engineering economy

Chapter 1
Basic Concept and Factors

1.1 Introduction of Engineering Economy

Engineering economy is at the heart of making decisions involve the fundamental
elements of:

• cash flows of money
• times of occurance of cash flows
• interest rates for time value of mpney
• Measure of economic worth for selecting an alternative

1.2 Interest Calculation

Interest is the manisfection of the time value of money. The diference between an
ending amount of money and beginning amount of money.

Interest = amount owed now – principal [1.1]

Interest rate (%)= interest accrued per time unit × 100% [1.2]
principal

Figure 1.1 : a) Interest paid over a time to lender b) Interest earned over a time by
investor.

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There are 2 types of interest which are a) interest paid over a time to lender and b)
Interest earned over a time by investor

Example 1.1: Interest paid over a time to lender

An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay
a total of $10,700 exactly 1 year later. Determine the interest amount and the
interest paid.

Solution:

Apply eq 1.1, Interest paid =$10,700 - $10,000 = $700

Apply eq 1.2 , Percent interest rate = $700 × 100% = 7%
$10,000

DIY:

RKI instruments borrowed $3,500,000 from a private equity firm for expansion
of its manufacturing facility for making carbon dioxide monitors/controllers.
The company repaid the loan after 1 year with a single payment of $3,885,000.
What was the interest rate on the loan?
Answer : 11%

From the perspective of a saver, a lender or an investor, interest earned in fig 1.1b ;
Interest earned = total amount now – principal

Interest earned over a specific period of time is expressed as a percentage of the
original amount and is called rate of return (ROR). The term return on investment
(ROI) is used equivalently with ROR, especially where large capital funds are
commited to engineering-oriented program.

Rate of return (%) = × 100%


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Interest period, same as borrower’s perspective, commonly 1 year.

Example 1.2: Interest earned over a time to investor

a) Calculate the amount deposited 1 year ago to have $1000 now at an interest
rate of 5% per year.

b) Calculate the amount of interest earned during this time period.

Solution:

a) The total amount accrued ($1000) is the sum of the original deposit and the
earned interest. If the X is the original deposit,

Total accrued = deposit +deposit(interest rate)

$1000 = X + X(0.05) =1.05X

The original deposit is = 1000 = $952.38
1.05

b) The interest earned = $1000 - $952.38 = $47.62

DIY:
Which of the following 1 year investment has the highest rate of return?
a) $12,500 that yields $1125 in interest
b) $56,000 that yields $6160 in interest, or
c) $95,000 that yields $7600 in interest.
Answer : b

1.3 Terminology and Symbols

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P = value or amount of money at a time designated as the present time or time 0.
referred to as present worth (PW), present value (PV), net present value (NPV),
discounted casf flow (DCF), capitalized cost (CC); monetary units such as
dollars,RM
F = value or amount of money at some future time. Reffered as future worth (FW)

and future value (FV).
A = series of consecutive, equal, end of period amounts of money. Reffered as annual

worth (AW), equivalent uniform annual worth (EUAW); RM per year, RM per
month.
n = number of interest periods; years, months,days
i = interest rate per time period; percent per year, percent per month
t = time, stated in periods; years, months, days

Example 1.3: Terminology and Symbols
Today, Julie borrowed $5000 to purchase furniture for her new house. She can
repay the loan in either oft he two ways described below. Determine the
engineering economy symbols and their value for each option.
a) Five equal annual installments with interest based on 5% per year
b) One payment 3 years from now with interest based on 7% per year.
Solution
a) P = $5000 i = 5% per year n = 5 years A = ?
b) P = $5000 i = 7% per year n = 3 years F =?

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DIY:

Identify the four engineering economy symbols and their values from the
following problem statement. Use a question mark with the symbol whose value
ist o be determined.
a) A green algae can produce hydrogen when temporarily deprived of sulfur for
up to 2 days at a time. A small company needs to purchase equipment costing
$3.4 million to commercialize the process. If the company wants to earn a rate
of return of 10% per year and recover its investment in 8 years, what must be
the net value of the hydrogen produced each year? Ans : P = $3.4 million; A =
?; i = 10%; n = 8
b) Vision Tech is a small company that uses ultra-wideband technology to
develop devices that can detect objects (including people) inside of buildings,
behind walls or below ground. The company expects to spend $100,000 per year
for labor and $125,000 per year for supplies before a product can be marketed.
At an interest rate of 15% per year, what ist he total equivalent future amount
oft he company’s expenses at the end of 3 years? Ans : F = ?; A = $100,000 +
$125,000?; i = 15%; n = 3

1.4 Cash flow
Cash inflows are the receipts, revenues, incomes and savings generated by project
and business activity. A plus sign indicates a cash flow.

Cash outflows are costs, disbursements, expenses, and taxes caused by projects and
business activity. A negative or minus sign indicates a cash outflow. When a project
involves only costs, the minus sign may be ommited for some techniques, such as
benefit/cost analysis.

Cash Inflow Estimates

Income : +$150,000 per year from sales

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Savings : +$24,500 tax savings

Receipt : $750,000 received on large business loan plus accrued interest

Savings : $150,000 per year saved by installing more effiecient air cond

Revenue : +$50,000 to $75,000 per month in sales for extended battery life iphones.

Cash Outflow Estimates

Operating costs : -$230,000 per year annual operating cost for software services

First cost : -$800,000 next year to purchase replacement eartmoving equipment

Expense : -$20,000 per year for loan interest payment tob ank

Initial Cost : -$1.2 million in capital expenditures for a water recycling unit

Net Cash flow = cash inflows – cash outflows [1.3]

NCF = R – D [1.4]

Where NCF is net cash flow, R is receipt and D is disbursement.

Figure 1.2 : Typical cash flow time scale for 5 years

Figure 1.3 : example of positive and negative cash flows.

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Example 1.4: Cashflow
An electrical enginner wants to deposit an amount P now such that she can withdraw an
equal annual amount of A1 = $2000 per year for the first 5 years, starting 1 year after the
deposit, and a different annual withdrawal of A2 = $3000 per year fort he following 3
years. How would the cash flow diagram appear if i =8.5% per year?
Solution

DIY:
Construct a cash flow diagram for the following cash flows;
a) $25,000 outflow a time 0, $9000 per year inflow in years 1 through 5 at an
interest rate of 10% per year, and an unknown future amount in year 5.
b) Find the present worth in year 0 at interest rate of 15% per year for the
following situation
Year Cash Flow,$
0 -19,000
1-4 +8,100
c) Represents the amount of money that will be accumulated in 15 years from
an investment of $40,000 now at an interest rate of 8% per year.

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1.5 Economic equivalence

Economic equivalence is a combination of interest rate and time value of money to
determine the different amounts of money at different points in time that are equal in
economic value.

Figure 1.4 : Equivalence of money at 6% per year interest

Example 1.5: Economic Equivalence
At 8% interest, what is the equivalent worth of $2,042 now 5 years from now?
Solution
2,042

?

0 1 23 4 5

= 2042(1 + 0.08)5 = 3,000

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DIY:

You borrowed $5,000 from a bank and you have to pay it back in 5 years. There
are many ways the debt can be repaid. ( i = 0.08)
Plan 1: At end of each year pay $1,000 principal plus interest due.
Plan 2: Pay interest due at end of each year and principal at end of five years.
Plan 3: Pay in five end-of-year payments ($1,252).
Plan 4: Pay principal and interest in one payment at end of five years
Tabulated the repayment for each plan in the table. Which plan would you
choose and justify why?

1.6 Simple and Compound Interest
Simple interest = (principal)(number of periods)(interest rate)

=

Compund interest =(principal + all accrued interest)(interest rate)
Total due after n years = principal(1+interest rate)n years

= (1 + )

Example 1.6 : Simple and Compound Interest

If a company sets aside $1,000,000 now into a contigency fund, how much will
the company have in 2 years, i fit does not use any oft he money and the
account grows at a rate of 10% per year?
Solution
Simple Interest

F2 = 1,000,000+1,000,000(2)(0.10) = 1,200,000
Compound interest

F2 = 1,000,000(1+0.1)2= + 1,000,000(0.10) = 1,210,000

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DIY:
A solid waste disposal company borrowed money at 10% per year interest to
purchase new haulers and other equipment needed at company owned landfill
site. If the company got the loan 2 years ago and paid it off with a single payment
of $4,600,000, what was the principal amount P of the loan? Ans:P = $3,801,653
If interest is compounded at 20% per year, how long it will take for $50,000 to
accumulate to $86,400? Ans : n = 3 years

1.7 Minimum Attractive Rate of Return
The MARR is a reasonable rate of return established fort he evaluation and selection
of alternatives. A project is not economically viable unless it is expected to return at
least the MARR. Also reffered as the hurdle rate, cutoff rate, benchmark rate and
minimum acceptable rate of return. As example, if you want to purchase a widescreen
HDTV,but do not have sufficient money (capital), a few alternatives available
1) Obtain a bank loan with a cost of capital 9% per year and pay for the TV in cash
now.
2) Use credit card with 15% per year and pay off the balance on a monthly basis.
3) Use funds from savings account that earn 5% per year and pay cash.

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For a corporation, the established MARR used as a criterion to accept or reject an
investment alternative will usually be equal to or higher than weighted average cost
of capital (WACC) that the corporation must bear to obtain the necessary capital
fund. So the inequality

ROR ≥ MARR > WACC must be correct for an accepted project. Exceptions may be
government regulated requirements (safety, security, environmental,legal,etc).

Types of investment

• Pure Investment

Project cash flow balances are greater than 0 when evaluated at i* at any time
period. All simple investments are pure investments.

• Mixed investment

Project cash flow balances are both + and -
Balances are reinvested at MARR for analysis purposes in order to find i*.

Example

End of Year Cash flow Current Balance
0 -1000 -1000
1 263.8
2 263.8 263.8 + (-1000)(1+0.1)1 = -863.20
3 263.8 263.8 + (-863.2)(1+0.1)1 = -656.02
4 263.8 263.8 + (-656.02)(1+0.1)1 = -487.82
5 263.8 263.8 + (-487.82)(1+0.1)1 = -239.80

263.8 + (-239.80)(1+0.1)1 = 0

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All balances are less than 0.
If i was not equal to i*, the final balance at the end of year 5 would not be
equal to 0.
If we had used i = 5% then CB(5) = 65.12 which is greater than 0.

Hence i =5% would not be i*, since at i* the cash balance at the last period
must be 0.

• Simple investment

Assume MARR = 10% , is this a good investment

Solve the problem by trial and errot to know either the investment is good or not

i* 200 200 400 600 PW total

(1+i*) -1 (1+i*) -2 (1+i*) -3 (1+i*) 41

12% 178.57 159.44 287.41 381.31 4.03 > 0

13% 176.99 156.63 277.22 367.99 -21.17<0

Since we need a PW Total = 0, solve for i*.

By interpolation : i* = 12% + 1% ( 4.03 ) = 12.16%
[4.03-(-21.17)]

12.16%> MARR= 10%

Hence investment is acceptable compared to MARR.

Other method than trial and error is Rule of 72.

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An investment will double in size in N years at i% relative to the following
relationship: 72 / N = i
Hence if most of income is at yr N and about double the initial outlay i = 72 / N

= 72 = 18% start , i* = 24.1 % actual

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Chapter 2

Factors : How Time and Interest Affect Money

2.1 Single- Amount Factor (F/P and P/F)
F/P Factor ,

F = P (1 + ) [2.1]

= [(1+1 ) ] = (1 + )− [2.2]

Figure 2.1 : Cash flow diagrams for single payment factors : (a) Find F, given P (b)
Find P, given F.

Factor Name Find/Given Notation Formula Excell
Notation F/P equation Function
(F/P,i,n) Single F= F=P =FV(i%,n,,P)
payment P(F/P,i,n) (1 + )
(P/F,i,n) amount
compund P/F P= P = F(1 + =PV(i%,n,,P)
Single F(P/F,i,n) )−
payment
present worth

Table 2.1 : F/P and P/F Factors

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Example 2.1: Single- Amount Factor (F/P and P/F)
Sandy, a manufacturing engineer, just received a year end bonus of $10,000 that
will be invested immediately. With the expectations of earning at the rate of 8%
per year, find the amount of funds that will be available after 20 years (draw cash
flow diagram
Solution
(a) Factor formula: Apply Equation [2.2] to find the future value F. Rounding to
four deci- mals, we have

Standard notation and tabulated value: Notation for the F/P factor is (F/P,i%,n)

DIY:
Look up the numerical values
a) (P/F, 6%,8)
b) (F/P, 10%,12)
How much can Haydon Rheosystem to spend now on an energy management
system if the software will save the company $21,300 per year for the next 5
years?Use an interest rate 10% per year. Ans : P = 80,744
Red Valve Co. of Carnegie, Pennsylvania, makes a control pinch valve that
provides accurate, repeat- able control of abrasive and corrosive slurries, out-
lasting gate, plug, ball, and even satellite coated valves. How much can the
company afford to spend now on new equipment in lieu of spending $75,000 four
years from now? The company’s rate of return is 12% per year. Ans : P = 47,663

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2.2 Uniform Series Present Worth Factor and Capital Recovery Factor (P/A
and A/P)

= [(1 (+1 + ) )− 1] ≠ 0 [2.3]

= [(1 (+1 + ) )− 1] [2.4]

Figure 2.2 : Cash flow diagrams used to determine (a)P, given a uniform series A
(b)A, given present worth P

Factor Find/ Notation Formula Excell
Notation Name Given equation (1 + ) − 1 Function
P/A P=A(P/A,i,n) [ (1 + ) ] =PV(i%,n,,A)
(P/A,i,n) Uniform
(A/P,i,n) series A/P A=P(A/P,i,n) (1 + ) =PMT(i%,n,,P)
present [(1 + ) − 1]
worth
Capital
recovery

Example 2.2: Uniform Series Present Worth Factor/ Capital Recovery Factor
How much money should you willing to pay now for a guaranteed $600 per
year for 9 years starting next year, at a return of 16% per year?

Solution
A =$600, i =16%, and n= 9.
The present worth is, P = 600(P/A,16%,9) = 600(4.6065) =$2763.90

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= [(1+ ) −1] [2.5]

= [(1+ ) −1] [2.6]



The uniform series A begins at the end of year (period) 1 and continues through the
year oft he given F.

Figure 2.3 : Cash flow diagrams to (a) Find A, given F (b) Find F, given A

Factor Find/ Notation Formula Excell
Notation Name Given equation
F/A F=A(P/A,i,n)
) Function

(F/A,i,n) Uniform (1 + − 1]=FV(i%,n,,A)
(A/F,i,n) series [
Compund
amount A/F A=F(A/P,i,n) =PMT(i%,n,,
Sinking fund

(1 + ) − 1 F)

Example 2.3: Uniform Series Present Worth Factor/ Capital Recovery Factor

The president of Ford Motor wants to know the equivalent future worth of a $1
million capital investment each year for 8 years, starting 1 year from now. Ford
capital earns at a rate of 14% per year.
Solution

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The cash flow diagram (shows the annual investments starting at the end of year 1
and ending in the year the future worth is desired. In $1000 units, the F value in year
8 is found by using the F/A factor.

F =1000(F/A,14%,8) = 1000(13.2328) = $13,232.80
Recalibration of sensitive measuring devices costs $8000 per year. If the machine will
be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year
equivalent uniform series at 16% per year.
Solution

Present worth method.
Calculate PA' for the shifted series in year 2, followed by PT in year 0. There are 6
years in the A series.
P'A =8000(P/A,16%,6)

PT = PA' (P/F,16%,2) = 8000(P/A,16%,6)(P/F,16%,2)
= 8000(3.6847)(0.7432) = $21,907.75
The equivalent series A' for 8 years can now be determined via the A/P factor.
A' =PT (A/P,16%,8) = $5043.60

Future worth method.

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In summary as table below

Semiannual Annual rates

Bid Nominal, CP per 6 Effective, Nominal, CP per Effective,

r per 6 mth, mths, m i % r per year , m i %

% year, %

1 4.5 2 4.55 9 4 9.31

2 6.0 2 6.09 12 4 12.55

3 4.4 6 4.48 8.8 12 9.16

b) For the effective annual rate, the time basis in Equation is 1 year. For bid 1,

9%
= 9% = 4 = 2.5%

= 4

Effective rate for a year : (1 + 0.09)4 − 1 = 1.0931 − 1 = 9.31%

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c) Bid 3 includes the lowest effective annual rate of 9.16%, which is equivalent to
an effective semiannual rate of 4.48% when interest is compounded monthly.

DIY:
A dot-com company plans to place money in a new venture capital fund that
currently returns 18% per year, compounded daily. What effective rate is this
( a ) yearly and ( b ) semiannually?
Ans : 19.176%, 9.415%
An interest rate of 8% per 6 month, compounded monthly, is equivalent to what
effective rate per quarter. Ans : 0.0405 per quarter

For the past 7 years, Excelon Energy has paid $500 every 6 months for a software
maintenance contract. What is the equivalent total amount after the last payment, if

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these funds are taken from a pool that has been returning 8% per year, compounded
quarterly? Ans : F=9171.09
What is the future worth of a present cost of $285,000 to Monsato Inc. 5 yrs from
now at an interest rate of 2% per month? Ans : 935,805
A company expects to spend $50,000 for a certain machine 4 yrs from now. At
interest rate of 12% per year, cp quarterly, the present worth of machine is? Ans :
P=F(P/F,3%,16)

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Chapter 4
Basic Analysis Tools
4.1 Present Worth Method

Present Worth (PW) alternatives categorized into equal lives and different lives.

4.1.1 PW Analysis of equal life alternatives
If the alternatives have the same capacities for the same time period (life), the equal
service requirement is met. Calculate the PW value at the stated MARR for each
alternative.

For mutually exclusive (ME) alternatives, whether they are revenue or cost
alternatives, the following guidelines are applied.

One alternative: if PW ≥ 0, the requested MARR is met or exceeded and the
alternative is economically justified.

Two or more alternatives : Select the alternative with the PW is numerically largest,
that is, less negative or more positive. This indicates a lower PW of cost for cost
alternatives or a larger PW of net cash flows for revenue alternatives.The selection
below correctly apply the guideline for two alternatives A and B.

PWA PWB Selected Alternatives
$ -2300 $-1500 B
+1000 B
-500 +2000 A
+2500 -400 A
+4800

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For independent projects, each PW is considered separately, that is, compared with
the DN project, which always has PW = 0. The selection guideline is as follows:

One or more independent projects: Select all projects with PW ≥ 0 at the MARR.

The independent projects must have positive and negative cash flow to obtain a PW
value that can exceed zero; that is, they must be revenue projects

Example 4.1: PW analysis

A university lab is a research contractor to NASA for in space fuel cell system
that are hydrogen and methanol based. During lab research, three equal services
machines need to evaluated economically. Perform the present worth analysis
with the costs shown below. The MARR is 10% per year.

Electric - Gas- Powered Solar-Powered
Powered

First cost , $ -4500 3500 -6000

Annual operating -900 -700 -50
cost (AOC),
$/year

Salvage value S, 200 350 100

$

Life, yrs 8 8 8

Solution

These are cost alternatives. The salvage values are considered a “negative” cost,
so a - sign precedes them. (If it costs money to dispose of an asset, the estimated
disposal cost has a - sign.) The PW of each machine is calculated at i =10% for n
= 8 years. Use subscripts E, G, and S.

PWE = -4500 -900(P/A,10%,8) + 200(P/F,10%,8) = $-9208

PWG = -3500 -700(P/A,10%,8) +350(P/F,10%,8) = $-7071

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PWS = -6000 -50(P/A,10%,8) +100(P/F,10%,8) =$-6220

The solar-powered machine is selected since the PW of its costs is the lowest; it
has the numerically largest PW value.

DIY

A metallurgical engineer is considering two materials for use in a space vehicle. All
estimates are made. (a) Which should be selected on the basis of a present worth
comparison at an interest rate of 12% per year? (b) At what first cost for the material
not selected above will it become the more economic alternative?

Material X Material Y
-35,000
First cost, $ -15,000 -7,000
20,000
Annual M&O cost, $ per year -9,000 5

Salvage value, $ 2,000

Life, years 5

Ans : PWx = -46,308, PWy = -48,886

A company that manufactures magnetic membrane switches is investigating two
production options that have the estimated cash flows shown ($1 million units).
Which one should be selected on the basis of a present worth analysis at 10% per
year?.

First cost, $ In - house Contract
Annual cost, $ per year -30 0
Annual income, $ per year -5 -2
Salvage value, $ 14 3.1
Life, years 2 -
5 5

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Ans :In house = 5,359,000 Contract : 4,170,000

4.1.2 PW Analysis of Different Life Alternatives
The PW of the alternatives must be compared over the same number of years and
must end at the same time to satisfy the equal service requirement.

LCM : Compare the PW of alternatives over a period of time equal to the least
common multiple (LCM) of their estimated lives.

The assumptions when using LCM approach are that

1. The service provided will be needed over the entire LCM years or more.

2. The selected alternative can be repeated over each life cycle of the LCM in exactly
the same manner.

3. Cash flow estimates are the same for each life cycle.

Example 4.2: PW analysis for different lifes

National Homebuilders to purchase new cut and finish equipment. Two
manufacturers offered the estimates below

First cost , $ Vendor A Vendor B
-15,000 -18,000
Annual operating cost -3,500 -3,100
(AOC), $/year
1,000 2,000
Salvage value S, $ 6 9

Life, yrs

(a) Determine which vendor should be selected on the basis of a present worth
comparison, if the MARR is 15% per year.

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(b) National Homebuilders has a standard practice evaluating all options over a
5 year period. If a study period of 5 years is used and the salvage values are not
expected to change, which vendor should be selected?

Solution

(a) Since the equipment has different lives, compare them over the LCM of 18
years. For life cycles after the first, the first cost is repeated in year 0 of each
new cycle, which is the last year of the previous cycle. These are years 6 and 12
for vendor A and year 9 for B. The cash flow diagram is shown in Figure 5–2.
Calculate PW at 15% over 18 years.

PWA = -15,000 - 15,000(P/F,15%,6) + 1000(P/F,15%,6) -15,000(P/F,15%,12)
+1000(P/F,15%,12) +1000(P/F,15%,18) -3,500(P/A,15%,18)

=$-45,036
PWB = -18,000 - 18,000(P/F,15%,9) + 2000(P F,15%,9) + 2000(P/F,15%,18)
- 3100(P/A,15%,18)

= $-41,384

Cash flow diagram for both PW

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Vendor B is selected, since it cost less in PW terms.

(b) For a 5-year study period, no cycle repeats are necessary. The PW analysis
is

PWA = -15,000 - 3500(P/A,15%,5) + 1000(P/F,15%,5) = $-26,236

PWB = -18,000 -3100(P/A,15%,5) + 2000(P/F,15%,5) = $-27,397

Vendor A is now selected based on its smaller PW value. This means that the
shortened study period of 5 years has caused a switch in the economic decision.
In situations such as this, the standard practice of using a fixed study period
should be carefully examined to ensure that the appropriate approach, that is,
LCM or fixed study period, is used to satisfy the equal-service requirement.

DIY

Machines that have the following costs are under consideration for a robotized
welding process. Using an interest rate of 10% per year, determine which
alternative should be selected on the basis of a present worth analysis.

Machine X Machine Y
- 430,000
First cost, $ -250,000 -40,000
95,000
Annual M&O cost, $ per year -60,000 6

Salvage value, $ 70,000

Life, years 3

Ans : PWx =-607, 037, Pwy = - 550585

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Machines that have the following costs are under consideration for a robotized
welding process. Using an interest rate of 10% per year, determine which alternative
should be selected on the basis of a present worth analysis.

First cost, $ Machine X Machine Y
Annual operating cost, $ per year -250,000 - 430,000
Salvage value, $ -60,000 -40,000
Life, years 70,000 95,000
3 6
Ans : PWx = -607,037, PWy = -550,585

A chemical processing corporation is considering three methods to dispose of a non-
hazardous chemical sludge: land application, fluidized-bed incineration, and private
disposal contract. The estimates for each method are shown. Determine which has
the least cost on the basis of a present worth comparison at 10% per year for the
following scenarios: (a) The estimates as shown (b) The contract award cost
increases by 20% every 2-year renewal

Land application Incineration Contract
0
First cost, $ -130,000 -900,000 -120,000

Annual operating cost, $ per -95,000 -60,000 0
year 2

Salvage value, $ 25,000 300,000

Life, years 36

Ans : a) PWland = -608,528, PWinc = - 991,968, PWcontract = -522,636 b)
PWcontract = -619,615

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4.2 Future Worth Analysis (FW)

The selection guidelines for FW analysis are the same for PW analysis; FW ≥ 0 means
the MARR is met or exceeded. For two or more mutually exclusive alternatives, select
the one with the numerically largest FW value.

Example 4.3: FW analysis

An industrial engineer is considering two alternatives or purchase by a fiber-optic
manufacturing company. Robot X will have a first cost of $80,000, an annual
maintenance and operation (M&O) cost of $30,000, and a $40,000 salvage value.
Robot Y will have a first cost of $97,000, an annual M&O cost of $27,000, and a
$50,000 salvage value. Which should be selected on the basis of a future worth
comparison at an interest rate of 15% per year? Use a 3-year study period.

Solution

First cost X Y
M & O cost 80,000 97,000
Salvage value 30,000 27,000
40,000 50,000
n
3 3

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DIY

Compare the alternatives shown below on the basis of a future worth analysis, using
an interest rate of 8% per year.

First cost, $ P Q
Annual operating cost, $ per year -23,000 -30,000
Salvage value, $ -4,000 -2,500
Life, years 3000 1000
3 6

Ans : FWp = -88, 036, FWQ = -64,947

4.3 Annual Worth Analysis (AW)
One alternative : If AW ≥ 0, the requested MARR is met or exceeded and the
alternative is economically justified.

Two or more alternatives: Select the alternative with the AW that is numerically largest,
that is less negative or more positive. This indicates a lower AW of cost for cost
alternatives or a larger AW of net cash flows for revenue alternatives.

Example 4.4: AW analysis

Luby’s Cafeterias is in the process of forming a separate business unit that
provides meals to facilities for the elderly, such as assisted care and long-term
care centers. Since the meals are prepared in one central location and distributed
by trucks throughout the city, the equipment that keeps food and drink cold and
hot is very important. Michele is the general manager of this unit, and she wishes
to choose between two manufacturers of temperature retention units that are
mobile and easy to sterilize after each use. Use the cost estimates below to select

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the more economic unit at a MARR of 8% per year.
Solution

Initial cost P, $ Hamilton (H) Infinity Care (IC)
-15,000 -20,000
Annual operating cost, -6,000 -9,000
$ per year
0 -2,000 every 4 yrs
Refurbishment cost,$ 20 40

Trade in value S, % of 4 12
P

Life, years

Solution

The best evaluation technique for these different-life alternatives is the annual
worth method, where AW is taken at 8% per year over the respective lives of 4
and 12 years.
AWH = annual equivalent of P - annual M&O +annual equivalent of S
=-15,000(A/P,8%,4) - 6000 +0.2(15,000)(A/F,8%,4)

=-15,000(0.30192) - 6000 +3000(0.22192)


= $-9,863

AWIC =annual equivalent of P - annual M&O - annual equivalent of
refurbishment + annual equivalent of S

= -20,000(A/P,8%,12) - 9000 - 2000[(P/F,8%,4) + (P/F,8%,8)](A/P,8%,12) +
0.4(20,000)(A/F,8%,12)

= -20,000(0.13270) - 9000 - 2000[0.7350 + 0.5403](0.13270) + 8000(0.05270)

= $-11,571,The Hamilton unit is considerably less costly on an annual
equivalent basis

35























DIY

For the cash flows shown, determine the sum of the cumulative cash flows.

Year 0 1 23 4

Revenue 25,000 15,000 8,000 4,000

Cost -60,000 -30,000 -7,000 -6,000 -12,000

Ans : 1,000

Stan-Rite Corp of Manitowoc, Wisconsin, is a B to B company that manufactures
many types of industrial products, including portable measuring arms with absolute
encoders, designed to perform 3D inspections of industrial parts. If the company’s
cash flow (in millions) for one of its product divisions is as shown, determine
( a ) the number of possible i * values and
( b ) all rate of return values between 0% and 100%.

Year Expenses Revenues
0 -30 0
1 -20 18
2 -25 19
3 -15 36
4 -22 52
5 -20 38
6 -30 70

Ans : b) i = 28.3%

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Chapter 6

Tax

6.1 Income Tax

Income tax is the amount of the payment (taxes) or income or profit that must be
delivered to a federal (or lower-level) government unit.

Net operating income = revenue – operating expenses [6.1]

Taxable income = revenue – operating expenses – depreciation [6.2]

Operating revenue, R also called gross income GI , is the total income realized from all
revenue- producing sources.

Operating expenses OE include all cost inccured in the transaction of business. These
expenses are tax deductible for corporations. For after tax economic evaluations, the
AOC (annual operating costs) and M&O (maintenance and operating) cost are
applicable here. Depreciation is not included here since it is not an operating expense.

Net operating income NOI, often called EBIT (earnings before interest and income
taxes is the difference between gross income and operating expenses.

NOI = EBIT = GI –OE [6.3]

Taxable income TI is the amount of income upon which taxes are based. A corporation
is allowed to remove depreciation, depletion and amortization, and some other
deductibles from net operating income.

TI = gross income – operating expenses – depreciation

TI = GI – OE – D [6.4]

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Tax rate T is a percentage or decimal equivalent, of TI that is owed in taxes.

Income taxes = applicable tax rate x taxable income

= (T)(TI) [6.5]

Net operating profit after taxes NOPAT ist he amount remaining each year after taxes
are substracted from taxable income

NOPAT = TI – taxes = TI –(T)(TI)

= TI (1 – T) [6.6]

Average tax rate =total taxes paid = [6.7]
taxable income

Table 6.1 US Corporate Income tax Rate for 2010

Table 6.2 Malaysia income tax schedule rate for 2019

Chargeable Income Calculations (RM) Rate (%) Tax (RM)
0 – 5,000 On the first 2,500 0 0
5,001 -20,000 On the first 5000 1 0
Next 15,000 50

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20,001-35,000 On the first 20,000 3 150
35,001 – 50,000 Next 15,000 8 450
50,001 – 70,000 14
70,001-100,000 On the first 35,000 21 600
100,001-250,000 Next 15,000 24 1200

On the First 50,000 1800
Next 20,00 2800

On the First 70,000 4,600
Next 30,00 6,300

On the First 100,000 10,600
Next 150,00 36,000

Table 6.1 : Assesment year for 2016 (LHDN, Malaysia)

Figure 6.1 : Tax filling procedure in Malaysia
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