Engineering economy lorie m. Cabanayan francisco d. Cuaresma


Case 2. Useful lives are different among alternatives and at least one do not match the study


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COMPILED LECTURE IN ENGINEERING ECONOMY

 
Case 2. Useful lives are different among alternatives and at least one do not match the study 
period. 
 When the useful lives of mutually exclusive alternatives are different, the repeatability 
assumption and co-terminated assumption maybe used. 
The repeatability assumption involves two main conditions: 
1.
The study period over which the alternatives are being compared is either indefinitely long 
or equal to a common multiple of the lives of the alternatives. 
2.
The economic consequences that are estimated to happen in an alternative‟s initial useful life 
span will also happen in all succeeding life spans (replacements) 
The co-terminated assumption uses a finite (limited) and identical study period for all 
alternatives. This planning horizon, combined with appropriate adjustments to the estimated 
cash flows, puts the alternatives on a common and comparable basis. 
Guidelines: 
1.
Useful life < Study period 
a.
Cost alternatives: contracting for the service or leasing the needed equipment for the 
remaining years maybe appropriate or repeat part of the useful life of the original 
alternative, and then use an estimated market value to truncate it at the end of the study 
period. 
b.
Investment alternatives: Assumption used here is that all cash flows will be reinvested in 
other opportunities available to the firm at the MARR to the end of the study period. 
2.
Useful life > Study period: The most common technique is to truncate the alternative at the 
end of the study period using an estimated market value, that the disposable assets will be 
sold at the end of the study period at that value. 
Sample Problem. The following data have been estimated for two mutually exclusive investment 
alternatives, A and B, associated with a small engineering project for which revenues as well as 
expenses are involved. They have useful lives of four and six years, respectively. If the MARR = 
10% per year, show which alternative is more desirable by using equivalent worth methods. Use the 
repeatability assumption. 

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