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.

  • $\begingroup$ 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


$\delta = 0.02$ 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.

  • $\begingroup$ $A$ 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

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