Question

. Consider the following model:

C = 500 + .3(Y − T) I = 150 + .1Y − 1000i G = 350 T = 250 (M/P) d = 2Y − 8000i M/P = 1600

(a) Derive the IS relation

(b) Derive the LM relation

(c) Solve for equilibrium real output.

(d) Solve for the equilibrium interest rate. i = .131

(e) Solve for the equilibrium values of C and I and verify the value you obtained for Y by adding C, I, and G

Answer 1

C = 500 + .3(Y − T)

I = 150 + .1Y − 1000i

G = 350

T = 250

(M/P) d = 2Y − 8000i

M/P = 1600

a) IS relation:

Y= C+I+G

Y= 500 + .3(Y − T)+150 + .1Y − 1000i+350

Y= 1000+ 0.3Y-0.3(250)+0.1Y-1000i

Y-0.3Y-0.1Y = 925 - 1000i

i= (925-0.6Y)/1000 IS equation

b) LM relation:

Money demand = Money supply

2Y − 8000i = 1600

i = (2Y-1600)/8000 LM equation

c) For equilibrium:

IS = LM

(925-0.6Y)/1000 = (2Y-1600)/8000

(925-0.6Y)8= 2Y-1600

7400-4.8Y = 2Y-1600

9000= 6.8Y

Y= 1323.53 Equilibrium level of real output.

d) For equilibrium level of i:

Put value of Y= 1323.53 into LM

i = (2Y-1600)/8000= (2x1323.53-1600)/8000= 1047.06/8000 = 0.131

e)

C = 500 + .3(Y − T) = 500 + .3(1323.53 − 250) = 500+322.10= 822.10

I = 150 + .1Y − 1000i = 150+0.1(1323.53)-1000(0.131)= 151.43

G= 350

Y= 822.1+151.43+350= 1323.53