Mathematical Growth Models

Some of the more important ideas about economic growth are based on mathematical models. This lesson looks at some of these.

The Malthusian ModelBefore 1800, technological progress was relatively slow. The result was that output per worker hardly increased at all, but population grew. In 1798, Thomas Malthus wrote an essay on population that presented a pessimistic picture of economic growth. He said that when food is ample, population grows exponentially. Because there are diminishing returns to labor in food production, exponential population growth leads to starvation, and population falls again.

Here is a numerical example of a two-equation Malthusian model.

[food production] Y_{t} = 1000 + L_{t}

[population growth] L_{t} = 600 + 100*(Y_{t-1}/L_{t-1})^{2}

This generation's food production, Y_{t}, increases linearly with this generation's labor supply (population). However, the next generation's labor supply increases with the square of this generation's ratio of food to population.

You can solve these two equations for values of Y and L that will be stable. These are called the equilibrium values. In this case, they are 2000 for Y and 1000 for L. If L is 1000, then according to the food production equation, Y will be 2000. If Y is 2000, then population will grow to be 1000.

What happens if we start out with 2000 units of food, but disease causes the population to fall to 900? You can use the calculator below to see what happens if population starts out too low or too high. If you click on "calculate" the economy will move forward in time one generation. Keep clicking on "calculate" and you will see Y and L oscillate back and forth until they converge to their equilibrium values. You can try starting out with different values of Y and L and see the convergence process from different starting points.

Next, suppose that we get better technology in food production, so that the food production equation becomes

Y_{t} = 1200 + L_{t}

What happens to the equilibrium values of Y, L, and Y/L? Use the calculator below to find out. Keep clicking on calculate until the values stop changing.

What is the equilibrium level of food? What is the equilibrium level for the population? What is the equilibrium ratio of food to population?

At first, with the population at 1000, the technological improvement brings food production to 2200, and the ratio of food to population rises to 2.200. However, in the final equilibrium, because of population increases, the ratio of food to population is only 2.136. This is the Malthusian effect by which population growth dissipates technological advances. In fact, prior to 1800, the Malthusian effect was so strong that there was very little progress in average output per capita; instead, nearly all of the inventions and technological advances until 1800 served primarily to increase population.

Capital AccumulationWhen the Industrial Revolution broke out of the Malthusian trap, economies began to accumulate capital goods. In order to accumulate capital, you have to save. This means that you cannot consume all of your output.

Start with a constant level of output, Y, and no growth. Suppose that capital depreciates at a rate of 5 percent per year. In order to keep the level of capital constant we have to replace 5 percent of the capital stock each year. This means that saving, S, must equal 5 percent of the capital stock.

[1] S = .05K

We think of the saving rate, s, as the ratio of savings to income, S/Y. Writing equation [1] in terms of s, we have

[2] s = S/Y = .05(K/Y)

where all we did was divide the previous equation by Y on both sides. What this equation says is that in order to maintain constant output, we need a savings rate that equals the rate of depreciation times the capital/output ratio. If we want to have high labor productivity, we need a high ratio of capital to output, and therefore we need a high saving rate. Thus, we expect to find a relationship between countries with high saving rates and countries with high productivity, and this is indeed what Brad DeLong found when he indicated that countries with high productivity tend to have saving rates over 20 percent.

Suppose that we want the capital stock to grow at a rate of 2 percent per year. In that case, we need

[3] S = .05K + .02K

Or, in terms of s and K/Y, we need

[4] s = .05(K/Y) + .02(K/Y)

If we use the symbol d to represent depreciation, the symbol k to stand for the capital-output ratio, and the symbol x to stand for the growth rate of capital, then we can write

[5] s = dk + xk

To see how the saving rate affects the growth rate of capital, we can solve [5] for x, the growth rate of capital.

[6] x = s/k - d

If the labor force is growing at a rate n, then the capital/labor ratio will grow at the rate of x-n. For example, suppose that the saving rate s is .25 (i.e., 25 percent), the capital/output ratio k is 2.5, the depreciation rate d is .05, and the growth rate of the labor force n is .01. Then we have

[7] x - n = s/k - d - n = .25/2.5 - .05 -.01 = .04

which says that the capital/labor ratio grows at 4 percent per year. (In a moment, when I discuss balanced growth, I will argue that this is not a reasonable long-term growth rate for the capital/labor ratio.)

Labor ProductivityWe are interested in the growth rate of labor productivity, Y/L. To look at productivity, we return to the production function that we used in the growth accounting lesson.

[8] (Y/L) = (K/L)^{0.25}E^{0.75}

where E is the efficiency of labor. When we took logs of both sides, we obtained an equation for the growth rate of productivity. If y is the growth rate of output and n is the growth rate of the labor force, then the growth rate of productivity is y-n. Letting g be the symbol for the growth rate of E, the efficiency of labor, we have

[9] y - n = 0.25(x - n) + .75g

When we made numerical assumptions in equation [7], we found that x-n = .04. Plugging this into equation [9] and assuming that the growth rate of the efficiency of labor, g, is .02, we have

[10] y - n = 0.25(.04) + .75(.02) = .025

Thus, the assumptions about saving rate, depreciation, and so forth imply growth in labor productivity of 2.5 percent per year.

Balanced GrowthEconomists define a balanced growth path as a path along which capital and output grow at the same rate. The alternatives to a balanced growth path are not sustainable. If capital grows more slowly than output, then the capital stock will eventually drop to zero. If capital grows more quickly than output, then the share of output that you set aside for capital goods will increase until you reach the point where the amount available for consumption is zero.

Looking at equation [9], the only way that x and y can be equal is if

[11] g = x - n

That is, for balanced growth, the growth rate of the efficiency of labor must be matched by the growth rate of capital minus the growth rate of the labor force.

The requirement for balanced growth implies that there is only one sustainable ratio of capital to ouput. That is, there is only one ratio of capital to output, k that is consistent with a balanced growth path. Using equations [11] and [7] we have

[12] g = x - n = s/k - d - n

We can solve this equation for a balanced-growth value for k, given the other parameters. Using s = .25, g = .02, n = .01 and d = .05, we have

[13] k = s/(g + n + d) = .25/(.02+.01+.05) = 3.125

Therefore, the balanced-growth capital-output ratio is 3.125. If the capital-output ratio happens to be above this level, the savings rate is not high enough to maintain it, and the ratio will tend to fall back to 3.125. Conversely, if the capital-output ratio happens to start out below the balanced-growth level, the savings rate is high enough to generate capital accumulation until the ratio rises back to 3.125. Back at equation [10] when we computed labor productivity growth, we had assumed earlier an arbitrary capital-output ratio of 2.5. Now, we know that this is not a balanced-growth ratio given the saving rate, depreciation rate, and other assumed parameters. Using the balanced-growth ratio of 3.125 in equation [7] gives

[7'] x - n = .25/3.125 - .05 - .01 = .02

Putting this into [10], we have

[10'] y - n = .25(.02) + .75(.02) = .02

What we have found is that on a balanced growth path, output per worker and capital per worker grow at the same rate as the efficiency of labor. In our example, this is 2 percent per year.

SummaryLet us review what we have learned from mathematical growth models.

For the Malthusian model:

Whenever the food supply expands, population grows exponentially.

The economy has an equilibrium in which population stays constant.

If the equilibrium is disturbed, population will oscillate. A small generation enjoying a high ratio of food to population will reproduce excessively, leading to a large generation with a low ratio of food to population. This large generation will reproduce minimally, leading to a small generation, etc. As the magnitude of the oscillations diminishes (if indeed they do dampen), the economy goes back to its equilibrium.

Increases in productivity will lead to less than proportionate increases in output per worker. Instead, population expansion and diminishing returns will dissipate much of the technological improvement.

For the balanced-growth model of capital accumulation:

The growth rate of the economy is equal to the growth rate of the efficiency of labor. Capital per worker and output per worker both grow at this rate.

The savings rate affects the level of productivity and the level of the capital/output ratio. The higher the savings rate, the higher the capital/output ratio and the higher the level of productivity.