ECON 3012
PROBLEM 7
Answers
I.
 = [Yn/(Yn +f)] = [400/(400+1002)] = 400/600 = 2/3 = .6667
II.
A.
AD0 is Y =  A0 +  M0 –  P0 = 1.25800 + 550 – P0 = 1000 + 250 – P0 = 1250 –
0P0 .
You can find Y0 two ways. The best is to use the solution, since you will have to do this
later:
Yt = At + Mt + f]– Pe. Pe = Pt-1 . .: Y0 = 1.25800 + .8550 + .82501 – .82501
= 800 + 200 +200 – 200 = 1000.
To find the initial interest rate, i0, use the LM curve: i0 = .021000 – .1 (50/1.0) = 20 – 5 =
15.
The simplest way to find Y0 is to use the AD curve. This works here because P is known, so
you don’t need a simultaneous solution yet: Y0 =1250 – 250P0 = 1250 – 250 = 1000.
B. I’ll underline all changes from one period to the next.
1. t = 1:
You don’t actually need the AD curve because you MUST use the solution since the SR
equilibrium is simultaneous. But, for interest, the new AD curve, AD1 is: Y = 1.25900 +
550 – P0 = 1125 + 250 – P0 = 1357 – P0 .
Y1 = 1.25900 + .8550 + .82501 – .82501 = 900 + 200 +200 – 200 = 1100.
Use SAS to get P1: P1 = 1.0 + 1[(1100-1000)/1000] = 1.0 + 0.1 = 1.1.
Use LM to get i1: i1 = .021100 – .1 (50/1.1) = 22 – 4.5 = 17.5.
t = 2:
Again, use the solution:
Y2 = 1.25900 + .8550 + .82501 – .82501.1 = 900 + 200 +200 – 220 = 1080.
Use SAS to get P2: P2 = 1.1 + 1[(1080-1000)/1000] = 1.1 + 0.08 = 1.18.
Use LM to get i2: i2 = .021080 – .1(50/1.18) = 21.6 – 4.2 = 17.4.
t = 3:
And yet again, use the solution:
Y3 = 1.25900 + .8550 + .82501 – .82501.18 = 900 + 200 +200 – 236 = 1064.
Use SAS to get P3: P3 = 1.18 + 1[(1064-1000)/1000] = 1.18 + 0.064 = 1.244.
Use LM to get i3: i3 = .021064 – .1(50/1.244) = 21.3 – 4.0 = 17.3.
You can see where Y is going, but it’s difficult to see to just what values P and i are
converging. SSEx 3 shows these clearly though.
2. t = 1:
Period 1 is the same as with #1 above: AD1 is: Y = 1357 – 250P0 . Y1 = 1100; P1 = 1.1; i1 =
17.5.
t = 2:
M2 = 1.150 = 55. Use the solution:
Y2 = 1.25900 + .8555 + .82501 – .82501.1 = 900 + 220 +200 – 220 = 1100.
Use SAS to get P2: P2 = 1.1 + 1[(1100-1000)/1000] = 1.1 + 0.1 = 1.2.
Use LM to get i2: i2 = .021100 – .1(55/1.2) = 22 – 4.6 = 17.4.
t = 3:
M3 = 1.250 = 60. Again use the solution:
Y3 = 1.25900 + .8560 + .82501 – .82501.2 = 900 + 240 +200 – 240 = 1100.
Use SAS to get P3: P3 = 1.2 + 1[(1100-1000)/1000] = 1.2 + 0.1 = 1.3.
Use LM to get i3: i3 = .021100 – .1(60/1.3) = 22 – 4.6 = 17.4.
3. t = 1: Use the “old” AD curve, AD0 .
Y1 = 1.25800 + 550 – 2501.2 = 1000 + 250 – 300 = 950.
P1 = 1.2 (given).
Use LM to get i1: i1 = .02950 – .1(50/1.2) = 19 – 4.17 = 14.83.
a)
t = 2: M2 = 1.250 = 60. Use the new AD curve, AD2 .
Yr = 1.25800 + 560 – 2501.2 = 1000 + 300 – 2501.2 = 1300 – 2501.2 = 1300 – 300 =
1000.
Since Yr = Yn  Y2 = Yr = 1000, and P2 = P1 = 1.2.
Use LM to get i2: i2 = .021000 – .1(60/1.2) = 20 – 5 = 15.
b)
t = 2: M2 = 1. = 60. P1 = 1.2
Y2 = 1.25800 + .8560 + .82501 – .82501.2 = 800 + 240 +200 – 240 = 1000.
Use SAS to get P2: P2 = 1.2 + 1[(1000-1000)/1000] = 1.2 + 0 = 1.2.
Use LM to get i2: i2 = .021000 – .1(60/1.2) = 20 – 5 = 15.
NOTE: Your answers to 3.b) are the same as your answers to 3.a). The reason is that the
condition for the MAS curve, that prices can’t fall, is exactly the same as our simple – too
simple – version of price expectations. The world shown in Problem II , part B, number 3
would be a nice world, at least if we had smart planners. In both cases, a) and b) we return to
full employment and no inflation. Unfortunately the world doesn’t work this way. How it
actually does is shown in Economics 4012. (The problem is that we have only expectations of
Prices because we’re working in the Price, Output space – the P,Y space. In reality, in all
likelihood, what develop are expectations of inflation. If so, we don’t return nicely to full
employment and no inflation.