How to Calculate the Solow Steady State
With y = k^α, steady-state capital per worker is k* = (s ÷ (n + d))^(1/(1 − α)). Then output y* = (k*)^α and consumption c* = (1 − s) × y*.
The Solow Steady State formula
Calculator
With y = k^α, enter the capital share, saving rate, population growth and depreciation to get the steady state per worker.
As a decimal: 0.2 means 20% of output is saved.
As a decimal per year.
As a decimal per year.
Where saving s × k^α just covers (n + d) × k.
- Output per worker (y*)
- 4
- Consumption per worker (c*)
- 3.2
- Investment per worker
- 0.8
The share of output not saved.
Equals break-even investment (n + d) × k* = 0.8.
How to calculate Solow Steady State, step by step
- 1Set saving equal to break-even investment. Capital per worker stops changing when investment s × k^α exactly covers the capital lost to depreciation plus the capital needed to equip new workers, (n + d) × k.
- 2Solve for k*. Divide both sides by (n + d) × k^α to get s ÷ (n + d) = k^(1 − α), then raise both sides to the power 1 ÷ (1 − α).
- 3Compute output per worker. y* = (k*)^α. The shortcut y* = (s ÷ (n + d))^(α/(1 − α)) gives the same number and works as a cross-check.
- 4Compute consumption per worker. Workers consume the output they do not save: c* = (1 − s) × y*. Investment per worker is the remaining share, s × y*.
- 5Check the break-even condition. Compute s × y* and (n + d) × k*. They must be equal; if they are not, an arithmetic step went wrong.
Worked example: Solow Steady State
Let α = 0.5, s = 0.2, n = 0.01 and d = 0.04, so n + d = 0.01 + 0.04 = 0.05. Then s ÷ (n + d) = 0.2 ÷ 0.05 = 4, and the exponent is 1 ÷ (1 − 0.5) = 2, so k* = 4² = 16. Output per worker is y* = 16^0.5 = 4 (the shortcut gives 4^(0.5/0.5) = 4^1 = 4 as well). Consumption per worker is c* = (1 − 0.2) × 4 = 0.8 × 4 = 3.2, and investment per worker is 0.2 × 4 = 0.8. Check: break-even investment is (n + d) × k* = 0.05 × 16 = 0.8, equal to investment per worker, so capital per worker holds steady at 16.
How each parameter moves the steady state
A higher saving rate raises k* and y*, because more output goes back into capital each period. Raise s from 0.2 to 0.3 in the example and s ÷ (n + d) = 0.3 ÷ 0.05 = 6, so k* = 6² = 36, y* = 36^0.5 = 6, and c* = 0.7 × 6 = 4.2.
Faster population growth or faster depreciation works the other way. Either one raises n + d, so more investment is needed just to keep capital per worker where it is, and the steady state settles lower. With s = 0.2 and n + d = 0.08, s ÷ (n + d) = 2.5, so k* = 2.5² = 6.25 and y* = 6.25^0.5 = 2.5.
The exponent matters too. With α = 1/3 instead of 0.5, and s = 0.2 and n + d = 0.05 as before, the exponent on 4 becomes 1 ÷ (2/3) = 1.5, so k* = 4^1.5 = 8, y* = 8^(1/3) = 2 and c* = 0.8 × 2 = 1.6. With these parameters, a smaller α means capital runs into diminishing returns sooner, so the same saving rate supports less capital and output per worker (k* falls from 16 to 8, y* from 4 to 2).
None of these changes produce lasting growth in output per worker. They move the level the economy converges to, and once it arrives, y* stays put until a parameter changes again. That is why the model points to technological progress as the source of sustained growth.
The saving rate that maximizes consumption: the golden rule
More saving always raises y*, but it does not always raise consumption, because a larger share of a larger output is set aside. The saving rate that maximizes steady-state consumption is the golden rule, and with y = k^α it is s = α.
Run it with α = 0.5 and n + d = 0.05. At s = 0.4, s ÷ (n + d) = 8, so k* = 64, y* = 8 and c* = 0.6 × 8 = 4.8. At s = 0.5, s ÷ (n + d) = 10, so k* = 100, y* = 10 and c* = 0.5 × 10 = 5. At s = 0.6, s ÷ (n + d) = 12, so k* = 144, y* = 12 and c* = 0.4 × 12 = 4.8. Consumption peaks at s = 0.5.
The same point shows up in the marginal product of capital. At the golden rule, MPK = α × k^(α − 1) equals n + d. Here that is 0.5 × 100^(−0.5) = 0.5 × 0.1 = 0.05, which matches n + d = 0.05.
Solow Steady State questions
Why is the Solow steady state where investment equals break-even investment?
Capital per worker rises by investment s × y and falls through depreciation d × k and through spreading capital over a workforce growing at rate n. When the two sides match, the change in k is zero and k stops moving.
Does a higher saving rate raise long-run growth in the Solow model?
No. It raises the steady-state levels k* and y*, and growth is faster only during the transition to the new steady state. Lasting growth in output per worker comes from technological progress.
How does technological progress change the Solow steady state?
With labor-augmenting technology growing at rate g, the same algebra runs in capital per effective worker, and the break-even term becomes (n + d + g) × k. In that steady state, output per worker grows at rate g.
Why is the Solow model written in per-worker terms?
With constant returns to scale, Y = K^α × L^(1 − α) divided by L gives y = k^α. That turns a model with two inputs into one with a single variable to track, capital per worker k.
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