SOLUTIONS TO HOMEWORK #2

-Question #2

a)If i=.05 then Md=60,000*(.35-.05)=18,000, while Bd=50,000-18,000=32,000. If i=.1 then Md=15,000 and Bd=35,000
b) As the interest rate increases bonds become more attractive: holding W fixed, Md falls and Bd raises
c) It falls by 50%
d) It falls by 50%
e) Higher income implies higher Md. A given % percentage increase in income yields to an indentical % increase in Md.

-Question #3
a) i=100/(price today)-1. If p=75 i=.33, if p=85 i=.17, if p=95 i=.05.
b)The higher the price the lower the interest rate
c)pt=92.6

-Question #4
a) From Ms=20=Md=100*(.25-i), i=.05
b) Ms=100*(.25-.15)=10

-Dynamic Adjustment in the Goods Market.

At t=0.
Y(0)=Z(0)=1/(1-c1)*(I+G+c0-c1*T)=1/(1-3/4)*(15+15+2.5-3/4*10)=100.

The rule used by firms to choose output in each period t is Y(t)=Z(t-1).
At t=1, c0 permanently falls to .5

At t=1.
Y(1)=Z(0)=100
Z(1)=c0+c1*(Y(1)-T)+I+G=.5+3/4*(100-10)+30=98
Inventories=Iu(1)=100-98=2
Total Investment=TI(1)=2+15=17
Private Savings=S(1)=-c0+(1-c1)*(Y(1)-T)=-.5+1/4*90=22
Public Savings=PS(1)=T-G=10-15=-5
National Savings=NS(1)=17=Total Investment
C(1)=.5+3/4*90=68
Y(1)=Z(1)+Iu(1)=68+30+2=100

At t=2.
Y(2)=Z(1)=98. Z(2)=.5+3/4*(98-10)+30=96.5
Iu(2)=98-96.5=1.5
TI(2)=1.5+15=16.5
S(2)=-.5+1/4*88=21.5
PS(2)=-5
NS(2)=16.5=TI(2)
C(2)=.5+3/4*88=66.5
Y(2)=Z(2)+Iu(2)=66.5+30+1.5=98

At t=3.
Y(3)=Z(2)=96.5. Z(3)=.5+3/4*(96.5-10)+30=95.375
Iu(3)=96.5-95.375=1.125
TI(3)=1.125+15=16.125
S(3)=-.5+1/4*86.5=21.125
PS(3)=-5
NS(3)=16.125=TI(3)
C(3)=.5+3/4*86.5=65.375
Y(3)=Z(3)+Iu(3)=65.375+30+1.125=96.5

At t=infinity.
Y(inf)=Z(inf)
Y(inf)=1/(1-3/4)*(30.5-3/4*10)=92
Iu(inf)=0
TI(inf)=15
S(inf)=-.5+1/4*82=20
PS(inf)=-5
NS(inf)=15=TI(inf)
C(inf)=.5+3/4*82=62
Y(inf)=Z(inf)=62+30=92.