Solow Steady-State Calculation Workshop

Why Calculate the Solow Steady State?

A Solow steady-state calculation connects a production function to a long-run level of capital and output per worker. Given the saving rate, depreciation rate, and population growth rate, the central task is to find the capital stock per worker at which investment just keeps pace with the capital needed to equip a growing workforce and replace worn-out capital. Once that stock is found, output and consumption per worker follow from the production function and the saving rate.

This tutorial uses a discrete-time Solow model with a fixed technology level and no technological progress. Rates refer to one period and are expressed as fractions in calculations: for example, 5% is entered as 0.05. The model and its assumptions matter because a different treatment of population or technology growth changes the steady-state condition. These calculations are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

The Model and Steady-State Condition

Definition: Let \(K_t\) be aggregate capital, \(L_t\) the number of workers, and \(k_t=K_t/L_t\) capital per worker in period \(t\). Output per worker is \(y_t=f(k_t)\). A constant fraction \(s\) of output is saved and invested, so consumption per worker is \(c_t=(1-s)y_t\). The workforce grows at rate \(n\), and a fraction \(\delta\) of the capital stock depreciates each period. Assume \(0\leq s\leq1\), \(n\geq0\), and \(0\leq\delta\leq1\).

Aggregate production is \(Y_t=L_t f(k_t)\). With a constant saving rate, investment is \(I_t=sY_t\). The workforce grows according to \(L_{t+1}=(1+n)L_t\), while next period’s aggregate capital is \(K_{t+1}=(1-\delta)K_t+sY_t\). Dividing next period’s capital by next period’s labor gives the per-worker accumulation equation:

k_{t+1}=\frac{(1-\delta)k_t+s f(k_t)}{1+n}.

A steady state is a level \(k^*\) that remains constant from one period to the next, so \(k_{t+1}=k_t=k^*\). Rearranging the accumulation equation gives the condition below. The term \((n+\delta)k^*\) is break-even investment per worker: the amount of investment required to maintain \(k^*\) as the workforce grows and capital depreciates.

s f(k^*)=(n+\delta)k^*.

The addition \(n+\delta\) follows from this model’s discrete-time equation; it is not a claim that depreciation and population growth are the same process. It combines their contributions to the amount of investment needed at a steady state. Do not add a technological-growth term here: technology is fixed in the model used throughout this calculation workshop.

Solving for Capital, Output, and Consumption

A commonly used production function is Cobb–Douglas production per worker, \(f(k)=A k^\alpha\), where \(A>0\) is the technology level and \(0<\alpha<1\) is the capital exponent. Substituting this function into the steady-state condition yields an equation that can be solved directly for \(k^*\).

sA(k^*)^\alpha=(n+\delta)k^* \quad\Longrightarrow\quad (k^*)^{1-\alpha}=\frac{sA}{n+\delta} \quad\Longrightarrow\quad k^*=\left(\frac{sA}{n+\delta}\right)^{\frac{1}{1-\alpha}}.

After finding capital, calculate steady-state output and consumption in that order:

y^*=A(k^*)^\alpha, \qquad c^*=(1-s)y^*.

The capital formula requires \(n+\delta>0\). With positive \(A\), \(s>0\), and \(0<\alpha<1\), it gives a positive steady-state capital stock. If \(s=0\), there is no investment and the positive-capital steady-state formula does not apply. Under the standard assumptions of diminishing returns and positive break-even investment, a positive saving rate gives a unique positive steady state.

Why the Formula Works

The steady-state equation is not a separate rule to memorize without context. It follows from the requirement that capital per worker does not change. Starting from the accumulation equation, impose \(k_{t+1}=k_t=k^*\), multiply both sides by \(1+n\), and collect the capital terms:

k^*=\frac{(1-\delta)k^*+s f(k^*)}{1+n}

(1+n)k^*=(1-\delta)k^*+s f(k^*)

(n+\delta)k^*=s f(k^*).

Thus, at the steady state, actual investment per worker equals break-even investment per worker. For the Cobb–Douglas formula, divide the equation \(sA(k^*)^\alpha=(n+\delta)k^*\) by \((k^*)^\alpha\). This gives \((k^*)^{1-\alpha}=sA/(n+\delta)\). Raising both sides to the reciprocal power \(1/(1-\alpha)\) produces the stated expression for \(k^*\).

Identify the model inputs.
Record \(A\), \(\alpha\), \(s\), \(n\), and \(\delta\), converting percentage rates to decimals and keeping all rates in the same time unit.

Calculate the break-even rate.
Compute \(n+\delta\). This is the required investment per unit of steady-state capital in the specified discrete-time model.

Solve for steady-state capital.
For Cobb–Douglas production, evaluate \(sA/(n+\delta)\), then raise the result to \(1/(1-\alpha)\).

Calculate output and consumption.
Use \(y^*=A(k^*)^\alpha\) and then \(c^*=(1-s)y^*\). Consumption is the unsaved share of output.

Check the steady state.
Verify that \(s y^*=(n+\delta)k^*\). This catches errors in the capital calculation or in the application of rates.

Worked Calculations

Worked Example: A Square-Root Production Function

A hypothetical economy has output per worker \(y=1\sqrt{k}\), saving rate \(s=0.20\), population growth rate \(n=0.02\) per period, and depreciation rate \(\delta=0.03\) per period. Calculate steady-state capital, output, and consumption per worker. The common trap is to use the saving rate as the exponent or to omit one of the two break-even components.

Method: apply the steady-state condition and then check investment. Here \(A=1\) and \(\alpha=1/2\). First, combine the two rates:

n+\delta=0.02+0.03=0.05.

Next, use the Cobb–Douglas steady-state formula. The ratio inside the power is \(sA/(n+\delta)=0.20(1)/0.05=4\), and the exponent is \(1/(1-1/2)=2\). Therefore:

$$ k^*=4^2=16. $$

Output per worker is \(y^*=1\sqrt{16}=4\). Consumption is the fraction not saved:

$$ c^*=(1-0.20)(4)=0.80(4)=3.20. $$

Check. Saving per worker is \(s y^*=0.20(4)=0.80\). Break-even investment is \((n+\delta)k^*=0.05(16)=0.80\). The two amounts agree, so the calculated capital stock satisfies the steady-state condition.

Worked Example: A One-Third Capital Exponent

A hypothetical production function is \(y=8k^{1/3}\). The saving rate is 0.25, population grows by 5% per period, and 7.5% of capital depreciates each period. Calculate steady-state capital, output, and consumption per worker. The key trap is to use the reciprocal of \(1-\alpha\), not the reciprocal of \(\alpha\), as the exponent on the steady-state ratio.

Method: solve the Cobb–Douglas condition and verify the implied saving. The parameters are \(A=8\), \(\alpha=1/3\), \(s=0.25\), \(n=0.05\), and \(\delta=0.075\). The break-even rate is:

n+\delta=0.05+0.075=0.125.

The ratio in the capital formula is \(sA/(n+\delta)=0.25(8)/0.125=2/0.125=16\). The required exponent is:

\frac{1}{1-\alpha}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}.

Thus steady-state capital is \(k^*=16^{3/2}=(\sqrt{16})^3=4^3=64\). For output, \(64^{1/3}=4\), so \(y^*=8(4)=32\). Consumption is:

$$ c^*=(1-0.25)(32)=0.75(32)=24. $$

Check. Saving per worker is \(0.25(32)=8\). Break-even investment is \(0.125(64)=8\). The equality confirms the result. Notice that the power of \(3/2\) is applied to 16 to find capital; output is then calculated from the original production function.

Worked Example: Find the Saving Rate for a Target Steady State

A hypothetical economy produces \(y=3\sqrt{k}\), has population growth \(n=0.04\) and depreciation \(\delta=0.08\) per period, and is intended to have steady-state capital \(k^*=25\) per worker. Find the saving rate required for that target and calculate steady-state output and consumption per worker. The trap is to report the saving amount as the saving rate.

Method: rearrange the steady-state condition for \(s\). First calculate the break-even rate and output at the target capital:

n+\delta=0.04+0.08=0.12, \qquad y^*=3\sqrt{25}=3(5)=15.

At a steady state, \(s y^*=(n+\delta)k^*\). Substituting the known values and dividing by output gives:

s=\frac{(n+\delta)k^*}{y^*} =\frac{0.12(25)}{15} =\frac{3}{15} =0.20.

The saving rate is 0.20, or 20%. The amount saved per worker is \(0.20(15)=3\); these are different quantities. Consumption per worker is:

$$ c^*=(1-0.20)(15)=0.80(15)=12. $$

Check. Break-even investment is \(0.12(25)=3\), equal to saving per worker. Output is 15 and is divided into saving of 3 and consumption of 12, so \(s y^*+c^*=3+12=15=y^*\).

Common Calculation Traps and Interpretation

A numerical answer is more useful when its components and units are clear. Capital, output, and consumption in these examples are all measured per worker and in the same output units. Population growth affects the steady-state capital requirement, but it is not itself a percentage of output. The expression \((n+\delta)k^*\) is an investment amount per worker, not a rate of saving.

Using percentages as whole numbers. A 4% rate enters as 0.04, not 4. When rates have different time units, convert them to a common period before combining them.
Forgetting population growth or depreciation. In the model here, both appear in \(n+\delta\). Using only one understates break-even investment and changes the calculated capital stock.
Using the wrong Cobb–Douglas exponent. The exponent is \(1/(1-\alpha)\). For \(\alpha=1/3\), it is \(3/2\), not 3.
Calling saved output consumption. Saving is \(s y^*\); consumption is \((1-s)y^*\). They add to output under this model’s assumption that output is divided between consumption and investment.
Skipping the consistency check. After calculating the results, compare \(s y^*\) with \((n+\delta)k^*\). If they differ, revisit the rate conversion, exponent, or arithmetic.

The steady-state calculation identifies a long-run level under specified parameters; it does not say that an economy already at that level, or that the level is optimal. In particular, a higher saving rate affects the division of current output between consumption and investment, while steady-state consumption must be calculated after finding the new steady-state output. Comparing a steady state with a consumption-maximizing capital stock is a separate question.

Key takeaway: In the fixed-technology discrete-time Solow model, steady-state investment per worker equals \((n+\delta)k^*\). For Cobb–Douglas production, solve this condition for capital first, then calculate output and the unsaved share of output as consumption.

Check Your Understanding

For each problem, use the model and rate conventions stated above. Answers are given at the end of each question for checking.

An economy has \(y=2\sqrt{k}\), saving rate \(s=0.20\), population growth \(n=0.06\), and depreciation \(\delta=0.04\), all per period. Find steady-state capital, output, and consumption per worker. (Answer: \(k^*=16,\ y^*=8,\ c^*=6.4\).)
A hypothetical economy has \(y=5k^{1/3}\), \(s=0.30\), \(n=0.03\), and \(\delta=0.07\). Find steady-state capital, output, and consumption per worker. (Answer: \(k^*=15^{3/2}=15\sqrt{15},\ y^*=5\sqrt{15},\ c^*=3.5\sqrt{15}\).)
For \(y=4\sqrt{k}\), population growth is 0.04 per period and depreciation is 0.06 per period. If the saving rate is 0.25, calculate steady-state capital, output, and consumption, then verify break-even investment. (Answer: \(k^*=100,\ y^*=40,\ c^*=30\); saving and break-even investment are both 10.)
An economy produces \(y=2\sqrt{k}\), with \(n=0.02\) and \(\delta=0.08\) per period. If its target steady-state capital is 64 per worker, find the required saving rate, output, and consumption per worker. (Answer: \(s=0.40,\ y^*=16,\ c^*=9.6\).)
A hypothetical economy has \(y=6k^{1/3}\), saving rate \(s=0.20\), population growth \(n=0.05\), and depreciation \(\delta=0.07\) per period. Calculate \(k^*\), \(y^*\), and \(c^*\), then check that saving equals break-even investment. (Answer: \(k^*=10^{3/2}=10\sqrt{10},\ y^*=6\sqrt{10},\ c^*=4.8\sqrt{10}\); both investment amounts equal \(1.2\sqrt{10}\).)

What You'll Learn