<blockquote>
<p>Are you sure that equation (4)
is correct? In absence of depreciation we should have ΔK=Kt+1−Kt=It
. An equation as (4)
implies that the capital stock of the preceding period Kt
is 0
, so that in each period the existing stock of capital is destroyed. It could be, even if it s unusual.</p>
</blockquote>
<p>It must be that the depreciation rate is 100% indeed.</p>

CC BY-SA 4.0
2023-04-10T18:01:53.353Z
Yes, it is the same, capital 'disappears' for some reason. It is a possible assumption, all models are not realistic to some extent. Just, to answer the question, one must be sure that the text is correct and there aren't typos.
BakerStreet
CC BY-SA 4.0
2023-04-10T18:04:48.933Z
I thought so but I'm not sure if it's right, or how to go about solving it.
sffffii
CC BY-SA 4.0
2023-04-10T18:06:32.270Z
Looking  for a solution is possible even if $(4)$ is correct.
BakerStreet
CC BY-SA 4.0
2023-04-10T18:08:05.053Z
@BakerStreet How would I go about it?
sffffii
CC BY-SA 4.0
2023-04-10T18:09:06.413Z
I write my suggestions in an answer
BakerStreet
CC BY-SA 4.0
2023-04-10T18:19:03.427Z
Why the downvote? It is just an observation, correct, a possible formalization, it is not wrong.
BakerStreet
CC BY-SA 4.0
2023-04-10T17:53:46.533Z
2023-04-10T17:53:46.533Z
TeX_novice
<p>To find the <strong>rate of growth  of capital</strong> just subtract <span class="math-container">$K_t$</span> from both sides of <span class="math-container">$(4)$</span>, to obtain:</p>
<p><span class="math-container">$$\Delta K\equiv K_{t+1}-K_t= I_t-K_t \qquad (1)$$</span></p>
<p>from which, recalling equality of investment and saving</p>
<p><span class="math-container">$$\frac {\Delta K}{K_t}= \frac {I_t-K_t}{K_t}   =\frac {sA{K_t}^{1/2}- K_t}{K_t}\qquad (2)$$</span> .</p>
<p>You can find the <strong>steady state value of capital and output</strong>, defined as a state in which the stock of capital is constant,  setting the preceding equation equal to <span class="math-container">$0$</span>:</p>
<p><span class="math-container">$$\frac {\Delta K}{K_t}= \frac {sA{K_t}^{1/2}- K_t}{K_t}=0\qquad (3)$$</span></p>
<p>From <span class="math-container">$(3)$</span> you obtain the steady state value of <span class="math-container">$K_t$</span> and then you can derive the steady state value of output plugging this value of <span class="math-container">$K_t$</span>  into the production function as usual.</p>

CC BY-SA 4.0
2023-04-10T18:15:01.573Z
Thats very helpful, thanks! In the second line, why didn't you divide by the Kt to cancel out Kt/Kt and reduce Kt^1/2 to Kt^-1/2?
sffffii
CC BY-SA 4.0
2023-04-10T18:17:13.650Z
Yes, you can do it, it is the same. You can do the steps you prefer to find the solution.
BakerStreet
CC BY-SA 4.0
2023-04-10T18:18:06.190Z
Much appreciated, thanks again.
sffffii
CC BY-SA 4.0
2023-04-10T18:19:27.263Z
You are welcome!
BakerStreet
CC BY-SA 4.0
2023-04-10T18:47:51.397Z
Hey, one more question. How would I go about deriving the saving rate that maximises the steady state consumption?
sffffii
CC BY-SA 4.0
2023-04-10T20:53:18.787Z
Find the steady state consumption using equation $(3)$, in terms of the saving rate $s$, then maximize it with respect to $s$.
BakerStreet
CC BY-SA 4.0
2023-04-10T21:27:30.650Z
So I simply replace the LHS with the value of Y*(steady state), It with sYt, then substitute the Yt with the steady state value of Y?

Then differentiate with respect to s?
sffffii
CC BY-SA 4.0
2023-04-10T21:33:01.347Z
Yes, if I understand correctly what you said, I don't understand what is LHS.  From $(3)$ you have $C_t=Y_t-I_t$, then  sostitute saving for $I_t$, all considered at their steady state vaues.
BakerStreet
CC BY-SA 4.0
2023-04-10T21:35:18.970Z
LHS = Left hand side but yes that's what I referring to, many thanks again
sffffii
CC BY-SA 4.0
2023-04-10T21:35:54.723Z
You are welcome!
BakerStreet
CC BY-SA 4.0
2023-04-10T18:10:42.583Z
2023-04-10T18:10:42.583Z
BakerStreet
<p>I've been stuck on an exam question regarding long run growth</p>
<p>Consider the following model of economic growth</p>
<p><a href="https://i.sstatic.net/DCnmB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/DCnmB.png" alt="enter image description here" /></a></p>
<p>A) Derive the equilibrium growth rate of capital, and hence find the steady-state values
of capital and output.</p>
<p>If there is no <span class="math-container">$K_t$</span> term in (4) how does one find the growth rate? Since the growth is found using <span class="math-container">$\Delta K/K_t \equiv \frac {Kt+1 - Kt} {Kt}$</span>.</p>

CC BY-SA 4.0
2023-04-10T17:07:40.113Z
Are you sure that equation $(4)$ is correct? In absence of depreciation we should have $\Delta  K= K_{t+1}-K_t=I_t$. An equation as $(4)$ implies that the capital stock of the preceding period $K_t$ is $0$, so that in each period the existing stock of capital is destroyed. It could be, even if it s  unusual.
BakerStreet
CC BY-SA 4.0
2023-04-10T17:52:02.967Z
Moreover, how is the steady state defined here? In the usual Solow model it is the state in which the capital-labor ratio remains constant, but here we haven’t any information about population growth and per capita variables.  I suppose $\Delta K=0$
BakerStreet
CC BY-SA 4.0
2023-04-10T18:04:21.390Z
@BakerStreet Yes I copied it straight from the past paper, it's unusual as no other questions given to me have depreciation and/or Kt missing. 

The steady state is defined as DeltaKt+1 = 0.
sffffii
CC BY-SA 4.0
2023-04-10T18:08:30.153Z
Ok, the absence of $K_t$ is not a problem, I give my suggestions in an answer below.
BakerStreet
CC BY-SA 4.0
2023-04-11T18:27:20.830Z
Those equations could perhaps model a primitive agricultural economy with seed retained from one year to the next as the only capital, also output less than proportional to seed input because of a limited supply of land and/or labour.
Adam Bailey
<macroeconomics>
Solow long run growth model question
sffffii
# Solow long run growth model question

(Asked by: username_0 on 2023-04-10)

I've been stuck on an exam question regarding long run growth

Consider the following model of economic growth

[![enter image description here](https://i.sstatic.net/DCnmB.png)](https://i.sstatic.net/DCnmB.png)

A) Derive the equilibrium growth rate of capital, and hence find the steady-state values
of capital and output.

If there is no $K\_t$ term in (4) how does one find the growth rate? Since the growth is found using $\Delta K/K\_t \equiv \frac {Kt+1 - Kt} {Kt}$.

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Are you sure that equation $(4)$ is correct? In absence of depreciation we should have $\Delta  K= K_{t+1}-K_t=I_t$. An equation as $(4)$ implies that the capital stock of the preceding period $K_t$ is $0$, so that in each period the existing stock of capital is destroyed. It could be, even if it s  unusual.  
> **username_1**: Moreover, how is the steady state defined here? In the usual Solow model it is the state in which the capital-labor ratio remains constant, but here we haven’t any information about population growth and per capita variables.  I suppose $\Delta K=0$  
> **username_0**: @BakerStreet Yes I copied it straight from the past paper, it's unusual as no other questions given to me have depreciation and/or Kt missing.   The steady state is defined as DeltaKt+1 = 0.  
> **username_1**: Ok, the absence of $K_t$ is not a problem, I give my suggestions in an answer below.  
> **username_2**: Those equations could perhaps model a primitive agricultural economy with seed retained from one year to the next as the only capital, also output less than proportional to seed input because of a limited supply of land and/or labour.  

---

## 2 answers

### Answer by username_1 on 2023-04-10. Score: 4 | Accepted answer

To find the **rate of growth of capital** just subtract $K\_t$ from both sides of $(4)$, to obtain:

$$\Delta K\equiv K\_{t+1}-K\_t= I\_t-K\_t \qquad (1)$$

from which, recalling equality of investment and saving

$$\frac {\Delta K}{K\_t}= \frac {I\_t-K\_t}{K\_t} =\frac {sA{K\_t}^{1/2}- K\_t}{K\_t}\qquad (2)$$ .

You can find the **steady state value of capital and output**, defined as a state in which the stock of capital is constant, setting the preceding equation equal to $0$:

$$\frac {\Delta K}{K\_t}= \frac {sA{K\_t}^{1/2}- K\_t}{K\_t}=0\qquad (3)$$

From $(3)$ you obtain the steady state value of $K\_t$ and then you can derive the steady state value of output plugging this value of $K\_t$ into the production function as usual.

> **Comments:**
> **username_0**: Thats very helpful, thanks! In the second line, why didn't you divide by the Kt to cancel out Kt/Kt and reduce Kt^1/2 to Kt^-1/2?  
> **username_1**: Yes, you can do it, it is the same. You can do the steps you prefer to find the solution.  
> **username_0**: Much appreciated, thanks again.  
> **username_1**: You are welcome!  
> **username_0**: Hey, one more question. How would I go about deriving the saving rate that maximises the steady state consumption?  
> **username_1**: Find the steady state consumption using equation $(3)$, in terms of the saving rate $s$, then maximize it with respect to $s$.  
> **username_0**: So I simply replace the LHS with the value of Y*(steady state), It with sYt, then substitute the Yt with the steady state value of Y?  Then differentiate with respect to s?  
> **username_1**: Yes, if I understand correctly what you said, I don't understand what is LHS.  From $(3)$ you have $C_t=Y_t-I_t$, then  sostitute saving for $I_t$, all considered at their steady state vaues.  
> **username_0**: LHS = Left hand side but yes that's what I referring to, many thanks again  
> **username_1**: You are welcome!  

---

### Answer by username_3 on 2023-04-10. Score: 0

> Are you sure that equation (4)
> is correct? In absence of depreciation we should have ΔK=Kt+1−Kt=It
> . An equation as (4)
> implies that the capital stock of the preceding period Kt
> is 0
> , so that in each period the existing stock of capital is destroyed. It could be, even if it s unusual.

It must be that the depreciation rate is 100% indeed.

> **Comments:**
> **username_1**: Yes, it is the same, capital 'disappears' for some reason. It is a possible assumption, all models are not realistic to some extent. Just, to answer the question, one must be sure that the text is correct and there aren't typos.  
> **username_0**: I thought so but I'm not sure if it's right, or how to go about solving it.  
> **username_1**: Looking  for a solution is possible even if $(4)$ is correct.  
> **username_0**: @BakerStreet How would I go about it?  
> **username_1**: I write my suggestions in an answer  
> **username_1**: Why the downvote? It is just an observation, correct, a possible formalization, it is not wrong.  

---

