Question #269620

Consider the following IS–LM model: C = 150 + 1/2YD T = 300 G = 300 I = 150 + 1/3Y − 10 000ρ ρ = i + x (M/P) d = 2Y − 20 000i M/P = 2600 a. Imagine the external finance premium (x) is zero. Derive the IS relation. b. Derive the LM relation. c. Solve for the equilibrium real output and interest rate. d. What is the cost of bank loans and the equilibrium level of investment? e. Now suppose that firms’ capital drops following a severe slump in stock prices and banks charge an external finance premium (x) on loans to firms equal to 0.5%


Expert's answer

C=150+12YDC=150+\frac{1}{2}Y_D

T=300T=300

G=300G=300

I=150+13Y10000ρI=150+\frac{1}{3}Y-10000\rho

ρ=i+x\rho=i+x

(MP)d=2Y20000i(\frac{M}{P})^d=2Y-20000i

MP=2600\frac{M}{P}=2600

(a)

IS relation:

AD= Consumption+Investment+ Government Expenditure.

Y=C+I+GY=C+I+G

Y=150+12YD+150+13Y10000ρ+300Y=150+\frac{1}{2}Y_D+150+\frac{1}{3}Y-10000\rho +300

Y=150+0.5(Y300)+150+13Y10000ρ+300Y=150+0.5(Y-300)+150+\frac{1}{3}Y-10000\rho+300

Y=2,647.158,823.53ρY=2,647.1-58,823.53\rho

x=0    ρ=ix=0\implies \rho=i

So,

Y=2,647.158,823.53iY=2,647.1-58,823.53i


(b)

LM relation:

(MP)s=(MP)d(\frac{M}{P})^s=(\frac{M}{P})^d

2600=2Y20000i2600=2Y-20000i

20000i=2Y260020000i=2Y-2600

i=Y100000.13i=\frac{Y}{10000}-0.13


(c)

IS:Y=2,647.158,823.53iIS:Y=2,647.1-58,823.53i

LM:i=Y100000.13LM:i=\frac{Y}{10000}-0.13

Subst. ii in IS relation:

Y=2,647.158,823.53(Y100000.13)Y=2,647.1-58,823.53(\frac{Y}{10000}-0.13)

Y=1,562.51Y=1,562.51

Subst. Y in the LM relation:

i=1,562.5110,0000.13=0.026251i=\frac{1,562.51}{10,000}-0.13=0.026251

i=0.026251×100=2.6251i=0.026251\times100= 2.6251%


(d)

Investment:

I=150+0.33Y10000iI=150+0.33Y-10000i

I=150+0.33(1562.51)10000(0.026251)I=150+0.33(1562.51)-10000(0.026251)

I=403.1183I=403.1183


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