Economic growth · Lesson 6 of 14

Solow in motion: saving, population and technology

The Solow model has a steady state, and an economy that sits in it is a still picture. This lesson sets it moving. Change the saving rate, the population growth rate or depreciation, or destroy some capital, and watch the economy travel to a new resting point: which things change where it ends up, which change how it grows on the way, and how long the trip takes. Then technical progress is added, which is what finally gives the model growth per worker.

← Economic growth moduleLesson 6 of 14 · about 50 minutes

The short answer

How do saving, population growth and technical progress change the Solow steady state?

In the basic Solow model, parameters change levels, not long-run growth rates. A permanent rise in the saving rate speeds growth for a while and then growth returns to its old rate, with output per worker settled on a higher path. Faster population growth lowers income per head. Sustained growth in income per head needs technical progress.

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

Saving, population growth and depreciation move the level of the steady state. A higher s raises capital and output per worker; a higher n or δ lowers them.
Only technical progress, g, sets the long-run growth rate of output per worker, and the model takes it as given. Total output grows at n + g.
The transition takes time. The gap to the steady state closes at rate λ = (1 − α)(n + g + δ), about 5% a year for textbook values, and λ does not depend on s.
A shock that leaves the parameters alone (an earthquake, a war) is undone: the economy grows faster than usual until it is back on its old path.

A question to hold on to

In the Solow model an economy is in its steady state when an earthquake destroys a third of its capital stock. The saving rate, the depreciation rate and population growth do not change. Does growth of output rise or fall, and where does the economy end up?

Hyderabad 2020Solved at the end ↓

New to this? The few things you need first

The Solow steady state: k̇ = s f(k) − (n + δ)k. At k* saving per worker equals break-even investment, k stops changing, and total output grows at n.
Neutral technical progress: Technical progress is Harrod-neutral when it acts like extra labour, Y = F(K, AL). Only that form is consistent with steady growth when n and g are constant.
Growth rates of ratios: The growth rate of a ratio is the difference of the growth rates: g_K/L = g_K − g_L.

What changes what

Level effects and growth effects

In Lesson 5 the steady state was k* = (s/(n + δ))^1/(1−α). That one formula says what every parameter does. Read down the rows below by asking how k* moves, and what that does to output per worker, y* = (k*)^α.

A permanent rise in…	Capital and output per worker, k* and y*	Long-run growth of output per worker	Long-run growth of total output	On the way
the saving rate, s	Rise	0, unchanged	n, unchanged	Growth is faster for a time
population growth, n	Fall	0, unchanged	new n, higher	Growth per worker is slower for a time
the depreciation rate, δ	Fall	0, unchanged	n, unchanged	Growth per worker is slower for a time
the level of productivity, A	y* rises in proportion to A^1/(1−α)	0, unchanged	n, unchanged	Growth is faster for a time

Every row of the “long-run growth” columns is the same, and that is the point. These are level effects: the economy moves to a different resting point, and its output per worker grows faster or slower along the way, but in the long run it is again not growing. A growth effect would change the long-run rate itself. In the basic model nothing inside the model has one.

Two exam statements follow directly. A higher saving rate “raises the growth rate of output in the short run” (the transition) but not the long-run rate (UGC NET 2025). And a higher population growth rate lowers the steady-state capital per worker while raising the steady-state growth rate of aggregate output (JAM 2021, GATE 2026).

The capital–output ratio in the steady state

\frac{K}{Y} = \frac{k}{y} = \frac{s}{n + δ}

Because k/y = k^(1−α) and k*^(1−α) = s/(n + δ), the ratio depends on s, n and δ only. It rises with s and falls with n or δ.

This ratio matters for one frequently asked question. UGC NET 2025 lists the effects of faster population growth in the neoclassical model, and the key accepts a fall in capital per head and a rise in the growth rate of total output. The statement that the capital–output ratio decreases is also true by the formula above, since n is in the denominator, yet it is not in the key, so that question is left out of the practice sets below. If you meet a question with this combination, trust the algebra.

Pause and predict

In the Solow model without technical progress, depreciation falls permanently. In the new steady state, output per worker

Interactive

Hit an economy and watch it recover

Meridia is an invented economy that has sat at its steady state for decades: saving 25%, population growth 1%, depreciation 5%, technical progress 2% and α = 0.33. Choose a shock. The top chart shows the natural logarithm of output per worker, so a straight line is steady growth; the dashed line is where the old parameters would have taken Meridia and the solid line is where it goes now. The lower chart shows the growth rate.

Growth jumps and then fades back to 2%; output per worker ends up higher for good. A level effect, not a growth effect.

Notice the shape of each answer. After the rise in saving, the solid line leaves the dashed line and then runs parallel to it: a level effect. After the rise in technology growth the two lines have different slopes for ever: a growth effect, and the only one. After the earthquake the solid line starts below the dashed line, grows steeply, and rejoins it. The earthquake is a useful test of understanding: because it changes no parameter, it leaves the destination alone and only displaces the economy, so the economy grows faster than normal until the lost capital is rebuilt.

Hyderabad 2020 · standard

y = √k, saving rate 0.4, depreciation 0.1, no population growth. The economy is at its steady state when 75% of its capital is destroyed. Find the growth rates of capital and of output immediately afterwards.

The steady state.
$0.4 k = 0.1 k \Rightarrow k = 4, k^{*} = 16$
After the earthquake.
k = 25% of 16 = 4, so y = √4 = 2.
Net investment.
$k ˙ = 0.4 (2) - 0.1 (4) = 0.8 - 0.4 = 0.4$
Growth rates.
Capital grows at k̇/k = 0.4/4 = 10%. For y = k^0.5 output grows at half that rate, 5%.

Answer

Capital grows at 10% and output at 5% a year, far above the steady-state rate of zero, and both slow down as k approaches 16 again. The economy returns to the original steady state, because nothing that determines k* has changed.

The missing piece

Technical progress and effective labour

Without technical progress the model has no growth per worker, which contradicts the facts of Lesson 1. Add technology that raises the effectiveness of each worker, so that production is $Y = F (K, A L)$ and A grows at a constant rate g. The product AL is effective labour. This is Harrod-neutral technical progress, the only kind compatible with a steady state when growth is constant (Lesson 2).

Now measure everything per unit of effective labour: $k^= K / (A L)$ and $y^= Y / (A L) = f (k^)$ . The growth rate of effective labour is n + g, so exactly as in Lesson 5 the growth rate of the ratio is the growth rate of K minus n + g:

The Solow equation with technical progress

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

The same equation as before, with g added to the break-even rate: new capital must now also equip the extra effectiveness of each worker.

At the steady state k̂ and ŷ are constant. But the variables people care about are Y/L = Aŷ and K/L = Ak̂, and these grow because A grows. The whole table of growth rates follows:

Variable	Steady-state growth rate	Why
Capital per effective worker, k̂	0	The definition of the steady state
Output per worker, Y/L	g	Y/L = A·ŷ with ŷ constant
Capital per worker, K/L	g	K/L = A·k̂ with k̂ constant
Total output, Y, and total capital, K	n + g	Y = ŷ·AL, with AL growing at n + g
Capital–output ratio, K/Y	0	K/Y = k̂/ŷ
Real wage, w	g	w = A·(ŷ − k̂ f′(k̂)), with the bracket constant
Rental rate and profit rate, r	0	r = f′(k̂), constant

Compare this with the list of Kaldor’s stylised facts in Lesson 1: output per worker grows at a steady rate, the capital–output ratio is constant, the profit rate is constant, wages grow with productivity. The Solow model with Harrod-neutral progress reproduces every one of them, which is why it became the benchmark. The price is that the long-run growth rate g is not explained: it is assumed. Explaining it is the work of Lessons 10 and 11.

Standard

Y = K^0.5(AL)^0.5, s = 0.24, n = 0.01, g = 0.02, δ = 0.05. Find k̂*, ŷ*, and the long-run growth rates of output per worker and of total output.

Break-even rate.
n + g + δ = 0.01 + 0.02 + 0.05 = 0.08.
Steady-state capital per effective worker.
$k^^{*} = (\frac{0.24}{0.08})^{\frac{1}{0.5}} = 3^{2} = 9$
Output per effective worker.
ŷ* = √9 = 3.
Growth rates.
Output per worker grows at g = 2%. Total output grows at n + g = 3%.

Answer

k̂* = 9, ŷ* = 3; output per worker grows at 2% and total output at 3%. The saving rate 0.24 appears only in the level, and would change neither growth rate.

Pause and predict

In the Solow model with labour-augmenting technical progress at rate g and population growth n, the steady-state growth rate of output per worker is

The transition

How fast the economy converges

The steady state says where the economy is going; transition dynamics say how long it takes. Take the growth rate of k̂ and ask how fast it responds to the gap:

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

This is zero at k̂* and falls as k̂ rises. Differentiate the right-hand side with respect to ln k̂ at the steady state, using s k̂*^α−1 = n + g + δ: the slope is −(1 − α)(n + g + δ). So near the steady state the log gap shrinks at a constant rate:

The speed of convergence

λ = (1 - α) (n + g + δ) t_{1/2} = \frac{ln 2}{λ}

Near the steady state, d ln(k̂/k̂*)/dt = −λ ln(k̂/k̂*). The half-life of the gap is ln 2 / λ. For Cobb–Douglas the result is exact in the variable k̂^(1−α) (see the derivation page).

Three things to notice. First, λ does not depend on the saving rate: s decides where the economy is going, not how fast it gets there. Second, a larger α slows convergence, because capital then loses its productivity more slowly, so poor economies get a smaller boost from being poor. Third, the numbers: with α = 1/3, n = 1%, g = 2% and δ = 5%, λ = (2/3)(0.08) = 5.3% a year and half the gap closes in about 13 years. That is fast. Real economies converge much more slowly, about 2% a year in cross-country data. Lesson 8 returns to this gap between the model and the evidence. The exact solution of the equation, which proves the formula for λ, is in the Solow equation, derived and solved.

Standard

α = 0.4, n = 1%, g = 2%, δ = 7%. Find the speed of convergence and the half-life.

Break-even rate.
n + g + δ = 0.10.
Speed.
λ = (1 − 0.4)(0.10) = 0.06, or 6% a year.
Half-life.
ln 2 / 0.06 = 0.693 / 0.06 = 11.6 years.

Answer

6% a year; half of any gap to the steady state closes in about 11.6 years.

Why the transition matters for policy

If convergence took a few years the distinction between levels and growth would hardly matter, because a country that raised its saving rate would reach its new steady state before the next election. The half-life of a decade or more is what makes the transition, with its extra growth, a real part of the story of a country that raises its saving rate.

The passage from one steady state to another has its own name in growth theory: the traverse (Hicks’s term). Steady-state analysis compares the two end points; the traverse is the path between them, and it is analytically awkward because along it neither saving nor the capital–labour ratio sits at its long-run value. UOH 2018 asks for its name.

Two variants

Two variants the exams like

1. A model in periods, with all capital used up each period

In discrete time the capital stock evolves by K_t+1 = (1 − δ)K_t + sY_t. With labour growing at n this gives k_t+1 = [(1 − δ)k_t + s f(k_t)]/(1 + n), and setting k_t+1 = k_t = k* gives s f(k*) = (n + δ)k*: the same steady state as in continuous time. The special case δ = 1, in which capital is fully used up within a period, makes K_t+1 = sY_t, which is easy to iterate by hand.

Delhi School style

Y_t = K_t^1/2N^1/2, N = 100 (constant), saving rate 0.4, 100% depreciation each period. Find the steady-state capital stock, output and consumption.

Capital next period.
K_t+1 = sY_t = 0.4 × 10√K_t = 4√K_t.
Steady state.
K = 4√K, so √K = 4 and K* = 16.
Output.
Y* = √(16 × 100) = 40.
Consumption.
C* = (1 − 0.4) × 40 = 24.

Answer

K* = 16, Y* = 40, C* = 24. The Delhi School entrance papers set exactly this kind of economy several times; Lesson 14 collects them.

2. Different saving rates out of wages and profits

Suppose workers save a share s_w of wage income and capitalists a share s_r of profit income. With Cobb–Douglas production and competitive factor payments, profits are a share α of output and wages 1 − α. Total saving is therefore (s_w(1 − α) + s_rα)Y, an ordinary Solow economy with the average saving rate

s = s_{w} (1 - α) + s_{r} α

For α = 1/3, s_w = 0.1 and s_r = 0.4 this gives s = 0.1 × 2/3 + 0.4 × 1/3 = 0.2. Raising the capital share α now raises the saving rate when s_r > s_w, a link between distribution and growth that Kaldor and the Cambridge school made central (Lesson 12).

Solving the opening question

The opening question, solved

Hyderabad 2020 · standard

An economy in its Solow steady state loses part of its capital in an earthquake. Saving, depreciation and population growth are unchanged. What happens to growth and to the steady state?

The steady state.
It satisfies s f(k*) = (n + δ)k*, which involves only s, n, δ and the production function. None has changed, so k* is the same.
After the quake.
k is below k*, so saving per worker s f(k) exceeds break-even investment and k̇ > 0.
Growth.
With k rising, output per worker rises: growth is above its steady-state rate while the economy rebuilds. As k approaches k* it slows to the steady-state rate.

Answer

Growth of output rises, and the economy ends up at the original steady state. Option A. The earthquake changed the starting point, not the destination.

Watch out for

The mistakes examiners are fishing for

Saying a higher saving rate raises long-run growth. It raises the level and growth only on the way.
Forgetting g in the break-even line. With technical progress the break-even rate is n + g + δ, not n + δ.
Mixing per-worker and per-effective-worker. The steady state is for k̂; output per worker is Aŷ and keeps growing at g.
Thinking an earthquake changes the steady state. A shock to the stock is not a change in parameters.
Using λ = n + δ. The speed of convergence is (1 − α)(n + g + δ), with the factor 1 − α.

Revise

The lesson on one screen

Level effects: s, n, δ and the level of A change k* and y*; none changes long-run growth per worker.
Growth effects: Only g, the rate of technical progress, which the model takes as given.
Effective labour: Y = F(K, AL). k̂ = K/(AL) obeys k̂̇ = s f(k̂) − (n + g + δ)k̂.
Steady-state rates: Per worker: g. Total output and capital: n + g. K/Y, r: constant. Wage: g.
Population: Higher n lowers k*, y* and K/Y, and raises total growth.
Speed: λ = (1 − α)(n + g + δ); half-life ln 2 / λ; independent of s.
Shocks: A shock to capital with the parameters unchanged is reversed. The path between steady states is the traverse.

Practice

Check yourself

0 of 16 correct so far

A. Numbers

Problem 1

Y = K^0.5(AL)^0.5, s = 0.24, n = 0.01, g = 0.02, δ = 0.05. Find k̂*, and the long-run growth rates of output per worker and of total output, in per cent.

k̂*Per worker (%)Total (%)

Problem 2

α = 0.4, n = 1%, g = 2%, δ = 7%. Find the speed of convergence λ (per cent a year) and the half-life of a gap in years (one decimal).

λ (%)Half-life (years)

Problem 3

y = k^0.5, s = 0.2, δ = 0.05, no technical progress. Population growth rises from 1% to 3%. Find y* before and after (two decimals).

y* beforey* after

Problem 4

y = k^0.5, s = 0.4, δ = 0.1, n = 0. The economy is at its steady state when 75% of its capital is destroyed. Find k̇ and the growth rate of output (per cent) immediately afterwards.

k̇Growth of y (%)

Problem 5

Y_t = K_t^1/2N^1/2, N = 100, s = 0.4, 100% depreciation each period. Find the steady-state capital stock and output.

K*Y*

B. Concepts

1A permanent increase in the saving rate in the basic Solow model

2In a Solow economy with labour-augmenting technical progress at rate g and population growth n, total output in the steady state grows at

3The speed of convergence in the Solow model depends on

4An economy in a Solow steady state loses a third of its capital stock in a flood; its parameters are unchanged. Afterwards

5The movement of an economy from one steady state to another is called

Real questions

Past papers, by level

Questions on what changes in the Solow model, transitions, and the path between steady states. The Delhi School’s full-depreciation and class-saving economies are in Lesson 14 when it opens; the golden rule is the next lesson.

Standard

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

3 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 3 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 happens in the Solow model when the saving rate rises?

The steady-state capital per worker, output per worker and consumption per worker (up to the golden rule) all rise, so a higher saving rate has a level effect. It does not change the long-run growth rate of output per worker, which is set by technical progress. During the transition to the new steady state, growth is temporarily faster, which is why a question can say both that saving raises growth in the short run and that it does not in the long run.

How does population growth affect the Solow steady state?

Faster population growth raises break-even investment, (n + δ)k, because more capital is needed to equip each new worker. Capital per worker and output per worker in the steady state therefore fall. Total output, which grows at the rate of the labour force, now grows faster. So population growth lowers the level of income per head but raises the growth rate of aggregate output.

What is the steady-state growth rate of output per worker in the Solow model with technical progress?

With labour-augmenting (Harrod-neutral) technical progress at rate g, output per worker grows at g in the steady state and total output grows at n + g. Capital per effective worker, k̂ = K/(AL), is the variable that settles down, and it satisfies k̂̇ = s f(k̂) − (n + g + δ)k̂.

What is the speed of convergence in the Solow model?

Near the steady state, the gap between capital per effective worker and its steady-state value closes at the rate λ = (1 − α)(n + g + δ). For α = 1/3 and n + g + δ = 8% a year, λ is about 5.3% a year and half the gap closes in about 13 years. The saving rate does not appear in λ, so it changes where the economy is going but not how fast it gets there.

What is the difference between a level effect and a growth effect?

A level effect moves the economy to a higher or lower path for output per worker but leaves the long-run growth rate unchanged: the path shifts in parallel. A growth effect changes the long-run growth rate itself, so the path gets steeper or flatter for ever. In the Solow model, saving, population growth and depreciation have level effects. Only the rate of technical progress has a growth effect, and it is taken from outside the model.

Why does the Solow model use capital per effective worker?

If technology A grows at rate g, then output per worker and capital per worker both keep growing, so there is no constant to find. Dividing by effective labour, AL, gives k̂ = K/(AL) and ŷ = Y/(AL), which do settle down. The steady state is found for these, and the growth rates of the original variables are recovered by adding g back: per worker, g; in total, n + g.

← Lesson 5The Solow–Swan model: capital accumulation and the steady state Lesson 7 →The golden rule: how much should an economy save?

Solow in motion: saving, population and technology

Level effects and growth effects

Hit an economy and watch it recover

Technical progress and effective labour

How fast the economy converges

Two variants the exams like

1. A model in periods, with all capital used up each period

2. Different saving rates out of wages and profits

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

Standard

Delhi School level

Frequently asked questions

Read next

Solow in motion: saving, population and technology

Level effects and growth effects

Hit an economy and watch it recover

Technical progress and effective labour

How fast the economy converges

Two variants the exams like

1. A model in periods, with all capital used up each period

2. Different saving rates out of wages and profits

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

Standard

Delhi School level

Frequently asked questions

Read next