Why Measure Capital Per Effective Worker?
When technology makes each worker more productive, the number of workers alone is not a complete measure of the labor input used in production. In the Solow model with labor-augmenting technical progress, a worker supplied with technology level \(A\) provides \(A\) units of effective labor. The relevant denominator for comparing capital and output with productive labor is therefore \(AL\), not just \(L\).
This change of units matters for capital accumulation. Investment must replace depreciated capital, equip a growing labor force, and keep pace with growth in the effective labor supplied by each worker. As a result, the break-even investment rate in effective-worker terms includes three components: depreciation, population growth, and technology growth.
Model and Effective-Worker Production
Use the continuous-time Solow model. Assume a closed economy with a constant saving rate \(s\), a constant depreciation rate \(\delta\), population growing at constant rate \(n\), and labor-augmenting technology growing at constant rate \(g\). Thus \(\dot L/L=n\) and \(\dot A/A=g\), where a dot denotes a derivative with respect to time. Assume aggregate production is \(Y=F(K,AL)\), with constant returns to scale, and all saving is invested. Under these assumptions, aggregate capital changes according to \(\dot K=sY-\delta K\).
Constant returns to scale means that multiplying both inputs by the same positive factor multiplies output by that factor. Define \(f(\tilde{k})=F(\tilde{k},1)\). Dividing aggregate output by effective labor gives output per effective worker as a function of capital per effective worker:
The third equality follows by applying constant returns to scale with the factor \(1/(AL)\). This transformation lets us express production and investment in units of effective labor. It does not assume a particular production function such as Cobb-Douglas; that form can be used when specified.
Worked Example: Converting a Cobb-Douglas Production Function
Suppose aggregate production is \(Y=K^{1/3}(AL)^{2/3}\). At a particular date, \(K=216\) units of capital and \(AL=8\) units of effective labor. Find capital and output per effective worker, and verify the transformed production function.
Method. Divide capital and output by effective labor, then express output per effective worker in terms of \(\tilde{k}\). The common mistake to avoid is dividing by \(L\) instead of \(AL\).
First calculate capital per effective worker: \(\tilde{k}=K/(AL)=216/8=27\).
Next calculate aggregate output from the stated production function: \(Y=216^{1/3}\times 8^{2/3}=6\times4=24\). Thus output per effective worker is \(\tilde{y}=Y/(AL)=24/8=3\).
For this production function, the transformed function is \(\tilde{y}=\tilde{k}^{1/3}\). At \(\tilde{k}=27\), it gives \(27^{1/3}=3\), matching the direct calculation. Check: both methods give 3 units of output per effective worker.
Deriving the Effective Capital Equation
Capital per effective worker is a ratio whose denominator changes over time. Its change therefore depends both on the change in the aggregate capital stock and on growth in effective labor \(AL\). Since \(\dot{(AL)}/(AL)=n+g\), differentiating \(K/(AL)\) gives the key result.
The first line is the quotient rule applied to \(K/(AL)\). In the second line, aggregate capital accumulation is substituted in, and the proportional growth rate of \(AL\) is \(n+g\). In the third line, use \(Y/(AL)=\tilde{y}\) and \(K/(AL)=\tilde{k}\). Finally, substitute the effective-worker production function.
This derivation explains why the technology-growth term appears: even if \(K\) were unchanged, a growing \(A\) would increase \(AL\) and reduce \(K/(AL)\). The equation is exact under the stated continuous-time assumptions. If growth rates vary over time, the same derivation applies instantaneously with \(n\) and \(g\) replaced by their current growth rates. A discrete-time model instead requires discrete growth factors; the continuous-time equation should not be used as though it were an exact one-period difference formula.
The equation separates the change in effective capital into two flows. Saving finances gross investment \(sf(\tilde{k})\). Break-even investment is the amount needed just to keep \(\tilde{k}\) constant: \(\delta\tilde{k}\) replaces depreciated capital, \(n\tilde{k}\) equips the additional workers, and \(g\tilde{k}\) keeps capital in step with greater labor effectiveness. Hence:
These comparisons describe the instantaneous direction of change in capital per effective worker. They do not, by themselves, establish the long-run behavior of the whole economy. In particular, a zero value of \(\dot{\tilde{k}}\) at a given capital level means investment just covers break-even needs at that moment.
Worked Example: Checking Whether Effective Capital Rises
Assume \(f(\tilde{k})=\sqrt{\tilde{k}}\), a constant saving rate \(s=0.25\), depreciation rate \(\delta=0.05\) per year, population growth \(n=0.02\) per year, and technology growth \(g=0.03\) per year. At \(\tilde{k}=9\), calculate the annual change in effective capital per effective worker and determine its direction.
Method. Compare saving per effective worker with full break-even investment. All rates are continuous-time annual rates. The common mistake to avoid is leaving out the technology-growth term \(g\).
Output per effective worker is \(f(9)=\sqrt{9}=3\). Gross investment per effective worker is \(sf(9)=0.25\times3=0.75\).
The total break-even rate is \(\delta+n+g=0.05+0.02+0.03=0.10\) per year. Break-even investment at \(\tilde{k}=9\) is \(0.10\times9=0.90\).
Therefore, \(\dot{\tilde{k}}=0.75-0.90=-0.15\) units per effective worker per year. Effective capital is falling at that instant. Check: saving is \(0.15\) below the investment required to keep effective capital unchanged.
Using the Equation Without Mixing Units
The equation is written entirely in effective-worker units: \(\tilde{k}\), \(f(\tilde{k})\), and their rates of change all refer to capital or output per effective worker. Aggregate investment is \(sY\); dividing by \(AL\) converts it to \(s\tilde{y}\). Similarly, aggregate depreciation is \(\delta K\); dividing by \(AL\) gives \(\delta\tilde{k}\). Population growth and technology growth dilute effective capital at rates \(n\) and \(g\), respectively.
Worked Example: How Faster Technology Growth Changes Break-Even Investment
Suppose \(f(\tilde{k})=\sqrt{\tilde{k}}\), the saving rate is \(s=0.30\), depreciation is \(\delta=0.04\) per year, and population grows at \(n=0.01\) per year. At \(\tilde{k}=25\), compare effective capital accumulation when technology grows at \(g=0.01\) and when it grows at \(g=0.03\). Hold the current capital level and all other assumptions fixed.
Method. Calculate saving once, then recalculate break-even investment for each technology-growth rate. The common mistake to avoid is treating a change in \(g\) as if it directly changed current output \(f(\tilde{k})\) in this comparison; the stated current \(\tilde{k}\) and production function are held fixed.
At \(\tilde{k}=25\), output per effective worker is \(\sqrt{25}=5\), so saving per effective worker is \(0.30\times5=1.50\).
With \(g=0.01\), the break-even rate is \(0.04+0.01+0.01=0.06\) per year. Break-even investment is \(0.06\times25=1.50\), so \(\dot{\tilde{k}}=1.50-1.50=0\).
With \(g=0.03\), the break-even rate is \(0.04+0.01+0.03=0.08\) per year. Break-even investment is \(0.08\times25=2.00\), so \(\dot{\tilde{k}}=1.50-2.00=-0.50\) units per effective worker per year.
At the same current effective capital and saving, faster technology growth raises break-even investment and changes effective capital accumulation from zero to negative \(0.50\). Check: the break-even investment increase is \(2.00-1.50=0.50\), exactly the change in \(\dot{\tilde{k}}\).
Common Mistake: Omitting Technology Growth
A frequent error is to use \((\delta+n)\tilde{k}\) as break-even investment even when technology grows. That expression accounts for replacement of depreciated capital and capital dilution from population growth, but it does not equip the increase in effective labor due to \(A\).
For the previous example with \(g=0.03\), incorrectly omitting technology growth gives break-even investment \((0.04+0.01)\times25=1.25\), and the wrong calculation reports \(\dot{\tilde{k}}=1.50-1.25=0.25\). The correct calculation includes technology growth: \((0.04+0.01+0.03)\times25=2.00\), giving \(\dot{\tilde{k}}=-0.50\). The mistake reverses the conclusion about the direction of effective capital.
A useful check is to list the three sources of break-even needs before calculating: depreciation \(\delta\), population growth \(n\), and labor-augmenting technology growth \(g\). This result is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.
Check Your Understanding
For each problem, use the continuous-time effective-worker equation and state whether effective capital rises, falls, or is unchanged where requested.
- Let \(f(\tilde{k})=\sqrt{\tilde{k}}\), \(s=0.20\), \(\delta=0.04\), \(n=0.01\), and \(g=0.03\), with rates per year. At \(\tilde{k}=36\), calculate \(\dot{\tilde{k}}\). (Answer: \(-1.68\) units per effective worker per year; it falls.)
- Let \(f(\tilde{k})=\tilde{k}^{0.4}\), \(s=0.25\), \(\delta=0.05\), \(n=0.02\), and \(g=0.03\). At \(\tilde{k}=32\), use \(32^{0.4}=4\) to calculate \(\dot{\tilde{k}}\). (Answer: \(-2.20\) units per effective worker per year; it falls.)
- Suppose \(f(\tilde{k})=\sqrt{\tilde{k}}\), \(\tilde{k}=9\), \(\delta=0.03\), \(n=0.02\), and \(g=0.05\). What saving rate makes effective capital unchanged at this instant? (Answer: \(s=0.30\).)
- At \(\tilde{k}=25\), let \(f(\tilde{k})=\sqrt{\tilde{k}}\), \(s=0.20\), \(\delta=0.04\), and \(n=0.02\). Compare \(\dot{\tilde{k}}\) when \(g=0.01\) and \(g=0.02\). (Answer: \(-0.75\) and \(-1.00\) units per effective worker per year, respectively; the increase in \(g\) reduces \(\dot{\tilde{k}}\) by \(0.25\).)
- For \(Y=K^{1/2}(AL)^{1/2}\), state the effective-worker production function. If \(\tilde{k}=16\), \(s=0.25\), and \(\delta+n+g=0.10\) per year, calculate the net change in effective capital. (Answer: \(\tilde{y}=\tilde{k}^{1/2}\) and \(\dot{\tilde{k}}=0.25(4)-0.10(16)=-0.60\) units per effective worker per year.)