Consumption, Saving, and Investment 65
NUMERICAL PROBLEMS
1. First, a general formulation of the problem is useful. With income of Y1 in the first
year and Y2 in the second year, the consumer saves Y2 – C in the first year and Y2
– C in the second year, where C is the consumption amount, which is the same in
both years. 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. So the key equation is (Y1 – C)(1 + r) +
(Y2 – C) = T.
a. Y1 = Y2 = $25 000, r = 10%, T = $6300. The key equation gives ($25 000 – C)
1.1 + ($25 000 – C) = $6300. This can be simplified to $25 000 – C =
$6300/2.1 = $3000, which can be solved to get C = $22 000. Then S = Y – C
= $25 000 – $22 000 = $3000.
d. With the increase in wealth of W, the total amount invested for the second
period is W + Y1 – C, so the key equation becomes ($525 + $25 000 – C)1.1
+ ($25 000 – C) = $6300. This can be simplified to ($25 525 × 1.1) + $25 000