SME
Engineering Economics · Problem type
Uniform series: (F/A, i%, n) and (A/F, i%, n)
Use this for “$1,500 is deposited at the end of each year for 10 years at 5%. How much is there at the end?”, or the deposit a sinking fund needs.
Chart
Terms: 1,500, 1,575, 1,654, 1,736, …, 2,327 (multiply by g = 1.05 each time).
a₁₀ = 1,500 × 1.05⁹ = 2,327
F
=
1,500
×
1 −
1.05
10
1 − 1.05
=
18,870
This is an example. Type your own values to start a new problem.
$
F
=
$
A
(F/A,
i
%,
n
)
g
Growth factor (1 + i)
worked out (not typed)
(F/A)
Factor (F/A, i%, n)
calculated
(A/F)
Factor (A/F, i%, n)
calculated
Formulas
g
=
1
+
i
100
1.05
=
1
+
5
100
(
F
A
)
=
(
(1 +
i
100
)
n
−
1)
÷
(
i
100
)
12.5779
=
(
(1 +
5
100
)
10
−
1)
÷
(
5
100
)
(
A
F
)
=
1
(
F
A
)
0.079505
=
1
12.5779
F
=
A
×
((
F
A
))
18,867 = 1,500 × 12.5779
Assumptions
•
Deposits come at the end of each year; the last one, made the day F is counted, earns nothing.
•
The same amount A every year for n years, all at one rate i.
•
(A/F) is the sinking-fund factor: the yearly deposit per dollar wanted at the end.
Step-by-step
WE KNOW
A = $1,500, i = 5%, n = 10 yr
FIND
growth factor (1 + i) (g), factor (F/A, i%, n) ((F/A)), factor (A/F, i%, n) ((A/F)), future worth (F)
Step 1 · Find growth factor (1 + i): g
Use g = 1 + i ÷ 100
Each
year
a
deposit
is
multiplied
by
1
+
i
.
g
=
1
+
5
100
g
=
1
+
0.05
g
=
1.05
Step 2 · Find factor (F/A, i%, n): (F/A)
Use (F/A) = ((1 + i ÷ 100)ⁿ − 1) ÷ (i ÷ 100)
Add
up
$1
a
year
grown
to
year
n
:
1
+
(1
+
i
)
+
…
+
(1
+
i
)
ⁿ
-1
,
a
geometric
series.
(
F
A
)
=
(
(1 +
5
100
)
10
−
1)
÷
(
5
100
)
(
F
A
)
=
(
(1 + 0.05)
10
−
1)
÷
(
5
100
)
(
F
A
)
=
1.05
10
− 1
5 ÷ 100
(
F
A
)
=
1.629 − 1
0.05
(
F
A
)
=
0.629
0.05
(
F
A
)
=
12.5779
Step 3 · Find factor (A/F, i%, n): (A/F)
Use (A/F) = 1 ÷ ((F/A))
Spreading
F
back
over
the
years
undoes
adding
the
deposits
up.
(
A
F
)
=
1
12.5779
(
A
F
)
=
0.079505
Step 4 · Find future worth (F)
Use F = A × ((F/A))
Each deposit, times the factor.
F
=
1,500
×
12.5779
F
=
about
$18,867
Check
Put the numbers back in. Both sides should match.
1.05
=
1
+
5
100
✓
12.5779
=
(
(1 +
5
100
)
10
−
1)
÷
(
5
100
)
✓
0.079505
=
1
12.5779
✓
18,867 = 1,500 × 12.5779 ✓
← Previous lesson
Uniform series: (P/A, i%, n) and (A/P, i%, n)
Next lesson →
Arithmetic gradient: (P/G, i%, n)
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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