<p>You can use Karush-Kuhn-Tucker conditions. As you are considering a minmisation problem, this gives:
<span class="math-container">$$
\frac{\partial F'}{\partial L_i} \ge \lambda \text{ with equality if $L_i &gt; 0$}\\
\frac{\partial F'}{\partial C_i} \ge \mu \text{ with equality if $C_i &gt; 0$}
$$</span>
where <span class="math-container">$\lambda$</span> and <span class="math-container">$\mu$</span> are the Lagrange multipliers for the adding up constraintes. <a href="https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions" rel="nofollow noreferrer">Wikipedia entry for KKT.</a></p>

CC BY-SA 4.0
2022-01-18T08:32:01.183Z
I don't understand. How do I find $\lambda$ and $\mu$? Is this just saying that the derivatives need to be positive? How do I apply this in practice, eg in the example I gave?
Jsevillamol
CC BY-SA 4.0
2022-01-18T13:26:51.240Z
@Jsevillamol Do you know Lagrangians and optimisation? In general there's no easy way to solve this except to consider all possible cases where $L_i > 0$ or $L_i = 0$ and see if you get a solution (or a contradiction). There are many resources online (e.g. Youtube) that show you how to use Kuhn-Tucker conditions.
tdm
CC BY-SA 4.0
2022-01-17T17:21:46.720Z
2022-01-18T13:22:31.117Z
2022-01-18T13:22:31.117Z
tdm
<p>I am trying to solve an allocation problem for a nested CES production function with three factors.</p>
<p>The production function we posit is:</p>
<p><span class="math-container">$$
F(K, \mathbf L, \mathbf C) = [\alpha K^\rho + \sum_{i\not\in \text{AT}} \beta_i L_i ^\rho + \sum_{i\in \text{AT}} \beta_i (L_i + \eta_i C_i)^\rho]^{\frac{1}{\rho}}
$$</span></p>
<p>where <span class="math-container">$K$</span> is capital, <span class="math-container">$\mathbf L$</span> is labour, <span class="math-container">$\mathbf C$</span> is compute, <span class="math-container">$\rho$</span> is the labour-capital substitution coefficient, <span class="math-container">$\alpha + \sum_{0 \le i \le n} \beta_i = 1$</span> are the task-specific share parameters, <span class="math-container">$\text{AT}$</span> is the set of automatable tasks and <span class="math-container">$\eta_i &lt; 1$</span> are the task-specific substitution factors between labour and compute.</p>
<p>Assuming capital and labour are gross complements (ie <span class="math-container">$\rho &lt; 0$</span>) then the problem is equivalent to minimizing
<span class="math-container">$$
F'(\mathbf L, \mathbf C) = \sum_{i} \beta_i (L_i + \eta_i C_i )^\rho
$$</span>
subject to the budget constraints</p>
<p><span class="math-container">$$
\sum_i L_i = L
$$</span>
<span class="math-container">$$\sum_{i} C_i = C$$</span>
<span class="math-container">$$C_i = 0, \text{ if } i\not\in \text{AT}$$</span></p>
<p>The point of minimization happens when the marginal returns to labour and compute on each task are equal.</p>
<p>This gives rise to the optimality conditions <span class="math-container">$\frac{\partial F'}{\partial L_i} = \frac{\partial F'}{\partial L_j}$</span> for all <span class="math-container">$i,j$</span>; and <span class="math-container">$\frac{\partial F'}{\partial C_i} = \frac{\partial F'}{\partial C_j}$</span> for <span class="math-container">$i,j \in \text{AT}$</span>.</p>
<p><span class="math-container">$$
\frac{\partial F'}{\partial L_i} = 
\beta_i \rho (L_i + \eta_i C_i)^{\rho - 1} = 
\beta_0 \rho (L_0 + \eta_0 C_0)^{\rho - 1} =
\frac{\partial F'}{\partial L_0}
$$</span></p>
<p><span class="math-container">$$
\frac{\partial F'}{\partial C_i} = 
\beta_i \rho (L_i + \eta_i C_i)^{\rho -1} \eta_i = 
\beta_0 \rho (L_0 + \eta_0 C_0)^{\rho -1} \eta_0 =
\frac{\partial F'}{\partial C_0}
$$</span></p>
<p>where we assumed that task 0 is automatable ie <span class="math-container">$0\in \text{AT}$</span>.</p>
<p>After eliminating <span class="math-container">$\rho$</span> and resolving the exponent we have that:</p>
<p><span class="math-container">$$
\frac{1}{\beta_0^{\sigma}}L_0 - \frac{1}{\beta_i^{\sigma}}L_i + 
\frac{\eta_0}{\beta_0^{\sigma}}C_0 - \frac{\eta_i}{\beta_i^{\sigma}}C_i = 0
$$</span></p>
<p><span class="math-container">$$
\frac{1}{\beta_0^{\sigma}\eta_0^{\sigma}}L_0 - \frac{1}{\beta_i^{\sigma}\eta_i^{\sigma}}L_i + 
\frac{\eta_0^{1 - \sigma}}{\beta_0^{\sigma}}C_0 - \frac{\eta_i^{1 - \sigma}}{\beta_i^{\sigma}}C_i = 0
$$</span>
where <span class="math-container">$\sigma = \frac{1}{1 - \rho}$</span> is the coefficient of elasticity of the CES production function.</p>
<p>Together with the budget conditions, these equations form a uniquely determined system of linear equations. Since <span class="math-container">$F'$</span> is convex, the solution is neccessarily the minimum.</p>
<hr />
<p>So far so good, but I wanted to check that the solution works in an example.</p>
<p>I chose <span class="math-container">$C = 1, L=1, \beta_1 = 1/3, \beta_2=2/3, \rho = -0.5, \eta_0 = 1, \text{AT} = \{0\}$</span>.</p>
<p>The solution that I get using a lineal algebra solver is <span class="math-container">$L_0 = -0.22702358, L_1 = 1.22702358, C_0 = 1, C_1 = 0$</span>.</p>
<p>So clearly I have forgotten to account for the restriction that each individual factor must be non negative. The question is, how do I incorporate that in the solution?</p>

<production-function><optimization><resource-allocation>
Deriving factor allocation of production function
Jsevillamol
# Deriving factor allocation of production function

(Asked by: username_0 on 2022-01-14)

I am trying to solve an allocation problem for a nested CES production function with three factors.

The production function we posit is:

$$
F(K, \mathbf L, \mathbf C) = [\alpha K^\rho + \sum\_{i\not\in \text{AT}} \beta\_i L\_i ^\rho + \sum\_{i\in \text{AT}} \beta\_i (L\_i + \eta\_i C\_i)^\rho]^{\frac{1}{\rho}}
$$

where $K$ is capital, $\mathbf L$ is labour, $\mathbf C$ is compute, $\rho$ is the labour-capital substitution coefficient, $\alpha + \sum\_{0 \le i \le n} \beta\_i = 1$ are the task-specific share parameters, $\text{AT}$ is the set of automatable tasks and $\eta\_i < 1$ are the task-specific substitution factors between labour and compute.

Assuming capital and labour are gross complements (ie $\rho < 0$) then the problem is equivalent to minimizing
$$
F'(\mathbf L, \mathbf C) = \sum\_{i} \beta\_i (L\_i + \eta\_i C\_i )^\rho
$$
subject to the budget constraints

$$
\sum\_i L\_i = L
$$
$$\sum\_{i} C\_i = C$$
$$C\_i = 0, \text{ if } i\not\in \text{AT}$$

The point of minimization happens when the marginal returns to labour and compute on each task are equal.

This gives rise to the optimality conditions $\frac{\partial F'}{\partial L\_i} = \frac{\partial F'}{\partial L\_j}$ for all $i,j$; and $\frac{\partial F'}{\partial C\_i} = \frac{\partial F'}{\partial C\_j}$ for $i,j \in \text{AT}$.

$$
\frac{\partial F'}{\partial L\_i} =
\beta\_i \rho (L\_i + \eta\_i C\_i)^{\rho - 1} =
\beta\_0 \rho (L\_0 + \eta\_0 C\_0)^{\rho - 1} =
\frac{\partial F'}{\partial L\_0}
$$

$$
\frac{\partial F'}{\partial C\_i} =
\beta\_i \rho (L\_i + \eta\_i C\_i)^{\rho -1} \eta\_i =
\beta\_0 \rho (L\_0 + \eta\_0 C\_0)^{\rho -1} \eta\_0 =
\frac{\partial F'}{\partial C\_0}
$$

where we assumed that task 0 is automatable ie $0\in \text{AT}$.

After eliminating $\rho$ and resolving the exponent we have that:

$$
\frac{1}{\beta\_0^{\sigma}}L\_0 - \frac{1}{\beta\_i^{\sigma}}L\_i +
\frac{\eta\_0}{\beta\_0^{\sigma}}C\_0 - \frac{\eta\_i}{\beta\_i^{\sigma}}C\_i = 0
$$

$$
\frac{1}{\beta\_0^{\sigma}\eta\_0^{\sigma}}L\_0 - \frac{1}{\beta\_i^{\sigma}\eta\_i^{\sigma}}L\_i +
\frac{\eta\_0^{1 - \sigma}}{\beta\_0^{\sigma}}C\_0 - \frac{\eta\_i^{1 - \sigma}}{\beta\_i^{\sigma}}C\_i = 0
$$
where $\sigma = \frac{1}{1 - \rho}$ is the coefficient of elasticity of the CES production function.

Together with the budget conditions, these equations form a uniquely determined system of linear equations. Since $F'$ is convex, the solution is neccessarily the minimum.

---

So far so good, but I wanted to check that the solution works in an example.

I chose $C = 1, L=1, \beta\_1 = 1/3, \beta\_2=2/3, \rho = -0.5, \eta\_0 = 1, \text{AT} = \{0\}$.

The solution that I get using a lineal algebra solver is $L\_0 = -0.22702358, L\_1 = 1.22702358, C\_0 = 1, C\_1 = 0$.

So clearly I have forgotten to account for the restriction that each individual factor must be non negative. The question is, how do I incorporate that in the solution?

**Tags:** production-function, optimization, resource-allocation


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## 1 answers

### Answer by username_1 on 2022-01-17. Score: 2

You can use Karush-Kuhn-Tucker conditions. As you are considering a minmisation problem, this gives:
$$
\frac{\partial F'}{\partial L\_i} \ge \lambda \text{ with equality if $L\_i > 0$}\\
\frac{\partial F'}{\partial C\_i} \ge \mu \text{ with equality if $C\_i > 0$}
$$
where $\lambda$ and $\mu$ are the Lagrange multipliers for the adding up constraintes. [Wikipedia entry for KKT.](https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions)

> **Comments:**
> **username_0**: I don't understand. How do I find $\lambda$ and $\mu$? Is this just saying that the derivatives need to be positive? How do I apply this in practice, eg in the example I gave?  
> **username_1**: @username_0 Do you know Lagrangians and optimisation? In general there's no easy way to solve this except to consider all possible cases where $L_i > 0$ or $L_i = 0$ and see if you get a solution (or a contradiction). There are many resources online (e.g. Youtube) that show you how to use Kuhn-Tucker conditions.  

---

