How do we solve for initial consumption in the Ramsey model? How do we solve for the initial consumption level in the Ramsey model?
Assume we have the Euler equation $$\frac {\dot c_t}{c_t}=r_t-\rho$$
Then $$c_t=c_0e^{R_t-\rho t}$$
Putting this into the lifetime budget constraint:
$$\int_0^\infty e^{-R_t}(w_t-c_t)dt=k_0$$
gives $$\int_0^\infty e^{-R_t}w_t dt-\frac {c_0} {\rho}=-k_0$$
Therefore $$c_0=\rho\left(\int_0^\infty e^{-R_t}w_t dt+k_0\right)$$
Where we know that $w_t=f(k_t)-k_t r_t$ and that $r_t=f'(k_t)$, and $R_t=\int_0^t r_\tau d\tau$.
However, the problem is that the initial amount of consumption will influence the path of $k_t$, because $\dot k_t=f(k)-c_t$, and therefore of $w_t$ and $r_t$ as well. Therefore it seems to me that solving for $c_0$ is almost impossible here. 
So let's say we would have a computer that could help us solve this, how would we do it? e.g. how does Dynare/matlab do it? 
 A: An algorithm to solve for equilibrium in a steady state (so that $\dot k_t=0$) is as follows:


*

*Guess $k^*$, the steady state level of capital;

*Compute steady state prices, $r^*$ and $w^*$;

*Compute steady state consumption, $c^*$;

*Check the market clearing condition $f(k^*)-c^*=0$;

*Adjust your guess for $k^*$ and repeat until $f(k^*)-c^*$ is close enough to zero.


Under usual assumptions, the equilibrium is unique so the algorithm will converge to the only root of the market clearing condition. 
Out of steady state, the procedure is similar. You should first obtain the analogous formulas in discrete time and you have to start from some $k_0$ which is set exogenously. Then, compute $k^*$ following the algorithm above and guess a path for $\{k_t\}_{t=0}^\infty$ which converges to $k^*$ after $T$ periods. Compute prices for every period, consumption for every period, then check the market clearing at every period and use it to adjust the guess for $\{k_t\}_{t=0}^\infty$. Once this converges, increase $T$ and repeat until the path stops changing. 
