Consider the following IS-LM Model C = 189 + 0.52Yd I = 145 + 0.12Y - 1025i G =

Question

Consider the following IS-LM Model

C = 189 + 0.52Yd

I = 145 + 0.12Y - 1025i

G = 329

T = 263

(M/P)^d = 1.6Y - 7737i

M/P = 1579

The IS equation is determined to be Y=1461.78-2847.22i

The LM equation is determined as i=-0.20408+0.00021Y

Initial equilibrium values of Y,i,c, and I are calculated bellow.

Y=1278

i=6.43%

C=717

I=232

Suppose that money supply increases to (m/p)=1737

Calculate the equation for the new LM relation.

i=-0.22451+0.00021Y

1. giving the above IS-LM equation, what is the equilibrium output(Y) interest(i)?

what is the equilibrium value of C?

what is the equilibrium value of I ?

DETERMINE THE VALUE OF C+I+G? round answer to the nearest integer.

Explanation / Answer

Since the IS equation is Y=1461.78-2847.22i and new LM equation is i=-0.22451+0.00021Y.So for equilibrium both goods market and money market should be in equilibrium and the interest ratge should be same in both the markets.So, putting the i from LM equation into the Is equation we get

Y=1461.78-2847.22*(-0.22451+0.00021Y)

Y=1461.78+639.23-0.598Y

1.598Y=2101.01

Y=1314.77

hence i=-0.22451+0.00021Y=0.0516

Hence Equlibrium value of Y is 1314.77 and so

equilibrium value of C= 189+0.52(Y-T)

C=189+0,52(1314.77-263)

C=735.92

Equlibrium value of I=145+0.12Y-1025i

I=145+0.12*1314.77-1025*0.0516=249.8824

Hence equilibrium value of I is 249.8824

The value of C+I+G=735.92+249.8824+329=1314.80=1315

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