Engineering economy lorie m. Cabanayan francisco d. Cuaresma


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COMPILED LECTURE IN ENGINEERING ECONOMY

 
 
 
 
 
 
 
 
 
 
 
 
The extra benefits obtained by investing the additional $13,000 of capital in B, have a present 
worth of: 
$10,131 - $9,738 = $393 
That is,
PW (10%)
diff
= -$13,000 + $4,225 (P/A, 10%, 4) = $393 
And the additional capital invested in B is justified. 
 
2
nd
 example involves a cost project situation: 
Alternatives C and D are two mutually exclusive cost alternatives with estimated net cash flows 
as shown over a three-year life. MARR at 10% per year. Cost alternatives are those with all 
negative cash flows except for a possible positive cash flow element from disposal of assets at 
the end of the projects‟ useful life. It occurs when the organization must take some action and 
the decision involves the most economical way of doing it (e.g. the addition of environmental 
control capability to meet new regulatory requirements). 
End of Year 
Alternative 



-$380,000 
-$415,000 

-$38,100 
-$27,400 

-$39,100 
-$27,400 

-$40,100 
-$27,400 
3
a

$26,000 
a
Market Value 
 
At MARR = 10% per year 
$60,00




A = $22,000 
Alternativ
$73,00




A = $26,225 
Alternativ
$13,000 




A = $4,225 
Alt. B 
– Alt. A 


55 
PW (10%) C = - $477,077 
PW (10%) D = - 463,607 
Alternative D is preferred to C because it has the lesser PW of costs. Hence the lower 
annual expenses obtained by investing the additional $35,000 of capital in Alternative D has a 
present worth of - 463,607 
– (- $477,077) = $13,470. Therefore, the additional capital 
invested in Alternative D is justified. 

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