Question about Pareto Optimality

Question

I am looking at an exchange economy where I have two types of goods and n consumers. Half of the consumers have a utility function given by $U(x)= 5\ln{x} +m $ and the other half of the consumers have a $U(x) = 3\ln{x} + m$. All of the consumers have been given an initial endowment of $20$ of good $x$ and $10$ of good $m$.

I need to find what is the maximum amount of good $x$ that the first type of consumers can get at a Pareto Optimal allocation under the constraint $m>0$ for all consumers

To solve this, I began with creating the following equation: \begin{align*} \frac{\frac{\partial U_1}{\partial x_1}}{\frac{\partial U_1}{\partial m_1}} &= \frac{\frac{\partial U_2}{\partial x_2}}{\frac{\partial U_2}{\partial m_2}}\\ \frac{\frac{5}{x_1}}{1} &= \frac{\frac{3}{x_2}}{1}\\ \frac{5}{x_1} &= \frac{3}{x_2}\\ \frac{5}{x_1} &= \frac{3}{20-x_1} \end{align*} Then, I proceeded to solve for $x_1$ and then plugged into the equation $x_2 = 20-x_1$ to get get $x_2$

However, this approach does not give me the correct solution, nor does it include the initial endowment of $m$ so I can't solve for $m_1$ and $m_2$. Can someone please help me understand what I am missing?

Answer 1

This question reminds me of my old days studying advanced micro-economics, and staring at those simply described economics questions during exam without knowing how to start the modeling (lol).

Pareto optimality means a state that where one party's situation cannot be improved without making another party's situation worse.

WOLG, let's assume there are two consumers, customer A of type 1, and customer B of type 2. Each of them have $20$ x and $10$ m to start with.

$U_A(x, m)=5lnx + m$, and $U_B(x,m)=3lnx + m$

Notice that if we move $1$ m away from A and give it to B, A's utility will decrease by 1, and B's increase by 1. But then we need to see how many x we need to move from B to A while keeping both Utilities non-decreasing.

$$5ln(20 + \Delta) - 5ln(20) \ge 1$$ $$3ln(20) - 3ln(20 - \Delta) \le 1$$

We get: $4.4 \le \Delta \le 5.67$, thus we'll move $5$ x from B to A.

We could check if we move 2 m from A to B, and you could find you cannot compensate for 2 differences in utility caused by moving m with re-arranging x

Thus, the equilibrium would be at A having $25$ x and $9$ m, while B having $15$ x and $11$ m. At this state both sides have higher utility than initial assignment.

Answer 2

Let me try a simpler approach. (You can ignore this bracketed comment, but I am using the quasi-linearity of the utility functions, i.e. the fact that the $m$-component is additively separable.)

Wlog assume two consumers. The total amount of the first good is $X = 40$ and the total amount of money $M=20$. For a given level $k$ of utility of the second consumer, it is Pareto-optimal to maximise the utility utility of the first player. Hence, a Pareto-optimal allocation must satisfy $$\max_{(x_1,m_1)} u_1(x_1,m_1) \mbox{ s.t. } u_2(x_2,m_2)=k, x_1+x_2=40, m_1+m_2=20$$ Replace your data and this problem can be rewritten as $$\max_{(x_1,m_1)} 5 \ln x_1 + m_1 \mbox{ s.t. } 3 \ln (40-x_1) + (20-m_1)=k$$ Solve the constraint for $m_1$ and replace in the objective function to get $$\max_{x_1} 5 \ln x_1 + k - 3 \ln (40-x_1)$$ which is maximized at $x_1^* = 25$. Any Pareto-efficeint allocation has $x^*=25$. How you apportion money between the two consumers decides only how much utility each will get.