NATIONAL UNIVERSITY OF SINGAPORE
ECA5102 MACROECONOMICS
Sample Exam
Solution
Time Allowed: 2 Hours
Note: This sample exam contains three questions from the past
midterm and final exams. Before you attempt to solve these ques-
tions, please first try to understand the lecture notes, relevant
textbook materials, redo the practice questions, and assign-
ment questions.
1. (40 marks) (AY2018/19 Final Exam) This problem consid-
ers a variation on the Solow model that we learned in class.
Suppose that instead of the population growing at a constant
rate, that people have fewer children as they become wealthier.
In particular, as always suppose that output is produced via a
Cobb-Douglas production function Y = KαL1−α, where Y is
the aggregate real output, K is the aggregate capital stock, L is
the total labor force, and α ∈ (0, 1) is a parameter. Consumers
save a constant fraction s of their income, and the capital stock
depreciates at rate δ. The capital stock evolves according to
the following equation
dK
dt
≡ K˙ = sY − δK
Define the per-capita variables as k = K/L, y = Y/L, and c =
C/L. However, instead of being a constant, now the population
growth rate is a decreasing function of per-capita income y:
L˙
L
= ny
α−1
α
where n is a constant, and 0 < n < s. Assume α−1
α
< 0,
capturing the declining population growth rates of wealthier
nations.
ECA5102
(a) Derive the production function in intensive form, i.e., y as
a function of k. (2 marks)
Solution: y = Y
L
= K
αL1−α
L
=
(
K
L
)α
= kα
(b) Derive the equation of motion for per-capita quantities of
capital k˙ ≡ dk
dt
= d(K/L)
dt
. (6 marks)
Solution: L˙
L
= ny
α−1
α = n (kα)
α−1
α = nkα−1
Divide by L in the capital accumulation equation:
K˙
L
= sy − δk = skα − δk.
Then remember that k = K/L, so:
k˙ =
K˙
L
− K
L
L˙
L
= skα − δk − knkα−1 = (s− n)kα − δk
(c) Solve for the steady state per-capita quantities of capital,
output, consumption, and population growth, i.e., k∗, y∗, c∗
and
(
L˙
L
)
steady state
. (10 marks)
Solution: The steady state is at k˙ = 0. Solve for steady
state:
0 = (s− n) (k∗)α − δk∗ ⇒ k∗ =
(
s− n
δ
) 1
1−α
Then
y∗ = (k∗)α = (
s− n
δ
)
α
1−α
c∗ = (1− s) y∗ = (1− s)(s− n
δ
)
α
1−α(
L˙
L
)
steady state
= n (k∗)α−1 = n(
s− n
δ
)−1 =
δ
s/n− 1
(d) What are the growth rates of aggregate capital stock K,
aggregate output Y , and aggregate consumption C in the
steady state? (6 marks)
Solution: Because K = kL, and the steady state per-
capita quantities of capital k∗ is a constant, taking logs
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ECA5102
and differentiating with respect to time t, we have(
K˙
K
)
steady state
=
(
L˙
L
)
steady state
=
δ
s/n− 1 .
Similarly,(
Y˙
Y
)
steady state
=
(
L˙
L
)
steady state
=
δ
s/n− 1 ,
and (
C˙
C
)
steady state
=
(
L˙
L
)
steady state
=
δ
s/n− 1 .
(e) What are the short run (transitional dynamics) and long
run (steady state) effects of an increase in n on the per-
capita quantities of capital, output, consumption, and pop-
ulation growth? (16 marks)
Solution:
Long-run effect:
Clearly, long run (steady state) per-capita quantities of
capital, output and consumption are decreasing in n, but
steady state population growth is increasing.
Short-run effect:
Recall that the equation for capital evolution is:
k˙t = (s− n)kαt − δkt.
(s− n)kαt shifts downwards as n increases and δkt doesn’t
shift (Figure 1 below), thus kt falls in the short run and
converges downwards to the new steady state. yt and ct
have the same tendency. However, population growth has
a different one. Notice that L˙
L
= nkα−1, where α − 1 < 0.
The population growth will jump upwards because of an
increase in n initially, and will continue to grow up to its
steady state level as k converges downwards to the new k∗′
(Figure 2 below).
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ECA5102
Figure 1: Shifts of the Curves
Figure 2: Population Growth Rate
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ECA5102
2. (24 marks) (AY2016/17 Midterm Exam) Consider a Ramsey-
Cass-Koopman economy that we learned in class. The dynamic
equations for c(t) (consumption per unit of effective labor) and
k(t) (capital stock per unit of effective labor) are given by:
˙c(t)
c(t)
=
f ′[k(t)]− δ − ρ− θg
θ
˙k(t) = f [k (t)]− c(t)− (n+ g + δ) k(t)
where f [k (t)] is output per unit of effective labor, n is the
population growth rate, θ is the inverse of the intertemporal
elasticity of substitution, g is the growth rate of technology , ρ
is the discount rate, and δ is the depreciation rate of capital.
Let k˙ ≡ dk
dt
and c˙ ≡ dc
dt
. Parameters n, θ, g , ρ, and δ are all
greater than zero. Assume the production function in intensive
form is y(t) = f [k (t)] = k(t)α, where α ∈ (0, 1).
(a) Assume that the economy is on the balanced growth path
with k (t) = k∗. Solve for the value of k∗ . (8 marks)
Solution: The balanced-growth-path capital stock per unit
of effective labor is implicitly defined by:
f ′ (k∗)− δ − ρ− θg = 0
Because f(k) = kα, f ′(k) = αkα−1, this implies
k∗ =
(
α
δ + ρ+ θg
) 1
1−α
(b) Define the balanced-growth-path savings rate as s∗ ≡ (y∗ − c∗) /y∗.
Solve for s∗. (8 marks)
Solution: The balanced-growth-path consumption per unit
of effective labor is
c∗ = y∗ − (n+ g + δ) k∗
Because y∗ = (k∗)α and s∗ = (y∗ − c∗) /y∗, the savings rate
is
s∗ = (n+ g + δ) (k∗)1−α =
α (n+ g + δ)
δ + ρ+ θg
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ECA5102
(c) Solve for the Golden-Rule capital stock per unit of effective
labor kGR. (8 marks)
Solution: The golden-rule capital stock per unit of effec-
tive labor is:
kGR = argmax
k
{kα − (n+ g + δ) k}
The first order condition with respect to k is
αkα−1 − (n+ g + δ) = 0.
The solution to this problem is
kGR =
[
α
n+ g + δ
] 1
1−α
3. (36 marks) (AY2016/17 Midterm Exam) Consider an over-
lapping generations (OLG) economy that we learned in class in
which each individual lives for two periods. The optimization
problem of an individual born at time t is:
max
{C1,t,C2,t+1}
{lnC1,t + lnC2,t+1}
subject to
C1,t +
1
1 + rt+1
C2,t+1 = wt
where C1,t and C2,t+1 are her consumption levels in the two
periods, rt+1 is the interest rate in period t + 1 and wt is her
income in the first period of her life (she does not work in the
second period). The income is equal to the wage rate because
she supplies one unit of labor in the first period. There are
Lt such individuals who were born at time t. The production
function in this economy is given by
Yt = K
α
t (Lt)
1−α ,
where Yt is the aggregate real output, Kt is the aggregate capi-
tal stock, Lt is the total labor force, and α ∈ (0, 1) is a param-
eter. All markets are competitive and factors are paid at their
marginal products. There is no depreciation. The labor force
grows at an exogenous rate n, i.e. 1 + n = Lt+1/Lt.
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ECA5102
(a) Set up the Lagrangian for an individual born at t, write
down the first order conditions of her problem and solve
for the equilibrium C1,t and C2,t+1 as functions of prices,
wt and rt+1. (10 marks)
Solution: The Lagrangian function is:
L = lnC1,t + lnC2,t+1 + λ
[
wt − C1,t − 1
1 + rt+1
C2,t+1
]
The first order conditions are
1
C1,t
= λ
1
C2,t+1
=
1
1 + rt+1
λ
Thus, the optimality condition is
C2,t+1
C1,t
= 1 + rt+1
Combine with the budget constraint to get
C1,t =
wt
2
C2,t+1 =
wt
2
(1 + rt+1)
(b) Solve for the individual’s equilibrium savings rate s(rt+1).
(6 marks)
Solution: The equilibrium savings rate of the young is
given by
s(rt+1) =
wt − C1,t
wt
= 1− C1,t
wt
Substituting the value of C1,t and simplifying
s(rt+1) =
1
2
(c) How does an increase in the interest rate rt+1 affect the
individual’s equilibrium consumption choices and savings
rate? Interpret your answer in terms of income and sub-
stitution effects. (10 marks)
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ECA5102
Solution: As rt+1 increases, there are two effects at work:
income effect and substitution effect. From the income ef-
fect, both C1,t and C2,t+1 increase; from the substitution ef-
fect, C1,t decreases and C2,t+1 increases. Both effects work
in the same direction for C2,t+1, hence
∂C2,t+1
∂rt+1
> 0. How-
ever, they work in opposite directions for C1,t . If IE > SE,
C1,t will increase and s will decrease. If IE < SE, the
opposite happens. If IE = SE, C1,t and savings rate are
independent of rt+1. In this question, we use log utility, IE
and SE offset, and thus an increase in the interest rate rt+1
doesn’t affect C1,t and savings rate.
(d) Derive an expression for kt+1 (kt+1 = Kt+1/Lt+1) in terms
of kt (kt = Kt/Lt) and the parameters of the model. Next
suppose the economy is in a steady state. Solve for the
steady-state capital per capita k∗ in terms of the parame-
ters of the model. (10 marks)
Solution: The next period capital stock is given by the
savings of the young this period
Kt+1 = s(rt+1)wtLt
Substitute the expression for s(rt+1) in the equation for
Kt+1 to get
Kt+1 =
1
2
wtLt
Divide both sides by Lt+1 to get
Kt+1
Lt+1
= kt+1 =
1
2 (1 + n)
wt
Note that
wt =
∂Yt
∂Lt
= (1− α)Kαt L−αt = (1− α) kαt
We can then substitute for wt to obtain
kt+1 =
1− α
2 (1 + n)
kαt
When the economy in a steady state, kt+1 = kt = k∗. Use
this to substitute for kt+1 and kt in the last equation and
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ECA5102
solve for k∗. This gives
k∗ =
[
1− α
2 (1 + n)
] 1
1−α
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