Lesson M18.L02: The Solow Model: Full Mathematical Treatment

Module: Economic Growth Part I Level: intermediate Duration: 30 minutes Learning Objective: Derive the steady-state capital per effective worker from the fundamental Solow equation. Data as of: 2024 Provenance: MIT OCW 14.02 | ABS National Accounts

Explanation

The Solow-Swan model (Solow 1956, Swan 1956) is the foundation of modern growth theory. It explains long-run GDP per worker through capital accumulation and exogenous technological progress.

Production Function (aggregate):

Y = F(K, AL) = Kᵅ (AL)^(1−α)

Notation: - Y = real GDP; K = capital stock; L = labour (workers); A = level of technology (labour-augmenting) - AL = "effective labour" (efficiency units of labour) - α ∈ (0,1) = capital elasticity of output = capital income share (by competitive factor pricing); α ≈ 0.35 for Australia - 1−α = labour income share ≈ 0.65 - n = growth rate of L (population/labour force growth); n ≈ 0.015 for Australia - g = growth rate of A (technological progress); g ≈ 0.015 for Australia - δ = depreciation rate of capital; δ ≈ 0.05 for Australia - s = gross saving/investment rate; s ≈ 0.25 for Australia (rounded from ~24%; used throughout examples)

Intensive form (per effective worker): Define:

k = K / (AL)   [capital per effective worker]
y = Y / (AL)   [output per effective worker]

Substituting into the production function:

y = Y/(AL) = Kᵅ(AL)^(1−α) / (AL) = (K/(AL))ᵅ = kᵅ

Capital accumulation equation: Gross investment = sY. Capital per effective worker changes as:

k̇ = dk/dt = s·y − (δ + n + g)·k = s·kᵅ − (δ + n + g)·k

The term (δ + n + g)k is "break-even investment" — the investment needed to: - Replace depreciated capital (δk) - Equip new workers with existing capital per worker (nk) - Maintain k as A grows (gk, since more efficient workers "dilute" existing capital per effective worker)

Steady State: Set k̇ = 0:

s·k*ᵅ = (δ + n + g)·k*
s·k*^(α−1) = (δ + n + g)
k*^(1−α) = s / (δ + n + g)
k* = [s / (δ + n + g)]^(1/(1−α))

Steady-state output per effective worker:

y* = k*ᵅ = [s / (δ + n + g)]^(α/(1−α))

Key comparative statics: - Higher s → higher k* and y* (but not higher long-run growth rate — only level effect) - Higher δ, n, or g → lower k* and y* (more break-even investment required) - Higher α → more sensitive k* to changes in s

Worked Example

Australian parameters: s = 0.25, δ = 0.05, n = 0.015, g = 0.015, α = 0.35

Step 1 — Compute break-even investment rate:

δ + n + g = 0.05 + 0.015 + 0.015 = 0.080

Step 2 — Compute s/(δ+n+g):

s / (δ + n + g) = 0.25 / 0.080 = 3.125

Step 3 — Compute the exponent:

1/(1−α) = 1/(1 − 0.35) = 1/0.65 = 1.5385

Step 4 — Compute k:*

k* = 3.125^(1.5385)

Taking logs: ln(k*) = 1.5385 × ln(3.125)
ln(3.125) = 1.1394
ln(k*) = 1.5385 × 1.1394 = 1.7529
k* = e^1.7529 = 5.771

Step 5 — Compute y:*

y* = k*ᵅ = 5.771^0.35

Taking logs: ln(y*) = 0.35 × ln(5.771) = 0.35 × 1.7529 = 0.6135
y* = e^0.6135 = 1.847

Step 6 — Verify steady state (check k̇ = 0):

Investment: s·y* = 0.25 × 1.847 = 0.4618
Break-even: (δ+n+g)·k* = 0.080 × 5.771 = 0.4617  ✓  (small rounding difference)

Interpretation: In steady state, an economy with Australian parameters has k* ≈ 5.77 units of capital per effective worker and produces y* ≈ 1.85 units of output per effective worker. 25% of this output (≈ 0.46) is invested, exactly replacing depreciation and equipping new effective workers.

Step 7 — Output per actual worker (grows at rate g in steady state):

Y/L = y × A = k*ᵅ × A
Since A grows at rate g = 1.5%, Y/L grows at 1.5% per year in steady state.

This matches Australia's long-run GDP per capita growth of ~1.5–2%.

Common Misconception

Misconception: "A higher saving rate leads to permanently faster economic growth in the Solow model."

Correction: In the Solow model, a higher saving rate raises the level of k* and y* (a level effect), but not the long-run growth rate. In steady state, output per effective worker y* is constant regardless of s. Output per actual worker Y/L grows at rate g (technological progress), independent of s. The saving rate determines where on the growth path the economy sits, not how fast it travels along it. Permanent growth effects require endogenous growth models (Romer, Lucas) where knowledge/technology is itself produced by saving and investment.

Practice Prompts

1. Conceptual: Why must break-even investment include a term for technological progress (gk) in the capital accumulation equation? → Answer: Capital per effective worker k = K/(AL). As A grows at rate g, the denominator (AL) grows, so k would fall even if K were constant. To maintain k at its current level, investment must also cover this "dilution" effect. More precisely: d(K/AL)/dt = (K̇·AL − K·(Ȧ·L + A·L̇)) / (AL)² = K̇/(AL) − K·(n+g)/(AL) = (K̇/AL) − (n+g)k. Setting this equal to sk^α − (n+g)k gives the full equation. The gk term represents the investment needed to equip each unit of new technological capability with the existing capital-per-effective-worker ratio.

2. Numerical: An economy has s = 0.30, δ = 0.06, n = 0.02, g = 0.02, α = 0.40. Derive k* and y*. → Answer:
   δ + n + g = 0.06 + 0.02 + 0.02 = 0.10
   s/(δ+n+g) = 0.30 / 0.10 = 3.0
   1/(1−α) = 1/(1−0.4) = 1/0.6 = 1.6667
   k* = 3.0^(1.6667)
   ln(k*) = 1.6667 × ln(3.0) = 1.6667 × 1.0986 = 1.8310
   k = e^1.8310 = 6.238**
   y = k*^0.4 = 6.238^0.4
   ln(y*) = 0.4 × 1.8310 = 0.7324
   y = e^0.7324 = 2.080**
   Verify: s·y = 0.30 × 2.080 = 0.624; (δ+n+g)·k* = 0.10 × 6.238 = 0.624 ✓

3. Application: If Australia's saving rate rose from 0.25 to 0.30 (a 5 pp increase), by what percentage would the steady-state capital per effective worker (k*) increase? Use α = 0.35, δ = 0.05, n = 0.015, g = 0.015. → Answer:
   k*(s=0.30) = [0.30/0.080]^(1/0.65) = [3.75]^1.5385
   ln(k*_new) = 1.5385 × ln(3.75) = 1.5385 × 1.3218 = 2.0337
   k*_new = e^2.0337 = 7.637
   From worked example: k*_old = 5.771
   Percentage increase = (7.637 − 5.771)/5.771 = 1.866/5.771 = 0.323 = 32.3%
   A 20% rise in the saving rate (from 0.25 to 0.30) raises k* by approximately 32%, illustrating how the level effect is amplified through the exponent 1/(1−α) = 1.54.

Visual — Solow Mathematics in Phase-Diagram Form

Figure: The Solow steady state k is where the saving curve sk^α intersects the break-even line (δ+n+g)k. Left of k*, capital deepening occurs (k̇ > 0); right of k*, capital per effective worker shrinks (k̇ < 0).*

Further Resources

• 📺 Solow Growth Model Part 1: Model Intro & Solution — Intermediate Macro Series (18 min)
• 📺 Solow model #3 — Steady State (algebra) — Macro Algebra Series (12 min)
• 📚 MIT OCW 14.02 Lecture Notes on Growth — Full mathematical treatment of the Solow model