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Module 10
Dr Tareq Albahri 2016
Economic Service Life
Physical Life: the life where an asset is physically capable of doing the intended job
Economic Life: the life where an asset does the intended job with lowest annual cost possible.
Therefore, you do not keep an asset for its complete physical life, but rather you keep it for it
economic life to avoid losses.
Example: Economic Life of an asset
Find the economic life of an existing asset that has a present salvage value of $1,000 but its
salvage value is expected to be $400 if it is retired at any time in the future. Assume 12% interest
End of Year
Salvage Value When Asset Retired at
Year n
$1,000
400
400
400
400
0
1
2
3
4
Operating Costs During Year
n
$1,500
1,500
1,500
1,500
Solution:
To find this asset’s economic life, it is necessary to identify the relevant cash flows associated
with retaining the asset 1, 2, 3, or 4 years as shown below.
$400
0
1
n=1
AE(12)1= CR1 - $1,500
A/P,12%, 1
= ($400 - $1,000) ( 1.12 ) - $400(0.12) - $1,500
$1,000
= - $2,220
$1,500
Notice that for an existing asset we used the present salvage value of the machine as the initial
cost for the purpose if studying its economic life.
Engineering Economy - © 2015 Dr. Tareq Albahri – Kuwait University
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$400
0
1
2
n=2
AE(12)2 = CR2 - 1500
A/P,12%, 2
= ($400-$1,000) ( 0.5917 ) - $400(0.12) -$1,500
$1,000
= -$1,903
$1,500
$400
0
1
2
3
n=3
AE(12)3 = CR1 - $1,500
A/P,12%, 3
= ($400 - $1,500) ( 0.4163 ) - $400(0.12) - $1,500
$1,000
= - $1,798
$1,500
$400
0
1
2
3
4
AE(12)4 = CR1 - 1500
n=4
A/P,12%, 4
= ($400 - $1,000) ( 0.3292 ) - $400(0.12) - $1500
= - $1,745
$1,000
$1,500
Most economical option is to operate 4 years (cheaper)
Notice that in this case, it is better to operate the asset for as long as possible (physical life
allows) because the annual equivalent decreases with time.
Engineering Economy - © 2015 Dr. Tareq Albahri – Kuwait University
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Example: Economic life of an asset
Find the economic life of an asset whose first cost is $3,000, with decreasing salvage values and
with operating costs beginning at $1,000 and increasing by $700 each year for an interest rate of
12%.
End of Year
Salvage Value When Asset Retired at
Year n
$1,500
1,000
500
0
1
2
3
4
Operating Costs During Year
n
$1.000
1,700
2,400
3,100
Solution:
To find this asset’s economic life, it is necessary to identify the relevant cash flows associated
with retaining the asset 1, 2, 3, or 4 years as shown below.
$1500
0
1
AE(12)1= CR1 - $1,000
n=1
A/P,12,1
= ($1,500 - $3,000) ( 1.12 ) - $1,500(0.12) - $1000
$1000
= - $2,860
$3,000
$1000
n=2
0
1
2
A/P,12, 2
AE(12)2 = ($1,000 - $3,000) ( 0.5917 ) - $1,000(0.12)
A/G,12, 2
$1000
$1700
- [$1,000 + $700( 0.4717 )]
= - $2,633
$3,000
Engineering Economy - © 2015 Dr. Tareq Albahri – Kuwait University
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$500
0
1
2
3
A/P,12, 3
AE(12)3 = ($500 - $3,000) ( 0.4164 ) - $500(0.12)
n=3
A/G,12, 3
- [$1,000 + $700( 0.9246 )]
$1000
$1700
= - $2,748
$2400
$3,000
n=4
0
1
2
3
4
A/P,12, 4
$1000
$1700
$2400
$3,000
AE(12)4 = ($0 - $3,000) ( 0.3292 ) - $0 (0.12)
A/G,12, 4
- [$1,000 + $700( 1.3589 )]
= - $2,938
$3100
Most economical option is to operate 2 years (cheaper)
Notice that in this case, it is not advisable to operate the asset for as long as possible (physical
life allows) because the annual equivalent increases significantly with time.
General: just because the physical life allows you the operate the asset say for 20 years it does
not mean you have to do that but rather operate it for its economic life (2 years in case) then sell
it.
Engineering Economy - © 2015 Dr. Tareq Albahri – Kuwait University