Growth with Abundant Resources

Abstract

An examination of the different factors that can contribute to growth: increased capital, abundant resource supplies, a larger labor force, and innovation, though the effect of different types of innovation will depend on the ease with which resource supplies can be expanded. A distinction is made between the things that physically have to happen for growth to occur and the social arrangements that make those physical outcomes more likely. The question of convergence between rich countries and poor is considered by means of the diminishing marginal product of capital. The end of the chapter introduces “conditional equivalence,” describing what has to be true for the conventional model and this book’s model to provide the same view of growth. An appendix goes through the algebra of the steady state.

Appendices

Appendix: Growth and the Steady State

Given particular parameters for growth, what level of output will the economy reach over time? In other words, what is the “steady state” toward which it is heading? It turns out we can solve for the level of capital per worker in the steady state, and that in turn implies a particular level of steady-state output per worker.

We start with the observation that next year’s capital is this year’s capital, plus whatever we build over the course of this year, minus whatever depreciates over the course of the year.

If we use a subscript t for this year and t + 1 for next year, we can write:

$$\displaystyle{ K_{t+1} = K_{t} + \text{investment} -\text{depreciation}. }$$

Then we just have to get more specific about investment and depreciation.

Make a couple of simplifying assumptions: first, that investment equals saving, and that we save a constant portion of each year’s GDP. If that portion is s, we can write:

$$\displaystyle{ \text{(Investment)}_{t} = sY _{t}. }$$

We can then make a similar simplifying assumption about depreciation, namely, that a constant portion of capital wears out in any given year. If that portion is d, then we can write:

$$\displaystyle{ \text{(Depreciation)}_{t} = dK_{t}. }$$

Now we can describe the evolution of capital in algebraic terms:

$$\displaystyle{ K_{t+1} = K_{t} + sY _{t} - dK_{t}. }$$

Having done that, we bring K _t over to the left-hand side, and divide both sides by N _t , the labor force:

$$\displaystyle{ \frac{K_{t+1} - K_{t}} {N_{t}} = s\frac{Y _{t}} {N_{t}} - d\frac{K_{t}} {N_{t}}. }$$

To simplify the notation, remember from Sect. 4.7 that Y∕N = y, or output per worker, and K∕N = k, capital per worker:

$$\displaystyle{ \frac{K_{t+1}} {N_{t}} - k_{t} = sy_{t} - dk_{t}. }$$

Note that the first fraction can’t quite be gotten rid of the way the other fractions could, because the subscripts on K and N aren’t the same; next year’s capital stock divided by this year’s work force isn’t a meaningful quantity. But it would be nice to be able to do something about that fraction, which will require getting a term N _t+1 under there. We can do that by multiplying by $\frac{N_{t+1}} {N_{t+1}}$ and rearranging:

$$\displaystyle{ \frac{K_{t+1}} {N_{t+1}} \frac{N_{t+1}} {N_{t}} - k_{t} = sy_{t} - dk_{t}. }$$

We need to put some meaning to this new fraction that has appeared, $\frac{N_{t+1}} {N_{t}}$. If you look at the terms, it’s next year’s labor force divided by this year’s labor force. If we say that the labor force is growing at the rate n, then we can rewrite $\frac{N_{t+1}} {N_{t}}$ as (1 + n) and put this into our equation (also noting that K _t+1∕N _t+1 = k _t+1):

$$\displaystyle{ k_{t+1}(1 + n) - k_{t} = sy_{t} - dk_{t}. }$$

Now we isolate the k _t+1 term:

$$\displaystyle{ k_{t+1}(1 + n) = sy_{t} + (1 - d)k_{t}, }$$

and divide both sides by (1 + n):

$$\displaystyle{ k_{t+1} = \frac{sy_{t} + (1 - d)k_{t}} {(1 + n)}. }$$

The next step is to subtract k _t from both sides:

$$\displaystyle{ k_{t+1} - k_{t} = \frac{sy_{t} + (1 - d)k_{t}} {(1 + n)} - k_{t}, }$$

and bring k _t into the numerator on the right by multiplying it by (1 + n)∕(1 + n):

$$\displaystyle{ k_{t+1} - k_{t} = \frac{sy_{t} + (1 - d)k_{t} - (1 + n)k_{t}} {(1 + n)}. }$$

On the right, we have a (+1k _t ) and a (−1k _t ), so we can consolidate:

$$\displaystyle{ k_{t+1} - k_{t} = \frac{sy_{t} - dk_{t} - nk_{t}} {(1 + n)}. }$$

What we have on the left is the growth of capital per worker (next year’s level, minus this year’s level). To turn it into a growth rate in percentage terms, we have to divide by k _t , this year’s level, so of course we have to do the same to the right-hand side as well:

$$\displaystyle{ \frac{k_{t+1} - k_{t}} {k_{t}} = \frac{s\frac{y_{t}} {k_{t}} - d - n} {(1 + n)}. }$$

(7.1)

Now, it would be nice to be able to do something with that y _t ∕k _t term on the right-hand side, and it turns out we can. Remember (Sect. 4.7) that the per-worker form of the Cobb-Douglas function is

$$\displaystyle{ y_{t} = k_{t}^{\alpha }Z^{\delta }\rho ^{\gamma }, }$$

and if we divide both sides of that by k _t we get

$$\displaystyle{ \frac{y_{t}} {k_{t}} = k_{t}^{\alpha -1}Z^{\delta }\rho ^{\gamma }. }$$

(7.2)

So if we take this expression in Eq. 7.2 and substitute it into Eq. 7.1, we get:

$$\displaystyle{ \frac{k_{t+1} - k_{t}} {k_{t}} = \frac{sk_{t}^{\alpha -1}Z^{\delta }\rho ^{\gamma } - d - n} {(1 + n)}. }$$

(7.3)

Now, believe it or not, we’re almost there.

The steady state is defined by things not changing. Among the things that have to not change in the steady state is the level of capital per worker. In other words, we can define the steady state as the situation when the left-hand side of Eq. 7.3 is zero (we can also move the k _t ^α−1 around within its part of the right-hand side):

$$\displaystyle{ 0 = \frac{sZ^{\delta }\rho ^{\gamma }k_{t}^{\alpha -1} - d - n} {(1 + n)}. }$$

Our goal now is to solve for k _t , in order to be able to define steady-state capital per worker in terms of the other items in the equation. Except that now we’ll call it $\overline{k}$ (pronounced “k-bar”) instead of k _t , because the time subscript suggests something that’s changing from one period to the next, and the whole point of this exercise is that we’re looking for a value that’s fixed; the bar over the k will denote that fixed, steady-state level of k.

The first step is to multiply through by (1 + n), so it disappears from the denominator on the right; it doesn’t show up on the left, since it’s getting multiplied by zero:

$$\displaystyle{ 0 = sZ^{\delta }\rho ^{\gamma }\overline{k}^{\alpha -1} - d - n. }$$

Next we can move the things that are just added on (d and n), and also turn the equation around for notational convenience:

$$\displaystyle{ sZ^{\delta }\rho ^{\gamma }\overline{k}^{(\alpha -1)} = d + n. }$$

Now we divide both sides by all the stuff that $\overline{k}$ is getting multiplied by:

$$\displaystyle{ \overline{k}^{(\alpha -1)} = \frac{d + n} {sZ^{\delta }\rho ^{\gamma }}. }$$

If we take the exponent on $\overline{k}$ and turn it around (so it becomes [1 −α]), that’s the same as taking the inverse of the left-hand side of the equation (in other words, 1 divided by $\overline{k}^{(\alpha -1)}$, which means we have to flip over the right-hand side as well:

$$\displaystyle{ \overline{k}^{(1-\alpha )} = \frac{sZ^{\delta }\rho ^{\gamma }} {d + n}. }$$

And now we are actually, really, truly, almost there—we just need to get rid of the exponent on $\overline{k}$. If you raise both sides to the 1∕(1 −α), that shows up as an exponent on the right, while it cancels out the exponent on the left:

$$\displaystyle{ \overline{k} = \left ( \frac{sZ^{\delta }\rho ^{\gamma }} {d + n}\right )^{ \frac{1} {(1-\alpha )} }. }$$

(7.4)

A few pages later, and we’ve done it: we have an expression for capital-per-worker in the steady state, expressed as a function of the level of efficiency, resource intensity of labor, the savings rate, the depreciation rate, and the growth rate.

If you plug that back into the per-worker form of output, you get the level of output per worker in the steady sate, as a function of the same factors that determine steady-state capital per worker.

Problems

Problem 7.1

The table below shows eight fictitious countries (creatively named “A” through “H”). Each country is given with its CO₂ footprint per capita, its renewables footprint per capita, and its GDP per capita. For each country, say whether it fits the expected pattern in terms of the relationship between resource use and GDP per capita. If your answer is that it doesn’t fit, explain specifically how it’s “out of line.”

Table 2

Problem 7.2

Identify a social institution. Say whether it tends to foster economic growth or hinder it. Your answer should operate through one of the more tangible factors of economic growth (e.g., saving, investment, innovation, access to resources, population growth).

Problem 7.3

The table below gives another four (fantastically creatively named) imaginary countries, this time “J” through “K”. The data show population (in thousands) and GDP (in billions) for each country in two different periods (1 and 2).

For each country calculate GDP per capita in each of the two periods, as well as the percent change in GDP from period 1 to period 2, and the percent change in GDP per capita from period 1 to period 2.

Table 3

Problem 7.4

Equation 7.4 of the appendix defines steady-state capital per worker $\overline{k}$ as a function of various influences, including s, Z, ρ, n, and d. For each of these influences:

• (a) Say what the letter represents.

• (b) Give a mathematical explanation of how that factor influences $\overline{k}$. For example, “A larger value of s makes $\overline{k}$ get” (and here you have to say “larger” or “smaller”) “because …”, and your explanation should be based simply on the equation for steady-state capital per worker.

• (c) Provide the economic logic behind the impact you identified in the previous part of the question. That is, why does it “make sense” that the relationship should be what you said it was in that part of the question?

Problem 7.5

Explain the circumstances under which the resource-inclusive growth model of this book leads to a different conclusion than a model that neglects the role of resources.