SME
Engineering Economics · Problem type
Equivalent uniform annual cost
Use this for “A $45,000 machine costs $6,000 a year to run and sells for $5,000 after 8 years. At 10%, what is its EUAC?”
Chart
EUAC = CR + AOC
7,998 + 6,000 = 13,998
The flows in add and the flows out take away: EUAC = 13,998.
This is an example. Type your own values to start a new problem.
$
EUAC
=
$
C
(A/P,
i
%,
n
)
−
$
S
(A/F,
i
%,
n
)
+
$
AOC
S is the salvage value, what the machine sells for at the end; it lowers the capital recovery CR, the first cost spread over the years.
(A/P)
Factor (A/P, i%, n)
calculated
(A/F)
Factor (A/F, i%, n)
calculated
CR
Capital recovery
calculated · $
Formulas
(A/P)
=
(
i
100
)
÷
(1
−
(1 +
i
100
)
−
n
)
0.18744
=
(
10
100
)
÷
(1
−
(1 +
10
100
)
-8
)
(A/F)
=
(
i
100
)
÷
(
(1 +
i
100
)
n
−
1)
0.087444
=
(
10
100
)
÷
(
(1 +
10
100
)
8
−
1)
CR
=
C
×
((A/P))
−
S
×
((A/F))
7,998 = 45,000 × 0.18744 − 5,000 × 0.087444
EUAC = CR + AOC
13,998 = 7,998 + 6,000
Assumptions
•
Costs count positive and the salvage negative, so the lowest EUAC wins.
•
EUAC compares alternatives of different lives when each would be repeated alike at the end of its life.
•
C is paid now, AOC at the end of each year, and S comes back at the end of year n.
•
EUAC = PWc × (A/P, i%, n): the same machine’s present worth, spread over its life.
Step-by-step
WE KNOW
C = $45,000, S = $5,000, AOC = $6,000, i = 10%, n = 8 yr
FIND
factor (A/P, i%, n) ((A/P)), factor (A/F, i%, n) ((A/F)), capital recovery (CR), equivalent annual cost (EUAC)
Step 1 · Find factor (A/P, i%, n): (A/P)
Use (A/P) = (i ÷ 100) ÷ (1 − (1 + i ÷ 100)⁻ⁿ)
The
capital-recovery
factor
is
1
P/A
:
the
yearly
payment
that
repays
$1
now
with
interest.
(A/P)
=
(
10
100
)
÷
(1
−
(1 +
10
100
)
-8
)
(A/P)
=
10
100
÷
(1
−
(1 + 0.1)
-8
)
(A/P)
=
10
100
÷
(1
−
1.1
-8
)
(A/P)
=
10
100
÷
(1
−
0.46651)
(A/P)
=
10
100
÷
0.53349
(A/P)
=
0.1
0.53349
(A/P) = 0.18744
Step 2 · Find factor (A/F, i%, n): (A/F)
Use (A/F) = (i ÷ 100) ÷ ((1 + i ÷ 100)ⁿ − 1)
The
sinking-fund
factor
is
1
F/A
:
the
yearly
deposit
that
grows
to
$1
at
year
n
.
(A/F)
=
(
10
100
)
÷
(
(1 +
10
100
)
8
−
1)
(A/F)
=
10
100
÷
(
(1 + 0.1)
8
−
1)
(A/F)
=
10
100
÷
(
1.1
8
−
1)
(A/F)
=
10
100
÷
(2.1436
−
1)
(A/F)
=
10
100
÷
1.1436
(A/F)
=
0.1
1.1436
(A/F) = 0.087444
Step 3 · Find capital recovery (CR)
Use CR = C × ((A/P)) − S × ((A/F))
Spread the first cost over the life with interest, less the salvage spread as a sinking fund.
CR = 45,000 × 0.18744 − 5,000 × 0.087444
CR = 8,434.8 − 437.22
CR = about $7,998
Step 4 · Find equivalent annual cost (EUAC)
Use EUAC = CR + AOC
Capital recovery and the operating cost are both yearly amounts, so they add.
EUAC = 7,998 + 6,000
EUAC = about $13,998
Check
Put the numbers back in. Both sides should match.
0.18744
=
(
10
100
)
÷
(1
−
(1 +
10
100
)
-8
)
✓
0.087444
=
(
10
100
)
÷
(
(1 +
10
100
)
8
−
1)
✓
7,998 = 45,000 × 0.18744 − 5,000 × 0.087444 ✓
13,998 = 7,998 + 6,000 ✓
Topic check
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