Engineering Economics

Time value of money

Use this for “What does $8,000 grow to at 6% a year for 9 years?”, what a future amount is worth today, or the years it takes to grow.

Chart

F(n) = 8,000(1 + 6/100)n024681002,0004,0006,0008,00010,00012,00014,00016,000Years n (yr)Worth F ($)(9, 13,520)
F(9) ≈ 13,520
This is an example. Type your own values to start a new problem.
$
F
=
$
P
(1
+
%)
(F/P)
Factor (F/P, i%, n)
calculated
(P/F)
Factor (P/F, i%, n)
calculated

Formulas

(
F
P
)
=
(1 +
i
100
)
n
1.6895
=
(1 +
6
100
)
9
(
P
F
)
=
1
(
F
P
)
0.5919
=
1
1.6895
F
=
P
×
((
F
P
))
13,516 = 8,000 × 1.6895

Assumptions

•
i is the interest rate per period, and n counts the same periods (years here).
•
One payment: P now, F at the end of year n.
•
Interest compounds once each period, so each year’s interest earns interest after it.

Step-by-step

WE KNOW
P = $8,000, i = 6%, n = 9 yr
FIND
factor (F/P, i%, n) ((F/P)), factor (P/F, i%, n) ((P/F)), future worth (F)
Step 1 · Find factor (F/P, i%, n): (F/P)
Use (F/P) = (1 + i ÷ 100)ⁿ
Each
year
multiplies
the
amount
by
1
+
i
,
so
n
years
multiply
it
by
(1
+
i
)ⁿ.
(
F
P
)
=
(1 +
6
100
)
9
(
F
P
)
=
(1 + 0.06)
9
(
F
P
)
=
1.06
9
(
F
P
)
=
1.6895
Step 2 · Find factor (P/F, i%, n): (P/F)
Use (P/F) = 1 ÷ ((F/P))
Bringing
money
back
n
years
undoes
carrying
it
forward.
(
P
F
)
=
1
1.6895
(
P
F
)
=
0.5919
Step 3 · Find future worth (F)
Use F = P × ((F/P))
Carry
the
present
worth
forward
n
years.
F
=
8,000
×
1.6895
F
=
about
$13,516
Check
Put the numbers back in. Both sides should match.
1.6895
=
(1 +
6
100
)
9
✓
0.5919
=
1
1.6895
✓
13,516 = 8,000 × 1.6895 ✓
Topic check
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