Economic growth · Lesson 5 of 14

The Solow–Swan model: capital accumulation and the steady state

Harrod and Domar left growth balanced on a knife edge. Robert Solow and Trevor Swan changed one assumption, letting capital and labour substitute, and the knife edge turned into a stable resting point. This lesson builds the model from its assumptions to its one equation and its one picture, which is probably the most-used diagram in growth theory.

← Economic growth moduleLesson 5 of 14 · about 55 minutes

The short answer

What is the Solow growth model, and what is its steady state?

The Solow–Swan model is the standard neoclassical growth model. A fixed share of output is saved and added to capital; some capital wears out; the workforce grows. Because each extra unit of capital adds less than the last, the economy settles in a steady state where capital per worker stops changing.

k ˙ = s f (k) - (n + δ) k s f (k^{*}) = (n + δ) k^{*} k^{*} = (\frac{s}{n + δ})^{\frac{1}{1 - α}}

One equation: capital per worker rises when saving per worker, s·f(k), exceeds the investment needed to stand still, (n + δ)k.
The steady state k* is where the two are equal. Capital and output per worker are constant there, and total output grows at the rate of labour-force growth, n.
It is stable: diminishing returns pull the economy to k* from any start, unlike Harrod–Domar’s knife edge.
Saving changes the level, not the long-run growth rate. Sustained growth per worker needs technical progress (Lesson 6).

A question to hold on to

An economy follows the Solow model with y = √k. The steady-state output per worker is y^ss = A^1/(1−α)(γ/δ)^α/(1−α), where γ is the investment share and δ the depreciation rate, both in per cent. Suppose A = 1, k = 400, γ = 50%, δ = 5% and α = 1/2. Is current output above, at, or below the steady-state level of output per worker?

IIT JAM 2022Solved at the end ↓

New to this? The few things you need first

The intensive form: With constant returns, output per worker depends only on capital per worker: y = f(k), with f′ > 0 and f″ < 0. For Cobb–Douglas, f(k) = k^α.
Depreciation and widening: Part of investment only replaces capital that wears out, and part equips new workers at the old capital per worker. Only the rest raises capital per worker.
Growth rates: The growth rate of a ratio is the difference of the growth rates: if K grows at g_K and L at n, then K/L grows at g_K − n.

The idea

The model in a handful of assumptions

The basic Solow model is a closed economy producing one good. It has five ingredients:

Production: $Y = F (K, L)$ with constant returns to scale and diminishing returns to each input, usually Cobb–Douglas (Lesson 2).
Saving: households save a constant fraction s of income, and everything saved is invested: $I = s Y$ .
Depreciation: a fraction δ of capital wears out each period.
Labour: the workforce grows at a constant rate n, from outside the model.
No technical progress, for now. It is added in Lesson 6.

Compare this with Harrod–Domar. There the capital–output ratio was fixed. Here it adjusts: if capital is plentiful relative to workers, its marginal product falls, and the economy uses it differently. That one change removes the knife edge.

The key step

From assumptions to one equation

Capital grows by what is invested and shrinks by what wears out:

K ˙ = s Y - δ K

Now convert to per worker, $k = K / L$ . Taking the growth rate of k, it is the growth rate of K minus the growth rate of L: $k ˙ / k = K ˙ / K - n$ . Substitute and use Y/K = y/k:

The fundamental equation of the Solow model

k ˙ = s f (k) - (n + δ) k

Saving per worker, minus break-even investment: the amount needed to replace depreciation (δk) and to give each new worker the same capital as the rest (nk).

Read it as a budget. The first term is how much capital is added per worker. The second, break-even investment, is how much must be added just to keep k constant: δk to replace what wore out and nk to spread the capital over the new workers. If saving exceeds break-even investment, k rises (capital deepening). If it falls short, k falls.

Derivation

Derive k̇ = s·f(k) − (n + δ)k from K̇ = sY − δK and Y = F(K, L) with constant returns.

Differentiate k = K/L.
$k ˙ = \frac{K ˙}{L} - \frac{K}{L} \frac{L ˙}{L} = \frac{K ˙}{L} - nk$
Substitute K̇.
$\frac{K ˙}{L} = \frac{s Y - δ K}{L} = s y - δ k$
Use y = f(k).
$k ˙ = s f (k) - δ k - nk = s f (k) - (n + δ) k$

Answer

That is the fundamental equation. Everything in this lesson, and the next two, comes from it. The same derivation with every symbol defined, and the exact solution of the equation, is on its own page: the Solow equation, derived and solved.

Interactive

The Solow diagram and the steady state

Draw the two terms as functions of k. The saving curve s·f(k) bends over, because of diminishing returns. The break-even line (n + δ)k is straight. They cross once, at the steady state k*. To the left of it saving is higher, so k rises; to the right the line is higher, so k falls. The lower chart follows k through time from the starting point you choose. Move the sliders and watch where the curves cross.

Capital per worker is below the steady state, so saving exceeds the investment needed to keep up and k rises.

The steady state is the value k* at which k̇ = 0:

The steady state

s f (k^{*}) = (n + δ) k^{*} ⟹ k^{*} = (\frac{s}{n + δ})^{\frac{1}{1 - α}}, y^{*} = (k^{*})^{α}, c^{*} = (1 - s) y^{*}

For Cobb–Douglas, f(k) = k^α. Raise saving and k* rises; raise population growth or depreciation and it falls.

CUET-PG 2023 · standard

y = k^0.5, depreciation 0.1, saving rate 0.4, no population growth. Find capital per capita in the steady state.

Set k̇ = 0.
$0.4 k^{0.5} = 0.1 k$
Divide both sides by the square root of k.
$k^{0.5} = \frac{0.4}{0.1} = 4$
Square.
k* = 16.

Answer

16. The same question with the saving rate chosen to maximise steady-state consumption (the “golden rule”, Lesson 7) gives 50%.

IIT JAM 2026 · hard

Y = K^αL^1−α, α = 0.5, saving rate 20%, depreciation 7% and labour growth 3%. Find the steady-state capital–labour ratio.

Break-even rate.
n + δ = 0.03 + 0.07 = 0.10.
Apply the formula.
$k^{*} = (\frac{0.20}{0.10})^{\frac{1}{1 - 0.5}} = 2^{2}$

Answer

GATE 2024 · hard

Output Y = 0.2K + 0.8L (a linear function), depreciation 0.2, saving rate 0.5, no population growth. Find steady-state capital per unit of labour.

Per worker.
y = 0.2k + 0.8, so saving per worker is 0.5(0.2k + 0.8) = 0.1k + 0.4.
k̇ = 0.
0.1k + 0.4 = 0.2k, so 0.1k = 0.4.
Solve.
k* = 4.

Answer

4. A steady state exists here because the break-even line has a steeper slope (0.2) than the saving line (0.1). The Inada conditions are violated, but the steady state is still found by setting k̇ = 0.

JNU 2011 · foundation

Two countries are identical in every respect except that A’s saving rate is lower than B’s. In the long run, which has the higher income per person?

Steady state.
k* = (s/(n + δ))^1/(1−α) is increasing in s.
Compare.
B has the larger k*, so the larger y*.

Answer

The higher-saving country has the higher level of income per person. JNU 2021 puts numbers on it: with α = 1/3, n = δ = 0.1, s_A = 0.2 and s_B = 0.4, k*_A = (0.2/0.2)^1.5 = 1 and k*_B = 2^1.5.

What the steady state means

What grows at what rate in the steady state

In the steady state k is constant, and so is y = f(k). But the workforce grows at n, so the totals must grow too:

Variable	Growth rate in the steady state	Why
Capital per worker, k	0	The definition of the steady state
Output per worker, y	0	y = f(k) and k is constant
Consumption per worker, c	0	c = (1 − s)y
Total capital, K	n	K = kL, with k constant
Total output, Y	n	Y = yL, with y constant

Three exam results follow. Output grows at the population growth rate: IIT JAM 2021 asks what happens to aggregate output growth when population growth rises to a new value: it settles at the new rate (and GATE 2026 gives the same answer, n). The saving rate has no effect on the long-run growth rate, only on the level (UGC NET 2025: a higher saving rate raises growth in the short run, during the transition). And growth per worker is zero without technical progress (JNU 2018: faster labour-force growth has no effect on the long-run growth rate of per-capita income, because that rate is zero whatever n is).

Pause and predict

In the Solow model without technical progress, the saving rate rises permanently. In the new steady state, output per worker

Pause and predict

Which statement about the steady state is NOT correct?

The contrast with Harrod–Domar

Why the economy settles: stability

Why does k converge to k* from any starting point? Look at the growth rate of capital per worker:

\frac{k ˙}{k} = s \frac{f ( k )}{k} - (n + δ)

Capital per worker grows when saving per unit of capital, $s f (k) / k$ , exceeds n + δ. With diminishing returns, f(k)/k falls as k rises: capital is spread over the same workers and each unit yields less. So when k is small, saving per unit of capital is high and k grows quickly; when k is large it is low and k falls. At k* the two sides are equal. A growth rate that falls as the variable rises is a force that always pulls it back to a steady state.

In Harrod–Domar a departure from the warranted path fed on itself. Here it dies away. That is what “stable” means, and it is why Solow’s model displaced the knife edge. JNU 2019 poses the contrast as an assertion and reason: Solow is a major improvement because it builds a long-run growth model without fixed proportions. And JNU 2016: capital accumulation drives long-term growth in Harrod–Domar but not in Solow.

The deepest point

Diminishing returns are what make the Solow economy settle. Remove them, so that capital never loses its productivity, and the economy never settles: that is the AK model of Lesson 10.

Solving the opening question

The opening question, solved

IIT JAM 2022 · standard

y = √k, k = 400, investment share γ = 50%, depreciation δ = 5%, α = 1/2, A = 1. Is current output above, at, or below the steady state?

Current output.
y = √400 = 20.
Steady-state output.
$y^{ss} = A^{\frac{1}{1 - α}} (\frac{γ}{δ})^{\frac{α}{1 - α}} = (\frac{50}{5})^{1} = 10$
Compare.
20 > 10.

Answer

Above the steady state. The economy has more capital per worker than it can maintain (k* = 100, but k = 400), so saving falls short of break-even investment, k will fall, and output will decline towards 10.

Watch out for

The mistakes examiners are fishing for

Steady state means no growth. Per-worker variables are constant; total output grows at n (and with technology, at n + g).
Forgetting n in the break-even line. Break-even investment is (n + δ)k, not δk. Population growth dilutes capital.
Mixing k* and y*. k* = (s/(n+δ))^1/(1−α); y* = (k*)^α. Check which one the question asks for.
Saving raises long-run growth. In the Solow model it raises the level only, and the growth only temporarily.
Applying δ to output. Depreciation is a fraction of the capital stock, not of output.

Revise

The lesson on one screen

Assumptions: CRS and diminishing returns; saving s; depreciation δ; labour growth n; no technology yet.
Fundamental equation: k̇ = s·f(k) − (n + δ)k. Saving per worker minus break-even investment.
Steady state: k̇ = 0: s·f(k*) = (n + δ)k*. Cobb–Douglas: k* = (s/(n+δ))^1/(1−α).
Growth rates: Per worker: 0. Totals K and Y: n.
Stability: Diminishing returns make k converge to k* from any start.
Saving: Raises k*, y* and c* (up to the golden rule) but not long-run growth.
Against Harrod–Domar: Substitutable inputs, no knife edge, saving affects level not growth.

Practice

Check yourself

0 of 10 correct so far

A. Numbers

Problem 1

y = k^0.5, s = 0.3, n = 0.01, δ = 0.05. Find k*, y* and c*.

k*y*c*

Problem 2

y = k^0.5, s = 0.25, n = 0.02, δ = 0.06, and k = 10. Find k̇, the change in capital per worker (to three decimal places).

k̇

Problem 3

In a Solow economy without technical progress the labour force grows 1.5% a year. What are the steady-state growth rates of output per worker and of total output, in per cent?

Per worker (%)Total (%)

B. Concepts

1In the Solow model, an increase in the population growth rate (n) leads to

2Two countries are identical except that A saves less than B. In the long run, according to the basic Solow model, income per person

3The condition for a steady state in the Solow model is that

4Why does the Solow economy converge while the Harrod–Domar economy does not?

Real questions

Past papers, by level

Questions on the steady state, its stability, and what grows at what rate, from CUET-PG, Hyderabad, JNU, IIT JAM and GATE. Questions on saving, population, transitions and technology are in Lesson 6.

Foundation

Definitions, one-step arithmetic and which-model questions. Most CUET and Hyderabad questions sit here.

3 questionsPractise →

Standard

Growth accounting, steady-state algebra, statement questions and matching, up to JNU, IIT JAM and UGC NET level.

12 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 14: The steady state and what moves it.

8 questionsLesson 14 →

All 15 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

Quick answers

Frequently asked questions

What is the Solow growth model?

The Solow–Swan model is the standard neoclassical growth model. Output is produced from capital and labour with constant returns to scale and diminishing returns to each input. A constant fraction s of output is saved and invested, capital depreciates at rate δ, and the labour force grows at rate n. The economy converges to a steady state in which capital per worker is constant.

What is the capital accumulation equation in the Solow model?

In per-worker terms, k̇ = s f(k) − (n + δ) k, where k is capital per worker, f(k) is output per worker, s is the saving rate, n the growth rate of the labour force and δ the depreciation rate. The first term is saving (investment) per worker; the second is the investment needed to replace depreciated capital and to equip new workers with the same capital per worker.

How do you find the steady state in the Solow model?

Set k̇ = 0, so that saving per worker equals break-even investment: s f(k*) = (n + δ) k*. For a Cobb–Douglas function y = k^α this gives k* = (s/(n + δ))^(1/(1 − α)). Output per worker in the steady state is y* = (k*)^α and consumption per worker is (1 − s) y*.

What is the steady-state growth rate in the Solow model without technical progress?

Capital per worker and output per worker are constant, so growth per worker is zero. Total capital and total output grow at the rate of labour-force growth, n. A change in the saving rate changes the level of output per worker but not its long-run growth rate, which is why the model needs technical progress to explain sustained growth per worker.

Is the Solow steady state stable?

Yes. Because of diminishing returns, saving per worker rises more slowly than the break-even line. Below k* saving exceeds break-even investment and capital per worker rises; above k* it falls. So the economy converges to k* from any positive starting point. This is the main difference from the Harrod–Domar knife edge.

What is the difference between the Solow and Harrod–Domar models?

Harrod–Domar has fixed proportions of capital and labour and growth equal to s/v, which makes growth depend permanently on saving and makes the warranted path unstable. Solow lets capital and labour substitute with diminishing returns to capital, so the economy settles in a stable steady state where a higher saving rate raises the level of income but not the long-run growth rate.

← Lesson 4The Harrod–Domar model and the knife edge Lesson 6 →Solow in motion: saving, population and technology

The Solow–Swan model: capital accumulation and the steady state

The model in a handful of assumptions

From assumptions to one equation

The Solow diagram and the steady state

What grows at what rate in the steady state

Why the economy settles: stability

The opening question, solved

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

A. Numbers

B. Concepts

Past papers, by level

Foundation

Standard

Delhi School level

Frequently asked questions

Read next

The Solow–Swan model: capital accumulation and the steady state

The model in a handful of assumptions

From assumptions to one equation

The Solow diagram and the steady state

What grows at what rate in the steady state

Why the economy settles: stability

The opening question, solved

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

A. Numbers

B. Concepts

Past papers, by level

Foundation

Standard

Delhi School level

Frequently asked questions

Read next