Why Capital per Worker Needs Break-Even Investment
Investment adds to the economy’s capital stock, but it does not automatically raise the amount of capital available to each worker. Some investment must replace capital that wears out. If the number of workers grows, additional capital is also needed to equip the new workers at the existing capital-per-worker level. The investment that covers these two needs is called break-even investment.
The key comparison is between investment per worker and break-even investment per worker. If investment is larger, capital per worker rises; if it is smaller, capital per worker falls. The comparison is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.
Model and Derivation
Use a closed-economy Solow model with one produced good, no technological progress, and a constant saving rate \(s\), where \(0\leq s\leq1\). Aggregate production is \(Y=F(K,L)\), with constant returns to scale, so output per worker can be written as \(y=f(k)\). A fraction \(s\) of output is invested, giving aggregate investment \(I=sY\). Capital depreciates at a constant rate \(\delta\), so depreciation per unit of time is \(\delta K\). The workforce grows continuously at rate \(n\), so \(\dot L=nL\). A dot denotes a change per unit of time.
These assumptions give the aggregate capital accumulation equation \(\dot K=sY-\delta K\). To find the change in capital per worker, differentiate \(k=K/L\). The quotient rule gives \(\dot k=\dot K/L-k(\dot L/L)\). Substituting the aggregate equations yields the per-worker accumulation equation.
This derivation explains why population growth appears alongside depreciation. Depreciation removes \(\delta k\) units of capital per worker, while population growth dilutes the existing capital stock across a growing workforce. Together, the two effects create a required investment of \((\delta+n)k\). The equation also gives the direction of change: \(\dot k\) is positive when investment per worker exceeds break-even investment, negative when it falls short, and zero when the two are equal.
- If \(sf(k)>(\delta+n)k\), then \(\dot k>0\) and capital per worker rises.
- If \(sf(k)<(\delta+n)k\), then \(\dot k<0\) and capital per worker falls.
- If \(sf(k)=(\delta+n)k\), then \(\dot k=0\) at that capital-per-worker level.
Reading the Investment Curves
A Solow diagram often plots capital per worker, \(k\), on the horizontal axis and investment per worker on the vertical axis. The production curve \(f(k)\) shows output per worker. Because only the fraction \(s\) is invested, the relevant investment curve is \(sf(k)\), not \(f(k)\) itself. The break-even investment curve is a straight line through the origin with slope \(\delta+n\), since \((\delta+n)k\) is proportional to \(k\).
At any chosen \(k\), compare the height of the saving-funded investment curve with the height of the break-even line. If \(sf(k)\) is higher, the vertical gap is positive and \(k\) increases. If it is lower, the gap is negative and \(k\) decreases. The production curve helps determine the investment curve, but comparing output directly with break-even investment would omit the saving rate.
The break-even line’s slope depends on both rates. A higher depreciation rate means more replacement investment is needed at each \(k\). A higher population growth rate means more capital must be provided for additional workers to maintain the same capital per worker. Either change raises break-even investment at a given \(k\). This comparison describes the direction of capital-per-worker change; it does not, by itself, determine a long-run equilibrium.
Worked Example: Depreciation and Population Growth Together
A hypothetical economy has production per worker \(f(k)=\sqrt{k}\), saving rate \(s=0.25\), depreciation rate \(\delta=0.05\) per year, population growth rate \(n=0.03\) per year, and current capital per worker \(k=16\). Using the continuous-time model, determine whether capital per worker rises or falls and calculate its rate of change.
Method. Calculate output per worker, multiply by the saving rate to find investment per worker, then compare that amount with break-even investment. This method avoids treating all output as investment.
First, output per worker is \(f(16)=\sqrt{16}=4\). Investment per worker is therefore \(sf(k)=0.25(4)=1\) unit per worker per year.
The combined rate is \(\delta+n=0.05+0.03=0.08\) per year. Break-even investment is \((\delta+n)k=0.08(16)=1.28\) units per worker per year.
The net change is \(\dot k=1-1.28=-0.28\) units per worker per year. Since investment is \(0.28\) below break-even investment, capital per worker falls. As a check, the accumulation equation gives \(0.25\sqrt{16}-(0.05+0.03)(16)=1-1.28=-0.28\), the same result.
Discrete-Time Timing and the Same Break-Even Amount
The expression \((\delta+n)k\) also has a useful discrete-time interpretation under a common timing convention. Suppose \(\delta\) is the fraction of beginning-of-period capital that depreciates, \(n\) is the workforce growth rate over the period, and \(i\) is investment per worker measured relative to the beginning-of-period workforce. Then capital per worker next period is \(\displaystyle k'=\frac{(1-\delta)k+i}{1+n}\). To keep capital per worker unchanged, set \(k'=k\). Multiplying by \(1+n\) gives \((1-\delta)k+i=(1+n)k\), so \(i=(\delta+n)k\). Under this stated timing convention, the break-even amount is the same. The continuous-time net-change equation and this discrete-time update are different equations; use the one that matches the model and timing specified in a problem.
Worked Example: Using the Production Curve to Compare Investments
A hypothetical economy has \(f(k)=2\sqrt{k}\), saving rate \(s=0.30\), depreciation rate \(\delta=0.04\) per year, population growth rate \(n=0.02\) per year, and \(k=25\). Under the continuous-time model, find output and investment per worker, break-even investment, and the direction and rate of change of capital per worker.
Method. Read output from the production function, scale it by the saving rate, and compare the result with the break-even line. This avoids comparing the unscaled production curve with the break-even line.
Output per worker is \(f(25)=2\sqrt{25}=2(5)=10\). Investment per worker is \(sf(k)=0.30(10)=3\) units per worker per year.
The break-even rate is \(\delta+n=0.04+0.02=0.06\) per year, so break-even investment is \(0.06(25)=1.5\) units per worker per year.
The net change is \(\dot k=3-1.5=1.5\) units per worker per year. Investment exceeds the amount needed to replace depreciated capital and equip the growing workforce, so capital per worker rises. Checking the full equation gives \(0.30(2\sqrt{25})-(0.04+0.02)(25)=3-1.5=1.5\).
When Population Growth Changes the Comparison
The same production level and saving rate can imply different changes in capital per worker if population growth differs. A rise in \(n\) increases the break-even line’s slope, so a given amount of investment may no longer be enough to maintain the same \(k\). To isolate this effect, compare economies at the same current capital per worker and with the same production and saving assumptions, changing only \(n\).
Worked Example: Testing Two Population Growth Rates
A hypothetical economy has \(f(k)=2\sqrt{k}\), saving rate \(s=0.25\), depreciation rate \(\delta=0.05\) per year, and current capital per worker \(k=9\). Compare the continuous-time change in capital per worker when \(n=0.05\) and when \(n=0.12\) per year.
Method. Hold output and saving fixed, calculate break-even investment separately for each population growth rate, and compare each amount with investment. This avoids assuming that a change in population growth also changes current production per worker.
At \(k=9\), output per worker is \(f(9)=2\sqrt{9}=6\). Investment per worker is \(0.25(6)=1.5\) units per worker per year in both cases.
When \(n=0.05\), the combined rate is \(0.05+0.05=0.10\), and break-even investment is \(0.10(9)=0.90\). Thus \(\dot k=1.5-0.90=0.60\), so capital per worker rises.
When \(n=0.12\), the combined rate is \(0.05+0.12=0.17\), and break-even investment is \(0.17(9)=1.53\). Thus \(\dot k=1.5-1.53=-0.03\), so capital per worker falls. The check is direct: the higher population growth raises the requirement by \(1.53-0.90=0.63\), changing the net rate from \(0.60\) to \(0.60-0.63=-0.03\).
Common Mistakes and Practical Checks
A frequent error is to count depreciation but leave out population growth. For example, at \(k=20\), \(\delta=0.04\), \(n=0.03\), and investment per worker of \(1.2\), the incorrect calculation using depreciation alone is \(1.2-0.04(20)=1.2-0.8=0.4\), suggesting that capital per worker rises. The correct break-even amount is \((0.04+0.03)(20)=1.4\); the correct net change is \(1.2-1.4=-0.2\), so capital per worker falls. Investment replaces some worn-out capital, but does not meet the additional requirement from workforce growth.
Other checks help prevent errors:
- Apply the saving rate. With a saving rate \(s\), investment per worker is \(sf(k)\), not \(f(k)\).
- Use rates consistently. If \(\delta\) is annual, \(n\) must also be annual. The two rates are added before multiplying by \(k\).
- Keep units aligned. If \(k\) is measured in units of capital per worker, then \((\delta+n)k\) is units of capital per worker per unit of time.
- Check the sign. Investment above break-even investment implies a positive change in \(k\); investment below it implies a negative change.
Check Your Understanding
For the first three questions, use the continuous-time model. For the fourth, use the discrete-time timing convention stated in the question.
- A hypothetical economy has \(k=12\), \(f(k)=8\), \(s=0.30\), \(\delta=0.04\) per year, and \(n=0.02\) per year. Calculate investment per worker, break-even investment, and the direction and rate of change of capital per worker. (Answer: investment \(=2.40\), break-even investment \(=0.72\), so \(\dot k=1.68\) and capital per worker rises.)
- A hypothetical economy has \(f(k)=3\sqrt{k}\), \(k=16\), \(s=0.20\), \(\delta=0.06\) per year, and \(n=0.04\) per year. Find output and investment per worker, then determine the net change in capital per worker. (Answer: output \(=12\), investment \(=2.40\), break-even investment \(=1.60\), so \(\dot k=0.80\) and capital per worker rises.)
- At \(k=20\), a hypothetical economy has output per worker \(f(k)=10\), saving rate \(s=0.25\), and depreciation rate \(\delta=0.03\) per year. What population growth rate makes capital per worker just remain constant? (Answer: \(n=0.095\), or \(9.5\%\) per year.)
- Under the discrete-time convention, let \(k=10\), \(\delta=0.10\), \(n=0.05\), and investment per worker \(i=1.20\). Calculate break-even investment and \(k'\), then state whether capital per worker rises or falls. (Answer: break-even investment \(=1.50\); \(k'=[(1-0.10)(10)+1.20]/1.05=9.7143\), so capital per worker falls.)
- A hypothetical economy has \(k=25\), \(f(k)=2\sqrt{k}\), \(s=0.20\), \(\delta=0.04\) per year, and \(n=0.04\) per year. If population growth rises to \(0.08\) while the other quantities stay fixed, calculate the new break-even investment and net change in capital per worker. (Answer: investment \(=0.20(10)=2.00\); new break-even investment \(=(0.04+0.08)(25)=3.00\); \(\dot k=-1.00\), so capital per worker falls.)