<p><span class="math-container">$\delta = 0.02$</span> is depreciation.</p>
<p><span class="math-container">$p = 0.02$</span> is population growth.</p>
<p><span class="math-container">$g = 0.03$</span> is technological growth.</p>
<p><span class="math-container">$s = 0.14$</span> is the savings rate.</p>
<p><span class="math-container">$Y=0.5\cdot K^{\frac{1}{3}}\left(AN\right)^{\frac{2}{3}}$</span> is the production function.</p>
<p>The equation of motion for capital is:
<span class="math-container">$$ K_{t+1}=I_{t}+K_{t}\left(1-\delta\right)$$</span>
<span class="math-container">$$ =s \cdot Y_{t}+(1-\delta)*K_{t}$$</span></p>
<p>Normalize both sides by <span class="math-container">$A_t \cdot N_t$</span>
<span class="math-container">$$ \frac{K_{t+1}}{A_t \cdot N_t} = s \cdot \frac{Y_{t}}{A_t \cdot N_t} +(1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$</span></p>
<p>Note that <span class="math-container">$ A_{t+1} \cdot N_{t+1} = A_t \cdot N_t \cdot (1+p)\cdot(1+g)$</span></p>
<p><span class="math-container">$$ \Rightarrow \frac{K_{t+1} \cdot (1+p)\cdot(1+g)}{A_{t+1} \cdot N_{t+1}} = s \cdot \frac{Y_{t}}{A_t \cdot N_t} + (1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$</span></p>
<p><span class="math-container">$$\Rightarrow \frac{K_{t+1} \cdot (1+p)\cdot(1+g)}{A_{t+1} \cdot N_{t+1}} = s \cdot \frac{0.5\cdot K^{\frac{1}{3}}\left(A_t \cdot N_t\right)^{\frac{2}{3}}}{A_t \cdot N_t} + (1-\delta)*\frac{K_{t}}{A_t \cdot N_t}$$</span></p>
<p>Define <span class="math-container">$\ell_t = \frac{K_{t}}{A_t \cdot N_t }$</span> and <span class="math-container">$\ell_{t+1} = \frac{K_{t+1}}{A_{t+1} \cdot N_{t+1} }$</span>
this is the capital per unit of effective worker.</p>
<p><span class="math-container">$$\Rightarrow \ell_{t+1} \cdot (1+p) \cdot(1+g)= 0.5 \cdot s \cdot \ell_t^{\frac{1}{3}} + (1-\delta)*\ell_t$$</span></p>
<p>Recognize that in the steady state:
<span class="math-container">$$ \ell_{t+1} = \ell_{t}$$</span>
, meaning capital per effective unit of labor is constant.</p>
<p><span class="math-container">$$\Rightarrow \ell \cdot (1+p) \cdot(1+g)= 0.5 \cdot s \cdot \ell^{\frac{1}{3}} + (1-\delta)*\ell$$</span></p>
<p><span class="math-container">$$\Rightarrow 2 \cdot \ell \frac{(1+p) \cdot (1+g) - (1-\delta)}{s} = \ell^{\frac{1}{3}} $$</span></p>
<p><span class="math-container">$$ \Rightarrow \ell^{2/3}=\frac{s}{2\cdot[(1+p) \cdot (1+g) - (1-\delta)]}$$</span></p>
<p><span class="math-container">$$ \Rightarrow \ell= \left\{\frac{s}{2\cdot[(1+p) \cdot (1+g) - (1-\delta)]}\right\}^\frac{3}{2}$$</span></p>
<p><span class="math-container">$$ \approx \left\{\frac{s}{2\cdot[p + g + \delta]}\right\}^\frac{3}{2} $$</span></p>
<p>Which equals 1 (1 = 0.14 / (2 * (.02 + .03 + .02)), so capital per effective worker is 1.</p>
<p>Output per unit of effective worker is:</p>
<p><span class="math-container">$$ \frac{Y_t}{A_t \cdot N_t} = \frac{0.5\cdot K_t^{\frac{1}{3}}\left(A_tN_t\right)^{\frac{2}{3}}}{A_t \cdot N_t} $$</span></p>
<p><span class="math-container">$$  = \frac{0.5\cdot K_t^{\frac{1}{3}}}{\left(A_t N_t\right)^{\frac{1}{3}}} = 0.5 \cdot \left(\frac{K_t}{A_t N_t }\right)^{\frac{1}{3}}$$</span>
<span class="math-container">$$ 0.5 \cdot \ell_t^{\frac{1}{3}}  $$</span></p>
<p>So output per effective worker is 0.5 when <span class="math-container">$s=0.14$</span> and therefore <span class="math-container">$\ell=1$</span>.</p>
<p>Given this setup, we can substitute any constant savings rate we want in for <span class="math-container">$s$</span>, giving us a new value of <span class="math-container">$\ell$</span> and output per effective worker.</p>

CC BY-SA 4.0
2022-02-13T20:41:54.680Z
$A$ is the state of technology, so did you use $g$ to represent $A$ instead?
user40459
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2022-02-13T21:32:10.407Z
In hindsight that does seem like an inferior choice.
BKay
CC BY-SA 4.0
2022-02-13T23:25:13.783Z
does is matter which one you choose though? Like would using $g$ instead of $A$ change the results?
user40459
CC BY-SA 4.0
2022-02-14T01:52:56.880Z
I don't think so. I'm assuming that $A_{t+1}=A_t \cdot (1+g)$ it could just as easily be $A_{t+1}=A_t \cdot (1+a)$
BKay
CC BY-SA 4.0
2022-02-14T01:55:41.510Z
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?
user40459
CC BY-SA 4.0
2022-02-14T01:59:33.773Z
also, it doesn't matter if you chose $g$ right or $a$, because they're just variables and it doesn't change the derivation in any way?
user40459
CC BY-SA 4.0
2022-02-14T02:10:10.030Z
Also if savings rate increases then to $15$%, then I would just need to input $0.15$ instead of $0.14$ right, since $l_{t+1}=l_t$?
user40459
CC BY-SA 4.0
2022-02-13T20:07:00.470Z
2022-02-14T11:01:35.987Z
2022-02-14T11:01:35.987Z
BKay
<p>I'm having trouble with the steady-state savings rate type of problems.</p>
<p>Here is the problem I'm stuck on:</p>
<p>The production is <span class="math-container">$Y = 0.5*K^{1/3}(AN)^{2/3}$</span>.</p>
<ol>
<li><p>If savings is <span class="math-container">$s$</span>%, what are the steady-state values of capital per unit of effective worker and output per unit of effective worker?</p>
</li>
<li><p>Now, suppose the savings rate increases to <span class="math-container">$s_1$</span>% from <span class="math-container">$s$</span>%, what will the capital per unit of effective worker be one year after the change in savings rate?</p>
</li>
</ol>
<p>Here's what I'm thinking:</p>
<p><span class="math-container">$A$</span> is the state of technology, so <span class="math-container">$AN$</span> is the amount of effective labor. And, output per effective worker is a function of capital per effective worker: <span class="math-container">$Y/(AN) = f(K/(AN))$</span>.</p>
<p>My solving attempt:</p>
<p>Want <span class="math-container">$K/(AN)^* = f(Y/(AN))^*$</span>. The <span class="math-container">$Y$</span> function is cobb-douglas. In a steady-state, saving per worker must be equal to depreciation per worker.</p>
<p>At steady state, <span class="math-container">$K_{t+1}/AN -K_t/AN = s(K_t/AN)^{1/3}-(K_t/AN)$</span></p>
<p>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.</p>
<p>So, from the formula I derived, capital per worker is <span class="math-container">$K^*/N=(S/)^3$</span></p>
<p>So using that, I get <span class="math-container">$K^*/AN=(14/2)^3=343$</span> So, <span class="math-container">$K^*=AN(343)=4(343)=1372$</span></p>
<p>This seems off...</p>
<p>And, steady-state of output per worker is <span class="math-container">$Y^*/AN=(K^*/AN)^{1/3}=(S/)^3=S/$</span>.</p>
<p>So using this, formula, <span class="math-container">$S/=14/2=7$</span>%</p>
<p>In the long run, does this mean output per worker doubles when the saving rate doubles?</p>
<p>Now, looking at the savings rate increase to <span class="math-container">$15$</span>%, capital per unit of effective worker after one year will be given by <span class="math-container">$K_{t+1}$</span>? I'm not too sure how to set this increase in percentage problem up.</p>

CC BY-SA 4.0
2022-02-13T17:48:55.080Z
Can you show us your attempt at finding the solution?
1muflon1
CC BY-SA 4.0
2022-02-13T17:57:40.010Z
@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
user40459
CC BY-SA 4.0
2022-02-13T18:01:18.650Z
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
1muflon1
CC BY-SA 4.0
2022-02-13T18:09:34.287Z
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...
Giskard
CC BY-SA 4.0
2022-02-13T18:14:29.647Z
@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
user40459
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2022-02-13T18:38:22.230Z
@1muflon1 I put it up, please provide me with any feedback you may have
user40459
CC BY-SA 4.0
2022-02-13T18:38:45.637Z
@Giskard I put up my solution, please let me know if I'm doing it right
user40459
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2022-02-13T18:49:59.817Z
How come you first write about $K/AN$ but then use $K/N$ in your calculations?
Giskard
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2022-02-13T18:50:39.480Z
Also, how do you derive "the formula"?
Giskard
CC BY-SA 4.0
2022-02-13T18:51:50.500Z
@Giskard is the formula wrong? I tried to derive it following steps for a function with power $1/2$ instead of $1/3$.
user40459
CC BY-SA 4.0
2022-02-13T19:21:17.683Z
Why do you think A and N do not have time subscripts?
erik
<macroeconomics><mathematical-economics><interest-rate><productivity><steady-state>
Steady-state savings rate
user40459
# Steady-state savings rate

(Asked by: username_0 on 2022-02-13)

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, $K\_{t+1}/AN -K\_t/AN = s(K\_t/AN)^{1/3}-(K\_t/AN)$

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 $K^\*/N=(S/)^3$

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 $Y^\*/AN=(K^\*/AN)^{1/3}=(S/)^3=S/$.

So using this, formula, $S/=14/2=7$%

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.

**Tags:** macroeconomics, mathematical-economics, interest-rate, productivity, steady-state
> **Comments:**
> **username_1**: Can you show us your attempt at finding the solution?  
> **username_0**: @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  
> **username_1**: 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  
> **username_2**: 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...  
> **username_0**: @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  
> **username_0**: @1muflon1 I put it up, please provide me with any feedback you may have  
> **username_0**: @Giskard I put up my solution, please let me know if I'm doing it right  
> **username_2**: How come you first write about $K/AN$ but then use $K/N$ in your calculations?  
> **username_2**: Also, how do you derive "the formula"?  
> **username_0**: @Giskard is the formula wrong? I tried to derive it following steps for a function with power $1/2$ instead of $1/3$.  
> **username_3**: Why do you think A and N do not have time subscripts?  

---

## 1 answers

### Answer by username_4 on 2022-02-13. Score: 3

$\delta = 0.02$ 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.

> **Comments:**
> **username_0**: $A$ is the state of technology, so did you use $g$ to represent $A$ instead?  
> **username_4**: In hindsight that does seem like an inferior choice.  
> **username_0**: does is matter which one you choose though? Like would using $g$ instead of $A$ change the results?  
> **username_4**: I don't think so. I'm assuming that $A_{t+1}=A_t \cdot (1+g)$ it could just as easily be $A_{t+1}=A_t \cdot (1+a)$  
> **username_0**: 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?  
> **username_0**: also, it doesn't matter if you chose $g$ right or $a$, because they're just variables and it doesn't change the derivation in any way?  
> **username_0**: Also if savings rate increases then to $15$%, then I would just need to input $0.15$ instead of $0.14$ right, since $l_{t+1}=l_t$?  

---

