# Typical Growth Model Social Planner Problem

Consider the following social planner's problem, hand-waving the usual assumptions on the preference, technology, endowment, and inelastic supply of labor:

$$V(k_o^*)= max_{\{c_t,k_{t+1}\}_{t=0}^{\infty}}\quad \sum\limits_{t=0}^\infty \beta^t U(c_t)$$

s.t. $$\quad c_t+k_{t+1}=f(k_t)+(1-\delta)k_t$$, $$c_t,k_{t+1}\geq0, k_o=k_o^*$$

$$\textbf{My Question}$$:

(Q1) Is the following a correct description of what happens in $$t=0,1$$?

(a) The economy starts with the given preference, technology, and endowment $$k_o^*$$.

(b) The planner chooses how much to put into the production and leave as to carried on as undepreciated capital to the next period. In other words:

(i) A fraction of $$k_o^*$$ is devoted to into the technology $$f(\cdot)$$.
(ii) The rest is simply carried onto the next period with the penalty of depreciation .

(c) Still in $$t=0$$ and with (b)-(i), the technology produces $$y_t$$.

(d) The consumer makes a choice as to how much to consume and invest from $$y_t$$ as $$c_t+i_t=y_t$$

(e) In $$t=1$$, the capital stock for tomorrow is defined as, $$k_1=(1-\delta)k_o^*+i_t$$, how much you left over from the initial capital stock to be carried over to $$t=1$$ plus how much you decided to leave out from the production as investment.

(Q2) In the social planner's problem, we have control vectors as a sequence of $$c_t$$ and $$k_{t+1}$$. These are two distinct control vectors, because the sequence $$\{c_t\}_{t=0}^\infty$$ requires the decision to be made after what you get from the production within the same period, but the decision of how much of the capital stock to have tomorrow involves how much of the capital stock today you will leave to be carried over to tomorrow in addition to the investment decision. For example, if for some reason, the capital stock is perishable, this $$\delta=1$$, our choice variable would be only $$c_t$$ as $$c_t$$ determines the level of $$i_t$$, which in turn determines the level of $$k_{t+1}$$, correct? Is $$k_{t+1}$$ not a control variable?

Q1: Your bullet point (b) is not how we usually interpret capital, it is usually not assumed that $$k_0$$ is divided between $$k_0^p$$, and $$k_0^s$$ the capital we use to produce and the capital we store for next period. Instead, we assume all the capital is used for production, but using it for production does not delete it, it only depreciates it. For example, if you use your machines to produce pizza that does not mean that they are obsolete to be used next period.
So I do not agree with your interpretation in (e), typically $$k_1=(1-\delta)k_0+i_0$$.
Consider this simple example. If I say that you can choose to buy chocolate (at a price of \$2) or cheese (at a price of \$4), but you only have \$10 dollars (sorry this are rough times :P ), it is clear that if you are to spend all your money on these two goods, then choosing how much chocolate to buy, immediately pins down how much cheese you will buy. However, the fact that this is true does not change the nature that both amounts of chocolate and cheese are your choice variables. The budget constraint may allow you to re-formulate this problem as one where you only choose how much chocolate to buy (since cheese would be implicit), but this will only be a technique to simplify the original problem. Something similar happens here. So to be concise, both sequences $$\{c_t,k_{t+1}\}_{t=0}^\infty$$ are control variables even if they have a simple linear relationship between them through the law of motion. • Hi Regio, thanks for the response. On your response to my Q2, it became more clear! – Frank Swanton Jun 15 '19 at 0:32 • Regarding your point on my Q1, to further clarify: what you are saying is the initial endowment$k_0$is getting depreciated by the discount factor$\beta$each period? So, in your example, if we are born with factory machines, that$k_0$is set for the rest of the problem but those machines depreciate over time (e.g. wear and tear)? Is this correct? – Frank Swanton Jun 15 '19 at 0:45 • I'm glad your Q2 was more clear. I'm not sure I follow your last comment, capital gets depreciated by the factor$\delta$not$\beta$(maybe that was a typo). In any case, you should think of capital as a stock variable and investment as a flow variable. Like the water in a tub, the quantity of water at any given point in time is the stock (this is$k_t$, the level of capital at time$t$), if you add more water to the tub (this is the investment$i_t$) the level will go up, and if the tub leaks (this would be the depreciation$\delta k_t$,), then the level of water might go down. – Regio Jun 15 '19 at 23:46 • So$k_0$is just the level of capital you start with, you use it to produce and with that you either consume or invest. If you don't invest, the level will go down to$k_1=(1-\delta)k_0<k_0$and that's the capital you'll have for operating your production. To some extent when you are in period 1, it does not matter a lot what was$k_0$, rather the only important thing is how much capital you have that period to use for production. Clearly,$k_1$can also be larger than$k_0\$ if your investment was larger than how much capital got depreciated. – Regio Jun 15 '19 at 23:52