Question

Slow model:

A flood destroys half population and capital. Draw four graphs plotting Kt, Nt, kt, and Yt against time. Begin at steady state, Be prepaid with distance, times, and movements, using words to explain.

Answer 1

Consider the given problem here the population is growing at the rate “n”, => the time path of “Nt” is given by, “Nt = (1+n)^t*N0”, here “N0” is constant. Now, let’s assume that the production function is given by, Yt = Kt^a*Nt^1-a, => yt = kt^a, here “yt=output per worker” and “kt=capital stock per worker”. Now, in the steady state equilibrium the change in the capital stock per worker is “0”, => “kt+1=kt”.

Now, let’s assume that at “t=0”the economy is in steady state equilibrium and the corresponding steady state “kt=k0=constant” and the steady state output per worker is given by “yt=y0=constant”, the corresponding labor is “Nt=N0” and capital stock is “Kt=K0=k0*N0”.

Now, since “K” and “N” destroys by half of their initial level, => “k=K/N” remain same and “y” also remain same as we know that there are direct relationship between “y” and “k”. So, here the steady state “k” and “y” will not be affected. Consider the following fig showing level of “k” over time.

Now, as we know that “N” is growing at the rate “n”, => the time path of “Nt” is exponential function and is given by, “Nt=(1+n)^t*N0”. Now, as the “N” reduce by half, => there will be a break in the function. So, in the following fig, at “t=t0”, the “Nt=N0”, now it reduced by half, => if decreases “A1” to “B1” then further it increases.

Now, “Kt” is the product of “kt” and “Nt”. Now, at “t=0”, “K0=k0*N0” and “k0” is constant and “N0” decreases by half. So, here also “Kt” decreases to “B2” from “A2” and then it growing at the same rate as “Nt”.

Now, “Yt” is the product of “yt” and “Nt”. Now, at “t=0”, “Y0=y0*N0” and “y0” is constant and “N0” decreases by half. So, here also “Yt” decreases to “B3” from “A3” and then it growing at the same rate as “Nt”.