1. Building Economy ARE 431
Dr. Mohammad A. Hassanain 1
1
Life Cycle Costing
Present worth method:
In this case, all costs or savings (cash flows) which
occur during the life time of an investment are
discounted to present value of money.
P
n-120 1 3 n-2 n
F
n
Fn = P(1 + i)n or P = Fn
1
(1 + i)n
2
Life Cycle Costing
Present worth method:
P
n-120 1 3 n-2 n
F
n
An alterative method, instead of using the F and P
formulas, is to use interest tables with the following
notations:
(F/P, i%, n) Find F given P
(P/F, i%, n) Find P given F

2. Building Economy ARE 431
Dr. Mohammad A. Hassanain 2
3
Life Cycle Costing
Example: Make a present-worth comparison of the equal-service life
projects for which costs are shown below, if i = 10%. Which project
would you select?
55Project service life (years)
$350,000$200,000Salvage value, F
$700,000$900,000Annual operating cost, A
$3,500,000$2,500,000First cost, P
Project BProject A
Present Worth Comparison of Equal-Lived Alternatives
4
Life Cycle Costing
A = $900,000
42
0
1 3
5
i = 10%
$2,500,000
$200,000
Project A
A = $700,000
42
0
1 3
5
i = 10%
$3,500,000
$350,000
Project B
PA = $2,500,000 + $900,000 (P/A, 10%, 5) - $200,000 (P/F, 10%, 5)
= $5,788,000
PB = $3,500,000 + $700,000 (P/A, 10%, 5) - $350,000 (P/F, 10%, 5)
= $5,936,000
Project A should be selected since PA < PB

3. Building Economy ARE 431
Dr. Mohammad A. Hassanain 3
5
Life Cycle Costing
Example: Two machines are under consideration for a carpentry
workshop. Machine A will have a first cost of $5,000, an annual
operation and maintenance cost of $300, and a $1,200 salvage
value. Machine B will have a first cost of $7,000, an annual
operation and maintenance cost of $200, and a $1,300 salvage
value. If both machines are expected to last for ten years, determine
which machine should be selected on the basis of their present-
worth values using an interest rate of 10%.
1010Service life (years), n
$1,300$1,200Salvage value, F
$200$300Operation and Maintenance cost, A
$7,000$5,000First cost, P
Machine BMachine A
Present Worth Comparison of Equal-Lived Alternatives
6
Life Cycle Costing
Example: Make a present-worth comparison of the different-service
life machines for which costs are shown below, if i = 15%. Which
machine would you select?
96Service life (years)
$2,000$1,000Salvage value
$3,100$3,500Annual operating cost
$18,000$11,000First cost
Machine BMachine A
Present Worth Comparison of Different-Lived Alternatives
The cash flow for one cycle of an alternative must be duplicated
for the least common multiple of years, so that service life is
compared over the total life for each alternative.

4. Building Economy ARE 431
Dr. Mohammad A. Hassanain 4
7
Life Cycle Costing
Since the machines have different lives, they must be compared
Over the least common multiple of years, which is 18 years.
A = $3,500
42
0
1 3 5 7 8 9 10 11 13
$11,000
14 15 16 17
$1,000 $1,000 $1,000
18
$11,000 $11,000
Machine A
A = $3,100
42
0
1 3 5 7 8 10 11 13
$18,000
14 15 16 17
$2,000 $2,000
18
Machine B
6 12
$18,000
8
Life Cycle Costing
PA = $11,000 + $11,000 (P/F, 15%, 6) + $11,000 (P/F, 15%, 12) +
$3,500 (P/A, 15%, 18) - $1,000 (P/F, 15%, 6) - $1,000 (P/F, 15%, 12) -
PB = $18,000 + $18,000 (P/F, 15%, 9) + $3,100 (P/A, 15%, 18) -
= $41,384
Machine A should be selected since PA < PB
$1,000 (P/F, 15%, 18)
= $38,559
$2,000 (P/F, 15%, 9) - $2,000 (P/F, 15%, 18)

5. Building Economy ARE 431
Dr. Mohammad A. Hassanain 5
9
Life Cycle Costing
Example: A company is trying to decide between two different
garbage disposals. A regular (RS) disposal has an initial cost of $65
and a life of 4 years. The alternative is a corrosion-resistant
disposal constructed of stainless steel (SS). The initial cost of the
SS disposal is $110, but it is expected to last 10 years.
The SS disposal is expected to cost $5 per year more than the
RS disposal. If the interest rate is 6%, which disposal should be
selected, assuming both have no salvage value?
104Service life (years)
--Salvage value
$5-Additional cost per year
$110$65Initial cost
SSRS
Present Worth Comparison of Different-Lived Alternatives
10
Life Cycle Costing
Since the disposals have different lives, they must be compared
Over the least common multiple of years, which is 20 years.
$65
420 1 3 5 7 8 9 10 11 13 14 15 16 17 18
RS Disposal
19 206
$65 $65$65 $65
12
Extra Cost (A) = $5
42
0
1 3 5 7 8 10 11 13
$110
14 15 16 17 18
SS Disposal
6 12 19 209
$110

6. Building Economy ARE 431
Dr. Mohammad A. Hassanain 6
11
Life Cycle Costing
$65 (P/F, 6%, 16)
PRS = $65 + $65 (P/F, 6%, 4) + $65 (P/F, 6%, 8) + $65 (P/F, 6%, 12) +
PSS = $110 + $110 (P/F, 6%, 10) + $5 (P/A, 6%, 20)
= $229
Disposal RS should be selected since PRS < PSS
= $215
12
Life Cycle Costing
Example: Two machines are under consideration for a carpentry
workshop. Machine A will have a first cost of $5,000, an annual
operation and maintenance cost of $300, and a $1,200 salvage
value. Machine B will have a first cost of $7,000, an annual
operation and maintenance cost of $200, and a $1,300 salvage
value. If machine A is expected to last for 10 years, while Machine B
would last for 15 years, determine which machine should be
selected on the basis of their present-worth values using an interest
rate of 10%.
1010Service life (years), n
$1,300$1,200Salvage value, F
$200$300Operation and Maintenance cost, A
$7,000$5,000First cost, P
Machine BMachine A
Present Worth Comparison of Different-Lived Alternatives