In the Solow model capital per worker k_t changes each year by
where A\,k_t^{\alpha} is output per worker (0 < \alpha < 1), s the share of output saved and invested, delta the depreciation rate and n the growth rate of the workforce (all as fractions: 5 % is 0.05).
A steady state is a level k^* > 0 at which capital per worker stops changing: k_{t+1} = k_t = k^*. There, each year's new capital exactly replaces what wears out and equips the newcomers. At the steady state, output per worker is y^* = A\,(k^*)^{\alpha} and consumption per worker is what is not saved, c^* = (1 - s)\,y^*.
The planning office of the hill country of Velm wants all three numbers. Write steady_state(A, alpha, s, delta, n) that returns the tuple (k_star, y_star, c_star). If s = 0, nothing is ever invested and the only resting point is k^* = 0: return (0.0, 0.0, 0.0) (the formula you derive gives this too).
Check your answer by running a long simulation of the rule from any positive start (Run with plot): it should settle at your k^*.
Examples
Input: A = 1, alpha = 0.5, s = 0.2, delta = 0.05, n = 0.05
Output: (4.0, 2.0, 1.6)
Explanation: at k = 4, output is 2, saving is 0.2 * 2 = 0.4 and the loss is
0.1 * 4 = 0.4: they balance. Consumption is 0.8 * 2 = 1.6.
Input: A = 2, alpha = 0.3, s = 0.25, delta = 0.06, n = 0.01
Output: (16.58888101, 4.64488668, 3.48366501)
Constraints
- answers are compared with a tolerance of
1e-6; do not round
Goals
- Find the steady state by setting next year's capital equal to this year's
- Solve $s A k^{\alpha} = (\delta + n) k$ for $k$ with exponents
- Compute steady-state output and consumption per worker