Capital Accumulation in Per-Worker Terms

Why Capital per Worker Changes

Aggregate capital can increase while capital per worker falls. The reason is that the workforce may grow at the same time as the capital stock: new workers share the available capital. To study productive capacity per worker, the Solow model therefore tracks not only aggregate capital, but also the number of workers over whom it is distributed.

The key equation combines three ideas. Saving finances investment, investment adds to the capital stock, and depreciation wears some of that stock out. Meanwhile, labor growth changes the denominator in capital per worker. The equation makes these effects explicit and is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Model, Variables, and Assumptions

Use a continuous-time model. Let $K(t)$ denote the aggregate capital stock, $L(t)$ the number of workers, and $Y(t)$ aggregate output. Define capital and output per worker as $k(t)=K(t)/L(t)$ and $y(t)=Y(t)/L(t)$. The production relationship established in the earlier material on output per worker is $y=f(k)$: output per worker is determined by capital per worker under the model’s production assumptions.

Assume a constant saving rate $s$, a constant depreciation rate $\delta$, and a constant labor-force growth rate $n$. Saving is a fixed share of output, so gross investment is $I=sY$. Capital depreciates continuously at rate $\delta$, and the labor force grows continuously at rate $n$. There is no technological change in this equation, and the model treats investment as the addition to the aggregate capital stock after accounting for depreciation.

Definition: Under these assumptions, aggregate capital changes according to $\dot K=sY-\delta K$, and labor changes according to $\dot L=nL$. Capital per worker is $k=K/L$. A dot denotes a change per unit of time; for example, $\dot k$ is the change in capital per worker per year when time is measured in years.

The rates and time units must be consistent. If $s$, $\delta$, and $n$ are annual rates, then the flow quantities $\dot K$, $\dot L$, and $\dot k$ are annual changes. The rate $s$ is a share of output, while $\delta$ and $n$ have units of inverse time.

Deriving the Per-Worker Equation

The derivation follows from differentiating a ratio. Since capital per worker is aggregate capital divided by labor, its change depends both on the change in the numerator and on the change in the denominator.

\dot k =\frac{d}{dt}\left(\frac{K}{L}\right) =\frac{\dot K L-K\dot L}{L^2} =\frac{\dot K}{L}-\frac{K}{L}\frac{\dot L}{L}.

Substitute the model’s aggregate equations. First, $Y/L=y=f(k)$ and $K/L=k$. Second, $\dot L/L=n$, by the assumed labor-growth equation. Therefore,

\dot k =\frac{sY-\delta K}{L}-k n =s\frac{Y}{L}-\delta\frac{K}{L}-nk =sf(k)-(\delta+n)k.

Per-worker capital equation: In the continuous-time Solow model with a constant saving rate, depreciation rate, and labor-force growth rate, and no technological change, $$ \dot k=sf(k)-(\delta+n)k. $$ The term $sf(k)$ is gross investment per worker. The term $\delta k$ is depreciation per worker, and $nk$ is the reduction in capital per worker associated with labor-force growth.

The $nk$ term follows from the changing denominator, not from physical wear on machines or structures. It is often called capital dilution: when the labor force grows, a given aggregate capital stock is spread across more workers. Both depreciation and labor growth reduce capital per worker, so their effects appear together as $(\delta+n)k$. The equation describes a level change in capital per worker, not a percentage growth rate.

Interpreting the Three Components

Term	Meaning per worker	Effect on $\dot k$
$sf(k)$	Gross investment financed by saving from output	Raises capital per worker
$\delta k$	Capital lost to depreciation	Reduces capital per worker
$nk$	Capital needed to maintain the existing capital-per-worker ratio as the workforce grows	Reduces capital per worker when $n>0$

The signs matter. A higher saving rate raises the investment term, holding output per worker fixed. A higher depreciation rate makes the loss from depreciation larger at a given $k$. A higher labor-growth rate makes the dilution term larger at a given $k$. These are direct implications of the equation; they do not require capital per worker to be constant or to be at any particular level.

Worked Applications

Worked Example: Verify the Equation From Aggregate Data

A hypothetical economy has $K=1{,}200$ units of capital, $L=400$ workers, and $Y=800$ units of annual output. Its saving rate is $s=0.25$, its annual depreciation rate is $\delta=0.04$, and its annual labor-force growth rate is $n=0.02$. Calculate the aggregate changes and the change in capital per worker, then verify the per-worker equation.

Method. Calculate aggregate investment and depreciation first, then use the quotient rule. This avoids treating aggregate capital growth as if it were the change in capital per worker.

Gross investment is $I=sY=0.25(800)=200$ units per year. Depreciation is $\delta K=0.04(1{,}200)=48$ units per year. Thus,

\dot K=200-48=152.

Labor grows by $\dot L=nL=0.02(400)=8$ workers per year. Current capital per worker is $k=K/L=1{,}200/400=3$, and output per worker is $y=Y/L=800/400=2$. Applying the quotient rule gives

\dot k=\frac{152}{400}-3\left(\frac{8}{400}\right) =0.38-0.06 =0.32

units of capital per worker per year. Now verify with the per-worker equation: $sf(k)=0.25(2)=0.50$, and $(\delta+n)k=(0.04+0.02)(3)=0.18$. Therefore, $\dot k=0.50-0.18=0.32$ units per worker per year. Both methods agree.

Check and common mistake. The aggregate change $\dot K=152$ is not the per-worker change. The labor-growth adjustment is $nk=0.06$, giving $\dot k=152/400-0.06=0.32$, not $0.38$.

Worked Example: Capital per Worker Can Fall While Investment Is Positive

A hypothetical economy has $K=5{,}000$, $L=1{,}000$, and annual output $Y=1{,}500$. Its saving rate is $s=0.18$, annual depreciation rate is $\delta=0.06$, and annual labor growth rate is $n=0.01$. Find $\dot K$ and $\dot k$, and explain why their signs differ.

Method. Compare investment with aggregate depreciation to determine $\dot K$, then account separately for workforce growth when finding $\dot k$. Positive investment alone does not guarantee that capital per worker rises.

Investment is $sY=0.18(1{,}500)=270$ units per year. Depreciation is $\delta K=0.06(5{,}000)=300$ units per year, so $\dot K=270-300=-30$ units per year. The labor force grows at $\dot L=nL=0.01(1{,}000)=10$ workers per year.

Capital per worker is $k=5{,}000/1{,}000=5$, and output per worker is $y=1{,}500/1{,}000=1.5$. The per-worker equation gives

\dot k=0.18(1.5)-(0.06+0.01)(5) =0.27-0.35 =-0.08.

The change is $-0.08$ units of capital per worker per year. To check using aggregate values, $\dot K/L=-30/1{,}000=-0.03$, while $k(\dot L/L)=5(10/1{,}000)=0.05$. Thus $\dot k=-0.03-0.05=-0.08$, as required. In this case aggregate capital itself falls, and growth in the workforce further reduces capital per worker.

Check and common mistake. The saving-funded investment of 270 is positive, but it is less than the 300 of depreciation. Reporting a positive change in capital per worker merely because investment is positive would ignore both depreciation and labor growth.

Worked Example: Check the Per-Worker Equation With a Production Function

A hypothetical economy has production per worker $f(k)=2\sqrt{k}$, current capital per worker $k=4$, saving rate $s=0.30$, annual depreciation rate $\delta=0.05$, and annual labor growth rate $n=0.025$. There are $800$ workers. Find annual output, $\dot K$, $\dot L$, and $\dot k$, and verify the per-worker calculation using aggregate changes.

Method. Evaluate production per worker at the current $k$, scale it up to aggregate output, and compute the aggregate flows. This checks that the production-function route and the aggregate accounting route use the same quantities.

Since $f(4)=2\sqrt{4}=2(2)=4$, output per worker is $y=4$. With $L=800$, aggregate output is $Y=yL=4(800)=3{,}200$. Also, $K=kL=4(800)=3{,}200$. Gross investment is $sY=0.30(3{,}200)=960$, and depreciation is $\delta K=0.05(3{,}200)=160$. Hence $\dot K=960-160=800$ units per year. Labor grows by $\dot L=nL=0.025(800)=20$ workers per year.

Using the per-worker equation, investment per worker is $sf(k)=0.30(4)=1.2$. The combined depreciation and labor-growth term is $(0.05+0.025)(4)=0.075(4)=0.3$. Therefore,

\dot k=1.2-0.3=0.9

units per worker per year. From the aggregate quantities, the quotient rule gives $\dot k=800/800-4(20/800)=1-0.1=0.9$, confirming the result.

Check and common mistake. The per-worker investment term is $1.2$, not aggregate investment of $960$. Dividing aggregate flows by the $800$ workers gives the appropriate per-worker units; the labor-growth adjustment must then still be included.

Common Exam Traps and Dimensional Checks

Omitting labor growth. The expression $sf(k)-\delta k$ leaves out the change in the denominator. When $n\ne0$, the full equation includes the additional term $-nk$.
Using the wrong sign for depreciation. Depreciation removes capital, so it enters as $-\delta k$, not $+\delta k$.
Confusing a level change with a growth rate. $\dot k$ is measured in units of capital per worker per unit of time. The percentage growth rate of $k$, when $k>0$, is $\dot k/k$.
Mixing time units. If $n$ is a quarterly rate but $\delta$ and $\dot K$ are annual, the calculation is inconsistent until the rates are converted to the same time unit.
Forgetting the production relationship. The investment term can be written $sY/L$ or, using $y=f(k)$, as $sf(k)$. Do not substitute aggregate $Y$ into a per-worker equation without dividing by $L$.

Find output per worker.
Use $y=Y/L=f(k)$, keeping aggregate and per-worker quantities distinct.

Compute investment and losses.
Per worker, investment is $sy$; depreciation is $\delta k$; labor-growth dilution is $nk$.

Combine with signs and units.
Subtract both losses from investment and report $\dot k$ as a change per worker per unit of time.

Key takeaway: The per-worker capital equation is a ratio-growth result applied to aggregate capital accumulation. Saving adds capital per worker; depreciation and labor-force growth both reduce it. Checking the equation against $\dot k=\dot K/L-k(\dot L/L)$ is an effective way to catch omitted terms and unit errors.

Check Your Understanding

For each problem, use consistent annual units and distinguish aggregate changes from per-worker changes.

A hypothetical economy has $K=900$, $L=300$, and $Y=600$, with $s=0.20$, $\delta=0.05$ per year, and $n=0.01$ per year. Find $k$, $\dot K$, and $\dot k$. (Answer: $k=3$, $\dot K=75$ units per year, and $\dot k=0.22$ units per worker per year.)
A hypothetical economy has $K=2{,}400$, $L=600$, $Y=1{,}200$, $s=0.15$, $\delta=0.03$ per year, and $n=0.02$ per year. Calculate investment, depreciation, $\dot K$, and $\dot k$, checking the last result by both methods. (Answer: investment $=180$, depreciation $=72$, $\dot K=108$ units per year, and $\dot k=0.10$ units per worker per year.)
A hypothetical economy has $K=1{,}500$, $L=500$, $Y=1{,}000$, $s=0.20$, $\delta=0.04$ per year, and labor growth of 15 workers per year. Find $n$, $\dot K$, and $\dot k$. (Answer: $n=0.03$ per year, $\dot K=140$ units per year, and $\dot k=0.19$ units per worker per year.)
At a particular time, $k=6$, $f(k)=3$, $\delta=0.04$ per year, and $n=0.01$ per year. If capital per worker is increasing at $0.15$ units per worker per year, find the saving rate $s$. (Answer: $s=0.15$.)
A hypothetical economy has annual output per worker $y=5$, saving rate $s=0.24$, annual depreciation rate $\delta=0.08$, and annual labor growth rate $n=0.02$. If current capital per worker is $k=5$, calculate gross investment per worker, the combined capital loss per worker, and $\dot k$. (Answer: investment per worker $=1.20$, combined loss $=0.50$, and $\dot k=0.70$ units per worker per year.)

Term	Meaning per worker	Effect on \(\dot k\)
\(sf(k)\)	Gross investment financed by saving from output	Raises capital per worker
\(\delta k\)	Capital lost to depreciation	Reduces capital per worker
\(nk\)	Capital needed to maintain the existing capital-per-worker ratio as the workforce grows	Reduces capital per worker when \(n>0\)

What You'll Learn