Future Worth Comparisons
We will not delve far into the analysis methods using Future Worth comparisons. Suffice it to say that F-W comparisons:
Engineering economy analysis is more apt to be concerned with present $.
Equivalent Annual Worth
Many comparisons are better suited to the use of a periodic worth comparison.
Consider the following cash flow:
Thus investing $1000 initially
and then receiving $400 at year 1 end and $900 at year 2 end is equivalent to
receiving $61.92 every year for as long as the process repeats.
EAW can be used to compare alternatives the same as PW and FW analyses.
EAW =
EAW(Benefits) - EAW(Costs)
Benefits:
Income, Savage
value
Costs: Annual operating & maintenance, Original investment/Purchase
cost
Example: Which option should be chosen if one must be chosen?
i = 7% / year , N =
5yrs
Option
A
B
1st
cost $15,000
$12,000
Maint. & Op
$3,000/yr
$4,000/yr
Salvage
$2,000
$1,000
EAW(A) =
-15,000(A/P,7,5)-3000 +
2000(A/F,7,5)
= -15,000(0.2439)-3000 +
2000(0.1739)
= -$6310/yr
EAW(B) =
-12,000(A/P,7,5)-4000 +
1000(A/F,7,5)
=
-$6753/yr
Choose Option A, since it has the least negative annual worth.
Note that this analysis would look better as an Equivalent Annual Cost. EAC = -EAW. Hence worths are negative costs.
Thus: EAC(A) = $6310/yr and EAC(B) = $6753/yr
Choose the lower cost option, which again is option A.
In cases where options aren't required
to be chosen, one alternative
is:
Do nothing ----- PW, FW, EAW =
0
Proposed alternatives must exceed Do
nothing.
Care must be taken in how costs and benefits are assigned.
Example: A sales increase of $20,000/year will occur if a mobile demonstration unit is built.
2
options
i = 9%
N=5yrs
Large unit
--- Cost $97,000 Salvage
$9,700
Small unit
--- Cost $63,000 Salvage
$3,500
The large unit has sleeping quarters and saves $11,000/year lodging, over the small unit. The large unit costs $3,100/year more to operate than the small unit.
Should a demo unit be built? And if so, which?
Wrong way : Let's first look at the incorrect method with which to analyze the two options.
EAW(Large):
Increase in profit
=
$20,000/yr
1st cost = -$97,000(A/P,9,5)=97,000(0.2571)
= -$24,939/yr
Lodging
savings
= $11,000/yr
Operating
costs
= -$3,100/yr
Salvage = $9,700(A/F,9,5) = 9,700(0.1671)
=
$1,621/yr
$4,583/yr
EAW(Small):
Increase in
profit
=
$20,000/yr
1st cost = -$63,000(A/P,9,5)=63,000(0.2571)
= -$16,197/yr
Salvage = $3,500(A/F,9,5)
=
3,500(0.1671) = $585/yr
$4,388/yr
The results show that both are acceptable options, (EAW>0) and to choose the larger unit. But this is not correct!!
When choosing between two or more options when the Do nothing option is available, assign costs as costs to option to which they apply, not as benefits to competitive option.
Correct way: Let's look at how the analysis should have been performed
EAW(Large):
Profit
= $20,000
1st
cost
= -$24,939
Operating
cost
= $3,100
Salvage =
$1,621
-$6,419/yr
EAW(Small):
Profit
= $20,000
1st
cost
= -$16,197
Lodging
cost
= -$11,000
Salvage =
$585
-$6,612/yr
EAW< 0. Neither is profitable. Choose the Do nothing option.
Note that any alternative can be made to look acceptable, by assigning phantom benefits to it, as in the wrong way analysis.
Unequal Life Options
Analysis over like periods is assumed to occur with
Recall our previous PW example that had different length lives.
Method
1: Repeat Replacement
If Repeat Replacement is assumed, EAC can be determined for each sub-multiple option.
EAC(A1) =
-$7,000 - $15,000(A/P,7,5) = -$7,000 - $15,000(0.38105)
EAC(A1) =
-$10,658/yr
EAC(A2) = -$2,000 - $20,000(A/P,7,3) = -$2,000 -
$20,000(0.38105)
EAC(A2) = -$9,621/yr
As with PW cost comparison A2 is the preferred
option, since it is a lower annual
cost.
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 http://cset.sp.utoledo.edu/engt3600/ |