7.2 equivalent uniform annual cost method

1. Building Economy ARE 431
Dr. Mohammad A. Hassanain 1
1
Life Cycle Costing
‰Uniform annual worth method:
‰In this case, all costs or savings (cash flows) which occur
during the life time of an investment are discounted to a
uniform annual series of cash flows over the life time of the
investment.
P = A []
i (1 + i)
n
(1 + i)
n
-1
‰An alternative method, instead of using the above formula, is
to use interest tables with the following notations.
(P/A, i%, n) Find P given A
(A/P, i%, n) Find A given P
(F/A, i%, n) Find F given A
(A/F, i%, n) Find A given F
P = ?
n-12
0
1 3 n-2n
A
n
AAAAA
2
Life Cycle Costing
‰Uniform annual worth method:
‰The major advantage of this method is that it
is not necessary to make the comparison
over the same number of years when the
alternatives have different lives.
‰The reason for that, is that this method is an
equivalent annual cost over the life of the
project.

2. Building Economy ARE 431
Dr. Mohammad A. Hassanain 2
3
Life Cycle Costing
Example:If the minimum required rate of return is 15%,
which project should be selected
$2,600,000Extra income taxes
$3,000,000$2,000,000Salvage value
106Project service life (years)
$7,000,000$11,000,000Annual labor cost
$300,000$800,000Annual maintenance cost
$36,000,000$26,000,000First cost
Project BProject A
Uniform Annual Worth Method
4
Life Cycle Costing
$11,800,000 = $11,000,000 + $800,000
42
0
1 3 5
6
$26,000,000
$2,000,000Project A
42
0
1 3 5 78
10
$36,000,000
$3,000,000Project B
6
$9,900,000 = 300,000 + 7,000,000 + 2,600,000
9

3. Building Economy ARE 431
Dr. Mohammad A. Hassanain 3
5
Life Cycle Costing
EUAC
A
= $26,000,000 (A/P, 15%, 6) - $2,000,000 (A/F, 15%, 6) +
$11,800,000
EUAC
B
= $36,000,000 (A/P, 15%, 10) - $3,000,000 (A/F, 15%, 10) +
= $16,925,000
Project B should be selected since EUAC
B
< EUAC
A
= $18,442,000
$9,900,000