SME
Engineering Economics · Problem type
Two alternatives of equal life
Use this for “Machine A costs $30,000, $8,000 a year and sells for $4,000; B costs $42,000, $5,000 a year and sells for $6,000. Both last 6 years. At 12%, which is cheaper?”
Chart
This is an example. Type your own values to start a new problem.
$
PWA
=
$
CA
+
$
AOCA
(P/A,
i
%,
n
)
−
$
SA
(P/F,
i
%,
n
)
$
PWB
=
$
CB
+
$
AOCB
(P/A,
i
%,
n
)
−
$
SB
(P/F,
i
%,
n
)
For each machine, C is its first cost, AOC what it costs to run each year and S what it sells for at the end; the bars show each one’s present worth.
Formulas
PWA
=
CA
+
AOCA
×
(1
−
(1 +
i
100
)
−
n
)
÷
(
i
100
)
−
SA
×
(1 +
i
100
)
−
n
60,865
=
30,000
+
8,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
4,000
×
(1 +
12
100
)
-6
PWB
=
CB
+
AOCB
×
(1
−
(1 +
i
100
)
−
n
)
÷
(
i
100
)
−
SB
×
(1 +
i
100
)
−
n
59,517
=
42,000
+
5,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
6,000
×
(1 +
12
100
)
-6
Assumptions
•
Both alternatives last n years and do the same job.
•
Costs count positive and the salvage negative, so the lower present worth of costs is chosen.
•
One MARR i for both; C now, AOC at each year’s end, S at the end of year n.
Step-by-step
WE KNOW
i = 12%, n = 6 yr, CA = $30,000, AOCA = $8,000, SA = $4,000, CB = $42,000, AOCB = $5,000, SB = $6,000
FIND
present worth of A (PWA), present worth of B (PWB)
Step 1 · Find present worth of A (PWA)
Use PWA = CA + AOCA × (1 − (1 + i ÷ 100)⁻ⁿ) ÷ (i ÷ 100) − SA × (1 + i ÷ 100)⁻ⁿ
Bring every amount to time 0: the yearly cost by (P/A), the salvage by (P/F).
PWA
=
30,000
+
8,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
4,000
×
(1 +
12
100
)
-6
(P/A,
12%,
6)
=
1 −
1.12
-6
0.12
=
4.1114
(P/F,
12%,
6)
=
1.12
-6
=
0.5066
PWA = 30,000 + 8,000 × 4.1114 − 4,000 × 0.5066
30,000
+
8,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
4,000
×
(1 +
12
100
)
-6
=
60,864.7341
PWA = about $60,865
Step 2 · Find present worth of B (PWB)
Use PWB = CB + AOCB × (1 − (1 + i ÷ 100)⁻ⁿ) ÷ (i ÷ 100) − SB × (1 + i ÷ 100)⁻ⁿ
Bring every amount to time 0 as for A; B’s present worth of costs is lower, so choose B.
PWB
=
42,000
+
5,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
6,000
×
(1 +
12
100
)
-6
PWB = 42,000 + 5,000 × 4.1114 − 6,000 × 0.5066
42,000
+
5,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
6,000
×
(1 +
12
100
)
-6
=
59,517.2499
PWB = about $59,517
Check
Put the numbers back in. Both sides should match.
60,865
=
30,000
+
8,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
4,000
×
(1 +
12
100
)
-6
✓
59,517
=
42,000
+
5,000
×
(1
−
(1 +
12
100
)
-6
)
÷
(
12
100
)
−
6,000
×
(1 +
12
100
)
-6
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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