ECONOMICS 4012
Problem Set 2
Answers
Notes on answers:
First, I will only write out detailed solutions once. If all you’re asked to do is to repeat an analysis
with different numbers, I will just give the answers.
Second, I have changed the order of Part V. The reason for this change is that V. C. doesn’t make
any sense the way it was written in the Problem 2 you have. I’m changing it so that it becomes
interesting. By making it interesting, I’m making it so that it sets up the introduction to the next unit:
the DAD curve. Our class on Monday will begin with me going over Part V again, but in the order
as appears here. Then on to the DAD curve.
I.
Given in the handout.
II.
IS curve only
A.
 = 1/(1-c) = 1/(1-.6) = 1/.4 = 2.5
B.
IS: Y = A – bi. A = ([a – cT) + h + G] You are given that a = 340, h = 300, G = 400, T =
400 and b = 2000. A = (340–240+300+400) = 800. IS: Y = 2.5800 – 2.52000i  Y =
2000 – 5000i
C.
Y0 = 2000 – 5000.10 = 2000 – 500 = 1500; I0 = 300 – 2000.10 = 300 – 200 = 100
D.
Y1 = 1750; I1 = 200
E.
New A  New IS: Y = 2.5900 – 5000i = 2250 – 5000i; Y2 = 1750
III.
IV.
LM curve only
A.
LM: You are given: d = .1; f = 500; M = 100; P = 1.00.
LM: i = (.1/500)Y – (1/500)(100/1.00)  i = .0002Y – .002(M/P)  i = .0002Y – .20
B.
i0 = .00021600 – .20 = .32 –.20 = .12
C.
i1 = .10
D.
New LM: i = .0002Y – (1/500)(120/1.00)  i = .0002Y – .24; i2 = .30 – .24 = .06.
E.
New LM: i = .0002Y – (1/500)(120/1.10)  i = .0002Y – (1/500)(109.1) 
= 0002Y – .218; i3 = .30 – .218 = .082
AD curve
A.
Substitute LM into IS: Y = 2000 – 5000(.0002Y – .20) = 2000 –1Y + 1000  2Y = 3000
 Y0 = 1500 (Surprise?); i0 = .10 (Another surprise?); I0 = 200
B.
First find AD: Substitute LM into IS: Y = 2000 – 5000(.0002Y – (1/500)(M/P)) 
2Y = 2000 + 10(M/P)  AD: Y = 1000 + 5(M/P)
A = 1000, A = 800,  = 1.25;  = 5
C.
Y = 1000 + 500(1/P)
D.
Y0 = 1000 + 500 = 1500 (Surprised again?); i0 = .10 (Or Again?) ; I0 = 100
E.
Y1 = 1000 + 500(1/P) = 1000 + 500(1/1.25) = 1000 + 400 = 1400;
Get i from LM: i1 = .00021400 – (.002)(100/1.25) = .28 – .16 = .12; I1 = 60
F.
Y2 = 1000 + 500(1/P) = 1000 + 500(1/0.80) = 1000 + 625 = 1625;
Get i from LM: i2 = .00021625 – (.002)(100/0.80) = .325 – .25 = .075; I2 = 150
i
V.
AD and MAS
A.
Begin withY0 = 1500 and P0 = 1.00. G increases by 100 to 500 so A increases by 100 to
900.
Get new AD: Y = 1.25900 + 5(100/1.0). Find Y-hat1 = 1125 + 500 = 1625 > 1500.
Y1 = 1500.
Now find P1 from AD: 1500 = 1125 + 500/P1  375 = 500/P1  375P1 = 500 
P1 = 500/375 = 1.33; Ms1 = 100/1.33 = 75;
Back to LM: i1 = .00021500 – (.002)(75) = .30 – .15 = .15 ; I1 = 0.
B.
Begin with withY1 = 1500 and P1 = 1.33. M increases to 120.
Get new AD: Y = 1.25900 + 5(120/1.33). Find Y-hat2 = 1125 + 450 = 1575 > 1500.
Y2 = 1500.
Now find P2 from AD: 1500 = 1125 + 600/P2  375 = 600/P2  375P2 = 600 
P2 = 600/375 = 1.60; Ms2 = 120/1.60 = 75; (Why is this the same as Ms1 ?
Back to LM: i2 = .00021500 – (.002)(120/1.60) = .30 – .15 = .15 ; I2 = 0.
C.
To hold the interest rate at .10, the real money stock must be held at 100. With a price
level of 1.6, this means setting M at 160. But if this happens in period 3, then the new
AD curve is: Y-hat = 1125 + 5(160/1.60) = 1625. We’re right back to the beginning of A.
above. New Y3 must be 1500, so new P3 is (5160)/375 = 2.13. But if P rises to 2.13, i3
rises to (.30 – .002(160/2.13)) = (.30 – .15) = .15. Same as i2!
So to hold the interest rate at .10 the government must set M at 213. But if this happens
in period 4, then the new AD curve is: Y-hat = 1125 + 5(213/2.13) = 1625. Again, we’re
right back to the beginning of B above. New Y4 must be 1500, so new P4 is (5213)/375
= 2.84. But if P rises to 2.84, i4 rises to (.30 – .002(213/2.84)) = (.30 – .15) = .15. Same
as i2! Again! Still!
So if the government tries to hold the interest rate at .10 by using what is called an
accommondating monetary policy, it ends up chasing its tail. It can not keep the interest
rate lower than .15 using monetary policy. All that happens as it tries is that prices keep
rising. In this case, what remain constant are Y, at 1500, and inflation,  .
{2 = (1.60-1.20)/1.20 = .33; 3 = (2.13-1.60)/1.60 = .33; 4 = (2.84-2.13)/2.13 = .33; … }
Thus we are led to the next step, to the DAD curve in the Y space. In that space, this
particular case is an equilibrium. But we need some new theory to get there.