Solow Model with Exogenous Technological Progress

Now assume that there is an exogenous rate of technological progress of 2% per year. To capture this, we need to slightly modify the production function as follows: \[Y(t)=10[K(t)]^{0.5}[T(t)L(t)]^{0.5}\]

In the equation above, \(T(t)\), is the state of technology and measures how “effective” our workers are. To capture exogenous technological progress, assume that in our initial period, \(T(0)=10\) and that \(T(t)\) grows exogenously by 2% every year for every country.

Assume that both \(North\) and \(South\) have a 10% savings rate and that initial capital stocks and all other exogenous parameters are the same as in problem 3 (\(n = 0.02\); \(\delta = 0.02\)).

Part A

What are the steady state levels of capital per effective worker and income per effective worker?

\(\Delta k_{e}(t)=sy_{e}(t)-(\delta +n+\theta)k_{e}(t)\) \(E(t)=T(t)L(t); k_{e}(t)=\frac{K(t)}{E(t)}; y_{e}(t)=\frac{Y(t)}{E(t)}\) \(\frac{Y(t)}{E(t)}=y_{e}(t)=\frac{10[K(t)]^{0.5}}{[T(t)L(t)]^{0.5}}=\frac{10[k_{e}(t)E(t)]^{0.5}}{E(t)^{0.5}}=10[k_{e}(t)]^{0.5} \equiv k_{e}(t)=[\frac{y_{e}(t)}{10}]^{2}\) \(\Delta k_{e}(t)=sy_{e}(t)-(\delta +n+\theta)k_{e}(t)=0 \equiv s10[k_{e}(t)]^{0.5}=(\delta +n+\theta)k_{e}(t) \equiv (0.10)10[k_{e}(t)]^{0.5}=(0.02+0.02+0.02)k_{e}(t)\equiv [k_{e}(t)]^{0.5}=(0.06)k_{e}(t) \equiv \frac{1}{0.06}=[k_{e}(t)]^{0.5} \equiv k_{e}(t)=(\frac{1}{0.06})^{2}=277.7778\) \(y_{e}(t)=10[277.7778]^{0.5}=166.6667\)

Part B

On one graph plot the per capita income levels for the two countries over 200 years under exogenous technological change. Title this graph “Figure 4A”. On a separate graph, plot the per-capita income growth rates over the same time period for both countries. Title this graph “Figure 4B”.