Question

Question 2 In the Keynesian cross, assume that the consumption function is given by C = 150 +0.7 (Y-T) Planned investment is: I = 100 – 10 *r Government purchases and taxes are both 50. a. Graph consumption as function of income. b. Graph investment as function of the real interest rate. c. Suppose that the real interest rate is 5. Write the equation of the planned expenditure. d. Suppose that the real interest rate is 5. What is the equilibrium level of income? e. Using the equation of the planned expenditure in question c', graph planned expenditure as a function of income. Draw a 45° degree line. Show the equilibrium level of income on the Keynesian cross. f. Use the equation of the planned expenditure in question c to answer this question: If government purchases increase to 125, what is the new equilibrium income? g. Calculate the government purchases multiplier.

Answer 1

(a) The consumption function is given as
C 1500.7(Y-T)
, and since T=G=50, we have
C 1500.7(Y-50)
or
150+0. 35
or
=115+0.Y
.

The graph would be as below.

1000 800 600 400 Consumption Function 200 Y 1000 400 200 600 800

(b) The investment function is given as . The graph is as below.

r 10 Investment 5- 50 00 LO

(c) For r=5, the investment would be
100 105=50
. The equation of PE would be as
+G PE C
or or .

(d) For r=5, we have PE as . The equilibrium level of income would be where or or or or . This would be the equilibrium level of income.

(e) The graph would be as below.

1000 РЕ 800 1E 600 400 PE PE=Y 200 Y*=716.67 Y 1000 400 200 800 600

(f) The PE would be
+G PE C
, and for new G, we have or . The new equilibrium income would be where or or or or .

(g) The government purchase multiplier would be $g = \frac{\Delta Y}{\Delta G}$ or $g = \frac{966.67 - 716.67}{125 - 50}$ or $g = \frac{250}{75}$ or .

We can also calculate this theoretically as for the equilibrium income be where , we have or or or $Y = \frac{1}{1 - 0.7}[165 + G]$ . For change in G as $\Delta G = G_2 - G_1$ , we have the change in equilibrium as $\Delta Y = Y_2 - Y_1$ or $\Delta Y = \frac{1}{1 - 0.7}[165 + G_2] - \frac{1}{1 - 0.7}[165 + G_1]$ or $\Delta Y = \frac{1}{1 - 0.7}[165 + G_2 - 165 - G_1]$ or $\Delta Y = \frac{1}{1 - 0.7}[\Delta G]$ , and hence $\frac{\Delta Y}{\Delta G} = \frac{1}{1 - 0.7}$ , ie $g = \frac{1}{1 - 0.7}$ or $g = \frac{1}{0.3} = 3.33$ .