Depreciation and the Steady-State Capital Stock

Why Depreciation Matters for Capital per Worker

A machine, building, or other productive asset does not remain fully productive forever. In the Solow model, depreciation represents the loss of productive capital over time. When depreciation rises, more investment is needed just to maintain a given amount of capital per worker. That changes the steady-state capital stock, and therefore affects steady-state output and consumption per worker.

The analysis here uses the basic Solow model without technological progress. Population grows at a constant rate, the saving rate is fixed, and production per worker depends on capital per worker. We compare steady states while holding the production function, saving rate, and population growth rate fixed. This isolates the effect of a change in depreciation.

Definition: Let \(k=K/L\) denote capital per worker and \(y=f(k)\) output per worker. Let \(s\) be the constant saving and investment share of output, \(n\) the constant population growth rate, and \(\delta\) the constant depreciation rate. Break-even investment per worker is \((n+\delta)k\): the investment needed to offset depreciation and the dilution of capital per worker as the population grows. A steady state \(k^*>0\) satisfies \(s f(k^*)=(n+\delta)k^*\).

Capital Accumulation and Break-Even Investment

Assume continuous time, a fixed saving rate \(0<s<1\), and a constant population growth rate \(n\geq 0\). Aggregate investment equals \(sY\), and depreciation removes capital at rate \(\delta K\). Since the number of workers grows at rate \(n\), capital per worker changes according to

\dot{k}=s f(k)-(n+\delta)k.

The first term is actual investment per worker. The second is break-even investment per worker. If \(s f(k)>(n+\delta)k\), capital per worker rises; if \(s f(k)<(n+\delta)k\), it falls. In a steady state, the two quantities are equal, so capital per worker does not change.

This equation follows from the aggregate capital equation \(\dot K=sY-\delta K\) and \(k=K/L\). Differentiating \(K/L\) gives \(\dot k=\dot K/L-k(\dot L/L)\). With \(Y/L=f(k)\) and \(\dot L/L=n\), substitution yields \(\dot k=s f(k)-\delta k-nk\), which is the stated equation. The population-growth term is important: replacing \(n+\delta\) with depreciation alone would omit capital dilution.

For a given \(k\), an increase in \(\delta\) raises \((n+\delta)k\) in direct proportion to \(k\). Thus, in a Solow diagram, the break-even investment line becomes steeper while the saving-investment curve \(s f(k)\) stays fixed. Under the usual assumptions that \(f(k)\) is increasing and concave, with \(f(0)=0\) and diminishing marginal product, the positive intersection moves left. This means a lower steady-state capital stock per worker.

A Direct Comparison of Steady States

The steady-state condition makes the comparative-static result clear: a higher depreciation rate requires a larger amount of investment per worker at any fixed positive \(k\). The saving-investment curve must therefore intersect the new break-even line at a lower \(k\), provided the production function and saving rate are unchanged.

The direction can also be verified by differentiating the steady-state condition. This argument assumes a positive steady state and a stable intersection, so \(s f'(k^*)<n+\delta\). Holding \(s\), \(n\), and \(f\) fixed gives

s f'(k^*)\frac{d k^*}{d\delta} =k^*+(n+\delta)\frac{d k^*}{d\delta}, \qquad \frac{d k^*}{d\delta} =-\frac{k^*}{n+\delta-s f'(k^*)}<0.

The derivative is negative because the denominator is positive at a stable intersection and \(k^*>0\). Since output per worker is \(y^*=f(k^*)\), an increasing production function implies \(d y^*/d\delta=f'(k^*)\,d k^*/d\delta<0\). With the saving rate fixed, steady-state consumption per worker is \(c^*=(1-s)f(k^*)\), so it falls as well.

Worked Example: Comparing Two Depreciation Rates

Suppose output per worker is \(f(k)=\sqrt{k}\), the saving rate is \(s=0.25\), and population grows at \(n=0.02\) per year. Compare the steady state when depreciation is \(\delta=0.03\) per year with the steady state when it rises to \(\delta=0.08\) per year. Find capital, output, and consumption per worker in each steady state.

Method. Solve \(s f(k^*)=(n+\delta)k^*\) separately for each depreciation rate. The annual population growth rate and depreciation rate must be added before solving.

At \(\delta=0.03\), total break-even investment per unit of capital is \(n+\delta=0.02+0.03=0.05\). Thus

0.25\sqrt{k^*}=0.05k^*, \qquad \sqrt{k^*}=5, \qquad k^*=25.

Output per worker is \(y^*=\sqrt{25}=5\). Consumption per worker is the share not saved: \(c^*=(1-0.25)(5)=0.75(5)=3.75\). As a check, saving is \(0.25(5)=1.25\), and break-even investment is \(0.05(25)=1.25\).

At \(\delta=0.08\), total break-even investment per unit of capital is \(0.02+0.08=0.10\). The new steady state satisfies

0.25\sqrt{k^*}=0.10k^*, \qquad \sqrt{k^*}=2.5, \qquad k^*=6.25.

Then \(y^*=\sqrt{6.25}=2.5\) and \(c^*=0.75(2.5)=1.875\). Saving and break-even investment both equal \(0.25(2.5)=0.625\), also \(0.10(6.25)=0.625\). The higher depreciation rate reduces steady-state capital, output, and consumption per worker. Common mistake avoided: using \(\delta k\) alone as break-even investment would omit population dilution; the correct term is \((n+\delta)k\).

Cobb–Douglas Production

For Cobb–Douglas production per worker, \(f(k)=k^\alpha\), with \(0<\alpha<1\), the steady-state condition can be solved explicitly. Divide both sides of \(s(k^*)^\alpha=(n+\delta)k^*\) by \((k^*)^\alpha\), then raise both sides to the power \(1/(1-\alpha)\):

k^*=\left(\frac{s}{n+\delta}\right)^{\frac{1}{1-\alpha}}, \qquad y^*=(k^*)^\alpha, \qquad c^*=(1-s)y^*.

These expressions apply to the positive steady state under the stated model assumptions. The formula for \(k^*\) shows that a larger \(n+\delta\) lowers steady-state capital when \(s\) is unchanged. The effect on output and consumption follows because both are increasing in \(k^*\) for a fixed production function and saving rate.

Worked Example: Cobb–Douglas Steady-State Changes

Let \(f(k)=k^{1/3}\), \(s=0.32\), and \(n=0.02\) per year. Calculate the steady-state capital, output, and consumption per worker when \(\delta=0.02\), then recalculate them when \(\delta=0.06\).

Method. Use the Cobb–Douglas steady-state formula with \(\alpha=1/3\), so the exponent on \(s/(n+\delta)\) is \(1/(1-1/3)=3/2\).

Initially, \(n+\delta=0.02+0.02=0.04\), and \(s/(n+\delta)=0.32/0.04=8\). Therefore

k^*=8^{3/2}=8\sqrt{8}=16\sqrt{2}\approx22.627.

Output is \(y^*=(k^*)^{1/3}=2\sqrt{2}\approx2.828\). Consumption is \(c^*=(1-0.32)y^*=0.68(2\sqrt{2})=1.36\sqrt{2}\approx1.923\). Check the steady-state condition: saving is \(0.32(2\sqrt{2})=0.64\sqrt{2}\), while break-even investment is \(0.04(16\sqrt{2})=0.64\sqrt{2}\).

After depreciation rises to \(0.06\), \(n+\delta=0.02+0.06=0.08\), and \(s/(n+\delta)=0.32/0.08=4\). Hence

k^*=4^{3/2}=8, \qquad y^*=8^{1/3}=2, \qquad c^*=0.68(2)=1.36.

The check is \(s y^*=0.32(2)=0.64\) and \((n+\delta)k^*=0.08(8)=0.64\). The reported approximate values are rounded to three decimal places. Common mistake avoided: the exponent is \(3/2\), not \(1/2\); it comes from \(1/(1-\alpha)\), not from \(\alpha\) itself.

The Transition Is Not the New Steady State

A change in depreciation does not instantly replace the existing capital stock with its new steady-state value. At the moment the rate changes, \(k\) is still the amount already in place. The changed break-even investment alters the direction and rate of capital accumulation; capital per worker then adjusts over time. The following calculation separates the immediate flow effect from the long-run comparison.

Worked Example: The Immediate Effect of Higher Depreciation

Suppose \(f(k)=\sqrt{k}\), \(s=0.30\), and \(n=0.02\) per year. The economy has \(k=16\) when depreciation rises from \(0.03\) to \(0.08\) per year. Calculate the direction and rate of change in capital per worker immediately before and after the increase. Also find the two steady-state capital stocks.

Method. Evaluate \(\dot{k}=s f(k)-(n+\delta)k\) at the unchanged current \(k=16\). Here \(f(16)=4\), so actual investment per worker is \(0.30(4)=1.20\).

Before the change, \(n+\delta=0.02+0.03=0.05\), and break-even investment is \(0.05(16)=0.80\). Thus \(\dot{k}=1.20-0.80=0.40\): capital per worker is rising at that moment.

After the change, \(n+\delta=0.02+0.08=0.10\), and break-even investment is \(0.10(16)=1.60\). Thus \(\dot{k}=1.20-1.60=-0.40\): capital per worker is falling at that moment.

For the old steady state, solve \(0.30\sqrt{k^*}=0.05k^*\). This gives \(\sqrt{k^*}=6\), so \(k^*=36\). For the new steady state, solve \(0.30\sqrt{k^*}=0.10k^*\), giving \(\sqrt{k^*}=3\) and \(k^*=9\). The current stock \(k=16\) is below the old steady state but above the new one, consistent with the calculated signs before and after the change. Common mistake avoided: substituting the new steady-state capital immediately into the transition equation would answer a different question; the immediate change is evaluated at the capital stock that exists when depreciation changes.

Interpretation and Common Traps

In this model, a higher depreciation rate means a larger replacement-investment requirement at every positive capital stock. With the saving rate fixed, the economy cannot sustain its former steady-state capital per worker, so it moves toward a lower one. Lower capital per worker means lower output per worker, and the fixed consumption share \(1-s\) then implies lower steady-state consumption per worker.

These conclusions are comparative statics: they describe the effect of changing \(\delta\) while holding \(s\), \(n\), and the production function fixed. They do not assert that every observed change in depreciation must lower consumption if other factors, such as the saving rate, also change. Nor does the steady-state comparison say that output or consumption jumps immediately when \(\delta\) changes; only the capital accumulation equation changes immediately.

Omitting population growth. Break-even investment is \((n+\delta)k\), not just \(\delta k\), when population grows.
Confusing a rate with an amount. The rate \(n+\delta\) multiplies \(k\); break-even investment per worker is the amount \((n+\delta)k\).
Mixing time units. If \(n\) and \(\delta\) are annual rates, use both as annual rates in the same equation.
Calling all investment consumption. In this model, consumption per worker is \((1-s)y\); the saved share finances investment, including the amount required to maintain capital per worker.
Equating a transition with a steady state. A new depreciation rate changes the target and the direction of accumulation, not the existing capital stock instantaneously.

Key takeaway: With the saving rate, population growth, and production function fixed, a higher depreciation rate raises break-even investment, lowers steady-state capital per worker, and therefore lowers steady-state output and consumption per worker.

Check Your Understanding

For each problem, use the stated model and keep all growth rates in the same time units.

Let \(f(k)=\sqrt{k}\), \(s=0.20\), \(n=0.01\), and \(\delta=0.04\) per year. Find steady-state capital, output, and consumption per worker. (Answer: \(k^*=16\), \(y^*=4\), and \(c^*=3.2\).)
Let \(f(k)=k^{1/3}\), \(s=0.25\), \(n=0.02\), and \(\delta=0.03\) per year. Find steady-state capital and consumption per worker, rounding to three decimals. (Answer: \(k^*=5\sqrt{5}\approx11.180\) and \(c^*=0.75\sqrt{5}\approx1.677\).)
Let \(f(k)=\sqrt{k}\), \(s=0.40\), and \(n=0.02\). Compare steady-state capital, output, and consumption per worker at \(\delta=0.03\) and \(\delta=0.08\). (Answer: At \(\delta=0.03\), \(k^*=64\), \(y^*=8\), \(c^*=4.8\); at \(\delta=0.08\), \(k^*=16\), \(y^*=4\), \(c^*=2.4\).)
Let \(f(k)=\sqrt{k}\), \(s=0.25\), \(n=0.01\), and current \(k=9\). If \(\delta\) rises from \(0.04\) to \(0.09\), calculate \(\dot{k}\) immediately before and after the change. (Answer: Before, \(\dot{k}=0.75-0.45=0.30\); after, \(\dot{k}=0.75-0.90=-0.15\) per year.)
Let \(f(k)=\sqrt{k}\), \(s=0.30\), and \(n=0.02\). What depreciation rate makes \(k^*=25\) a steady state? Find the corresponding steady-state output and consumption per worker. (Answer: \(\delta=0.04\) per year, \(y^*=5\), and \(c^*=3.5\).)

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