engineering fundamentals ch.20

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


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


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


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


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.


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.

engineering fundamentals ch.20

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