Numerical Problems
1.
The price level in the West is PW  5 guilders per ordinary soap bar. The price level in the East is
PE  100 florins per deluxe soap bar. The real exchange rate is 2 ordinary soap bars per deluxe
soap bar.
(a) Use Eq. (13.6) to get enom  ePFOR /P  2 ordinary soap bars per deluxe soap bar  5 guilders per
ordinary soap bar/100 florins per deluxe soap bar  0.10 guilders per florin, or 10 florins per
guilder.
(b) Inflation in the West is W  10%. Inflation in the East is  E  20%. The real exchange rate is
constant. Use Eq. (13.3) to get enom/enom  (e/e)  FOR –   0  10% – 20%  –10%. So the
nominal exchange rate (guilders per florin) depreciates at a 10% rate. The East Bubble florin
depreciates, the West Bubble guilder appreciates.
2.
(a) Japan imports 64 barrels of oil, worth 16 cameras at 4 barrels of oil per camera. It exports
40 cameras, so the real value of its net exports is 24 cameras.
(b) Japan now imports 60 barrels of oil, worth 20 cameras at 3 barrels of oil per camera. It exports
42 cameras, so the real value of its net exports declines to 22 cameras.
(c) In the long run, Japan imports 54 barrels of oil, worth 18 cameras at 3 barrels of oil per camera.
It exports 45 cameras, so the real value of its net exports rises to 27 cameras.
(d) This illustrates the J curve, as a real depreciation leads to an initial decline in net exports
followed by a later rise in net exports.
3.
Begin by writing the equation for the IS curve, which is S d – I d  NX.
S d  Y – C d – G  Y – (300  0.5Y – 200r) – G.
NX  150 – 0.1Y – 0.5e  150 – 0.1Y – 0.5(20  600r)  140 – 0.1Y – 300r.
Using these in the IS curve equation gives:
(0.5Y – 300  200r – G) – (200 – 300r)  140 – 0.1Y – 300r.
Rearranging terms and simplifying gives the IS curve:
800r  640 – 0.6Y  G.
(a) With G  100 and Y  900, the IS curve gives 800r  640 – 540  100  200, so r  .25.
Then e  20  600r  170, NX  150 – 90 – 85  –25, C  300  450 – 50  700, and I 
200 – 75  125.
(b) With Y  940, the IS curve gives 800r  640 – 564  100  176, so r  .22.
Then e  20  600r  152, NX  150 – 94 – 76  –20, C  300  470 – 44  726, and I 
200 – 66  134. The rise in domestic output reduces the real interest rate and real exchange rate,
and increases net exports, consumption, and investment.
(c) With G  132, the IS curve gives 800r  640 – 564  132  208, so r  .26.
Then e  20  600r  176, NX  150 – 94 – 88  –32, C  300  470 – 52  718, and I 
200 – 78  122. The rise in government spending increases the real interest rate and the real
exchange rate, and decreases net exports, consumption, and investment.
4.
(a) Begin by writing the equation for the IS curve, which is S d – I d  NX.
NX  150 – 0.08Y – 500r.
S d  Y – C d – G  Y – {200  0.6[Y – (20  0.2Y)] – 200r} – G  0.52Y – (188  G)  200r.
Using these in the IS curve equation gives:
0.52Y – (188  G)  200r – (300 – 300r)  150 – 0.08Y – 500r.
Rearranging terms and simplifying gives the IS curve:
1000r  (638  G) – 0.6Y.
The LM curve comes from the expression M/P  L, which is 924/P  0.5Y – 200r. In the long
run we’ll use this equation to find the price level, so we’ll write this as P  924/(0.5Y – 200r). In
the short run we’ll combine the LM curve with the IS curve to find equilibrium, so we’ll write it
as 200r  0.5Y – 924/P.
With G  152 and Y  Y  1000, the IS curve gives 1000r  790 – 600  190, so r  .19.
From the LM curve, P  924/(500 – 38)  2. Then NX  150 – 80 – 95  –25, C  200 
0.6(1000 – 220) – 38  630, and I  300 – 57  243.
(b) In the short run with G  214 and P  2, the IS curve now gives 1000r  (638  214) – 0.6Y, and
the LM curve is 200r  0.5Y – 462. Take five times the LM equation and subtract it from the IS
equation to get (852  2310) – 3.1Y  0, or Y  1020. Plug this in the LM equation to get r  .24.
Then NX  150 – 81.6 – 120  –51.6, C  200  0.6(1020 – 224) – 48  629.6, and I  300 –
72  228.
In the long run, using Y  1000 in the IS curve gives 1000r  (638  214) – (0.6  1000)  252,
so r  .252. From the LM curve, P  924/(500 – 50.4)  2.055. Then NX  150 – 80 – 126  –56,
C  200  468 – 50.4  617.6, and I  300 – 75.6  224.4.
(c) With G  152 and an increase in net exports of 62, the IS curve is the same as in part (b). The
only difference in the results is the amount of G and NX. In the short run, NX is 62 higher, or
10.4, while G is 62 lower, or 152. In the long run, NX is 62 higher than before (NX  6), while
G is 62 lower at 152.
5.
(a) AD intersects AS at 1000  400  50M/P, which means M/P  12. With M  48 francs, P  4
francs/bottle. Use the formula enom  ePFor /P  5 wedges/bottle  20 crowns/wedge/4 francs/
bottle  25 crowns/franc.
(b) At 50 crowns/franc, the franc is overvalued as the official rate exceeds the fundamental value. The
domestic central bank will lose reserve assets over time if it tries to maintain the official rate.
(c) To get enom  50 crowns/franc for the fundamental value, use the formula enom  ePFor/P to find
the necessary price level. This is 50 crowns/franc  5 wedges/bottle  20 crowns/wedge/P
francs/bottle, so P  2 francs/bottle. Since M/P  12, you need M  24 francs to get P  2
francs/bottle.
6.
(a) c0  200, cY  0.6, t0  20, t  0.2, cr  200
i0  300, ir 300
x0 150, xY  0.08, xYF  0, xr  500, xrF 0
r  ’IS – ’ISY
’IS  (c0  i0  G – cYt0  x0  xYFYFor  xrFrFor)/(cr  ir  xr)
 (200  300  152 – (0.6  20)  150  0  0)/(200  300  500)
 790/1000  0.79
’IS  [1 – (1 – t)cY  xY]/(cr  ir  xr)
 [1 – (1 – 0.2)0.6  0.08]/1000  0.0006
r  LM – (1/ r)M/P  LMY
LM  ( 0/ r) – e, LM  Y/ r
' 0  0, r  200, Y  0.5, e  0
LM  0
 LM  0.5/200  0.0025
(c) In general equilibrium, Y  1000, so
IS: r  0.79 – (0.0006  1000)  0.19
LM: r  0 – (1/200)924/P  (0.0025  1000), so 4.62/P  2.31, so P  2.
(d) AD: Combine IS and LM in terms of P and Y:
LM: r  – 0.005M/P  0.0025Y
IS: r  0.79 – 0.0006Y
–0.005M/P  0.0025Y  0.79 – 0.0006Y, so
0.0031Y  0.79  0.005M/P  0.79  (0.005  924/P), so Y  254.84  1490.3/P.
If NX rises by 62, x0 rises, which changes ’IS by 62/1000, or 0.062, increasing the constant term
in the equation for ’IS from 0.79 to 0.852. So the AD curve becomes:
0.0031Y  0.852  0.005M/P  0.852  (0.005  924/P), so Y  274.84  1490.3/P.
(e) With P  2, Y  274.84  1490.3/2, so Y  1020.
(b)