ENGINEERING ECONOMICS
LECTURE 4
PART 1
PRESENT WORTH ANALYSIS
Present Worth Analysis
It is further noted that when the alternatives are service type, the selection can only be
mutually exclusive with only one selected and with no consideration for DN.
In evaluating alternatives, the following criteria are used:
Present Worth Analysis of Equal-Life Alternatives:
One alternative – Present worth, PW, is calculated at the MARR. If PW ≥ 0, which
indicates
that the MARR is met or exceeded, the alternative is acceptable. If PW < 0, alternative is not
acceptable and we opt for DN.
It should be obvious from this that one alternative evaluation is relevant only for the revenue
type alternatives. (PW of service type alternatives are less than zero as they only include
costs).
Two or more alternatives – Calculate the PW values of all the alternatives at the MARR.
For mutually exclusive alternatives, whether service or revenue type, select the alternative
with the PW value that is numerically largest.
For independent alternatives (applicable only for revenue types), select all the alternatives
that have PW ≥ 0.
Example 1
Alternative A has an initial cost of $20,000, an operating cost of $9,000 per year, and
a$5,000 salvage value after 5 years. Alternative B will cost $35,000 with an operating cost
of $4,000 per year and a salvage value of $7,000 after 5 years. At an MARR of 12% per
year, which should be selected?
Solution: Find PW at MARR and select numerically larger PW value
(A) (B) 7,000
5,000
0 1 2 3 4 0 1 2 3 4
9,000 9,000 9,000 9,000 9,000 4,000 4,000 4,000 4,000 4,000
20,000 35,000
PWA = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5) = -$49,606
PWB = -35,000 - 4000(P/A,12%,5) + 7000(P/F,12%,5) = -$45,447
Select alternative B
Example 2
Perform annual worth analysis for equal - service devices with the costs shown below, if the MARR is 10 % per year.
Machine (A) Machine (B) Machine (C)
First Cost $30,000 $35,000 $40,000
Taxes ------------------------- $20,000 $23,000
AOC $3,500 $1,500 $1,000
Salvage value $1,000 $2,000 $5,000
Estimated Life 6 6 6
Solution: Find PW at MARR and select numerically larger PW value (A) 1,000
0 1 2 3 4 5 6
PWA = -30,000 – 3,500(P/A,10%,6) + 1,000(P/F,10%,6)
3,500 3,500 3,500 3,500 3,500 3,500
= -30,000 – 3,500(4.3553) + 1,000(0.5645) = -$44,679.1
30,000
2,000
(B)
1
PWB = -35,000 – 21,500 (P/A,10%,6) + 2,000(P/F,10%,6) 0 2 3 4 5 6
= -35,000 – 21,500 (4.3553) + 2,000(0.5645)= -$127,509.95
21,500 21,500 21,500 21,500 21,500 21,500
35,000
Machine (A) Machine (B) Machine (C)
First Cost $30,000 $35,000 $40,000
Taxes ------------------------------------- $20,000 $23,000
AOC $3,500 $1,500 $1,000
Salvage value $1,000 $2,000 $5,000
Estimated Life 6 6 6
(C) 5,000
PWC = -40,000 – 24,000(P/A,10%,6) + 5,000(P/F,10%,6) 0 1 2 3 4 5 6
= -40,000 – 24,000(4.3553) + 5,000(0.5645)
= -144,527.2+ 2,822.5 = -$ 141704.7 24,000 24,000 24,000 24,000 24,000 24,000
40,000
PWA PWB PWC
-$44,679.1 -$127,509.95 -$ 141704.7
Keep Machine (A) . Less cost.
Exercise
Perform annual worth analysis for equal - service devices with the costs shown below, if the MARR is 18 % per year.
(1) performing a present worth (PW) analysis.
(2) drawing three cash flows of the machines if the MARR is 18% per year.
Electric powered Gas powered Solar powered
First Cost, L.E. -50,000 -63,000 -60,000
Annual Operating Cost, L.E./year -7,000 -5,000 -3,000
Salvage Value, L.E. 20,000 24,000 28,000
life years 5 5 5
PW- Electric powered = - 50,000 – 7,000(P/A,18%,5) + 20,000(P/F,18%,5)
= - 50,000 – 7,000 (3.12717) + 20,000 (0.43711) = LE -63152.99
Gas powered = - 63,000 – 5,000(P/A,18%,5) + 24,000 (P/F,18%,5)
= - 63,000 – 5,000(3.12717) + 24,000 (0.43711) = LE -68145.21
Solar powered = - 60,000 – 3,000(P/A,18%,5) + 28,000 (P/F, 18%,5)
= - 60,000 – 3,000(3.12717) + 28,000 (0.43711) = LE -57142.43
Solar powered device is the most economic, since the PW of its Cost is the Lowest
