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Rate of Return Analysis Lecture

This document provides instructional material for an engineering economics lecture on rate of return analysis. It introduces rate of return and internal rate of return, and provides examples of calculating rate of return for projects. It outlines the steps to calculate rate of return, which include drawing a cash flow diagram, setting up the rate of return equation, selecting trial interest rates, and interpolating to find the internal rate of return. An example problem demonstrates calculating the internal rate of return of 10.5511% for an investment in green manufacturing techniques that provides annual savings and a one-time savings after 10 years.

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0% found this document useful (0 votes)
189 views20 pages

Rate of Return Analysis Lecture

This document provides instructional material for an engineering economics lecture on rate of return analysis. It introduces rate of return and internal rate of return, and provides examples of calculating rate of return for projects. It outlines the steps to calculate rate of return, which include drawing a cash flow diagram, setting up the rate of return equation, selecting trial interest rates, and interpolating to find the internal rate of return. An example problem demonstrates calculating the internal rate of return of 10.5511% for an investment in green manufacturing techniques that provides annual savings and a one-time savings after 10 years.

Uploaded by

Aziezah Palinta
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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College of Engineering

Civil Engineering Department


Instructional Material for

BES 125 ENGINEERING ECONOMICS


Lecture 8: Rate of Return Analysis
October 13- 16, 2020

Prof. Jesssica Maria Paz S. Casimiro, CE, EnP, DiSDS

Note: I shall greatly appreciate if you can take time to call my attention to any error(s) that you may discover in the lecture slides.
THANK YOU.
Introduction

The most commonly quoted measure of economic worth for a


project or alternative is its rate of return (ROR). Whether it is
an engineering project with cash flow estimates or an
investment in a stock or bond, the rate of return is a well-
accepted way of determining if the project or investment is
economically acceptable.
Objectives of this lecture:
1. To calculate Rate of Return (RoR) given the
exercises. (Homework No. 3)
2. To state the difficulties of using the RoR
method
Rate of Return (ROR)
 ROR is technically known as the internal rate of return (IROR) and
return on investment (ROI).
 Rate of return (ROR) – from the perspective of a borrower,
ROR is the rate paid on the unpaid balance of borrowed money, or
from the perspective of a lender, ROR is the rate earned on the
uncovered balance of an investment, so that the final payment or
receipt bring the balance to exactly zero. See example 1(a).
 It is stated as a percent (%) per period. The numerical value of i
can range from -100% to ∞. In terms of investment, a return of -
100% means the entire amount is lost.
 The ROR is used as a relative measure, while PW and AW are
absolute measures. Since the resulting interest rate depends only
on the cash flows themselves, the correct term is internal rate of
return (IROR). However, the term ROR is used interchangeably.
Example 1.
To get started in a new telecommuting position with AB Hammond
Engineers, Jane took out a $1000 loan at i=10% per year for 4 years to
buy a home office equipment. From the lender’s perspective, the
investment in this young engineer is expected to produce an equivalent net
cash flow of $315.47 for each of 4 years.
A=1000(A/P, 10%,4) = 1000(0.31547)=$315.47
This represents a 10% per rate of return on the unrecovered balance.
Compute the amount of unrecovered investment for each of the 4 years
using (a) the rate of return on the unrecovered balance (the correct basis)
and (b) the return on the initial $1000 investment. (c) Explain why all of
the initial $1000 amount is not recovered by the final payment in part (b).
Solution: A=$315.47

0 1 2 3 4 year

$1000

(a) Unrecovered balances using a Rate of Return of 10% on the unrecovered balance.
(1) (2) (3)=0.10x(2) (4) (5) =(4)-(3) (6)=(2)+(5)
Year Beginning Interest on Cash flow Recovered Ending
unrecovered unrecovered amount unrecovered
balance balance balance
0 - - 1,000 - 1,000
1 -1000 100 +315.47 $215.47 -784.53
2 -784.53 78.45 +315.47 237.02 -547.51
3 -547.51 54.75 +315.47 260.72 -286.79
4 -286.79 28.68 +315.47 286.79 0
$261.88 $1000
(b) Unrecovered balances using 10% Return on the Initial Amount
(1) (2) (3)=0.10x(2) (4) (5)=(4)-(3) (6)=(2)+(5)
Beginning Interest on initial Cash flow Recovered Ending
Year uncovered amount amount Unrecovered
balance Balance
0 - $-1,000 $1,000
1 $-1,000 $100 315.47 215.47 -784.53
2 -784.53 100 315.47 215.47 -569.06
3 -569.03 100 315.47 215.47 -353.59
4 -353.69 100 315.47 215.47 -138.12
$ 400 $ 861.88
Clearly, the interest rate applied only on the initial amount (principal) represents a higher
rate than is stated. In practice, a so-called add-on interest rate is frequently based on
principal only, as in part (b). This is sometimes referred to as the installment financing
problem.
The above example is not ROR method because the ending balance is not exactly
equal to zero.
Installment financing:

 Installment financing can be discovered in many forms in


everyday finances, e.g. purchases on installment, PAG-IBIG
multipurpose loans, etc.
 “No-interest” or “Zero-Interest” programs are variations of
installment financing commonly offered by retail stores on
the sale of major appliances, computers, furniture, etc.
 If the purchase is not paid in full by the time the promotion
is over, usually 6 months to 1 year later, finance charges are
assessed from the original date of purchase.
Exercise No. 1 Try this:

1. Assume you borrow $50,000 at 10% per year interest and


you agree to repay the loan in five equal annual payments.
What is the amount of the unrecovered balance
immediately after you make the third payment? Tabulate
your calculations.
2. A shrewd investor loaned $1,000,000 to a start-up
company at 10% per year interest for 3 years, but the
terms of the agreement were such that interest would be
charged on the principal rather than on the unpaid balance.
How much extra interest did the company pay?

Note: Follow the QL rubric guide, if possible, when answering the problems.
Rate of Return
 To determine the rate of return, develop the ROR equation using either
a PW or AW relation, set it equal to 0 and solve for the interest rate.
0=PW
 Alternatively, the present worth of cash outflows (costs and
disbursements) may be equated to the present worth of cash inflows
(revenues and savings).
PWoutflows = PWinflows
 Annual Worth approach
0=AW
AW outflows = AW inflows
 The i value that makes these equations numerically correct is called i*. It
is the root of the ROR relation. To determine if the investment project’s
cash flow series is viable, compare i* with the established MARR.
 If i*≥MARR, accept the project as economically viable.
 If i*<MARR, the project is not economically viable.
Steps for calculation of ROR

1. Draw a cash flow diagram.


2. Set up the rate of return equation in the form 0=PW or
PWinflow = PWoutflow
3. Select the values of i by trial and error until the equation is
balanced.
a) Convert all disbursements (or receipts) into either single or
uniform values.
b) Use P/F, P/A or A/F to find the approximate interest rate at
which the P/F, P/A, or A/F value is satisfied.
c) Use this i to estimate ROR for the first trial.
Example 2.
Applications of green, lean manufacturing techniques coupled with value stream
mapping can make large financial differences over future years while placing
greater emphasis on environmental factors. Engineers with Monarch Paints have
recommended to management an investment of $200,000 now in novel methods
that will reduce the amount of wastewater packaging materials, and other solid
waste in their consumer paint manufacturing facility. Estimated savings are
$15,000 per year for each of the next 10 years and an additional savings of
$300,000 at the end of 10 years on facility and equipment upgrade costs.
Determine the ROR.
Step 1: Draw the cash flow diagram. 300,000

A=15,000

0 1 2 3 4 5 6 7 8 9 10

200,000
Step 2: 0 = −200,000 + 15,000(P/A, i,10)+300,000(P/F,i,10)
Example 2. Continued
 200,000=450,000(P/F,i,10)
(P/F,i,10)=0.444
The roughly estimated i is between 8% and 9%. Use 9% as the first trial because
this approximate rate for the P/F factor will be lower than the true value when the
time value of money is considered.
0=−200,000+15,000(P/A, 9%,10)+300,000(P/F, 9%,10)
=−200,000+15,000(6.4177)+300,000(0.4224)
= 22,985
The result is not equal to 0 and positive indicating that i greater than 9%. 2nd trial.
Try 10%
0=−200,000+15,000(P/A,10%,10) + 300,000(P/F, 10%,10)
= −200,000+15,000(6.1146)+300,000(0.3855)
= 7369
3rd try: i=11%
=−200,000+15,000(P/A,11%,10)+300,000(P/F,11%,10)
=−200,000+15,000(5.8892)+300,000(0.3522)
=−6002.
Therefore, i is between 10% and 11%.
Interpolate the value of i
i* PW (differences) By ratio and proportion:
10 7369
y 7369 𝑦 7369
= 13371 y=0.5511%
1% i* 0 13,371 1

11 -6002 i*=y+0.55 = 10.00 +0.5511


i*=10.5511

Rate of return = 10.6%


Substitute in the equations to check 0= PW

1.105511 10 −1
0 = −200,000 + 15,000 + 300,000(1.105511)−10
0.1055 1.105511 10
0 ≈ 0.00
Special considerations when using
ROR
 Multiple i* values.
 Reinvestment at i*.
 Different procedure for multiple alternative evaluations.
 If possible from an engineering economic study
perspective, the AW or PW method at a stated
MARR should be used in lieu of the ROR method.
 When it is important to know the exact value of i*, a good
approach is to determine PW or AW at MARR, then
determine the specific i* for the selected alternative.
Exercise No. 2: Try this
1. Determine the rate of return for the cash flows shown in the diagram.
5000

0 1 2 3 4 5 6 7 8 year

963 963 963 963 963


2. In 2010, the city of Houston, Texas, collected $24,112.054
in fines from motorists because of traffic violations caught by
red-light cameras. The cost of operating the system was
$8,432,372. The net profit, that is, after operating costs, is split
equally (that is, 50% each) between the city and the operator of
the camera system. What will be the rate of return over a 3-
year period to the contractor that paid for, installed, and
operates the system, if its initial cost was $9,000,000 and the
profit for each of the 3 years is the same as it was in 2010?

Note: Always follow the QL rubric guide when solving your problems.
3. Water damage from a major flood in a Midwestern city
resulted in damages estimated at $108 million. As a result
of the claimant payouts, insurance companies raised
homeowners’ insurance rates by an average of $59 per year
for each of the 160,000 households in the affected city. If a
20-year study period is concerned, what was the rate of
return on the $108 million paid by he insurance
companies?
4. For each of the cash flows shown, determine the rate of
return:
Year 0 1 2 3 4 5
Expense, $ -17,000 -2500 -2500 -2500 -2500 -2500
Revenue, $ 0 5,000 6,000 7,000 8,000 12,000
5. Rubber sidewalks made from ground up tires are said to be
environmentally friendly and easier on people’s knees.
Rubbersidewalks Inc. of Gardena California, manufactures the
small rubberized squares that are being installed where tree
roots, freezing weather, and snow removal have required
sidewalk replacement or major repairs every 3 years. The
District of Columbia spent $60,000 for a rubber sidewalk
replace broken concrete in a residential neighborhood lines
with towering willow oaks. If a concrete sidewalk costs
$28,000 and lasts only 3 years versus a 9-year life for the
rubber sidewalks, what rate of return does this represent?
Thank you.

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