P = $1,000 __1___ +
$1,000 __1___ + $1,000 __1___ + $1,000 __1___
(1 + 0.06)1
(1 + 0.06)2
(1 + 0.06)3
(1 + 0.06)4
which is
Ž SPPW
Ž SPPW
Ž SPPW
Ž SPPW
P = $1000 | i = .06 | + $1,000 | i = .06 | +
$1000 | i = .06 | +
$1000 | i = .06 |
‘
n = 1 ž
‘
n = 2 ž
‘
n = 3 ž
‘
n = 4ž
or its equivalent
P =
$1000 (1 +0.06)4 - 1
0.06(1 + 0.06)4
= $1000 (3.465) = $3,465, which
is said to be the present
value of a 4-year stream of $1000
amounts. The factor, 3.465, is called the
Òuniform series
present worthÓ factor for n = 4, i =0.06 (see Table 2, next slide).