Consider the following dynamic IS-LM model. C(t) = 20+ 0.8Y(t-1) Ya(t) = Y(t)- Tx(t) Tx(t) = 5 +0.25Y(t) I(t) = 20 2r(t) G = 50 E (t) = C(t)+1(t) + G AY(t+ 1) = 0.05 [E(t) - Y(t)] Ma(t) = 10+ 0.25Y(t) - 0.5r(t) M₂(t) = 55 Ar(t+1) = 0.8[Ma(t) - M₂(t)] (i) What is the equilibrium level of Y and r? (ii) Show that dynamic IS and LM equations are the recursive equations for Y(t+1) and r(t+1). That is, Y(t + 1) = 86a +(1− a)Y(t) + 0.6aY(t-1) - 2ar(t) r(t + 1) = -45ß +0.25BY(t) + (1 - 0.5B)r(t) where a = 0.05 is the speed of good market adjustment and ß= 0.8 is the speed of money market adjustment. [Hint: Substitute all the relationships in each of the adjustment equations in turn. The algebra can be somewhat tedious but not intellectually difficult.] (iv) Use the spreadsheet this policy change? Suppose the Reserve Bank reduces the money supply (Ms) to 53.6 in period 2, given Y and r are at their equilibrium values in periods 0 and 1. (iii) Use the model set up in the spreadsheet to calculate the new equilibrium output and interest rate. [Hint: The LM curve shifts left so you need to re-calculate the dynamic LM curve] to plot the trajectory of the economy resulting from

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Consider the following dynamic IS-LM model.
C(t) = 20+ 0.8Ya (t-1)
Ya(t) = Y(t) - Tx(t)
Tx(t) = 5 +0.25Y(t)
I(t) = 20 2r(t)
G = 50
E(t) = C(t) + 1(t) + G
AY(t + 1) = 0.05 [E(t) - Y(t)]
Ma(t) = 10+ 0.25Y(t) - 0.5r(t)
M, (t) = 55
Ar(t + 1) = 0.8[Ma(t)- M¸(t)]
(i) What is the equilibrium level of Y and r?
(ii) Show that dynamic IS and LM equations are the recursive equations for Y(+1) and
r(t+1). That is,
Y(t + 1) = 86a+ (1 -a)Y(t) + 0.6aY(t-1) - 2ar(t)
r(t + 1) = 45ß +0.25BY(t) + (1 - 0.5B)r(t)
where a = 0.05 is the speed of good market adjustment and ß = 0.8 is the speed of
money market adjustment. [Hint: Substitute all the relationships in each of the
adjustment equations in turn. The algebra can be somewhat tedious but not
intellectually difficult.]
(iv) Use the spreadsheet
this policy change?
Suppose the Reserve Bank reduces the money supply (Ms) to 53.6 in period 2, given Y and r
are at their equilibrium values in periods 0 and 1.
(iii) Use the model set up in the spreadsheet
to calculate the new equilibrium
output and interest rate. [Hint: The LM curve shifts left so you need to re-calculate the
dynamic LM curve]
to plot the trajectory of the economy resulting from
Transcribed Image Text:Consider the following dynamic IS-LM model. C(t) = 20+ 0.8Ya (t-1) Ya(t) = Y(t) - Tx(t) Tx(t) = 5 +0.25Y(t) I(t) = 20 2r(t) G = 50 E(t) = C(t) + 1(t) + G AY(t + 1) = 0.05 [E(t) - Y(t)] Ma(t) = 10+ 0.25Y(t) - 0.5r(t) M, (t) = 55 Ar(t + 1) = 0.8[Ma(t)- M¸(t)] (i) What is the equilibrium level of Y and r? (ii) Show that dynamic IS and LM equations are the recursive equations for Y(+1) and r(t+1). That is, Y(t + 1) = 86a+ (1 -a)Y(t) + 0.6aY(t-1) - 2ar(t) r(t + 1) = 45ß +0.25BY(t) + (1 - 0.5B)r(t) where a = 0.05 is the speed of good market adjustment and ß = 0.8 is the speed of money market adjustment. [Hint: Substitute all the relationships in each of the adjustment equations in turn. The algebra can be somewhat tedious but not intellectually difficult.] (iv) Use the spreadsheet this policy change? Suppose the Reserve Bank reduces the money supply (Ms) to 53.6 in period 2, given Y and r are at their equilibrium values in periods 0 and 1. (iii) Use the model set up in the spreadsheet to calculate the new equilibrium output and interest rate. [Hint: The LM curve shifts left so you need to re-calculate the dynamic LM curve] to plot the trajectory of the economy resulting from
t
Y(t + 1) = 86a + (1 − a)Y(t) + 0.6aY(t− 1) −2ar(t)
r(t+1)= -xx.xß +0.25Y(t)+ (1 - 0.5p)r(t)
0
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12345678 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 4142434456F48 49
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Y(t)
190
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r(t)
5
5
5
5
5
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5
5
5
5
5
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5
5
5
5
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5
5
5
6
r3
1
0
20
40
Monetary contraction
60
80
alpha=
beta =
100
Y
120
140
160
0.05
0.8
180
200
Transcribed Image Text:t Y(t + 1) = 86a + (1 − a)Y(t) + 0.6aY(t− 1) −2ar(t) r(t+1)= -xx.xß +0.25Y(t)+ (1 - 0.5p)r(t) 0 1 2 3 4 5 6 7 8 9 10 12345678 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 4142434456F48 49 11 47 Y(t) 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 190 r(t) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 r3 1 0 20 40 Monetary contraction 60 80 alpha= beta = 100 Y 120 140 160 0.05 0.8 180 200
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