Question

Consider an economy in which the production function is given by Y = 6K1/2N1/2.(a) Derive the per-worker production function: Y/N = f(K/N). (Hint: A short-cut to get the per-worker production function is to keep the constant, write the ratio (K/N) and then raise the ratio by the same exponent on capital (K).Assume that the depreciation rate is 15% per year and the savings rate is 10% per year.(a) Solve for the steady-state level of capital per worker (K/N).(b) Solve for the steady-state level of output per worker (Y/N)(c) Solve of the steady-state level of consumption per worker (C/N)(d) Solve for the steady-state level of investment per worker (I/N)(e) Suppose that thedepreciationrate in the economy decreases to 12%, while the savingsrate remains constant at 10%Recalculate the steady state values of (K/N), (Y/N), (C/N)and (I/N)(f)Graphically illustrate the effect of a decrease of the depreciationrate on the steady-state level of capital. (Hint: The graph will have the investment per worker (savings per worker) curve and the depreciation curve).

Answer 1

Production function is given by Y = 6 K1/2 N1/2 ---------------- (1)

Where Y = Output, K = Capital Stock, N = Labor Units

a) Production function expressed in terms of Output per worker is known as per worker production function. In this we divide the production function Y = f(K,N) by N units, that gives :

y = f (k)   {such that, Y/N = y = output per worker ; K/N = k = capital per worker}  

Dividing both sides of the production function in equation (1) by N, we get :

Y/N = 6 K1/2 N1/2 / N

y = 6 K1/2 N-1/2  

y = 6 (K/N) 1/2

y = 6 (k)1/2 , which is the per worker production function.

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a) Given δ = depreciation rate = 15 % (0.15) and saving rate = s = 10% (0.10),

As per the solow growth model, change in capital stock is given by by the difference of the amount of investment and amount of depreciation :

sf(k) - δk  

In the above equation, Capital growth, sf(k) = Investment,  δk = Depreciation amount.

At steady state, capital growth = 0, which means sf(k) = δk -------------------- (2)

Using the given values and the derived output per worker production function in equation (2) :

0.10 * 6(k)1/2 = 0.15 * k

Solving for k*, which is the steady state level of capital we get,

0.6(k)1/2 = 0.15 * k

0.6/0.15 = k / k1/2

4 = k1/2

Squaring both sides, we get k* = 16

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b) Using k* , now we can find steady state output per worker y* from equation (1) :

y* = 6 (k*)1/2

y* = 6 (16)1/2

y* = 6 * 4 = 24

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c ) In an economy, out of their income every individual save a fraction "s" and consume a fraction "1-s", thus the consumption per worker at steady state level is given by:

c* = (1 - s)y*

c* = (1 - 0.10) * 24

c* = 0.9 * 24 = 21.6

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d ) Steady state level of investment per worker is given by :

sf(k) = 0.10 * 6 (k*)1/2 = 0.6 * (16)1/2 = 0.6 * 4 = 2.4

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e ) If depreciation dicreases to δ1 =12% and savings rate remains the same:

(we use the same output per worker production function as above and follow the same process and formulas to calculate all the steady states)

• Let the new Steady state capital per worker be k1* such that ;

sf(k1*) = δ(k1*)

0.10 * 6 (k1*)1/2 = 0.12 (k1*)

0.6 (k1*)1/2 = 0.12 (k1*)

0.6 / 0.12 = k1* / (k1*)1/2

5 = (k1*)1/2

Thus squaring both sides we get  k1* = 25  

• New Steady state level of output per worker " y1* " can be calculated as :

y1* = 6 (k1*)1/2

y1* = 6 (25)1/2

y1* = 6 * 5 = 30

• New Steady state level of consumption per worker " c1* " can be calculated as :

c1* = (1 - s)y1*

c1* = (1 - 0.10) * 30

c1* = 0.9 * 30

c1* = 27

• New Steady state level of investment per worker can be calculated as :

sf(k1*) = 0.10 * 6 (k1*)1/2  = 0.10 * 30 = 3

Thus a fall in the rate of depreciation, leads to a rise in the steady state level of capital, output, consumption and investment.

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e ) Graphical representation of a fall in depreciation

In the given below diagram,

y = f(k) = output per worker production function

i = sf(k) is the Investment curve

δk = Initial depriciation curve

k* = Initial Steady State level of capital, where investment is equal to depriciation. At this point, the initial deprecitaion curve and the investment curve intersects each other at Point A. At this point, the output per worker is y*, represented on Y axis.

As there is a fall in depreciation rate from δ to δ1, there is a fall in the slope of the depreciation curve which makes it rotate towards right. The new depriciattion curve is given by δ1(k).

The new depreciation curve intersects the investment curve at point B. There is a rise in steady stae level of capital to K1* and steady state level of income to y1*.