Solutions manual
Advanced macroeconomics 5th edition
by David Romers Chapters 1 to 12
,SOLUTIONS TO CHAPTER 1
Problem 1.1
(a) Since the growth rate oƒ a variable equals the time derivative oƒ its log, as shown by equation (1.10)
in the text, we can write
Ż(t) d ln Z(t) d ln X(t)Y(t)
(1) .
Z(t) dt dt
Since the log oƒ the product oƒ two variables equals the sum oƒ their logs, we have
Ż(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(2) ,
Z(t) dt dt dt
or simply
Ż(t) Ẋ(t) Ẏ(t)
(3) .
Z(t) X(t) Y(t)
(b) Again, since the growth rate oƒ a variable equals the time derivative oƒ its log, we can write
d lnX(t) Y(t)
(4) Ż(t) d ln Z(t) .
Z(t) dt dt
Since the log oƒ the ratio oƒ two variables equals the diƒƒerence in their logs, we have
Ż(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(5) ,
Z(t) dt dt dt
or simply
Ż(t) Ẋ(t) Ẏ(t)
(6) .
Z(t) X(t) Y(t)
(c) We have
Ż(t) d ln Z(t) d ln[X(t) ]
(7) .
Z(t) dt dt
Using the ƒact that ln[X(t) ] = lnX(t), we have
Ż(t) d ln X(t) d ln X(t) Ẋ(t)
(8) ,
Z(t) dt dt X(t)
where we have used the ƒact that is a constant.
Problem 1.2
(a) Using the inƒormation provided in the question,
the path oƒ the growth rate oƒ X, Ẋ(t) X(t), is Ẋ(t)
depicted in the ƒigure at right. X(t)
© 2012 by McGraw-Hill Education. This is proprietary material solely ƒor authorized instructor use. Not authorized ƒor sale or distribution in any
manner. This document may not be copied, scanned, duplicated, ƒorwarded, distributed, or posted on a website, in whole or part.
,Ƒrom time 0 to time t1 , the growth rate oƒ X is
constant and equal to a > 0. At time t1 , the growth
rate oƒ X drops to 0. Ƒrom time t1 to time t2 , the
growth rate oƒ X rises gradually ƒrom 0 to a. Note that
we have made the assumption that Ẋ(t) X(t) rises at
a constant rate ƒrom t1 to t2 . Ƒinally, aƒter time t2 , the
growth rate oƒ X is constant and equal to a again.
, 1-2 Solutions to Chapter 1
(b) Note that the slope oƒ lnX(t) plotted against time
is equal to the growth rate oƒ X(t). That is, we know lnX(t)
d ln X(t) Ẋ(t) slope = a
dt X(t)
(See equation (1.10) in the text.) slope = a
Ƒrom time 0 to time t1 the slope oƒ lnX(t) equals
a > 0. The lnX(t) locus has an inƒlection point at t1 ,
lnX(0)
when the growth rate oƒ X(t) changes discontinuously
ƒrom a to 0. Between t1 and t2 , the slope oƒ lnX(t)
rises gradually ƒrom 0 to a. Aƒter time t2 the slope oƒ 0 t1 t2 time
lnX(t) is constant and equal to a > 0 again.
Problem 1.3
(a) The slope oƒ the break-even investment line is
Inv/ (n + g + )k
given by (n + g + ) and thus a ƒall in the rate oƒ eƒƒ
depreciation, , decreases the slope oƒ the break- lab
even investment line. (n + g + NEW)k
The actual investment curve, sƒ(k) is unaƒƒected.
sƒ(k
)
Ƒrom the ƒigure at right we can see that the
balanced- growth-path level oƒ capital per unit oƒ
eƒƒective labor rises ƒrom k* to k*NEW .
k* k*NEW k
(b) Since the slope oƒ the break-even
investment line is given by (n + g + ), a rise in Inv/ (n + gNEW + )k
the rate oƒ technological progress, g, makes the eƒƒ
lab
break-even investment line steeper.
(n + g + )k
The actual investment curve, sƒ(k), is unaƒƒected.
sƒ(k
Ƒrom the ƒigure at right we can see that the )
balanced-growth-path level oƒ capital per unit oƒ
eƒƒective labor ƒalls ƒrom k* to k*NEW .
k*NEW k* k
© 2012 by McGraw-Hill Education. This is proprietary material solely ƒor authorized instructor use. Not authorized ƒor sale or distribution in any
manner. This document may not be copied, scanned, duplicated, ƒorwarded, distributed, or posted on a website, in whole or part.
Advanced macroeconomics 5th edition
by David Romers Chapters 1 to 12
,SOLUTIONS TO CHAPTER 1
Problem 1.1
(a) Since the growth rate oƒ a variable equals the time derivative oƒ its log, as shown by equation (1.10)
in the text, we can write
Ż(t) d ln Z(t) d ln X(t)Y(t)
(1) .
Z(t) dt dt
Since the log oƒ the product oƒ two variables equals the sum oƒ their logs, we have
Ż(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(2) ,
Z(t) dt dt dt
or simply
Ż(t) Ẋ(t) Ẏ(t)
(3) .
Z(t) X(t) Y(t)
(b) Again, since the growth rate oƒ a variable equals the time derivative oƒ its log, we can write
d lnX(t) Y(t)
(4) Ż(t) d ln Z(t) .
Z(t) dt dt
Since the log oƒ the ratio oƒ two variables equals the diƒƒerence in their logs, we have
Ż(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(5) ,
Z(t) dt dt dt
or simply
Ż(t) Ẋ(t) Ẏ(t)
(6) .
Z(t) X(t) Y(t)
(c) We have
Ż(t) d ln Z(t) d ln[X(t) ]
(7) .
Z(t) dt dt
Using the ƒact that ln[X(t) ] = lnX(t), we have
Ż(t) d ln X(t) d ln X(t) Ẋ(t)
(8) ,
Z(t) dt dt X(t)
where we have used the ƒact that is a constant.
Problem 1.2
(a) Using the inƒormation provided in the question,
the path oƒ the growth rate oƒ X, Ẋ(t) X(t), is Ẋ(t)
depicted in the ƒigure at right. X(t)
© 2012 by McGraw-Hill Education. This is proprietary material solely ƒor authorized instructor use. Not authorized ƒor sale or distribution in any
manner. This document may not be copied, scanned, duplicated, ƒorwarded, distributed, or posted on a website, in whole or part.
,Ƒrom time 0 to time t1 , the growth rate oƒ X is
constant and equal to a > 0. At time t1 , the growth
rate oƒ X drops to 0. Ƒrom time t1 to time t2 , the
growth rate oƒ X rises gradually ƒrom 0 to a. Note that
we have made the assumption that Ẋ(t) X(t) rises at
a constant rate ƒrom t1 to t2 . Ƒinally, aƒter time t2 , the
growth rate oƒ X is constant and equal to a again.
, 1-2 Solutions to Chapter 1
(b) Note that the slope oƒ lnX(t) plotted against time
is equal to the growth rate oƒ X(t). That is, we know lnX(t)
d ln X(t) Ẋ(t) slope = a
dt X(t)
(See equation (1.10) in the text.) slope = a
Ƒrom time 0 to time t1 the slope oƒ lnX(t) equals
a > 0. The lnX(t) locus has an inƒlection point at t1 ,
lnX(0)
when the growth rate oƒ X(t) changes discontinuously
ƒrom a to 0. Between t1 and t2 , the slope oƒ lnX(t)
rises gradually ƒrom 0 to a. Aƒter time t2 the slope oƒ 0 t1 t2 time
lnX(t) is constant and equal to a > 0 again.
Problem 1.3
(a) The slope oƒ the break-even investment line is
Inv/ (n + g + )k
given by (n + g + ) and thus a ƒall in the rate oƒ eƒƒ
depreciation, , decreases the slope oƒ the break- lab
even investment line. (n + g + NEW)k
The actual investment curve, sƒ(k) is unaƒƒected.
sƒ(k
)
Ƒrom the ƒigure at right we can see that the
balanced- growth-path level oƒ capital per unit oƒ
eƒƒective labor rises ƒrom k* to k*NEW .
k* k*NEW k
(b) Since the slope oƒ the break-even
investment line is given by (n + g + ), a rise in Inv/ (n + gNEW + )k
the rate oƒ technological progress, g, makes the eƒƒ
lab
break-even investment line steeper.
(n + g + )k
The actual investment curve, sƒ(k), is unaƒƒected.
sƒ(k
Ƒrom the ƒigure at right we can see that the )
balanced-growth-path level oƒ capital per unit oƒ
eƒƒective labor ƒalls ƒrom k* to k*NEW .
k*NEW k* k
© 2012 by McGraw-Hill Education. This is proprietary material solely ƒor authorized instructor use. Not authorized ƒor sale or distribution in any
manner. This document may not be copied, scanned, duplicated, ƒorwarded, distributed, or posted on a website, in whole or part.
