Economic growth · Lesson 14 of 14

The hard problems: the Delhi School set

Q: What kind of growth questions does the Delhi School entrance ask?

The DSE entrance asks the same models as other exams, but in general form and in sets. A single economy is described once (a production function, a saving rule, a labour-supply rule) and several questions follow: the steady state, its comparative statics, the golden rule, the effect of changing a parameter. The tools are the Solow equation k̇ = sf(k) − (n + δ)k, the golden-rule condition f′(k) = n + δ, and the habit of writing the model in per-worker terms first.

Q: How do you solve a Solow steady-state question with a general production function?

Write output per worker as y = f(k), set saving per worker equal to break-even investment, s f(k) = (n + δ)k, and solve for k. With Cobb–Douglas y = A k^α this gives k* = (sA/(n + δ))^(1/(1 − α)). Remember that total output grows at the rate of labour-force growth n in the steady state, and that a change in the level of productivity A or in the saving rate changes levels but not that growth rate.

Q: How do you find the golden-rule capital stock?

Maximise steady-state consumption per worker, f(k) − (n + δ)k, which gives f′(k) = n + δ. For Cobb–Douglas, f′(k) = αAk^(α−1), so k_gold = (αA/(n + δ))^(1/(1 − α)), and the golden-rule saving rate is α. If the question gives aggregate production, convert to per-worker form first and check whether it wants capital per worker or total capital.

Q: What if the production function is linear in capital?

Then returns to capital are not diminishing and the Solow convergence argument fails. With y = αk + β, the accumulation equation is k̇ = (sα − n − δ)k + sβ. A steady state k* = sβ/(n + δ − sα) exists only if sα < n + δ. Otherwise capital per worker grows without bound, which is the AK case.

Q: How does the Solow model change with separate saving rates for wages and profits?

With competitive factor pay and Cobb–Douglas production, profits are a share α of output and wages 1 − α, so the economy saves the average rate s = s_w(1 − α) + s_r α, and the steady state is k* = (s/(n + δ))^(1/(1 − α)). A transfer from wage earners to profit earners raises the average saving rate when s_r > s_w, and so raises capital per worker.

The Delhi School entrance asks about growth in a form no other exam does: a model described once, then four to ten questions about it, in letters as often as in numbers. Every tool was built in Lessons 2 to 12. This lesson shows how to use them on a whole set, grouped by technique, and lists the few questions whose keys we could not reproduce.

← Economic growth moduleLesson 14 of 14 · about 60 minutes

A question to hold on to

Y = 20 K^1/2L^1/2, no depreciation, a quarter of output saved, labour growing at 10% a year. What is output per worker in the steady state?

Delhi School of Economics entrance, 2011Solved at the end ↓

New to this? The few things you need first

The Solow equation: k̇ = s f(k) − (n + δ)k; steady state where saving per worker equals break-even investment.
The golden rule: f′(k) = n + δ; for Cobb–Douglas the saving rate is α.
Class saving: With wage and profit saving rates the average rate is s_w(1 − α) + s_rα.

The approach

How these questions differ

Three habits solve most of them.

Convert to per-worker form before anything else. Divide the production function by L; replace K by k; write k̇ = s f(k) − (n + δ)k. The options are often traps for a reader who works with totals.
Name what the question varies. A parameter in the production function (A), the saving rate, the population growth rate: the first changes levels, the second changes levels, the third changes levels and the growth of the total.
Check the assumption the standard model needs. Diminishing returns, constant returns, a continuous production function, a fixed saving rate, full employment of labour. When a question changes one of them, expect the answer to change.

Part 1

Part 1: the steady state and what moves it

DSE 2011 · six questions on one economy

Y = A K^αL^1−α, A = 20, α = 1/2, s = 1/4, n = 1/10, no depreciation. Find output per worker in the steady state, the growth of total output, the golden-rule capital and consumption, and what happens when A or n rises.

Per worker.
y = 20 k^1/2, and k̇ = 0.25 × 20k^1/2 − 0.1k.
Steady state.
5k^1/2 = 0.1k gives k^1/2 = 50, so k* = 2,500 and y* = 20 × 50 = 1,000.
Growth of the total.
n = 1/10: with y constant, Y grows with L.
A rises.
k* = (sA/n)² rises, so y* rises; growth of the total is still n.
n rises.
k* falls, y* falls, and total output grows faster (at the new n).

Answer

y* = 1,000; total output grows at 1/10; a higher A raises y* and leaves growth unchanged; a higher n lowers y* and raises growth. The golden-rule values for the same economy are in Part 2.

DSE 2008 and 2009: in the Solow model an increase in the saving propensity leaves the steady-state growth rate unchanged, and raises growth only on the way to the steady state (Lesson 6). DSE 2021 compares two countries with the same technology and starting position, saving rates of 30% and 25%, population growth of 3%, depreciation of 5% and technical progress of 3%: the higher-saving country grows faster at first, and in the steady state both grow at n + g = 6% (the break-even rate is 11%, so k̂* = (s/0.11)^1/(1−α) is higher at 30%). DSE 2021 also asks whether “balanced growth” in the Solow model means identical growth rates in different sectors (no: it is a one-sector model) or constant factor shares (yes: with Cobb–Douglas the shares are constant).

Part 2

Part 2: the golden rule

Golden rule, in two forms

k_{g o l d} = (\frac{α A}{n + δ})^{\frac{1}{1 - α}} c_{g o l d} = f (k_{g o l d}) - (n + δ) k_{g o l d} s_{g o l d} = α

Per worker: f′(k) = n + δ. For Y = K^α L^(1−α) with total capital K and labour L, the same condition is α K^(α−1) L^(1−α) = n + δ.

DSE 2011 · the same economy

Y = 20K^1/2L^1/2, δ = 0, n = 1/10. Find the golden-rule capital per worker and consumption per worker.

Golden rule.
f′(k) = 10k^−1/2 = n = 0.1, so k^1/2 = 100 and k = 10,000.
Consumption.
c = 20 × 100 − 0.1 × 10,000 = 2,000 − 1,000 = 1,000.

Answer

k_gold = 10,000 and c_gold = 1,000. The economy with s = 1/4 had k* = 2,500 and y* = 1,000: saving 1/4 is below the golden rule s = 1/2.

Four more from the same technique. DSE 2018: with Y = K^1/2N^1/2, full depreciation and N = 100, the saving rate that maximises steady-state consumption is α = 1/2 and the capital stock at that rate is 25 (Part 3). DSE 2019: Y = √(KL), with K_t+1 = (1 − δ)K_t + sY_t and constant L: the answer is again s = 1/2; the options that bring in δ are distractors. DSE 2020: Y = (KL)^1/2, L = Γ: steady state sY = δK gives K* = s²Γ/δ², Y* = sΓ/δ and consumption per capita (1 − s)s/δ. DSE 2021: F = L^2/3K^1/3, δ = 10%. Per worker f(k) = k^1/3; f′(k) = (1/3)k^−2/3 = 0.1 gives k^2/3 = 10/3 and k = (10/3)^3/2.

DSE 2009 asked what the golden rule means in the language of wages and profits. At s = α, saving equals the income of capital, so all wages are consumed and all profits invested. It also asked for the definition: the saving rate that maximises consumption per capita in the steady state, not in the transition.

Part 3

Part 3: capital that lasts one period

DSE 2018 · five questions on one economy

Y_t = K_t^1/2N^1/2, N = 100, initial capital 9, half of income saved, 100% depreciation within a period. Find the steady-state capital, output and consumption.

Next period's capital is last period's saving.
K_t+1 = 0.5 × K_t^1/2 × 10 = 5K_t^1/2.
Steady state.
K = 5K^1/2, so K^1/2 = 5 and K* = 25.
Output.
Y* = √(25 × 100) = 50.
Consumption.
C* = Y* − K* = 50 − 25 = 25. (Half of income.)

Answer

K* = 25, Y* = 50, C* = 25. The initial capital of 9 is a distraction: the path 9 → 15 → 19.36 → … converges to 25. With full depreciation all of saving is replacement, so consumption is output minus capital.

Part 4

Part 4: linear technology and the Inada conditions

DSE 2013 · four questions on one economy

Y = αK + βL (a linear function), wages β and rental α, a share s saved, no depreciation, labour growing at n. Which neoclassical property does it violate? What is the equation of motion for k, and the steady state for α = 1/2, β = 12, s = 1/4, n = 1/2?

What it violates.
Diminishing returns to each factor: the marginal product of capital is the constant α.
Per worker.
y = αk + β.
Accumulation.
k̇ = s(αk + β) − nk = (sα − n)k + sβ.
Numbers.
k̇ = (1/8 − 1/2)k + 3 = 3 − (3/8)k = 0 gives k* = 8.

Answer

It violates diminishing returns; k̇ = sαk + sβ − nk; k* = 8. In general k* = sβ/(n − sα), which exists only if sα < n; otherwise k grows without bound. (The follow-up question on a rise in s is among those left out below.)

DSE 2021 posed the same point with exponential output: if the economy produces e^k per capita from k units of capital, then k̇ = s e^k − (n + δ)k. Since s e^k/k has a minimum of s·e at k = 1, if s·e > n + δ then saving always exceeds break-even investment and capital grows without bound, with no steady state. If s·e < n + δ there are two steady states, one stable and one unstable. So “there may not be a steady state” is the correct option.

Part 5

Part 5: saving out of wages and profits

The average saving rate

s = s_{w} (1 - α) + s_{r} α k^{*} = (\frac{s}{n})^{\frac{1}{1 - α}} (no depreciation)

With Cobb–Douglas and competitive factor payments, the wage bill is (1 − α)Y and profits are αY, so saving is a weighted average of the two rates.

DSE 2015 · three of the questions on one economy

Y = K^αL^1−α, households save s_w of wages and s_r of interest, labour grows at n, no depreciation. (a) With s_r = 0, what does a rise in n do? (b) With both positive and s_w < s_r, to what does k converge? (c) Wages are taxed at τ and the revenue goes to capital owners: effect of a rise in τ?

(a) s_r = 0.
s = s_w(1 − α), and k* = (s_w(1 − α)/n)^1/(1−α) falls as n rises.
(b) The limit.
k* = [((1 − α)s_w + αs_r)/n]^1/(1−α).
(c) Tax and transfer.
Saving is s_w(1 − τ)(1 − α)Y + s_r(α + τ(1 − α))Y. Its derivative with respect to τ is (1 − α)(s_r − s_w) > 0.

Answer

(a) k* falls, unambiguously; (b) the expression in (b); (c) the average saving rate rises with τ because income is moved to the group that saves more, so k* rises, unambiguously.

Part 6

Part 6: a labour market inside a growth model

DSE 2006. Y = min{K/2, L/4}, 20% of output saved, no depreciation, labour grows by 4 units a period; K = 200 and L = 420. Output is limited by the scarcer factor: K/2 = 100 and L/4 = 105, so Y = 100 and employment is the labour needed, 4 × 100 = 400 (20 workers are unemployed). Investment of 20 makes K = 220 and L = 424; now K/2 = 110 and L/4 = 106, labour is the scarce factor, Y = 106 and employment is 424: the economy goes from capital-limited unemployment to full employment of labour with idle capital. The answer is 400 and 424.

DSE 2014 · a short-run labour market with endogenous supply

N identical firms with Y = K^αL^1−α in total. H households, each with 1 unit of labour, supply l to maximise wl − l^δ (δ > 1). Find labour demand, labour supply and the market-clearing wage in the short run.

Labour demand.
w = (1 − α)K^αL^−α, so L^d = [(1 − α)/w]^1/αK.
Labour supply.
w = δl^δ−1, so l = (w/δ)^1/(δ−1) while this is below 1, that is for w < δ; above that all of the unit is supplied. L^s = H(w/δ)^1/(δ−1) for w < δ, and H for w ≥ δ.
Clearing, with w below δ.
[(1 − α)/w]^1/αK = H(w/δ)^1/(δ−1) gives w* = [K(1 − α)^1/αδ^1/(δ−1)/H]^{α(δ−1)/(α+δ−1)}.
Full employment region.
Once K ≥ K̄ = H(δ/(1 − α))^1/α, L = H and w* = (1 − α)(K/H)^α.

Answer

Demand [(1 − α)/w]^1/αK; supply as above; wage w* below K̄ and (1 − α)(K/H)^α above it. The number of firms N does not appear: with identical competitive firms and the aggregate technology given, N does not change the wage (DSE 2014 Q58).

The fixed-disutility variant (DSE 2014): each household gets wl − D if it works (l = 1) and 0 if it does not. It works if w ≥ D, so aggregate supply is 0 for w < D and H for w ≥ D. The market-clearing wage is D for small K (the wage cannot fall below D, and employment is limited by demand) and (1 − α)(K/H)^α once labour demand at D reaches H.

Part 7

Part 7: growth accounting and research

DSE 2021 and 2018

(a) The wage bill is 40 and output 100; output grows 10%, capital 10% and labour 5%. Find TFP growth. (b) Y = A(1 − λ)L₀ with Ȧ = λL₀A and a fixed λ = λ₀. Find the long-run growth rate.

(a) Shares.
Labour share 40/100 = 0.4, capital share 0.6.
(a) Residual.
10 − 0.6 × 10 − 0.4 × 5 = 2%.
(b) Growth of A.
Ȧ/A = λ₀L₀. Output Y is proportional to A with λ and L₀ fixed, so it grows at the same rate.

Answer

(a) 0.02. (b) λ₀L₀: the Romer result of Lesson 11 with no capital.

Honesty

Questions we left out, and why

Twelve Delhi School growth items are not in the practice sets, because the key does not follow from the question as printed. We list them so you can see the arithmetic, and so you do not lose marks by doubting a correct derivation. In each case, if you met the question in an exam, solve it and mark the answer the key would want if it is clear which one that is.

DSE 2013 Q60. The linear economy above with α = n = 1/2. The steady state is k* = sβ/(n − sα) = 24s/(1 − s), which exists for every s below 1 and rises with s. The key says the effect is ambiguous because a steady state may not exist for large s, which is true only at s = 1.
DSE 2015 Q54. With s_w = 0 and s_r = 1, k* = (α/n)^1/(1−α). Its derivative with respect to α has the sign of (1 − α)/α + ln(α/n), which is positive for every n between 0 and 1. The key says “increases if α > n”.
DSE 2015 Q58 and Q59. Q58 is void in the bank: no printed option is correct. With labour growth n = Ak below a threshold the steady state is k = (s/A)^1/(2−α), which is not the constant printed in any option. Q59 depends on it.
DSE 2014 Q54. The keyed constant for equilibrium output does not match the wage of Q53 (checked numerically against the market-clearing solution). Q57 and Q59 print the capital threshold K̃ with exponent α where the wage formula in the same option needs 1/α. Q55 and Q60 key “output approaches a constant in the long run”, but with no depreciation capital never stops growing (k̇ = ½Y > 0), so output grows without bound, ever more slowly.
DSE 2018 Q32. With Y = A(1 − λ)L₀ and Ȧ = λL₀A at a price of one, the value of the marginal product of labour is A in both sectors for every λ, so the allocation is not determined. The key is 1/2.
DSE 2019 Q46 and DSE 2020 Q11. Both ask about growth of income per head at two capital–labour ratios below the steady state. One key says “none of the above” and the other g₁ > g₂. For Cobb–Douglas, g₁ > g₂; for a general production function the ordering can depend on the shape of f.

Solving the opening question

The opening question, solved

DSE 2011 · standard

Y = 20K^1/2L^1/2, s = 1/4, n = 1/10, δ = 0. Steady-state output per worker?

Per worker.
y = 20k^1/2.
k̇ = 0.
0.25 × 20k^1/2 = 0.1k, so k^1/2 = 50.
Output.
y* = 20 × 50 = 1,000.

Answer

1,000.

Watch out for

The mistakes examiners are fishing for

Answering for capital when asked for output. k* = 2,500 here, y* = 1,000: different options.
Forgetting n in total-output growth. In the steady state total output grows at n (and at n + g with technical progress).
Using aggregate capital when per-worker is asked. Check the units in the options.
Assuming a steady state exists. With linear or exponential output it may not.
Averaging saving rates without shares. The weights are factor shares of output.

Revise

The lesson on one screen

Steady state: s f(k) = (n + δ)k. Cobb–Douglas k* = (sA/(n + δ))^1/(1−α). Total output grows at n.
Golden rule: f′(k) = n + δ; s = α; all wages consumed, all profits invested.
Full depreciation: K_t+1 = sY_t; iterate or solve K = sY(K).
Linear technology: k̇ = (sα − n − δ)k + sβ; k* = sβ/(n + δ − sα) if sα < n + δ.
Class saving: s = s_w(1 − α) + s_rα; moving income to the higher saver raises k*.
Labour market: Demand [(1 − α)/w]^1/αK; supply from w = δl^δ−1; clear.
Accounting, research: Residual = g_Y − αg_K − (1 − α)g_L; Ȧ/A = λL₀.

Real questions

The Delhi School questions

Every DSE growth question above Easy that we could verify, 35 in all, grouped by technique, moderate before hard within each part.

Part 1 · from Lessons 5 and 6

The steady state and what moves it

Solow economies with letters and numbers: the steady-state output per worker, the growth of the total, and the effect of a change in saving, productivity or population.

8 questionsPractise →

Part 2 · from Lesson 7

The golden rule

Capital, consumption and the saving rate at the golden rule, in per-worker and in aggregate form, and the wage-and-profit reading of it.

9 questionsPractise →

Part 3 · from Lesson 6

Capital that lasts one period

A Solow economy with full depreciation, iterated by hand to its steady state.

3 questionsPractise →

Part 4 · from Lessons 2 and 5

Linear technology and the Inada conditions

What breaks when output is linear in capital and labour, or grows exponentially in capital: no diminishing returns, and perhaps no steady state.

4 questionsPractise →

Part 5 · from Lessons 6 and 12

Saving out of wages and out of profits

The Solow economy with different saving rates for wage and profit income, a transfer between the two, and what it does to the long-run capital–labour ratio.

3 questionsPractise →

Part 6 · from Lesson 5

A labour market inside a growth model

Fixed proportions with unemployment, and a labour market with endogenous supply, cleared period by period while capital accumulates.

6 questionsPractise →

Part 7 · from Lessons 3 and 11

Growth accounting and research

The Solow residual from a wage share, and the balanced growth rate of an economy with an idea-producing sector.

2 questionsPractise →

All 35 DSE questions, moderate before hard within each part: 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 kind of growth questions does the Delhi School entrance ask?

The DSE entrance asks the same models as other exams, but in general form and in sets. A single economy is described once (a production function, a saving rule, a labour-supply rule) and several questions follow: the steady state, its comparative statics, the golden rule, the effect of changing a parameter. The tools are the Solow equation k̇ = sf(k) − (n + δ)k, the golden-rule condition f′(k) = n + δ, and the habit of writing the model in per-worker terms first.

How do you solve a Solow steady-state question with a general production function?

Write output per worker as y = f(k), set saving per worker equal to break-even investment, s f(k) = (n + δ)k, and solve for k. With Cobb–Douglas y = A k^α this gives k* = (sA/(n + δ))^(1/(1 − α)). Remember that total output grows at the rate of labour-force growth n in the steady state, and that a change in the level of productivity A or in the saving rate changes levels but not that growth rate.

How do you find the golden-rule capital stock?

Maximise steady-state consumption per worker, f(k) − (n + δ)k, which gives f′(k) = n + δ. For Cobb–Douglas, f′(k) = αAk^(α−1), so k_gold = (αA/(n + δ))^(1/(1 − α)), and the golden-rule saving rate is α. If the question gives aggregate production, convert to per-worker form first and check whether it wants capital per worker or total capital.

What if the production function is linear in capital?

Then returns to capital are not diminishing and the Solow convergence argument fails. With y = αk + β, the accumulation equation is k̇ = (sα − n − δ)k + sβ. A steady state k* = sβ/(n + δ − sα) exists only if sα < n + δ. Otherwise capital per worker grows without bound, which is the AK case.

How does the Solow model change with separate saving rates for wages and profits?

With competitive factor pay and Cobb–Douglas production, profits are a share α of output and wages 1 − α, so the economy saves the average rate s = s_w(1 − α) + s_r α, and the steady state is k* = (s/(n + δ))^(1/(1 − α)). A transfer from wage earners to profit earners raises the average saving rate when s_r > s_w, and so raises capital per worker.

Why do some Delhi School growth questions have no valid answer?

A few have a key that does not follow from the question as printed: a typo in a formula, an option set with no correct choice, or a claim that fails at a boundary. This lesson lists the ones we have set aside, with the arithmetic, so that you can see why. If you meet such a question, work the model out and say what follows from it.

← Lesson 13Growth in practice: East Asia, India and institutions

The hard problems: the Delhi School set

How these questions differ

Part 1: the steady state and what moves it

Part 2: the golden rule

Part 3: capital that lasts one period

Part 4: linear technology and the Inada conditions

Part 5: saving out of wages and profits

Part 6: a labour market inside a growth model

Part 7: growth accounting and research

Questions we left out, and why

The opening question, solved

The mistakes examiners are fishing for

The lesson on one screen

The Delhi School questions

The steady state and what moves it

The golden rule

Capital that lasts one period

Linear technology and the Inada conditions

Saving out of wages and out of profits

A labour market inside a growth model

Growth accounting and research

Frequently asked questions

Read next

The hard problems: the Delhi School set

How these questions differ

Part 1: the steady state and what moves it

Part 2: the golden rule

Part 3: capital that lasts one period

Part 4: linear technology and the Inada conditions

Part 5: saving out of wages and profits

Part 6: a labour market inside a growth model

Part 7: growth accounting and research

Questions we left out, and why

The opening question, solved

The mistakes examiners are fishing for

The lesson on one screen

The Delhi School questions

The steady state and what moves it

The golden rule

Capital that lasts one period

Linear technology and the Inada conditions

Saving out of wages and out of profits

A labour market inside a growth model

Growth accounting and research

Frequently asked questions

Read next