Chapter 4
Numerical Problems
1.
General formulation of the problem:
With income of Y1 in the first year and Y2 in the second year, the consumer saves S1=Y1 - C1 in the first year and S2=Y2
- C2 in the second year, where C1, C2 are the consumption amounts.
Saving in the first year earns interest at rate r, where r is the real interest rate. And the consumer needs to accumulate just
enough after two years to pay for college tuition, in the amount T.
As S2 =0 and W3=0,
C2+T=Y2+(W1+S1)(1+r) where W1 is the level of wealth at the beginning of this period, W2=W1(1+r) will be the level of
assets at the beginning of next period.
So C2+T=Y2+(W1+Y1-C1)(1+r) or C1+ C2(1+r)+T= Y2+(W1+Y1)(1+r) or C1/(1+r)+C2+T/(1+r)= Y2/(1+r)+W1+Y1
As C2=C1=C we have C+ C (1+r)= Y2+(W1+Y1)(1+r)-T
So C= (Y2+(W1+Y1)(1+r)-T)/(2+R) and S1=Y1-C
a.
Y1 = Y2= $50,000
r = 10 %,
T = $12,600.
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C= $44,000.
And then S1 = Y1 - C = $50,000 - $44,000 = $6000.
b.
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C = $46,200.
S1 = Y1 - C = $54,200 - $46,200 = $8000.
This illustrates that a rise in current income increases savings.
c.
Y2 = $54,200.
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C = $46,000.
S1 = Y1 - C = $50,000 - $46,000 = $4000.
This illustrates that a rise in future income decreases savings.
d.
With the increase in wealth of W1, the total amount invested for the second period is W1 + Y1 - C,
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C = $44,550.
S1 = Y1-C =$50,000 - $44,550 = $5450.
This illustrates that a rise in wealth decreases savings.
e.
T = $14,700.
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C = $43,000.
Then S1 =Y1 - C = $50,000 - $43,000 = $7000.
The rise in targeted wealth needed in the future raises current saving.
f.
r = 25%.
C= (Y2+(W1+Y1)(1+r)-T)/(2+R) implies C = $44,400.
Then S1 = Y1 - C = $50,000 - $44,400 = $5600.
The rise in the real interest rate, with a given wealth target, reduces current saving.
2.
a.