Common Errors in Solow Growth Rates and Units

Why Solow Growth Calculations Go Wrong

Many errors in Solow calculations are not difficult algebra mistakes. They come from applying a correct calculation to the wrong variable: confusing aggregate capital with capital per effective worker, treating a stock as a growth rate, or reading a steady-state condition as if every quantity in the economy stopped growing. A reliable check begins by identifying the variable, its units, and the time period to which each rate refers.

This tutorial uses a continuous-time Solow model with labor-augmenting technological progress. The population growth rate is $n$, the growth rate of technology is $g$, the depreciation rate is $\delta$, and the saving rate is $s$. These rates are constant. The distinction between a level and a rate remains important even if a problem presents data in discrete annual periods; in that case, use the discrete model specified by the problem rather than treating a finite-period change as an instantaneous rate.

Definition: Let $K$ be the aggregate capital stock, $L$ the number of workers, and $A$ labor-augmenting technology. Effective labor is $AL$. Capital per effective worker is $k=K/(AL)$, and output per effective worker is $y=Y/(AL)$. Aggregate output is $Y$; output per worker is $Y/L$. A growth rate of a positive variable $X$ is $\gamma_X=\dot X/X$, where $\dot X$ is its change per unit of time.

Keep the Production Function and Capital Units Straight

Assume the aggregate production function $F(K,AL)$ has constant returns to scale: scaling both inputs by the same positive factor scales output by that factor. Define the intensive production function by $f(k)=F(k,1)$. Constant returns to scale then gives $Y=ALf(k)$, with $k=K/(AL)$. The function $f$ takes capital per effective worker as its input; it is not the aggregate production function in $K$ and $AL$.

This distinction prevents a common units error. $K$ is an aggregate stock, measured, for example, in dollars of capital. $k$ is a stock per effective worker, measured in dollars per effective worker. A growth rate such as $\dot k/k$, by contrast, is a rate per unit of time, often reported as a percentage per year. A capital stock, a change in a stock, and a percentage growth rate are different objects.

Model equation: With a constant saving rate and depreciation, aggregate capital accumulates according to $\dot K=sY-\delta K$. Since effective labor grows at rate $n+g$, capital per effective worker changes according to $$ \dot k=s f(k)-(n+g+\delta)k. $$ The first term is investment per effective worker. The second is break-even investment: the amount needed to offset depreciation and to provide capital for a growing number of effective workers.

The equation assumes that saving is a constant fraction $s$ of output, that capital depreciates at rate $\delta$, and that $L$ and $A$ grow at constant rates $n$ and $g$. In this continuous-time model, the terms $n$, $g$, and $\delta$ are rates per unit of time. They must be expressed using the same time unit before being added.

Deriving the Per-Effective-Worker Equation

The key conversion is from the aggregate capital law to the growth of capital per effective worker. The derivation also shows why population and technology growth dilute capital per effective worker.

Start from the aggregate stock.
The capital law is $\dot K=sY-\delta K$. Divide by $AL$ to express the change in aggregate capital in units per effective worker: $\dot K/(AL)=sf(k)-\delta k$.

Differentiate the definition of $k$.
Because $k=K/(AL)$, the quotient rule gives $\dot k=\dot K/(AL)-k\bigl(\dot A/A+\dot L/L\bigr)$. The growth rate of $AL$ is $g+n$.

Substitute and collect the dilution terms.
Using $\dot K/(AL)=sf(k)-\delta k$, obtain $\dot k=sf(k)-\delta k-(n+g)k=sf(k)-(n+g+\delta)k$. Thus all three rates reduce the amount of capital per effective worker, holding investment fixed.

Dividing by the level $k$ gives the growth rate of capital per effective worker:

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

This is not the growth rate of aggregate capital. Since $K=ALk$, the product rule for growth rates gives

\gamma_K=n+g+\gamma_k.

A positive $\dot k$ means that capital per effective worker is increasing; it does not by itself tell you whether aggregate capital is increasing. Aggregate capital can grow while $k$ falls, provided $AL$ grows sufficiently quickly.

Worked Example: Investment, Break-Even Investment, and a Sign Check

In a hypothetical economy, output per effective worker is $f(k)=\sqrt{k}$. The saving rate is $s=0.25$, population grows at $n=0.01$ per year, technology grows at $g=0.02$ per year, depreciation is $\delta=0.03$ per year, and current capital per effective worker is $k=16$. Find $\dot k$ and the percentage growth rate of $k$. Use the continuous-time model.

Method. Calculate investment per effective worker and break-even investment separately, then subtract. This avoids changing the sign of the net change or confusing a level change with a growth rate.

First, output per effective worker is $f(16)=\sqrt{16}=4$. Investment per effective worker is therefore $sf(k)=0.25(4)=1$ unit per effective worker per year.

The combined break-even rate is $n+g+\delta=0.01+0.02+0.03=0.06$ per year. Break-even investment is $0.06(16)=0.96$ units per effective worker per year.

Subtracting gives $\dot k=1-0.96=0.04$ units per effective worker per year. To obtain the growth rate, divide this level change by the current level: $\gamma_k=0.04/16=0.0025$, or $0.25\%$ per year.

Check. The change is positive because investment, $1$, exceeds break-even investment, $0.96$. The rate is $0.25\%$, not $0.04\%$: $0.04$ is a change in units of capital per effective worker, not a percentage.

Population Dilution and Aggregate Growth

Population growth enters the growth rate of $k$ because more workers require more capital to keep capital per worker unchanged. Technological progress also increases effective labor, so maintaining $k$ requires capital to keep pace with both workers and their efficiency. Depreciation is a separate drain: it replaces capital that wears out. These terms have distinct economic meanings even though they appear together in the break-even rate.

Growth rates of related variables are connected by their definitions. Since $Y=ALf(k)$, aggregate output grows at rate $n+g+\gamma_y$, where $\gamma_y=\dot y/y$ is the growth rate of output per effective worker. Output per worker satisfies $Y/L=A f(k)$, so its growth rate is $g+\gamma_y$. These identities do not require a Cobb-Douglas production function. To turn $\gamma_k$ into $\gamma_y$, however, the shape of $f$ matters.

For example, if $f(k)=k^\alpha$ for a constant $0<\alpha<1$, then $y=k^\alpha$ and differentiating logarithms gives $\gamma_y=\alpha\gamma_k$. This relation applies to output per effective worker, not directly to aggregate output or output per worker.

Worked Example: Converting Aggregate Capital Growth to Output Growth

A hypothetical economy has a Cobb-Douglas intensive production function $f(k)=k^{1/3}$. Aggregate capital grows at $5\%$ per year, population at $2\%$ per year, and labor-augmenting technology at $1\%$ per year. Find the growth rates of capital per effective worker, output per effective worker, output per worker, and aggregate output.

Method. First use $K=ALk$ to find the growth rate of $k$. Then apply the production function's elasticity to find output growth. This avoids treating the aggregate capital growth rate as if it were the growth rate of $k$.

Capital per effective worker grows at $\gamma_k=\gamma_K-n-g=5\%-2\%-1\%=2\%$ per year. Since $y=k^{1/3}$, output per effective worker grows at $\gamma_y=(1/3)(2\%)=0.6667\%$ per year.

Output per worker is $Y/L=A y$. Its growth rate is $g+\gamma_y=1\%+0.6667\%=1.6667\%$ per year. Aggregate output is $Y=L(Y/L)$, so its growth rate is $n+1.6667\%=2\%+1.6667\%=3.6667\%$ per year.

Check. Aggregate output also has growth rate $n+g+\gamma_y=2\%+1\%+0.6667\%=3.6667\%$. The two calculations agree. The final rates are shown to four decimal places; the calculations use the unrounded fraction $2\%/3$.

Steady-State Levels Are Not Zero Growth Everywhere

A steady state in the Solow model with labor-augmenting progress means that capital and output per effective worker are constant. It does not mean that the aggregate capital stock or aggregate output is constant. At a steady state, $\dot k=0$, so investment per effective worker exactly equals break-even investment. Aggregate capital and output can still grow with effective labor.

On a balanced growth path, $k$ and $y$ are constant. The growth rates are therefore $\gamma_K=\gamma_Y=n+g$, while capital and output per worker grow at $g$. This follows from $K=ALk$, $Y=ALy$, $K/L=Ak$, and $Y/L=Ay$. These statements presume a steady state in effective-worker terms and constant $n$ and $g$.

Worked Example: Checking a Steady-State Level and Its Growth Rates

A hypothetical economy has $f(k)=k^{1/3}$, saving rate $s=0.24$, population growth $n=0.02$ per year, technology growth $g=0.01$ per year, and depreciation $\delta=0.03$ per year. Find the steady-state capital and output per effective worker, verify the steady-state condition, and state the balanced-growth rates of aggregate capital, aggregate output, and output per worker.

Method. Set investment per effective worker equal to break-even investment to solve for the steady-state level. Then use the definitions of aggregate and per-worker variables to determine growth rates. This avoids concluding that a stationary $k$ implies stationary aggregate capital.

The break-even rate is $n+g+\delta=0.02+0.01+0.03=0.06$. The steady-state equation is $0.24k^{1/3}=0.06k$. For positive $k$, divide by $0.06k^{1/3}$ to get $4=k^{2/3}$, so $k^*=4^{3/2}=8$.

Output per effective worker is $y^*=f(8)=8^{1/3}=2$. Investment is $s y^*=0.24(2)=0.48$. Break-even investment is $0.06(8)=0.48$, verifying $\dot k=0$.

At this steady state, $\gamma_k=0$ and $\gamma_y=0$. Aggregate capital and output each grow at $n+g=0.02+0.01=3\%$ per year. Output per worker grows at $g=1\%$ per year.

Check. Constant $k$ and $y$ are consistent with positive growth in aggregate $K$ and $Y$, because effective labor $AL$ grows at $3\%$ per year. The steady-state condition concerns quantities per effective worker, not aggregate levels.

Common Errors and Their Corrections

The following checks are useful before accepting a calculated Solow growth rate:

Using aggregate capital growth as growth in $k$. This omits the growth of effective labor. The correct identity is $\gamma_k=\gamma_K-n-g$.
Forgetting population dilution or depreciation. In the continuous-time model, the break-even rate is $n+g+\delta$. Do not omit one of these terms or count it twice.
Reporting $\dot k$ as a percentage. $\dot k$ is a level change per unit of time. The percentage growth rate is $\dot k/k$.
Calling a stationary per-effective-worker variable a stationary aggregate. If $k$ is constant, aggregate capital grows with $AL$ at rate $n+g$; it does not generally have zero growth.
Mixing units of time. If one rate is stated per month and another per year, convert them to a common time unit before adding. Also distinguish a percentage, such as $2\%$, from its decimal form, $0.02$, when calculating.
Applying an output-growth relation to the wrong quantity. Under $f(k)=k^\alpha$, $\alpha\gamma_k$ is the growth rate of output per effective worker. Add $g$ for output per worker and $n+g$ for aggregate output on a balanced path.

Key takeaway: Identify the variable before calculating its growth. Use $K=ALk$ to move between aggregate capital and capital per effective worker, and distinguish a stationary level per effective worker from the growth of aggregate quantities.

Check Your Understanding

For each problem, identify whether the requested result is a level change, a growth rate, or a growth rate of an aggregate or per-worker variable.

A hypothetical Solow economy has $f(k)=2\sqrt{k}$, $s=0.20$, $n=0.015$ per year, $g=0.005$ per year, $\delta=0.03$ per year, and $k=9$. Find $\dot k$ and the growth rate of $k$. (Answer: $\dot k=0.75$ units per effective worker per year; $\gamma_k=8.3333\%$ per year.)
Aggregate capital grows at $4.5\%$ per year, population at $1.5\%$, and technology at $1\%$. If $f(k)=k^{0.4}$, find the growth rates of capital per effective worker, output per effective worker, output per worker, and aggregate output. (Answer: $2\%$, $0.8\%$, $1.8\%$, and $3.3\%$ per year, respectively.)
A hypothetical economy has $f(k)=\sqrt{k}$, $s=0.30$, $n=0.02$, $g=0.01$, and $\delta=0.02$, all rates per year. Find steady-state $k$ and $y$, then state the balanced-growth rates of aggregate capital and output. (Answer: $k^*=36$, $y^*=6$; aggregate capital and output each grow at $3\%$ per year.)
Aggregate capital is $800$ million dollars. Effective labor grows at $4\%$ per year, while capital per effective worker falls at $2\%$ per year. Find aggregate capital's growth rate and its instantaneous dollar increase per year. (Answer: $2\%$ per year and $16$ million dollars per year.)
In a hypothetical economy, population grows at $2\%$ per year, technology at $1.5\%$, and aggregate capital at $3\%$. If $f(k)=k^{0.3}$, find the growth rates of capital per effective worker, output per effective worker, output per worker, and aggregate output. (Answer: $-0.5\%$, $-0.15\%$, $1.35\%$, and $3.35\%$ per year, respectively.)

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