Module 14 · Chapters 19, 20

Long-Run Growth: Solow

Capital, savings, and the steady-state gap between rich and poor.

“Why capital accumulation alone cannot deliver perpetual growth.”

~35 min· 4 sub-skills·5 exercises0% mastered

The Solow growth model strips macro down to two ingredients: a production function with diminishing returns to capital and an exogenous saving rate. The result is the cleanest long-run growth result in macro: in the absence of technological progress, per-capita growth converges to zero.
Production function (intensive form)
$y \;=\; A\,f(k) \;=\; A\,k^{\alpha}$
$y = Y/N$
output per worker
$k = K/N$
capital per worker
$A$
total factor productivity
$\alpha$
capital share, 0 < α < 1
Diminishing returns: ∂²y/∂k² < 0. The first €1 of capital matters more than the millionth.
Steady-state condition
$s\,A\,k^{*\alpha} \;=\; (n + d)\,k^*$
$s$
saving rate
$n$
population growth rate
$d$
depreciation rate
At steady state, investment per worker = capital widening. Solving: $k^* = (sA/(n+d))^{1/(1-\alpha)}$.
Steady-state k* and y*
$k^* = \left(\frac{sA}{n+d}\right)^{\frac{1}{1-\alpha}}, \quad y^* = A\,(k^*)^{\alpha}$
Higher s, A → higher k*, y*. Higher n, d → lower k*, y*. **In steady state, per-capita growth = 0.** Total output grows at n.
Figure · Solow steady-state diagram
Solow Lab
Drag sliders to alter steady-state k* and y*.
k* = 6.03 · y* = 1.81
Per-worker. Curve: s·A·k^α (saving). Line: (n + d)·k (break-even).
Saving rate s0.20
TFP A1.00
Capital share α0.33
Population growth n1.0%
Depreciation d5.0%
Curve: s·A·f(k). Line: (n+d)·k. Intersection at k*.
Insight·Why technology matters
If g_A is the rate of technological progress, the steady-state growth rate of output per worker equals g_A. Solow's punchline: capital accumulation alone cannot sustain long-run growth — you need ongoing innovation. This is why the endogenous growth literature (Romer, Aghion-Howitt) opens the black box of A.
Common misconception
A higher saving rate raises long-run growth.
ActuallyHigher s raises the *level* of y* — but the steady-state growth rate is still g_A. The transition to higher k* is faster growth temporarily; eventually growth returns to g_A.
Exercise · true false · +8 XP
Diminishing returns
In the Cobb-Douglas y = k^α with α = 0.33, doubling k less than doubles y.
"In the Cobb-Douglas y = k^α with α = 0.33, doubling k less than doubles y."
Exercise · numerical · +14 XP
Steady-state k* numerical
s = 0.2, A = 1, α = 0.5, n = 0.01, d = 0.05. Compute k*.
Your answer
Exercise · numerical · +12 XP
Steady-state y* numerical
Continuing: same parameters, with k* ≈ 11.11. Compute y*.
Your answer
Exercise · multiple choice · +12 XP
Conditional convergence
The Solow model predicts that countries with similar s, n, A converge to:
Exercise · multi step · +18 XP
Tech progress and steady-state growth
n = 0.01, g_A = 0.02, d = 0.05.
Context: With Harrod-neutral technical progress at rate g_A, what is the steady-state growth rate of (a) Y, (b) Y/N, (c) Y/(A·N)?
(a)g(Y) — total output growth rate (in %):
(b)g(Y/N) — output per worker (%):
(c)g(Y/(A·N)) — output per effective worker (%):

Mastery check

5 questions · pass with 80%

Answer all five to confirm you've internalised the module. A passing run unlocks the next module.

Q1
Population growth rate n rises. Effect on k*?
Δn > 0; s, A, d, α unchanged.
Q2
In Solow with technical progress, the steady-state per-capita growth rate equals:
Q3
"Countries far below their steady-state k* grow faster than countries near their steady-state k*."
Q4
Cobb-Douglas with α = 0.33. The capital share of income is approximately:
Q5
Endogenous growth models (Romer, Lucas) differ from Solow primarily by:

0 / 5 answered

Exam pitfalls

Saying higher saving rate gives higher long-run growth. It only raises the *level* of y*.
Forgetting that without technical progress, per-capita growth is zero in steady state.
Mixing α (capital exponent) with s (saving rate). Both critical, different roles.
Computing k* with the wrong exponent. The exponent is 1/(1−α), not α.