Question

# Problem 2. Consider the following example of the IS-LM model: C = 340 + 0.5(Y–T) I...

Problem 2. Consider the following example of the IS-LM model:

C = 340 + 0.5(Y–T)
I = 400 – 1500(r + x) G = 150 T = 100 x=0.02

r = 0.04 ?e = 0.02

(1) Derive the IS equation.

(2) Find the equilibrium value of Y.

(3) Write down the zero lower bound constraint. Does the real interest rate of r=0.04 satisfy the constraint?

(4) Suppose that the risk premium x has increased to x=0.09.

(a) Derive the new IS equation and the new equilibrium value of Y.

(b) Suppose that the Fed wants to maintain the equilibrium value of Y at the value of

Y in part (2). What is the real interest rate required to achieve this goal?

(c) Does this real interest rate satisfy the zero lower bound constraint?

Solution:

(1). IS equation;.

We know, Y= C+I+G

Now C= 340+0.5(y-100)

I= 400-1500(r+k)

G= 150

Therefore,

Y = 340+0.5(y-100) + 400 - 1500(r+k) + 150

Y = 340 + 0.5y - 50 + 400 - 1500(r+k) + 150

Y = 0.5y + 840 - 1500(r+k)

Y - 0.5Y = 840 - 1500(r+k)

Y(1- 0.5) = 840 - 1500(r+k)

1/2 Y = 840 - 1500(r+k)

Y= 1680 - 3000(r+k). This is our IS equation.

(2) .Equilibrium value of Y

By putting the values of r=0.04 and k= 0.02 in the IS equation we can find equilibrium value of Y.

We know in IS equation Y=1680- 3000(r+k)

Y= 1680- 3000(0.04 + 0.02)

Y = 1680 - 3000(0.06)

Y = 1680 - 180

Y = 1500. This is the equilibrium value of Y.

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