Chapter 6, Problem 6NP

a

To determine

To calculate:Steady state values of Output per worker, Capital per worker, Consumption per worker, Investment per worker

Answer to Problem 6NP

the steady-state value of the capital per worker is $k=16$ .

the steady-state value of output per worker is $y=24$ .

the steady-state value of consumption per worker is $c=21.6$ .

Explanation of Solution

Given Information:

Households save 10% of income, so savings, $s=0.1$ . The population growth rate is given as $n=0.01$ and the depreciation rate is given as $d=0.14$ .

At steady stae:

Output per worker

Capital per worker

Consumption per worker are equal.

To determine the steady-state value of the capital per worker, use the equation $sf(k)=(n+d)k$ .

Households save 10% of income, so savings, $s=0.1$ . The population growth rate is given as $n=0.01$ and the depreciation rate is given as $d=0.14$ .

Substitute the given values $s=0.1$ , $y_t=f(k_t)=6\sqrt{k}$ into the equation and solve for k.

  $sf(k)=(n+d)k_t$

  $0.1\times 6\sqrt{k}=(0.01+0.14)k$

  $4\sqrt{k}=k$

Continue solving for k by dividing both sides by $\sqrt{k}$ .

  $\frac{4\sqrt{k}}{\sqrt{k}}=\frac{k}{\sqrt{k}}$

  $4=\sqrt{k}$

  $4^2={(\sqrt{k})}^2$

  $16=k$

Therefore, the steady-state value of the capital per worker is $k=16$ .

To find the steady-state value of output per worker, substitute the calculated value for capital-labor ratio from the previous step into the given per-worker production function, $y=6\sqrt{k}$ .

  $y=6\sqrt{k}$

  $=6\sqrt{16}$

  $=24$

Hence, the steady-state value of output per worker is $y=24$ .

To find the steady-state value of consumption per worker, use the equation $c=y-(n+d)k$ , substituting the values of steady-state output per worker from the previous step, $y$ , steady-state value of the capitaI-labor ratio $k$ and the given values of population growth rate and depreciation rate,

  $n=0.01$ and $d=0.14$

  $c=y-(n+d)k$

  $c=24-(0.01+0.14)(16)$

  $c=21.6$

Therefore, the steady-state value of consumption per worker is $c=21.6$ .

To fund the steady-state value of investment per worker, use the equation $I=(n+d)k$ , substituting the values of steady-state output per worker from above, $y$ , steady-state value of the capital-labor ratio $k$ and the given values of population growth rate and depreciation rate, $n=0.01$ and $d=0.14$ .

  $I=(n+d)k$

  $=(0.01+0.14)(16)$

  $=2.4$

Hence, the steady-state value of investment per worker is $I=2.4$ .

Economics Concept Introduction

Introduction:

Steady state is a situation at which investment is equal to depreciation. It implies all the investment done is used to replace the depreciating capital.

b)

To determine

Value of capital-labor ratio required to double steady state value of output per capita, Amount of savings to achieve output per worker.

Answer to Problem 6NP

The steady-state value of capital-labor ratio needed is $k=64$ .

Households would need to save $20\%$ of income.

Explanation of Solution

Given Information:

Households save 10% of income, so savings, $s=0.1$ . The population growth rate is given as $n=0.01$ and the depreciation rate is given as $d=0.14$ .

To determine the steady-state value of the capitallabor ratio needed to double the steady-state value of output per capita, double the value for y in part a, $y=24$ and set it equal to the production function, $y=6\sqrt{k}$ , so $48=6\sqrt{k}$ . Solve for k.

  $48=6\sqrt{k}$

  $\frac{48}{6}=\frac{6\sqrt{k}}{6}$

  $64=k$

So to double the steady-state value of output per capita, the steady-state value of capital-labor ratio needed is $k=64$ .

To determine the fraction of income households would need to save to have a double steady-state value of output per capita than in part a, use the steady-state value of capitaI-Iabor ratio from the previous step, $k=64$ . Substitute this value, double the value of output, $y=48$ and the values of population growth rate and depreciation rate from part a., $n=0.01$ and $d=0.14$ in the equation $sf(k)=(n+d)k$ .

Solve for s.

  $sf(k)=(n+d)k$

  $s(48)=(0.01+0.14)(64)$

  $48s=(0.15)(64)$

  $48s=9.6$

Continue solving for $s$ ,

  $48s=9.6$

  $\frac{48s}{48}=\frac{9.6}{48}$

  $s=0.2$

So households would need to save $20\%$ of income.

Economics Concept Introduction

Introduction:

Steady state is a situation at which investment is equal to depreciation. It implies all the investment done is used to replace the depreciating capital.