3
$\begingroup$

I'm having trouble with the steady-state savings rate type of problems.

Here is the problem I'm stuck on:

The production is $Y = 0.5*K^{1/3}(AN)^{2/3}$.

  1. If savings is $s$%, what are the steady-state values of capital per unit of effective worker and output per unit of effective worker?

  2. Now, suppose the savings rate increases to $s_1$% from $s$%, what will the capital per unit of effective worker be one year after the change in savings rate?

Here's what I'm thinking:

$A$ is the state of technology, so $AN$ is the amount of effective labor. And, output per effective worker is a function of capital per effective worker: $Y/(AN) = f(K/(AN))$.

My solving attempt:

Want $K/(AN)^* = f(Y/(AN))^*$. The $Y$ function is cobb-douglas. In a steady-state, saving per worker must be equal to depreciation per worker.

At steady state, $K_{t+1}/AN -K_t/AN = s(K_t/AN)^{1/3}-𝛿(K_t/AN)$

I'm not sure if that's the correct formula and if I derived it correctly. This should describe the evolution of capital over time.

So, from the formula I derived, capital per worker is $K^*/N=(S/𝛿)^3$

So using that, I get $K^*/AN=(14/2)^3=343$ So, $K^*=AN(343)=4(343)=1372$

This seems off...

And, steady-state of output per worker is $Y^*/AN=(K^*/AN)^{1/3}=(S/𝛿)^3=S/𝛿$.

So using this, formula, $S/𝛿=14/2=7$%

In the long run, does this mean output per worker doubles when the saving rate doubles?

Now, looking at the savings rate increase to $15$%, capital per unit of effective worker after one year will be given by $K_{t+1}$? I'm not too sure how to set this increase in percentage problem up.

$\endgroup$
11
  • $\begingroup$ Can you show us your attempt at finding the solution? $\endgroup$
    – 1muflon1
    Feb 13, 2022 at 17:48
  • $\begingroup$ @1muflon1 I don't really know how to get is started. The explanation I have below the question is what I came up with so far $\endgroup$
    – user40459
    Feb 13, 2022 at 17:57
  • $\begingroup$ on this site we have rule that homework/self-study questions need to showcase attempt at solution. The attempt does not need to be correct but you should at least try $\endgroup$
    – 1muflon1
    Feb 13, 2022 at 18:01
  • $\begingroup$ It seems like this is not "what I'm thinking" but rather "what is in my lecture notes". Spend some time trying to apply the concepts, ask help from a peer or consult with your professor. You can probably find similar problems on this very site if you put some effort into it... $\endgroup$
    – Giskard
    Feb 13, 2022 at 18:09
  • $\begingroup$ @1muflon1 Okay, I'll put up the solution I tried, I didn't want to post it because I think it's wrong, but I get what you're saying and I'll put up what I tried $\endgroup$
    – user40459
    Feb 13, 2022 at 18:14

1 Answer 1

3
$\begingroup$

$\delta = 0.02$ is depreciation.

$p = 0.02$ is population growth.

$g = 0.03$ is technological growth.

$s = 0.14$ is the savings rate.

$Y=0.5\cdot K^{\frac{1}{3}}\left(AN\right)^{\frac{2}{3}}$ is the production function.

The equation of motion for capital is: $$ K_{t+1}=I_{t}+K_{t}\left(1-\delta\right)$$ $$ =s \cdot Y_{t}+(1-\delta)*K_{t}$$

Normalize both sides by $A_t \cdot N_t$ $$ \frac{K_{t+1}}{A_t \cdot N_t} = s \cdot \frac{Y_{t}}{A_t \cdot N_t} +(1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$

Note that $ A_{t+1} \cdot N_{t+1} = A_t \cdot N_t \cdot (1+p)\cdot(1+g)$

$$ \Rightarrow \frac{K_{t+1} \cdot (1+p)\cdot(1+g)}{A_{t+1} \cdot N_{t+1}} = s \cdot \frac{Y_{t}}{A_t \cdot N_t} + (1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$

$$\Rightarrow \frac{K_{t+1} \cdot (1+p)\cdot(1+g)}{A_{t+1} \cdot N_{t+1}} = s \cdot \frac{0.5\cdot K^{\frac{1}{3}}\left(A_t \cdot N_t\right)^{\frac{2}{3}}}{A_t \cdot N_t} + (1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$

Define $\ell_t = \frac{K_{t}}{A_t \cdot N_t }$ and $\ell_{t+1} = \frac{K_{t+1}}{A_{t+1} \cdot N_{t+1} }$ this is the capital per unit of effective worker.

$$\Rightarrow \ell_{t+1} \cdot (1+p) \cdot(1+g)= 0.5 \cdot s \cdot \ell_t^{\frac{1}{3}} + (1-\delta)*\ell_t$$

Recognize that in the steady state: $$ \ell_{t+1} = \ell_{t}$$ , meaning capital per effective unit of labor is constant.

$$\Rightarrow \ell \cdot (1+p) \cdot(1+g)= 0.5 \cdot s \cdot \ell^{\frac{1}{3}} + (1-\delta)*\ell$$

$$\Rightarrow 2 \cdot \ell \frac{(1+p) \cdot (1+g) - (1-\delta)}{s} = \ell^{\frac{1}{3}} $$

$$ \Rightarrow \ell^{2/3}=\frac{s}{2\cdot[(1+p) \cdot (1+g) - (1-\delta)]}$$

$$ \Rightarrow \ell= \left\{\frac{s}{2\cdot[(1+p) \cdot (1+g) - (1-\delta)]}\right\}^\frac{3}{2}$$

$$ \approx \left\{\frac{s}{2\cdot[p + g + \delta]}\right\}^\frac{3}{2} $$

Which equals 1 (1 = 0.14 / (2 * (.02 + .03 + .02)), so capital per effective worker is 1.

Output per unit of effective worker is:

$$ \frac{Y_t}{A_t \cdot N_t} = \frac{0.5\cdot K_t^{\frac{1}{3}}\left(A_tN_t\right)^{\frac{2}{3}}}{A_t \cdot N_t} $$

$$ = \frac{0.5\cdot K_t^{\frac{1}{3}}}{\left(A_t N_t\right)^{\frac{1}{3}}} = 0.5 \cdot \left(\frac{K_t}{A_t N_t }\right)^{\frac{1}{3}}$$ $$ 0.5 \cdot \ell_t^{\frac{1}{3}} $$

So output per effective worker is 0.5 when $s=0.14$ and therefore $\ell=1$.

Given this setup, we can substitute any constant savings rate we want in for $s$, giving us a new value of $\ell$ and output per effective worker.

$\endgroup$
7
  • $\begingroup$ $A$ is the state of technology, so did you use $g$ to represent $A$ instead? $\endgroup$
    – user40459
    Feb 13, 2022 at 20:41
  • $\begingroup$ In hindsight that does seem like an inferior choice. $\endgroup$
    – BKay
    Feb 13, 2022 at 21:32
  • $\begingroup$ does is matter which one you choose though? Like would using $g$ instead of $A$ change the results? $\endgroup$
    – user40459
    Feb 13, 2022 at 23:25
  • $\begingroup$ I don't think so. I'm assuming that $A_{t+1}=A_t \cdot (1+g)$ it could just as easily be $A_{t+1}=A_t \cdot (1+a)$ $\endgroup$
    – BKay
    Feb 14, 2022 at 1:52
  • $\begingroup$ I'm trying to do this on my own, the deriving part, but I get stuck at the last part where $0.5*l^{1/3}_t$. Where does this $0.5$ come from? Is this relation of $0.5$ and $l^{1/3}_t$ somehow connected to production function? $\endgroup$
    – user40459
    Feb 14, 2022 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.