# How to solve this aggregate production problem?

Consider a continuum of firms indexed by $$a\in [0,1]$$ and that $$a$$ is uniformaly distributed. a firm of type $$a$$ has a production set $$Y^n$$ as depicted in the figure below. note that good 1 is used as an input to produce good 2. suppose that the price of the input (good 1) is fixed at $$w=1$$ while the price of output is $$p>0$$.

How should I get the input demand and output supply of a firm of type $$a$$ when the price of output is $$p$$? And what is the aggregate or average input demand and output supply when the output price is $$p$$? And what is the aggregate (or average) production set?

Observing the highest possible iso-profit curve passing through the feasible set, we obtain the supply of $$y_2$$ and demand for $$y_1$$ as follows. Here iso-profit curve is set of $$(y_1,y_2)$$ pairs that yields the same level of profit given $$p$$.

So, $$\begin{eqnarray*} (y_1^d,y_2^s)(p)\in\begin{cases}\{(0,0),(1+a,1)\} & \text{if } p = 1+a \\ \{(1+a,1)\} & \text{if } p > 1+a \\ \{(0,0)\} & \text{if } p < 1+a \end{cases} \end{eqnarray*}$$

Corresponding optimal profit is

$$\begin{eqnarray*} \pi(p)=\max(0,p-(1+a)) \end{eqnarray*}$$

Given that $$a\sim\text{Unif}(0,1)$$, aggregate input demand is $$\begin{eqnarray*}Y_1^d(p) = \int_0^1y_1^d(p)da = \begin{cases} \int_0^{1}(1+a)da=\frac{3}{2} & \text{if } p \geq 2 \\ 0 & \text{if } p \leq 1 \\ \int_0^{p-1}(1+a)da=\dfrac{(p-1)(p+1)}{2} & \text{if } 1 < p < 2\end{cases} \end{eqnarray*}$$

Aggregate output supply is $$\begin{eqnarray*}Y_2^s(p) = \int_0^1y_2^s(p)da = \begin{cases} \int_0^{1}da=1 & \text{if } p \geq 2 \\ 0 & \text{if } p \leq 1 \\ \int_0^{p-1}da=p-1 & \text{if } 1 < p < 2\end{cases} \end{eqnarray*}$$

• Hi - What is the average production set in this case? Commented Dec 8, 2023 at 15:27
• Boundary of the aggregate Production Set is given by $g(y_1)=\begin{cases} 1 & \text{for } y_1<-2\\ \int_0^1 \min(-y_1-a,1)da & \text{for } -2\leq y_1\leq -1 \\ \int_0^1 \max(-y_1-a,0)da & \text{for } -1<y_1\leq 0 \end{cases}$ and aggregate production set is simply $\{(y_1,y_2)|0\leq y_2\leq g(y_1)\}$
– Amit
Commented Dec 11, 2023 at 2:34

Let's first look at the profit of a single firm with parameter $$a$$. Let us denote by $$x = -y_1$$ the input and $$f(x) = y_2$$ the output.

The production function is the following: $$f(x) = \begin{cases} 0 &\text{ if } x < a,\\ x - a &\text{ if } a \le x \le 1+a,\\ 1 &\text{ if } x > 1+a\end{cases}$$

The firm optimizes profits: $$x^\ast = \arg\max_x p f(x) - x.$$

You can solve this by looking at various scenarios and taking the best (profit maximising) one.

1. if $$x^\ast < a$$ then profits are $$p 0 - x^\ast$$, so here it is best to put $$x^\ast = 0$$.
2. if $$a \le x^\ast \le 1+a$$ then profits are $$p(x^\ast - a) - x^\ast= (p-1)x^\ast - pa$$. If $$p > 1$$ then profits increase in $$x^\ast$$ so it is best to put $$x^\ast = 1+a$$ and profits will be $$p - 1 - a$$. if $$p < 1$$, then it is best to set $$x^\ast = a$$ and profits will be $$-a$$ (which is negative). In the latter case, it will be best to produce nothing.
3. if $$x^\ast \ge 1+a$$ then profits are $$p - x^\ast$$ so here it is optimal to put $$x = 1+a$$ and profits are given by $$p - 1 - a$$.

As such, we see that profits are either equal to $$p-1-a$$ with optimal demand $$x^\ast = 1+a$$ or equal to zero with optimal demand $$x^\ast = 0$$.

The first will be optimal if $$a \le p-1$$, while the second is optimal if $$a > p-1$$. This gives the following optimal demand for a firm with parameter $$a$$: $$x^\ast = \begin{cases} 1+a &\text{ if } a\le p-1 \\ 0 &\text{ if } a > p-1\end{cases}$$

Now let us derive optimal aggregate demand. If $$p < 1$$ then $$a \ge 0 > p-1$$ so no firm will demand anything: aggregate demand equals zero.

If $$p > 2$$ then $$a \le 1 \le p - 1$$ so the firm of type $$a$$ will demand $$1+a$$. Aggregate demand is then given by: $$\int_0^1 (1+a) da = [a + a^2/2]^1_0 = \frac{3}{2}.$$

If $$1 \le p \le 2$$ then firms with $$a \le p-1$$ will demand $$1+a$$ and firms with $$a > p-1$$ will not demand anything. As such, aggreagate demand is: \begin{align*} \int_0^{p-1} (1+a) da &= [a + a^2/2]^{(p-1)}_0 \\ &= (p-1) + \frac{(p-1)^2}{2} \\ &= p-1 + \frac{p^2}{2} - \frac{2p}{2} + \frac{1}{2},\\ & = \frac{p^2}{2} - \frac{1}{2}. \end{align*}

We get $$\text{aggregate demand} = \begin{cases} 0 &\text{ if } p < 1\\ \frac{p^2}{2} - \frac{1}{2} &\text{ if } 1 \le p \le 2\\ \frac{3}{2} &\text{ if } p > 2.\end{cases}$$

Aggregate demand is given below.

• According to the "production function" (<<a production set Yn as depicted in the figure below>>), the firm cannot produce more than 1 unit of output. Commented Nov 3, 2023 at 9:37
• @escaiguolquer thanks. I made some corrections (supply->demand)
– tdm
Commented Nov 3, 2023 at 12:48
• Hi - What is the average production set in this case? Commented Dec 8, 2023 at 15:17