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Lesson M09.L02: Capital Accumulation and Depreciation

Module: Savings, Capital Formation, and Economic Growth Level: intro Duration: 30 minutes Learning Objective: Apply the capital accumulation equation ΔK = I − ÎīK to determine the conditions for steady-state capital. Data as of: 2024 Provenance: OpenStax Macro 3e | MIT OCW 14.02

Explanation

Capital (K) refers to the stock of physical assets used in production — machines, buildings, infrastructure, and equipment. The capital stock changes over time through two opposing forces: investment (I) adds to it, and depreciation (ÎīK) erodes it.

Depreciation (Îī, the Greek letter "delta") is the fraction of the capital stock that wears out each period. If Îī = 0.10, then 10% of capital deteriorates every year.

The capital accumulation equation is:

ΔK = I − ÎīK

where ΔK is the change in the capital stock over one period.

Steady-state capital (K*) occurs when ΔK = 0 — investment exactly replaces worn-out capital:

I = ÎīK*

Because investment equals saving times output (I = sY in the Solow framework, where s is the saving rate), at steady state:

sY = ÎīK*

Two key marginal productivity formulas describe how capital and labour contribute to output, using a Cobb-Douglas production function Y = AK^α L^(1−α):

  • Marginal product of capital: MPK = Îą(Y/K)
  • Marginal product of labour: MPL = (1−α)(Y/L)

MPK tells us how much extra output one additional unit of capital produces. As K rises (holding L constant), MPK falls — diminishing returns to capital. This is crucial: it means simply adding more capital yields smaller and smaller gains, eventually driving the economy to a steady state.

For Australia, the ABS estimates the national capital stock at over $5 trillion, with an average depreciation rate of roughly 5–6% per year across all asset types.

Worked Example

Scenario: Australia's economy has: - Capital stock K = $5,000b - Investment I = $440b - Depreciation rate Îī = 0.05 (5% per year) - Output Y = $2,200b, capital share Îą = 0.35 (so labour share = 1−α = 0.65)

Step 1 — Calculate depreciation this period:

ÎīK = 0.05 × 5,000 = $250b

Step 2 — Calculate the change in the capital stock:

ΔK = I − ÎīK = 440 − 250 = +$190b

Capital is growing — the economy is still below its steady state.

Step 3 — Calculate MPK:

MPK = Îą × (Y/K) = 0.35 × (2,200 / 5,000) = 0.35 × 0.44 = 0.154

Each additional $1b of capital adds approximately $154m of output.

Step 4 — Calculate MPL (labour share = 1 − α = 0.65, assume L = 14 million workers):

MPL = (1−α) × (Y/L) = 0.65 × (2,200 / 14) = 0.65 × 157.14 = $102,143 per worker

Each additional worker adds roughly $102,000 of output.

Step 5 — Find steady-state capital (assuming saving rate s = 0.20):

At steady state: sY* = ÎīK* We need sAK*^Îą = ÎīK* (in per-worker terms, but simplified here) Using the ratio: K*/Y* = s/Îī = 0.20/0.05 = 4 So K* = 4 × Y* — the steady-state capital-to-output ratio is 4.

If current K/Y = 5,000/2,200 ≈ 2.27, which is below 4, we confirm capital is still accumulating.

Common Misconception

Misconception: "Depreciation is always bad for the economy — it reduces capital, so the government should try to eliminate it."

Correction: Depreciation is an unavoidable physical reality. More importantly, it plays a crucial role in the Solow model: without depreciation, capital would accumulate indefinitely and the model would have no steady state. Depreciation ensures old capital is replaced, making room for newer, more productive assets. The policy goal is not to eliminate depreciation but to maintain investment above ÎīK to grow (or sustain) the capital stock.

Practice Prompts

  1. In the capital accumulation equation ΔK = I − ÎīK, what economic conditions lead to ΔK = 0, and what is the significance of this point? → Answer: ΔK = 0 when investment exactly equals depreciation (I = ÎīK). This is the steady state — the capital stock stops changing. It is significant because it represents the long-run equilibrium where the economy settles; beyond this point, adding more capital (without improving technology or saving) produces no further growth in per-capita output.

  2. NUMERICAL CALCULATION: An economy has K = $3,000b, Îī = 0.08, I = $280b, Y = $1,500b, Îą = 0.40. Calculate (a) depreciation, (b) ΔK, and (c) MPK. → Answer: (a) Depreciation = ÎīK = 0.08 × 3,000 = $240b (b) ΔK = I − ÎīK = 280 − 240 = +$40b (capital still accumulating) (c) MPK = Îą × (Y/K) = 0.40 × (1,500 / 3,000) = 0.40 × 0.50 = 0.20 Each $1b of capital adds $200m of output.

  3. Australia increases its national saving rate from 18% to 22% of GDP. Using the steady-state condition sY = ÎīK*, explain what happens to the long-run capital stock. → Answer: At the new, higher saving rate, investment (sY) exceeds depreciation (ÎīK) at the old steady state, so ΔK > 0 and capital accumulates. The economy moves to a new, higher steady-state capital stock K*. In the long run, both K* and output Y* are higher. However, the growth rate of per-capita output eventually returns to zero — sustained growth requires technological progress (TFP), not just more saving.

Further Resources