SME
Engineering Economics
Present worth and annual worth
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 present worth of costs?”
Chart
PWc = C + PWA − PWS
45,000 + 32,010 − 2,333 = 74,677
The flows in add and the flows out take away: PWc = 74,677.
This is an example. Type your own values to start a new problem.
$
PWc
=
$
C
+
$
AOC
(P/A,
i
%,
n
)
−
$
S
(P/F,
i
%,
n
)
AOC is the annual operating cost, what the machine costs to run each year; the bar PWA is what those payments are worth today.
(P/A)
Factor (P/A, i%, n)
calculated
(P/F)
Factor (P/F, i%, n)
calculated
PWA
Worth of annual costs
calculated · $
PWS
Worth of salvage
calculated · $
Formulas
(P/A)
=
(1
−
(1 +
i
100
)
−
n
)
÷
(
i
100
)
5.3349
=
(1
−
(1 +
10
100
)
-8
)
÷
(
10
100
)
(P/F)
=
(1 +
i
100
)
−
n
0.46651
=
(1 +
10
100
)
-8
PWA = AOC × ((P/A))
32,010 = 6,000 × 5.3349
PWS
=
S
×
((P/F))
2,333 = 5,000 × 0.46651
PWc
=
C
+
PWA
−
PWS
74,677 = 45,000 + 32,010 − 2,333
Assumptions
•
Costs count positive, the salvage negative (FE’s signed form is −PWc); of alternatives with one life, the lowest PWc wins.
•
C is paid now, AOC at the end of each year, and S comes back at the end of year n.
•
One interest rate i (the MARR) per year, compounded once a year.
Step-by-step
WE KNOW
C = $45,000, AOC = $6,000, S = $5,000, i = 10%, n = 8 yr
FIND
factor (P/A, i%, n) ((P/A)), factor (P/F, i%, n) ((P/F)), worth of annual costs (PWA), worth of salvage (PWS), present worth of costs (PWc)
Step 1 · Find factor (P/A, i%, n): (P/A)
Use (P/A) = (1 − (1 + i ÷ 100)⁻ⁿ) ÷ (i ÷ 100)
Add the present worth of $1 at the end of each year: the series factor sums them.
(P/A)
=
(1
−
(1 +
10
100
)
-8
)
÷
(
10
100
)
(P/A)
=
(1
−
(1 + 0.1)
-8
)
÷
(
10
100
)
(P/A)
=
1 −
1.1
-8
10 ÷ 100
(P/A)
=
1 − 0.4665
0.1
(P/A)
=
0.5335
0.1
(P/A) = 5.33493
Step 2 · Find factor (P/F, i%, n): (P/F)
Use (P/F) = (1 + i ÷ 100)⁻ⁿ
Discount
$1
back
n
years:
divide
by
(1
+
i
)
once
for
each
year.
(P/F)
=
(1 +
10
100
)
-8
(P/F)
=
(1 + 0.1)
-8
(P/F)
=
1.1
-8
(P/F) = 0.46651
Step 3 · Find worth of annual costs (PWA)
Use PWA = AOC × ((P/A))
The yearly cost times the series factor brings every year’s cost to now.
PWA = 6,000 × 5.33493
PWA = about $32,010
Step 4 · Find worth of salvage (PWS)
Use PWS = S × ((P/F))
The
salvage
comes
back
at
year
n
,
so
discount
it
to
now.
PWS = 5,000 × 0.46651
PWS = about $2,333
Step 5 · Find present worth of costs (PWc)
Use PWc = C + PWA − PWS
Add the costs’ present worths and take off the salvage’s: every amount now sits at time 0.
PWc = 45,000 + 32,010 − 2,333
PWc = about $74,677
Check
Put the numbers back in. Both sides should match.
5.33493
=
(1
−
(1 +
10
100
)
-8
)
÷
(
10
100
)
✓
0.46651
=
(1 +
10
100
)
-8
✓
32,010 = 6,000 × 5.33493 ✓
2,333 = 5,000 × 0.46651 ✓
74,677 = 45,000 + 32,010 − 2,333 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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