Econ 1110 Spring 2008
Assignment 2 Solutions
1.
Exercise 1 in [AB] (Numerical Problems)
First, a general formulation of the problem is useful. With income of
Y
1
in the first year and
Y
2
in
the second year, the consumer saves
Y
1
–
C
in the first year and
Y
2
–
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 (
Y
1
–
C
)(1
+
r
)
+
(
Y
2
–
C
)
=
T
.
(a)
Y
1
=
Y
2
=
$50,000,
r
=
10%,
T
=
$12,600. The key equation gives ($50,000 –
C
)1.1
+
($50,000
–
C
)
=
$12,600. This can be simplified to $50,000 –
C
=
$12,600/2.1
=
$6000, which can be
solved to get
C
=
$44,000. Then
S
=
Y
–
C
=
$50,000 – $44,000
=
$6000.
(b)
Y
1
=
$54,200. The key equation is now ($54,200 –
C
)1.1
+
($50,000 –
C
)
=
$12,600. This can
be simplified to ($54,200
×
1.1)
+
$50,000 – $12,600
=
2.1
C
, or $97,020
=
2.1
C
, so
C
=
$46,200. Then
S
=
Y
1
–
C
=
$54,200 – $46,200
=
$8000. This illustrates that a rise in current
income increases saving.
(c)
Y
2
=
$54,200. The key equation is now ($50,000 –
C
)1.1
+
($54,200 –
C
)
=
$12,600. This can
be simplified to ($50,000
×
1.1)
+
$54,200 – $12,600
=
2.1
C
, or $96,600
=
2.1
C
, so
C
=
$46,000. Then
S
=
Y
1
–
C
=
$50,000 – $46,000