient allocations. There can be more.</p>

CC BY-SA 4.0
2024-11-02T09:38:26.733Z
2024-11-02T10:14:06.667Z
2024-11-02T10:14:06.667Z
Amit
<p>I'm struggling with the following General Equilibrium exercise:</p>
<p>Find the Pareto Efficient set for this Pure Exchange Economy;</p>
<p>The consumers are <span class="math-container">$i = 1,2,3$</span> with these Leontief utilities:</p>
<p><span class="math-container">$u_i(x_{1i},x_{2i}) = \min(ix_{1i},x_{2i})$</span></p>
<p>Normally I would approach Leontiefs graphically but I don't know how to visualize it for 3 consumers.</p>
<p>I remember from previous courses the approach of maximizing the sum of all the utilities to obtain the efficient bundles. Is this valid here?</p>
<p>I would appreciate any help with this.</p>

<microeconomics><optimization><general-equilibrium><pareto-efficiency><leontief>
Find the Pareto Efficient set for 3 Leontiefs
Nicolas Torres
# Find the Pareto Efficient set for 3 Leontiefs

(Asked by: username_0 on 2024-11-01)

I'm struggling with the following General Equilibrium exercise:

Find the Pareto Efficient set for this Pure Exchange Economy;

The consumers are $i = 1,2,3$ with these Leontief utilities:

$u\_i(x\_{1i},x\_{2i}) = \min(ix\_{1i},x\_{2i})$

Normally I would approach Leontiefs graphically but I don't know how to visualize it for 3 consumers.

I remember from previous courses the approach of maximizing the sum of all the utilities to obtain the efficient bundles. Is this valid here?

I would appreciate any help with this.

**Tags:** microeconomics, optimization, general-equilibrium, pareto-efficiency, leontief


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## 1 answers

### Answer by username_1 on 2024-11-02. Score: 3

Maximizing the sum of utilities or more generally weighted sum of utilities will give you efficient allocations as solutions.

To get efficient allocations, we can solve the following problem:

$\max\_{((x\_1,y\_1),(x\_2,y\_2),(x\_3,y\_3))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+\times\mathbb{R}^2\_+}\alpha\_1\min(x\_1,y\_1)+\alpha\_2\min(2x\_2,y\_2)+\alpha\_3\min(3x\_3,y\_3)$

subject to $x\_1+x\_2+x\_3=\omega\_X, \ y\_1+y\_2+y\_3=\omega\_Y$

where $(\alpha\_1,\alpha\_2,\alpha\_3)\in\mathbb{R}^3\_+\setminus (0,0,0)$ and $\omega\_X>0$ and $\omega\_Y>0$ are given.

Solution to the above problem is necessarily Pareto efficient. As we vary $(\alpha\_1,\alpha\_2,\alpha\_3)$ and consider the union of set of solutions generated in the process, we get the set of all Pareto efficient allocations.

For example: maximising the sum of utilities for the case where $\omega\_X=\omega\_Y=\omega>0$,

$\max\_{((x\_1,y\_1),(x\_2,y\_2),(x\_3,y\_3))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+\times\mathbb{R}^2\_+}\min(x\_1,y\_1)+\min(2x\_2,y\_2)+\min(3x\_3,y\_3)$

subject to $x\_1+x\_2+x\_3=\omega, \ y\_1+y\_2+y\_3=\omega$

will give the following set of feasible allocations as solutions to the above problem:
$\{((x\_1,y\_1),(x\_2,y\_2),(x\_3,y\_3))\in\mathcal{F}|y\_1\leq x\_1 \wedge y\_2 \leq 2x\_2 \wedge y\_3 \leq 3x\_3\}$

Here $\mathcal{F}$ is the set of feasible allocations i.e.
$\mathcal{F}=\{((x\_1,y\_1),(x\_2,y\_2),(x\_3,y\_3))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_1 +x\_2+x\_3=y\_1+y\_2+y\_3=\omega\}$

These solutions are Pareto efficient, but not necessarily always the only efficient allocations. There can be more.


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