Question #149862
C: Short-run impacts of fiscal policy The economy of Wonderland has the following characteristics: C d = 100 + 0.5(Y − T) − 1000r I d = 400 − 1500r Md P = 0.5Y − 2000(r + π e ) π e = 0 where Y is output, T is a lump-sum tax, G is government spending, r is the real rate of interest, P is the price level. Questions 19-28 relate to this information.

C2: Fiscal policy in a small open economy with flexible exchange rate. Now suppose Wonderland trades with one large open economy which has a price level of PF or = 200. The behaviour of Wonderland can still be represented by the equations in part C1. Plus its net exports follows NX = 50 − 0.1Y − 50e where e is the real exchange rate. Questions 22-26 relate to this information
Q24. What are the implied levels of nominal exchange rate and net exports?
(a) enom = 21.1; NX = −120
(b) enom = 2; NX = −100
(c) enom = 8; NX = −180
(d) enom = 21.1; NX = −180
1
Expert's answer
2020-12-10T14:40:24-0500

Solution

C: Given the following characteristics: C d=100+0.5(YT)1000r I d=4001500r Md P=0.5Y2000(r+πe)πe=0C\ d = 100 + 0.5(Y − T) − 1000r\ I\ d = 400 − 1500r\ Md\ P = 0.5Y − 2000(r + π e ) π e = 0

where

Y = output,

T = lump-sum tax,

G = government spending,

r = real rate of interest,

P is the price level.



AD(4)=Cd+Id+Nx+GAD(4)=C^d+I^d+Nx+G

AD curve; relationship between p and 4 and keep other variables as they are


4=100+0.5(YT)1000r+4001500r+500.1450e+G=04=550+0.5y0.5T2500r0.1y50e+G40.5y+0.1y=5500.5T2500r50e+G0.6y=5500.5T2500r50e+G4αP=0.5y(2000)(r+re)πe=0,NP=0.5T2000r    2000r=0.5TMP    r=0.5T2000MP×120000.6y=5500.5T2000(0.5yMP2000)50e+G    0.6y=5500.5T1.25(6.5yMP)50e+G    0.6y=5500.5T0.625y+1.25MP)50e+G    0.6y+0.625y5500.5T+1.25MP50e+G1.25MP=1.205y550+0.5T+50e+G4= 100 + 0.5(Y − T) − 1000r + 400 − 1500r + 50-0.14 − 50e+G = 0 \\ 4=550+0.5y-0.5T-2500r-0.1y-50e+G\\ 4-0.5y+0.1y=550-0.5T-2500r-50e+G\\ 0.6y=550-0.5T-2500r-50e+G\\ \therefore \frac{4^{\alpha}}{P}=0.5y-(2000)(r+r^e)\\ \therefore \pi^e=0,\\ \frac{N}{P}=0.5T-2000r\\ \implies 2000r=0.5T-\frac{M}{P}\\ \implies r=\frac{0.5T}{2000}-\frac{M}{P}\times\frac{1}{2000}\\ \therefore 0.6y=550-0.5T-2000(\frac{0.5y-\frac{M}{P}}{2000})-50e+G\\ \implies 0.6y=550-0.5T-1.25(6.5y-\frac{M}{P})-50e+G\\ \implies 0.6y=550-0.5T-0.625y+1.25\frac{M}{P})-50e+G\\ \implies 0.6y+0.625y-550-0.5T+\frac{1.25M}{P}-50e+G\\ \frac{1.25M}{P}=1.205y-550+0.5T+50e+G\\

Multiplying (4) both sides we get


5MP=4.9y2200+2T+200e+4G\frac{5M}{P}=4.9y-2200+2T+200e+4G

p=5M4.94+2200+2T4G+200ep=\frac{5M}{4.94+2200+2T-4G+200e}

For Q24, option d is correct


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