I have the following CES nested in a Cobb-Douglas production function:

$$y(i)=[l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}$$

$\beta \in (0,1)$, $\tilde{\gamma}$ is a productivity parameter, and $\epsilon$>1 is the EOS between low-skilled workers,$l(i)$, and machines $x(i)$. Also, $\alpha(i)$ is an indicator function taking value one if the product $i$ is automated and zero otherwise. Mathematically, $\alpha(i) \in \{0,1\}$. Finally, $h(i)$ are high-skilled workers.

The unit cost function of product $y(i)$ is:

$$c= \beta^{-\beta}(1-\beta)^{-(1-\beta)}(w_L^{1-\epsilon}+\gamma \alpha(i))^{\frac{\beta}{1-\epsilon}}w_H^{1-\beta}$$

where $w_L$ is the unit cost of the input $l(i)$, one is the unit cost of input $x(i)$ and $w_H$ is the unit cost of input $h(i)$. Finally $\gamma \equiv \tilde{\gamma^\epsilon}$

Could you tell me how to derive this unit cost function?

  • 1
    $\begingroup$ In this post, you can see the steps that are generally applicable, economics.stackexchange.com/a/5273/61 $\endgroup$ Sep 12, 2022 at 9:30
  • 1
    $\begingroup$ @John M. you can try to use the indicator function, first suppose $\alpha=0$ then find $c$ for that case, and then find $c$ for $\alpha=1$, the algebra is hellish for this case as I'm trying to do it myself. Will upload a solution if I am able to do it myself. Also, for the three input case you can try to solve the cost minimization problem in stages $\endgroup$
    – mynameparv
    May 1 at 5:54
  • $\begingroup$ @mynameparv can you show me your attempt? $\endgroup$
    – John M.
    May 1 at 21:14
  • $\begingroup$ I've written an answer but it's still in complete, the case for $\alpha=1$ is not at all simple. Do you want me to post the incomplete solution and you try the rest yourself? $\endgroup$
    – mynameparv
    May 2 at 4:05
  • $\begingroup$ @John M. The solution contains enough information and explanation so that you may attempt the problem on your own and I will get to completing it once I'm a bit more free. Let me know if I should post that attempt or not $\endgroup$
    – mynameparv
    May 2 at 4:19

2 Answers 2


I'm going to provide the steps to follow for solving this optimization problem:

The Cost Function for $y(i)$ is given by:

\begin{equation*} C(w_H,w_L,y(i))= w_H h(i)^{\ast} + w_L l(i)^{\ast} + x(i)^{\ast} \label{cost} \end{equation*}

The production function is homogenous of degree 1, that is,it exhibits constant return to scale. This makes average cost equal to the marginal cost.

For convenience, write the production function as

\begin{equation} y(i)=z(i)^{\beta}h^{1-\beta} , \text{$~~~$ with $~~$} z(i) \equiv [l(i)^{\frac{\epsilon -1}{\epsilon}} + (\alpha(i) \tilde{\gamma})^{\frac{\epsilon -1}{\epsilon}}]^{\frac{\epsilon}{\epsilon -1}} \end{equation} The cost function reads as

\begin{equation} C(w_H,w_z,y(i))= w_H h(i)^{\ast} + w_z z(i)^{\ast} \label{cost2} \end{equation}

In the first stage, you minimize:

\begin{equation} \begin{array}{rrclcl} min ~~~~~{ w_H h(i) + w_z z(i)}\\ \textrm{s.t.} & z(i)^{\beta}h^{1-\beta} \ge y(i) \end{array} \end{equation}

Solving this cost minimization problem you get $h(i)^{\ast},z(i)^{\ast}$. Then, plug these conditional factor input demands in the objective function. As a result, you get the cost function (which depends upon $w_H,w_Z,y(i))$.

In the second stage, you minimize

\begin{equation} \begin{array}{rrclcl} min ~~~~~{ w_L l(i) + x(i)}\\ \textrm{s.t.} & \left[ l(i)^{\frac{\epsilon -1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon -1}{\epsilon}} \right]^{\frac{\epsilon }{\epsilon -1}} = z(i) \end{array} \end{equation}

Solving this cost minimization problem, you find $x(i)^{\ast},l(i)^{\ast}, \lambda^{\ast}$

Now, since the composite good, $z(i)$, is priced at its marginal cost, you get $w_z=\lambda^{\ast}$. Using this result into the cost function coming from the first stage, you are done, you have just found the total cost function. Divide it by $y(i)$, and you get the unit cost of production


Since we are interested in finding the unit cost let the output needed to be produced be equal to 1 i.e., $y(i)\overset{set}=1$ We need to solve: $$\begin{align} \min_{l(i),x(i),h(i)\geq0} \quad & w_Ll(i)+x(i)+w_{H}h(i)\\ \textrm{s.t.} \quad & [l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}=1\\ \textrm{and} \quad & \alpha(i)=\begin{cases}1 & \text{if } i\text{ is automated}\\ 0 & \text{otherwise}\end{cases} \end{align}$$

I am going to use the indicator function and solve for two different cases and then try to combine them using the indicator.

Case 1: If $i$ is not automated i.e., $ \alpha(i)=0$

If $i$ is not automated, then the cost minimization problem becomes a standard one with a Cobb-Douglas  production function $$\begin{align} \min_{l,x,h\geq 0} \quad & w_Ll+x+w_{H}h\\ \textrm{s.t.} \quad & l^{\beta}h^{1-\beta}=1\\ \end{align}$$

This is just a standard cost minimization problem with a cobb-douglas production function, so I think you can verify yourself that the above problem  gives: $l(i)=\left(\frac{w_H\beta}{w_L(1-\beta)}\right)^{1-\beta}, \quad x(i)=0, \quad h(i)= \left(\frac{w_L(1-\beta)}{w_{H} \beta}\right)^{\beta}$

therefore, $$\begin{eqnarray} & c=w_L^\beta\left(\frac{w_H\beta}{1-\beta}\right)^{1-\beta}+w_H^{1-\beta}\left(\frac{w_L(1-\beta)}{\beta}\right)^{\beta}\\\\ \implies & {\boxed{c=w_L^\beta w_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}}}\tag{a} \end{eqnarray}$$

Case 2: If $i$ is automated $$\begin{align} \min_{l,x,h\geq 0} \quad & w_Ll+x+w_{H}h\\ \textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h^{1-\beta}=1\\ \end{align}$$

let us write the above problem as: $$\begin{align} {\min_{0\leq h \leq 1} \underbrace{\begin{pmatrix} \underset{l,x}{\min} \quad & w_Ll+w_Hh+x \\ \textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta} \end{pmatrix}}_{\text{auxiliary problem}}} \tag{1} \end{align}$$

We can solve the above problem in two stages:

Stage 1: first solve the auxiliary problem for $l$ and $x$ holding $h$ fixed

Stage 2: Substitute the solutions of the auxiliary problem$-$ $l(h)$ and $x(h)$ $-$ into $(1)$ and solve the problem to find optimal $h$

Stage 1: we need to solve: $$\begin{align} \underset{l,x}{\min} \quad & w_Ll+w_Hh+x \\ \textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta}\\ \text{gives:} \quad & {\left.\begin{matrix} l(h)=\frac{h^\frac{\beta -1}{\beta}}{[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}} \\ x(h)=\frac{w_L^\epsilon \tilde{\gamma}^{\epsilon -1} h^\frac{\beta -1}{\beta}}{[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}} \end{matrix}\right\}}\tag{2} \end{align}$$

Because we are given that $\epsilon>1$, You can verify that isoquants will be convex. Thus, I was able to solve the above by equating ratios of MP to MC for $l$ and $x$ and substituting them into the constraint.

Stage 2: substituting $(2)$ in $(1)$ we get the final problem solving which will give optimal $h$ $$\begin{align} \min_{0\leq h\leq 1} \quad & w_Hh+w_L[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon} h^\frac{\beta -1}{\beta}\tag{3} \end{align}$$

the above objective is convex in $h \; \because \beta<1$, so assuming the parametric condition gives us a stationary point that lies in the interval $(0,1)$ we can solve for the optimal h in the above problem and substitute it in the problem itself to get $c$ for $\alpha=1$:

$$h^*=\left(\frac{(1-\beta)w_L[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon}}{w_H\beta}\right)^\beta$$

substituting $h$ in $(3)$ give us: $$\boxed{c=w_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon}}\tag{b}$$

we can combine the solutions to case 1 and case 2 , i.e., $(a)$ and $(b)$ Therefore, we have

$c=\begin{cases} w_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}w_L^\beta & \text{if } \alpha(i)=0 \\ w_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon} & \text{if } \alpha(i)=1 \end{cases}$

using $\alpha(i)$, the above can also be written as: $$\boxed{c=w_H^{1-\beta}\beta^{-\beta}(1-\beta)^{\beta-1}\left(w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1} \alpha(i) \right)^\frac{\beta}{1-\epsilon}}$$

  • $\begingroup$ Thanks so much, I'm going to check the computations in a while and see if I can get useful hints to fully answer my question. $\endgroup$
    – John M.
    May 2 at 11:33
  • $\begingroup$ consider rewarding me the bounty if you do $\endgroup$
    – mynameparv
    May 2 at 12:15
  • 1
    $\begingroup$ For sure, don't you worry about that $\endgroup$
    – John M.
    May 2 at 12:20
  • $\begingroup$ How do you get $h$? I mean $h$ just after $(3)$ $\endgroup$
    – John M.
    May 2 at 15:29
  • $\begingroup$ I think that one problem with your procedure is that you omitted $\alpha(i)$ into the constraint on $(2)$ $\endgroup$
    – John M.
    May 2 at 16:21

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