Social Planner problem: two period

The production function is $F(K_t,N_t)=AK_t^alpha N_t^{1-alpha}$ and depreciation $(delta)$ is equal to 1. The given preferences are as follows: $$U(c_1,l_1,c_2,l_2)=gamma log(c_1)+(1-gamma)log(l_1)+beta[gamma log(c_2)+(1-gamma)log(l_2)]$$

From which $gammain(0,1)$,$betain(0,1)$. We assume $forall t:l_t + n_t ≤ 1$ and all non-negativities on all variables. How would we solve the Social Planner?

Since $delta=1$ we would have the constraint for period 1: $$c_1+k_2=AK_1^alpha N_1^{1-alpha}$$, from which $$c_1=AK_1^alpha N_1^{1-alpha} -k_{2}$$.
For period 2, $$c_2=AK_2^alpha N_2^{1-alpha}$$ since we don’t have anymore capital for next period.
Now we need to find the allocation ${c_1,c_2,l_1,l_2}$ by doing the Lagrangian.

$$U(c_1,l_1,c_2,l_2)=gamma log(c_1)+(1-gamma)log(l_1)+beta[gamma log(c_2)+(1-gamma)log(l_2)]-lambda_1[c_1-AK_1^alpha N_1^{1-alpha} -k_{2}]-lambda_2[c_2-AK_2^alpha N_2^{1-alpha}]$$

Finding F.O.Cs wrt $c_1,c_2,k_2:$

{$c_1$}: $frac{gamma}{c_1}=lambda_1$

{$c_2$}: $frac{betagamma}{c_2}=lambda_2$

{$k_2$}: $lambda_2Aalpha k_2^{alpha-1}N_2^{1-alpha}=lambda_1$

Generating $frac{lambda_1}{lambda_2}$ we would get $$c_2=Aalpha k_2^{alpha-1}N_2^{1-alpha}c_1beta$$

That we substitute in the original constraint to get $$k_2=c_1alphabeta$$ to which we substitute in $$k_2+c_1=Ak_1(1-l_1)^{1-alpha}$$
Some algebra and we get $$c_1=frac{Ak_1^alpha(1-l_1)^{1-alpha}}{alphabeta+1}$$
$$c_2=k_2^{alpha}N_2^{1-alpha}$$

For $l_t$ we substitute $c_1,c_2$ in the FOCs that equal each $lambda$ respectively along with the FOCs of $l_1,l_2$ and get the following, again after the algebra:
$$l_1=frac{1-gamma}{gamma(alphabeta+1)(1-alpha)+1-gamma}$$
$$l_2=frac{1-gamma}{gamma(1-alpha)(1-alpha)+1-gamma}$$

Phew! This was hella lot of algebra and I don’t know if I made a mistake or not. I hope this is correct but feel free to point my errors.

• Define the Arrow-Debreu equilibrium for this economy. Find the equilibrium prices and quantities.

Besides the market clearing and prices and allocations conditions how do we continue from here 🙂