SME
Engineering Economics · Problem type
Arithmetic gradient: (P/G, i%, n)
Use this for “Maintenance costs $3,000 in year 1 and rises $400 a year. At 8%, what are 10 years of it worth today?”
Chart
Terms: 3,000, 3,400, 3,800, 4,200, …, 6,600 (add G = 400 each time).
a₁₀ = 3,000 + 9 × 400 = 6,600
This is an example. Type your own values to start a new problem.
$
P
=
$
A₁
(P/A,
i
%,
n
)
+
$
G
(P/G,
i
%,
n
)
(P/A)
Factor (P/A, i%, n)
calculated
(P/G)
Factor (P/G, i%, n)
calculated
Formulas
(
P
/A)
=
(1
−
(1 +
i
100
)
−
n
)
÷
(
i
100
)
6.7101
=
(1
−
(1 +
8
100
)
-10
)
÷
(
8
100
)
(
P
G
)
=
(
(1 +
i
100
)
n
−
1)
÷
(
(
i
100
)
2
×
(1 +
i
100
)
n
)
−
n
÷
(
i
100
×
(1 +
i
100
)
n
)
25.9768
=
(
(1 +
8
100
)
10
−
1)
÷
(
(
8
100
)
2
×
(1 +
8
100
)
10
)
−
10
÷
(
8
100
×
(1 +
8
100
)
10
)
P
=
A₁
×
((
P
/A))
+
G
×
((
P
G
))
30,521 = 3,000 × 6.7101 + 400 × 25.9768
Assumptions
•
Year 1 holds A₁ alone; the gradient’s first G comes at the end of year 2.
•
Each year’s amount is G more than the year before; a falling series has G below 0.
•
(P/G) counts only the growing part; the base amount A₁ goes through (P/A).
Step-by-step
WE KNOW
A₁ = $3,000, G = $400, i = 8%, n = 10 yr
FIND
factor (P/A, i%, n) ((P/A)), factor (P/G, i%, n) ((P/G)), present worth (P)
Step 1 · Find factor (P/A, i%, n): (P/A)
Use (P/A) = (1 − (1 + i ÷ 100)⁻ⁿ) ÷ (i ÷ 100)
The
base
amount
A₁
comes
every
year:
add
the
present
worths
of
$1
a
year.
(
P
/A)
=
(1
−
(1 +
8
100
)
-10
)
÷
(
8
100
)
(
P
/A)
=
(1
−
(1 + 0.08)
-10
)
÷
(
8
100
)
(
P
/A)
=
1 −
1.08
-10
8 ÷ 100
(
P
/A)
=
1 − 0.4632
0.08
(
P
/A)
=
0.5368
0.08
(
P
/A)
=
6.7101
Step 2 · Find factor (P/G, i%, n): (P/G)
Use (P/G) = ((1 + i ÷ 100)ⁿ − 1) ÷ ((i ÷ 100)² × (1 + i ÷ 100)ⁿ) − n ÷ (i ÷ 100 × (1 + i ÷ 100)ⁿ)
The
gradient
pays
G
in
year
2,
2
G
in
year
3,
…,
(
n
−
1)
G
in
year
n
:
the
factor
adds
their
present
worths
per
dollar
of
G
.
(
P
G
)
=
(
(1 +
8
100
)
10
−
1)
÷
(
(
8
100
)
2
×
(1 +
8
100
)
10
)
−
10
÷
(
8
100
×
(1 +
8
100
)
10
)
(
P
G
)
=
(
(1 + 0.08)
10
−
1)
÷
(
0.08
2
×
(1 + 0.08)
10
)
−
10
÷
(
8
100
×
(1 + 0.08)
10
)
(
P
G
)
=
1.08
10
− 1
0.08
2
×
1.08
10
−
10
8 ÷ 100 ×
1.08
10
(
P
G
)
=
2.1589 − 1
0.0064 × 2.1589
−
10
8 ÷ 100 × 2.1589
(
P
G
)
=
1.1589
0.013817
−
10
0.08 × 2.1589
(
P
G
)
=
1.1589
0.013817
−
10
0.17271
(
P
G
)
=
83.875
−
57.901
(
P
G
)
=
25.9768
Step 3 · Find present worth (P)
Use P = A₁ × ((P/A)) + G × ((P/G))
Split the series into the base amount every year and the growing part, and add their present worths.
P
=
3,000
×
6.7101
+
400
×
25.9768
P
=
20,130.3
+
10,390.72
P
=
about
$30,521
Check
Put the numbers back in. Both sides should match.
6.7101
=
(1
−
(1 +
8
100
)
-10
)
÷
(
8
100
)
✓
25.9768
=
(
(1 +
8
100
)
10
−
1)
÷
(
(
8
100
)
2
×
(1 +
8
100
)
10
)
−
10
÷
(
8
100
×
(1 +
8
100
)
10
)
✓
30,521 = 3,000 × 6.7101 + 400 × 25.9768 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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