Finding the Steady State in the Solow Diagram

Why the Solow Diagram Has an Intersection

In the Solow model, capital per worker changes when investment adds to the capital stock and depreciation or population growth reduces capital available per worker. The steady state is the capital-per-worker level at which these opposing flows balance. In the Solow diagram, that balance appears where the investment-per-worker curve meets the break-even investment line.

Finding this intersection is more than a graph-reading exercise. Once steady-state capital per worker is known, the production function gives steady-state output per worker, and the saving rate determines how much of that output is consumed. The method applies to a range of production functions; it does not require a particular functional form.

Definition: Let $k$ be capital per worker, $y=f(k)$ output per worker, $s$ the constant saving rate, $\delta$ the depreciation rate, and $n$ the population growth rate. Investment per worker is $sf(k)$. Break-even investment is $(\delta+n)k$: the investment required to keep capital per worker unchanged. A steady state $k^*$ satisfies $$ sf(k^*)=(\delta+n)k^*. $$ At that point, output per worker is $y^*=f(k^*)$, and consumption per worker is $c^*=(1-s)f(k^*)$.

Reading the Intersection

The diagram places capital per worker $k$ on the horizontal axis and investment per worker on the vertical axis. The investment curve is $sf(k)$; it shows the amount invested from output at each capital level. The break-even line $(\delta+n)k$ starts at the origin and rises at a constant rate. Its slope is the combined rate of depreciation and population growth.

At an intersection, investment replaces depreciated capital and supplies the additional capital needed to keep pace with a growing workforce. This does not mean all output is invested: only the share $s$ is invested, while the remaining share $1-s$ is consumed. The intersection identifies a capital level, not an output level; output must still be evaluated using $f(k^*)$.

For a positive steady state, use the positive intersection. If $f(0)=0$, both curves also meet at $k=0$. That zero-capital point is a mathematical equilibrium, but it is not the positive steady state typically sought in the diagram. Under the standard neoclassical assumptions that $f$ is increasing and strictly concave, with $f(0)=0$, the ratio $f(k)/k$ falls as $k$ rises. A positive solution exists when saving-funded output per unit of capital is high enough at low $k$ to exceed the break-even requirement, and falls below it at larger $k$. Under those conditions, the positive intersection is unique.

Why the Steady-State Equation Holds

The intersection condition follows from the per-worker capital accumulation equation. Consider a discrete-time model with no technological progress. Let population grow by the rate $n$ each period, let fraction $\delta$ of capital depreciate each period, and let investment per worker be $sf(k_t)$. If $k_t=K_t/L_t$, then aggregate capital next period is $(1-\delta)K_t+I_t$, while the workforce is $(1+n)L_t$. Dividing next period’s capital stock by next period’s workforce gives

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

At a steady state, $k_{t+1}=k_t=k^*$. Multiplying the equation by $1+n$, then collecting the terms involving $k^*$, gives

(1+n)k^*=(1-\delta)k^*+sf(k^*) \quad\Longrightarrow\quad sf(k^*)=(n+\delta)k^*.

Thus the equality is not an arbitrary graphical rule: it is exactly the condition that next period’s capital per worker equal this period’s capital per worker in this model. The displayed break-even formula uses this discrete-time convention. If a problem specifies a different timing convention, use that model’s stated capital accumulation equation rather than silently mixing formulas.

Write the two curves.
Investment per worker is $sf(k)$; break-even investment is $(n+\delta)k$ under the model convention just stated.

Set them equal.
Solve $sf(k)=(n+\delta)k$ for the positive value of $k$. If the production schedule is tabulated, locate where the curves cross and interpolate only if the problem supports that method.

Recover output and consumption.
Evaluate $y^*=f(k^*)$, then calculate $c^*=(1-s)y^*$. Check that $sf(k^*)=(n+\delta)k^*$.

Worked Intersections and Steady-State Values

Worked Example: Solving a Smooth Intersection

A hypothetical economy has production per worker $f(k)=16k/(k+8)$, saving rate $s=0.25$, depreciation rate $\delta=0.05$ per year, and population growth rate $n=0.03$ per year. Find the positive steady-state capital, output, and consumption per worker. Assume the rates and production function are constant, and use the discrete-time convention above.

Method. Equate investment per worker with break-even investment, then evaluate output and consumption at the resulting capital level. The common mistake to avoid is setting output $f(k)$, rather than the saved share $sf(k)$, equal to break-even investment.

First, the break-even rate is $\delta+n=0.05+0.03=0.08$. Investment per worker is $0.25[16k/(k+8)]=4k/(k+8)$. The steady-state equation is

\frac{4k}{k+8}=0.08k.

For a positive solution, divide both sides by $k$: $4/(k+8)=0.08$. Therefore $k+8=4/0.08=50$, so $k^*=42$. Output is $f(42)=16(42)/(42+8)=672/50=13.44$. Investment is $sf(k^*)=0.25(13.44)=3.36$. Consumption is $c^*=(1-0.25)(13.44)=0.75(13.44)=10.08$.

Check. Break-even investment is $0.08(42)=3.36$, matching investment. The per-worker accounting also holds: output $13.44$ equals consumption $10.08$ plus investment $3.36$.

Worked Example: Finding an Intersection from a Production Schedule

A hypothetical economy has saving rate $s=0.20$, depreciation rate $\delta=0.04$ per year, and population growth rate $n=0.04$ per year. Its production schedule reports $f(10)=5$ and $f(30)=10$, with production assumed linear between those capital levels. Find the positive intersection within that interval, along with output and consumption per worker.

Method. Form the linear production schedule on the relevant interval, then equate saved output to break-even investment. The trap is to compare $f(k)$ directly with the break-even line, even though the investment curve is only the saved fraction of output.

The combined break-even rate is $0.04+0.04=0.08$. Between $k=10$ and $k=30$, the slope of the production schedule is $(10-5)/(30-10)=5/20=0.25$. Thus $f(k)=2.5+0.25k$ on this interval: at $k=10$, the expression gives $5$, and at $k=30$, it gives $10$.

Set investment equal to break-even investment:

0.20(2.5+0.25k)=0.08k \quad\Longrightarrow\quad 0.5+0.05k=0.08k \quad\Longrightarrow\quad k^*=\frac{0.5}{0.03}=\frac{50}{3}.

This is approximately $16.67$, which lies within the interval used for interpolation. Output is $f(50/3)=2.5+0.25(50/3)=20/3$, or approximately $6.67$. Investment is $0.20(20/3)=4/3$, or approximately $1.33$. Consumption is $0.80(20/3)=16/3$, or approximately $5.33$.

Check. Break-even investment is $0.08(50/3)=4/3$, matching investment. Values are rounded to two decimal places only for the approximate results.

Worked Example: Distinguishing the Saving Rate from Investment

A hypothetical economy has $f(k)=24k/(k+12)$, saving rate $s=0.25$, depreciation rate $\delta=0.04$ per year, and population growth rate $n=0.06$ per year. Find the positive steady-state capital, output, consumption, and investment per worker.

Method. Solve the intersection first, then split output into its saved and consumed shares. The common mistake to avoid is reporting the saving rate, $0.25$, as the amount of investment per worker.

The break-even rate is $0.04+0.06=0.10$, and investment per worker is $0.25[24k/(k+12)]=6k/(k+12)$. Equating the two curves gives

\frac{6k}{k+12}=0.10k \quad\Longrightarrow\quad \frac{6}{k+12}=0.10 \quad\Longrightarrow\quad k+12=60 \quad\Longrightarrow\quad k^*=48.

Output is $f(48)=24(48)/(48+12)=1152/60=19.2$. Investment is $0.25(19.2)=4.8$, while consumption is $(1-0.25)(19.2)=0.75(19.2)=14.4$.

Check. Break-even investment is $0.10(48)=4.8$, exactly the investment amount. Consumption plus investment is $14.4+4.8=19.2$, equal to output.

Common Traps and Interpretation

The steady-state calculation has a simple structure, but several mistakes can change the answer or its interpretation:

Using output instead of investment. The investment curve is $sf(k)$, not $f(k)$. Apply the saving rate before finding the intersection.
Confusing a rate with a quantity. The saving rate $s$ is a fraction. Investment per worker at a particular $k$ is the amount $sf(k)$.
Stopping at capital per worker. The intersection gives $k^*$. To report output and consumption, calculate $f(k^*)$ and $(1-s)f(k^*)$.
Treating steady-state investment as zero. Investment is generally positive at a positive steady state. It replaces depreciated capital and offsets the effect of population growth.
Ignoring the zero-capital intersection. If both curves start at the origin, zero is also an intersection. Check whether the question asks for a positive steady state.
Using incompatible timing conventions. Apply the break-even expression specified by the model. Here, with the stated discrete-time population and depreciation rates, the condition is $(n+\delta)k$.

The positive intersection matters because it links the capital accumulation mechanism to per-worker living standards in the model. It identifies the capital level at which capital per worker is unchanged, and therefore the output and consumption levels associated with that capital. The calculation alone does not say that every economy has the same steady state: the intersection depends on its production schedule, saving rate, depreciation rate, and population growth rate.

Key takeaway: Find the positive steady state by equating investment per worker, $sf(k)$, with break-even investment, $(n+\delta)k$. Then compute $y^*=f(k^*)$ and $c^*=(1-s)y^*$, and verify that investment equals break-even investment.

Check Your Understanding

Solve each problem using the stated model convention. Answers for approximate quantities are rounded to two decimal places.

A hypothetical economy has $f(k)=20k/(k+5)$, $s=0.25$, $\delta=0.06$, and $n=0.04$. Find positive $k^*$, output per worker, and consumption per worker. (Answer: $k^*=45$, $y^*=18$, $c^*=13.50$.)
A production schedule has $f(10)=5$ and $f(30)=10$, with linear interpolation between the points. If $s=0.20$, $\delta=0.05$, and $n=0.03$, find the intersection in that interval, output, and consumption. (Answer: $k^*=16.67$, $y^*=6.67$, $c^*=5.33$.)
A hypothetical economy has $f(k)=15k/(k+5)$, $s=0.30$, $\delta=0.03$, and $n=0.07$. Find positive steady-state capital, output, and investment per worker. (Answer: $k^*=40$, $y^*=13.33$, investment $=4.00$.)
At $k=20$, a hypothetical production schedule gives $f(20)=12$. If $\delta=0.04$ and $n=0.02$, what saving rate makes this capital level a steady state, and what is consumption per worker at that rate? (Answer: $s=0.10$, $c=10.80$.)
A schedule gives output per worker $f(25)=15$. At $k=25$, $s=0.20$, $\delta=0.05$, and $n=0.07$. Is this point a steady state, and what is consumption per worker? (Answer: Yes; investment and break-even investment are both $3.00$, and consumption is $12.00$.)

What You'll Learn