Efficiency Measures
So, now that you can develop business scenarios, estimate
cashflows, discount cashflows, etc. How do you know if this is a
"good" project? This is where efficiency measures come in.
We will not be focusing on typical engineering efficiency measures,
but instead financial efficiency measures. Such as, does the project
recover its costs and provide a suitable return on capital. There is not
one specific measure of efficiency that covers all cases, but there are
commonly used measures that taken together can go a long way
toward providing an answer to the question "is this a good project?"
To determine the answer to this question, we will cover Payback
Period, Present Worth, Rate of Return, and Net Present Value Index.
Efficiency Measures: Payback Period
The simplest measure of project financials is how long it takes to recover the
initial investment dollar for dollar on an undiscounted basis. This period of
time is called the
payback period
or simply,
payback
.
This is measured by creating a cumulative cashflow account where all
negative cashflows are accumulated to a total negative cashflow and then as
positive cashflows are earned they are accumulated until the initial negative
cashflow is recovered, the total accumulation account nets to zero.
The payback period is best illustrated by the following example.
Question:
Determine the payback time for each of the following simple investment
projects:
Project
Identifier
Required
Investment
Returns
Salvage
Value
A
$1000
$300/yr eoy for 6 years
$400
B
$1000
$150/yr eoy for 6 years
$400
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
C
$1000
$150/yr eoy for 6 years
$50
D
$1000
$300/yr continuous for 6
years
$400
E
$1000
$150/yr eoy forever

F
$1000
$150/yr continuous
forever

Answer:
Project A:
nA = 6(300) = $1800 > $1000 Y = é (1000/300) = é 3.33 =
4
years
Project B:
nA = 6(150) = $900.
nA + S = 900 +400 = $1300
Y = n =
6 years
Project C:
nA + S = 900 + 50 = $950 < I.
Y does not exist
Project D:
nA = 6(300) = $1800 > $1000.
Y = 1000/300 =
3.33 years
Project E:
Y = é (1000/150) = é 6.66 =
7 years
Project F:
Y = (1000.150) =
6.66 years