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7/27/2019 ENGINEERING ECONOMY CASE 2.docx
http://slidepdf.com/reader/full/engineering-economy-case-2docx 1/2
CASE 2: USEFUL LIVES ARE DIFFERENT AMONG THE ALTERNATIVES
1.) THE REPEATABILITY ASSUMPTION
Conditions:
1. Study period is either indefinitely long or equal to a common
multiple of the lives of the alternative.
2. The cash flows associated with an alternative's initial life span
are representative of what will happen in succeeding life spans.
EXAMPLE:
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. If the MARR = 10% per year,
show which alternative is more desirable using equivalent worth methods. Use the repeatability assumption.
Solution by the PW method:
PW(10%)A =-$3,500 - $3,500 [(P/F,10%,4)+(P/F,10%,8)]+($1,255)(P/A,10%,12) = $1,028
PW(10%)B =-$5,000 - $5,000(P/F,10%,6) + ($1,480)(P/A,10%,12) = $2,262
Based on the PW method, we would select Alternative B because it has the larger value ($2,262)
Solution by the AW Method:
(1.) The AW of each alternative over the 12-year analysis period based on the previous PW values.
AW(10%)A = $1,028(A/P,10%,12) = $151
AW(10%)B = $2,262(A/P,10%,12) = $332
(2) Determining the AW of each alternative over one useful life cycle.
AW(10%)A = -$3,500 (A/P,10%,4) + ($1,255) = $151
AW(10%)B = -$5,000(A/P,10%,6) = $332
When the repeatability assumption is applied, simply compare the AW amounts of each alternative over its own
useful life and select the alternative that maximizes AW. So we would select Alternative B because it has the largervalue ($332).
What if the study period is not a common multiple of the alternative lives or repeatability is not applicable? Use
coterminated assumption!
Alternative A Alternative B
Capital investment $3,500 $5,000
Annual cash flow 1,255 1,480
Useful life (years) 4 6
Market value at end of useful life 0 0
7/27/2019 ENGINEERING ECONOMY CASE 2.docx
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CASE 2: USEFUL LIVES ARE DIFFERENT AMONG THE ALTERNATIVES
2.) THE COTERMINATED ASSUMPTION
EXAMPLE:
Suppose that (example) is modified to 6 years instead of 12 years. Suppose that the responsible manager did not
agree with Repeatability Assumption and wanted 6 years analysis period.
Solution:
FW(10%)A = [-$3,500(F/P,10%,4)+($1,255)(F/A,10%,4)](F/P,10%,2) = $847
FW(10%)B = -$5,000 (F/P,10%,6)+($1,480)(F/A,10%,6) = $2,561
Based on the FW of each alternative at the end of the six-year study period, we would select Alternative B because
it has the larger value ($2,561).
Study Period > Useful Life
Procedure: The cash flows of the alternatives need to be adjustedto terminate at the end of the study period.
· Cost alternatives: Assuming repeatability, repeat part of the useful life of the
original alternative, and then use an estimated MV to truncate it at the end of
the study period. Without repeatability, we must purchase/lease the service/asset
for the remaining years.
· Investment alternatives: Assume all cash flows will be reinvested at the MARR
to the end of the study period (i.e., calculate FW at end of useful life and move
this to the end of the study period using the MARR).
Study Period < Useful Life
When the study period is explicitly stated to be shorter than the
useful life, use the cotermination assumption.
Procedure: The cash flows of the alternatives need to be adjustedto terminate at the end of the study period.
· Truncate the alternative at the end of the study period using
an estimated Market Value.