IS-LM Problem

1. The following equations describe an economy:
C = 10 + 0.5y
I = 190 – 20r 
Solution: IS curve describes the equation for product market equilibrium at various 
combinations of level of income and rate of interest.
Y = AD = C + I
Y = 10 + 0.5Y + 190 – 20r
Y – 0.5Y = 200 – 20r
Y (1- 0.5) = 200 – 20r
½ Y = 200 – 20r
Y = 400 – 40r.

2. The following equations describe an economy
C = 100 + 0.75Yd
I = 50 – 25r
T = G = 50
Where C is aggregate consumption, Yd is disposable income, I is aggregate investment. T is taxes, G is government purchases and r is the rate of interest. Derive the IS curve for the economy.
Solution: Y = C + I + G
Now, 
C = 100 + 0.75Yd = 100 + 0.75 (Y- T) = 100 + 0.75 (Y – 50) 
 I = 50 – 25r 
G = T = 50
Hence, Y = C + I + G
Y = 100 + 0.75 (Y – 50) + 50 – 25r + 50
Y = 200 + 0.75Y – 37.5 – 25r 
Y – 0.75Y = 162.5 – 25r
0.25Y = 162.5 – 25r 
Y = 650 – 100r

3. Given the following data about the monetary sector of the economy:
Md = 0.4Y – 80r
Ms = 1200 million
Where, Md is demand for money, Y is the level of income, r is the rate of interest and Ms is the supply of money.
Derive the equation for LM curve and give the economic interpretation of this curve.
Sol:-
     For money market to be in equilibrium, 
Md = Ms
0.4Y – 80r = 1200
80r = 0.4Y – 1200
r = 0.005Y - 15
Thus we get the following LM curve: 
r = 0.005Y – 15
LM curve means what would be rate of interest when money market is in equilibrium, given the level of income. Thus, if level of national income is SR 4000 million, then using LM equation, we have:
r = 0.005× 4000 – 15
r = 20 – 15 = 5%
Thus, at income of SR 4000 million, rate of interest will be 5 per cent when money market is in equilibrium.
Now, if level of income is SR 4400 million, equilibrium rate of interest will be:
r = 0.005Y - 15
r = 0.005× 4400 – 15
r = 22 – 15 = 7%
Thus, at income of SR 4400 million, rate of interest will be 7 per cent when money market is in equilibrium.

4.For an economy the following functions have been given:
C = 100 + 0.8Y
S = -100 + 0.2Y
I = 120 – 5r 
Ms = 120
Md = 0.2Y – 5r
Y=?
Sol:-
I=S
120-5r=-100+0.2y
0.2y=220-5r
Again,
Md=Ms
0.2y-5r=120
0.2y-120=5r
r=0.04y-24
Now,
0.2y=220-5r=220-5(0.04y-24)
0.2y=220-0.2y+120
0.4y=340
y=3400/4=850

5.For an economy the following functions have been given:
C = 100 + 0.8Y
S = -100 + 0.2Y
I = 120 – 5r 
G=500
Ms = 120
Md = 0.2Y – 5r
Y=?
Sol:-
Y=C+I+G
=100+0.8y+120-5r+500
=720+0.8y-5r
0.2y=720-5r
Again,
Md=Ms
0.2y-5r=120
0.2y-120=5r
r=0.04y-24
Now,
0.2y=720-5r=220-5(0.04y-24)
0.2y=720-0.2y+120
0.4y=840
y=8400/4=2100