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8/11/2019 Engineering Fundamentals Ch.20
1/20
EcoNoM I S
~ conomic
considerations play a
t
vital role in product
and
servrce
r:
'' development nd in
engineering
design
decision
making
process
97
8/11/2019 Engineering Fundamentals Ch.20
2/20
98
CHAPTER
20
ENGINEERING
Ec0Noi'l-11cs
s we explained in Chapter 3, economic factors always play important roles
in
engi-
neering design decision making. f
ou
design a product
that is
too expensive to
man-
ufacture, then
it
can not be sold at a price that consumers can ajfard and still be
profitable to your company. The fact is that companies design products
and
provide
services not only to make our lives better but also to make money
In
this section, we
will
discuss the basics
o
engineering economics. The information provided here
not
only applies
to
engineering projects
but
can also be applied
to
financing a car or a
house or borrowing from or investing money in banks. Some of ou may want to
apply the knowledge gained here
to
determine
your
student loan payments or your
credit card payments. Therefore, we advise you to develop a good understanding
of
engineering economics; the information presented here could help you manage
your
money more wisel.y.
' , : _ > _
' ,,-,
_
.
Figure
20.1
A ash flow
diaqram for
borrowed
money
and
the monthly payments.
Cash flow diagrains are visual aids that show the flow
of
costs and revenues over a period
of
rime. Cash flow diagrams show ivhen the cash flow occurs the
cash flow
rnagnitude, and whether
the cash flow i out ofyourpock et (cost) or into your pocket (revenue).
It
is
an important visual tool
rhat shows rhe timing, the magnirude, and the direction
of
cash
flo\-v.
To shed more light on
the concept of the cash flo\.v diagram, imagine that you are interested in-purchasing a new car.
Being a first-year engineering student, you may
not
have too much money in your savings
ac
count at this rime; for the sake
of
this example, er
us
say that you have $1200 to your name in
a savings account.
The
car that you are interested in buying costs $15,500; let
us
further assume
rhat inc luding the sales tax and other fees the total cost
of
the car would be
$16,880.
Assum
ing you can afford to
put
do .-vn $1000
as
a down payment for your new shiny car, you ask your
bank
for a loan.
The bank
decides to lend you the remainder, which is
$15,880
at 80/o interest.
You
will sign a contract that requires you to pay
$315.91
every month for the next
five years.
You
\vill soon learn ho\v to calculate these month ly payments, but for now let
us
focus on how
to draw the cash
Row
diagram.
The
cash flow diagram for this activity
is
shown in Figure 20.1.
Note
in Figure
20.1
the direction
of
the arrows representing the n1oney given to you by the
bank and rhe payments that you must make to the bank over the next
five
years
{60
months).
$15,880
1
2
3
4
:
7
58
59
6
0
I I
I
J
0
8/11/2019 Engineering Fundamentals Ch.20
3/20
fiqure 10.i
20.2
SIMPLE A.ND
CoMPol lND INTEREST
599
Draw the cash
flow
diagram for
an
investment that includes purchasing a machine that costs
$50,000 with a maintenance and operating cost of $1000 per year. l t is expected that the ma
chine will generate revenues of $15,000 per year for five years. The expected salvage value of
the machine at the end of five years is $8000.
he
cash flow diagram for the invesnnent is shown in Figure 20.2. Again, note the direc-
tions of arrows in the cash
flow
diagram. We have represented the initial cost of $50,000 and
the maintenance cost by arrows pointing down, while the revenue and the salvage value of the
machine are shown
y
arrows po inting up
2
l
0
$\S,000
3
4
$1000
$8000
5
Tne
cash now diagram
or
Example
10.\.
$50,000
Interest is rhe extra money in addition to the borrowed amount that one must pay fot the pur
= < O w f ~
='
0 , ,.,,,_, """'f _. '
vith
draw 1000
in
the next four years,
and 3000
in five years, and 5000 in seven years.
Up
to
this point, we have been discussing general relationships that deal -Vith rnoney, time, and
interest rates. Let us now consider the application of these relationships in an
engineerlng
set
ting.
Imagine
that you are assigned
the
task
of
choosing
which
air-conditioning unit ro pur
chase for
your
company. After
an
exhaustive search, you have narrowed your selection to rwo
alternatives, both
of which
have an
anticipated
10 years of
working
life. Assuming an 8o/o in
terest rate, find the best alternative. Addjtional information is given in Table 20.9.
The
cash
flow diagrams for each alternative are
shown
n Figure 2D.6.
Here we
will discuss three diff erent
methods that
you can use to choose
the
best
econom
ical
alternative from many opdons. The three methods are cominonly referred to as (1) pres
ent
worrh
(P\V) or
present
cost analysis, (2)
annual worth
(AW) or annual cost analysis, and
(3)
future
worth (F\Xl) or future cost analysis. When these: tnethods are applied
to
a
problem,
they all lead
t
the same conclusion. So in practice) you need only apply one of these
1nethods
to evaluate options; however, in
order
ro
show
you
the
details
of
these procedures, \Ve \-vill
ap
ply all of rhese n1ethods to the preceding problem.
Present Viorth or
Prtsent
Cost With this
approach
you con1pute rhe total
present
worth or
the present
cost of each alternative and then
pick
the alternative with
the
lo>vest pres
ent cost or choose rhe alternative 'ith
the
highest present V> orth or profit.
To
emp1oy rh[s
T BLE
208
Data
to
Be
Used in
Selection of
an
Air Conditioning Unit
Criteria
Inirial cost
Salvage value afcer 10 years
Operating cost per
year
Maintenance cost per
ye:ar
Alternative A
$100,000
S 0,000
52500
$1000
Alternative B
$85,000
$5000
$3400
$1200
8/11/2019 Engineering Fundamentals Ch.20
18/20
6 4
CH PTER
20 ENGINEERING c o N o ~ u c s
10,000
0
l
2
3 4
5
6
7
8
9
lO
3500
100,000
Alternative A
5000
0
l
2
3
4
5
6
7
8
9 lO
4600
85,000
Alternative B
Figure
10 6
The cash flow
diagrams
for
the
example
problem_
method, you begin
by
calculating the equivalent p resent value of all cash flow.
For
the example
problem mentioned, the application
of
the present
worth
analysis leads to:
AlternativeA
PW=
-100,000 - (2500
+ IOOO)(PIA, 8 ,
10)
+ 10,000(PIF,
8 , 10)
The interest-rime
factors for = 8o/o are given
in
Table 20.8.
PW=
-100,000 - (2500 + 1000)(6.71008140) + (10,000)(0.46319349)
PW=
-118,853.35
Alternative B
PW=
-85,000 - (3400
+
1200)(PIA, 8 , 10)
+ 5000(PIF,
8 , 10)
PW= -85,000 - (3400 + 1200)(6.71008140) + 5000(0.46319349)
PW=
-113,550.40
8/11/2019 Engineering Fundamentals Ch.20
19/20
Note
that we have determined the equivalent present worth of all future cash flow including
the yearly maintenance and operating costs and the salvage value of the air-conditioning unit.
ln the preceding analysis, the negative sign indicates cost, and because alternative B has a lower
present cost, we choose alternative B.
er
' : Using this approach, we compute the equivalent annual
worth or annual cost value of each alternative and
then
pick the alternative with the lowest an
nual cost or select rhc alternative with the highest annual
worth
or revenue. Applying the an-
nual
worth
analysis to our example problem, we have
Alternative
A
W = -(2500
+ \000) - 100,ooo AIP, 8 ,
\0)
+ \O,OOO A/F, 8 , 10)
A W=
-(2500
+ 1000) - (100,000)(0.14902949) + (10,000)(0.06902949)
A W=
-17,7\2.65
Alternative
B
W= -(3400 + 1200) - 35,ooo AIP, 8 , io) + 5ooo AIF, 8 , 10)
A W= -(3400
+ 1200) - 85,000(0.14902949) + 5000(0.06902949)
A W = -16,922.35
Note that using this method, we have determined the equivalent annual worth of all cash
flow
and because alternative B has a lower annual cost, we choose alternative B.
cr:c r \\rr:: :.:: .::
This
approach is based
on
evaluating the future worth or
future cost
of
each alternative.
Of
course,
you
will then choose
the
alternative with
the
lowest
future case or pick the alternative with the highest future worth
of
profit. The future worth
analysis o our example problem follows
Alternative
A
W = + lo,ooo - ioo,ooo FIP, so/a io) - (2500 + 1ooo)(F/A, 8 , 10)
F\Y f +10,000 - (100,000)(2.15892500) - (2500 + 1000)(14.48656247)
F\Y f = -256,595.46
Alternative
B
\Y f = +sooo
- .
35,ooo(F/P, 8 . 10) - (3400 + \200) F/A, 8 , 10)
F\Y f
+5000 - (85,000)(2.1ss92soo) - (3400 + 1200)(14.48656247)
F\Y f = -245,146.81
Because alternative B has a lower future cost, again we choose alternative B. Note that regard
less of which method we decide to use, alternative B is economically the better option. More
over, for each alternative, ail of the approaches discussed here are related to one another
through the interest-time relationships (factors). For example.
8/11/2019 Engineering Fundamentals Ch.20
20/20
61 6 CHAPTER
20
ENGINEERING ECONOMICS
Alternative
A:
PW= AW P/A,
8%,
IO =
(-17,712.65)(6.71008140)
-118,853.32
or
PW=
FW PIF,
8%,
IO)
= (-256,595.46)(0.46319349)
- l I 8,853.34
Alternative
:
PW= i\.W P/A, 8%, 10) = (-16,922.35)(6.71008140)
I
13,550.40
or
PW= FW PIF, 8%, 10) = (-245,I46.81)(0.46319349) = -113,550.40
Finally, it
is worth noting
that you can rake semester-long classes
in
engineering eco
nomics. Some
of
you will eventually
do
so.
You
will learn more
in depth about
the principles
of
money-rime
relationships,
including
rate-of-return analysis, benefit-cost ratio analysis,
general price inflation, bonds, depreciation me thods, evaluation of alternatives
on
an after-tax
basis, and risk
and
uncertainty
in
engineering economics. For now,
our intent
has been to
introduce you to engineering economics, but keep in
mind that
we have
just
scratched rhe sur
face
We cannot
resist
but
to
end
this section
with
definition
of
some
of
these
important
con
cepts that you will learn
more about
them
later.
States, counties,
and
cities issue bonds to raise
money
to pay for various projects, such as
schools,
highways, convent ion centers,
and
stadiums . Corporation s also issue bonds to raise money
to
expand
or
to modernize their facilities.
There
are
many
different types
of
bonds,
but
basically,
they are loans that investors
make
to
government
or corporations
in
return for some gain. When
a
bond is
issued, it will have a m turity d te (a year or less to 30 years or longer),parvalue the
amount
originally paid for
the bond and the
amount
that
will be repaid
at
maturity date), and
an interest rate (percentage of par value
that
is paid to bond holder at regular intervals).
Assets (such
as
machines, cars, and computer s) lose their value over a period of time. For ex
ample, a
computer
purchased today
by
a
company
for 2000
is not worth
as much in three or
four years. COmpanies use this
reduction in
value of an asset against their before-tax
i n c o m _ e ~ _
There
are rules
and
guidelines
that
specify what can be depreciated, by
how
much, and over
what period of rime. Examples of depreciation methods include the Straight Line and .c.;, ''
Modified Accelerated Cost Recovery System (MACRS).
u
In engineering, the term life cycle cost refers to
the
sum of
all
the costs that are
a.ssoda,ted,
vith a structure, a service,
or
a
product during
irs life span. For example,
if
you are
d e s i i g r t i n ~ ;
a bridge or a highway, you
need
to consider the costs that are related
-to
the initial
deliniiti?. >;;
and
assessment, conceptual design, detailed design, planning, construction, o p e r a n o n i 1 1 1 ~
tenance and disposal of the project at
the
end of its life span.