User:Alexnally/Currently Working On/Optimal Control Utility Example

Take a simplified version of the Ramsey–Cass–Koopmans model. We wish to maximize an agent's discounted lifetime utility achieved through consumption

$max\int _{0}^{\infty }e^{-\rho t}u(c(t))dt$

subject to the time evolution of capital per effective worker

${\dot {k}}={\frac {\partial k}{\partial t}}=f(k(t))-(n+\delta )k(t)-c(t)$

where $c(t)$ is period t consumption, $k(t)$ is period t capital per worker, $f(k(t))$ is period t production, $n$ is the population growth rate, $\delta$ is the capital depreciation rate, the agent discounts future utility at rate $\rho$, with $u'>0$ and $u''<0$.

Here, $k(t)$ is the state variable which evolves according to the above equation, and $c(t)$ is the control variable. The Hamiltonian becomes

$H(k,c,\mu ,t)=e^{-\rho t}u(c(t))+\mu (t){\dot {k}}=e^{-\rho t}u(c(t))+\mu (t)[f(k(t))-(n+\delta )k(t)-c(t)]$

The optimality conditions are

${\frac {\partial H}{\partial c}}=0\Rightarrow e^{-\rho t}u'(c)=\mu (t)$

${\frac {\partial H}{\partial k}}=-{\frac {\partial \mu }{\partial t}}=-{\dot {\mu }}\Rightarrow \mu (t)[f'(k)-(n+\delta )]=-{\dot {\mu }}$

If we let $u(c)=ln(c)$, then log-differentiating the first optimality condition with respect to $t$ yields

$-\rho {\frac {\dot {c}}{c(t)}}={\frac {\dot {\mu }}{\mu (t)}}$

Setting this equal to the second optimality condition yields

$\rho {\frac {\dot {c}}{c(t)}}=f'(k)-(n+\delta )$

This is the Keynes–Ramsey rule or the Euler–Lagrange equation, which gives a condition for consumption in every period which, if followed, ensures maximum lifetime utility.