Answer to Question #223018 in Macroeconomics for Amoa

Question

Instructions : Answer true or false and explain your answer

(1) If an economy can raise its annual real GDP growth rate from 3.8 percent to 4.5 percent, its real GDP doubling time is reduced by 15 years.

(2) Suppose that the government passes a law requiring households to increase savings 10% above previous levels. According to Solow's growth theory, in the long run output per capita will grow less rapidly.

(3) If an economy has a real GDP doubling-time of 48 years, this will be increased to 56 years if annual GDP growth is reduced by 3.2 percentage points.

(4) If K = 3000, n = 0.02, and depreciation, δ= 0.04 and g =0.03, then investment of 320 will hold (K/AL) constant.

Answer 1

1.

If economy is initially at the level (x) and grows at 3.8 %, number of years required to double are :

We use compounding formula :

2X = X (1 + 0.038)^T  , where T = number of years to double 

2 = 1.038^T

T = 19 years approx

when economy grows at 4.5 %, years to double are :

2 = 1.045^T

T = 16 years approx 

Therefore FALSE ,Because doubling time is reduced by 3 years 

2.

According to Solow Model , in steady state output / capita grows is at a constant rate .

Only change brought about by increasing saving rate is that the steady state output/capital level is increased to an higher level but the growth rate of it is similar to what it was before .

So growth rate of output /capita is unchanged 

Hence FALSE

3.

We will use compounding formula to answer this question ,

When economy doubles in 48 years , we find the rate of growth:

2X = X(1 + r)⁴⁸

2 = (1 + r )⁴⁸ ->  

$r=\sqrt[48]{2}-1=0.0139\\=13.9\%=14\% \space approx$ 

= 13.9 % = 14% approximate

When economy doubles in 56 years , we find the rate of growth:

2X = X(1+r)⁵⁶

2 = (1 + r )⁵⁶ ->

$r=\sqrt[56]{2}-1=0.0112\\=11.2\%$

Now difference between interest rates are : 14 % - 11.2% = 3.8%

Therefore, TRUE 

4.

It is known that at steady-state, $k =\frac{K}{AL}$

K = kAL

Taking log on both sides:

$Log K = Log k + Log A + Log L$

Differentiating in terms of time t:

$\frac{dLog K }{ dt} =\frac{d Log k }{dt} +\frac{ d Log A }{ dt} + \frac{dLog L }{ dt}$

Now, to bring the equation to concise form:

$\frac{d}{dK}[log(K)].\frac{d}{dt}[K]=\frac{d}{d\bar{k}}[log(\bar{k})].\frac{d}{dA}[\bar{k}]+\frac{d}{dA}[log(A)].\frac{d}{dt}[A]+\frac{d}{dL}[log(L)].\frac{d}{dt}[L]$

Now substituting all the log derivatives and use dot notation instead:

$\frac{1}{K}K^* =\frac{1}{\bar{k}}k^*+\frac{1}{A}A^*+\frac{1}{L}L^*=\frac{k}{\bar{k}}+g+n$

$\frac{k^*}{\bar{k}}=0$

so

$\frac{K^*}{K}=g+n$

$\frac{320}{ 3000} = 0.107$ this is not equal to $(g+n = 0.03+0.02)$  

Thus, an investment of 320 will not hold (K/AL) constant.

Answer- False