ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 1
OUTLINE
• Questions?
• News?
• Quiz Results
• Go over quiz
• Recommendations
• Go over homework
• New homework
• More on stocks
• IRR
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 2
Quiz Results
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 3
Quiz Results
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Quiz Results
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Recommendations
• Tate Britain – a few pictures (mostly Moore) during break
– Turner, Constable, Moore, Blake and many others
– Beautiful building
– Some modern stuff also
– Free, walk from your housing or tube stops Pimlico or
Vauxhall
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Recommendations
• National Portrait Gallery– Annual Competition of portrait
painting
– Wonderful paintings of all kinds with explanations by the
artists
– Selected from thousands of applicants
– Free, by Trafalgar square (we walked)
– No photos allowed
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Recommendations
• Serpentine Gallery– Open Tue- Thu to 6, all free
– Two venues, both in Kensington Gardens, take the tube
to Lancaster Gate walk South through the garden ( worth
the trip in itself)
– On the East side of the lake – life size sculptures by
Duane Hanson (pictures during the break). While there,
peak into the restaurant, The Magazine a fabulous design
by Zaha Hadid – the current darling of the architectural
design world.
– On the West side – Oil paintings by a British painter with
parents from Ghana – I found them very impressive
(pictures at break)
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Recommendations
• Victoria and Albert– free on Cromwell Road
– Take the 14 bus from South of Accent to V&A (1/2 hour)
– Lot’s of different things – mostly decorative arts – we
loved the glass
– Of special interest a small exhibit showing the new
addition in progress
– Special free exhibit : “What is Luxury?”
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 9
Selling short and options
• Definition – stock in street name – the broker holds the stock and
the company does not know that you are the owner
• Selling short – sell someone else’s stock hoping that it goes down
and you can buy it back cheaper and replace it. No actual transfer
of stock ownership happens
• Without dividend – trading a stock without the associated dividend
• Put - An option contract giving the owner the right, but not the
obligation, to sell a specified amount of an underlying security at a
specified price (strike price) within a specified time. This is the
opposite of a call option, which gives the holder the right to buy
shares.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Standard deviations
• If we have the average and the standard deviation of a
variable, we can calculate the probability of obtaining a
specific value
• To demonstrate what this means, what is the probability that
a member of this class is taller than 5 foot 7?
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 11Contemporary Engineering Economics, 4th edition, © 2007
Incremental Analysis
Lecture No. 13
Chapter 7
Contemporary Engineering Economics
Copyright © 2006
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Comparing Mutually Exclusive Alternatives Based on IRR
• Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR?n A1 A2
0
1
IRR
-$1,000 -$5,000
$2,000 $7,000
100% > 40%
$818 < $1,364PW (10%)
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 13Contemporary Engineering Economics, 4th edition, © 2007
Who Got More Pay Raise?
Bill Hillary
10% 5%
ENGINEERING ECONOMICS ISE460SESSION 15
Annual Equivalent, June 23, 2014
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Can’t Compare without Knowing Their Base Salaries
Bill Hillary
Base Salary $50,000 $200,000
Pay Raise (%) 10% 5%
Pay Raise ($) $5,000 $10,000
For the same reason, we can’t compare mutually exclusive projects based onthe magnitude of its IRR. We need to know the size of investment and its timingof when to occur.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Incremental Investment
• Assuming a MARR of 10%, you can always earn that rate from other investment source, i.e.,
$4,400 at the end of one year for $4,000 investment.
• By investing the additional $4,000 in A2, you would make additional $5,000, which is equivalent to earning at the rate of 25%. Therefore, the incremental investment in A2 is justified.
n Project A1 Project A2
Incremental Investment (A2 – A1)
0
1
-$1,000
$2,000
-$5,000
$7,000
-$4,000
$5,000
ROR
PW(10%)100%
$818
40%
$1,364
25%
$546
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Incremental Analysis (Procedure)
Step 1: Compute the cash flow for the difference between the projects (A,B) by subtracting the cash flow of the lower investment cost project (A) from that of the higher investment cost project (B).Step 2: Compute the IRR on this incremental
investment (IRR ).Step 3: Accept the investment B if and only if
IRR B-A > MARR
B-A
NOTE: Make sure that both IRRA and IRRB are greater than MARR.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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Example 7.10 - Incremental Rate of Return
n B1 B2 B2 - B1
0
1
2
3
-$3,000
1,350
1,800
1,500
-$12,000
4,200
6,225
6,330
-$9,000
2,850
4,425
4,830
IRR 25% 17.43% 15%
Given MARR = 10%, which project is a better choice?Since IRRB2-B1=15% > 10%, and also IRRB2 > 10%, select B2.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
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IRR on Increment Investment:Three Alternatives
n D1 D2 D3
0 -$2,000 -$1,000 -$3,000
1 1,500 800 1,500
2 1,000 500 2,000
3 800 500 1,000
IRR 34.37% 40.76% 24.81%
Step 1: Examine the IRR for each project to eliminate any project that fails to meet the MARR.
Step 2: Compare D1 and D2 in pairs. IRRD1-D2=27.61% > 15%, so select D1. D1 becomes the current best.
Step 3: Compare D1 and D3. IRRD3-D1= 8.8% < 15%, so select D1 again.
Here, we conclude that D1 is the bestalternative.
ENGINEERING ECONOMICS ISE460SESSION 15
Annual Equivalent, June 23, 2014
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Practice Problem
You are considering
four types of
engineering designs.
The project lasts 10
years with the
following estimated
cash flows. The
interest rate (MARR) is
15%. Which of the four
is more attractive?
Project
A B C D
Initial cost $150 $220 $300 $340
Revenues/Year
$115 $125 $160 $185
Expenses/Year
$70 $65 $60 $80
IRR (%) 27.32 24.13 31.11 28.33
ENGINEERING ECONOMICS ISE460SESSION 15
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Geza P. Bottlik Page 20
Items CMS Option FMS Option
Annual O&M costs:
Annual labor cost $1,169,600 $707,200
Annual material cost 832,320 598,400
Annual overhead cost
3,150,000 1,950,000
Annual tooling cost 470,000 300,000
Annual inventory cost
141,000 31,500
Annual income taxes 1,650,000 1,917,000
Total annual costs $7,412,920 $5,504,100
Investment $4,500,000 $12,500,000
Net salvage value $500,000 $1,000,000
Example 7.13 Incremental Analysis for Cost-Only Projects
ENGINEERING ECONOMICS ISE460SESSION 15
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n CMS Option FMS Option
Incremental
(FMS-CMS)
0 -$4,500,000 -$12,500,000 -$8,000,000
1 -7,412,920 -5,504,100 1,908,820
2 -7,412,920 -5,504,100 1,908,820
3 -7,412,920 -5,504,100 1,908,820
4 -7,412,920 -5,504,100 1,908,820
5 -7,412,920 -5,504,100 1,908,820
6 -7,412,920 -5,504,100
$2,408,820Salvage + $500,000 + $1,000,000
Incremental Cash Flow (FMS – CMS)
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 22
Solution:
PW i
P A i
P F i
IRR
FMS CMS
FMS CMS
( ) $8, ,
$1,908, ( / , , )
$2, , ( / , , )
.43%
000 000
820 5
408 820 6
0
12 15%,
select CMS.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 23
Summary Rate of return (ROR) is the interest rate earned on
unrecovered project balances such that an investment’s cash receipts make the terminal project balance equal to zero.
Rate of return is an intuitively familiar and understandable measure of project profitability that many managers prefer to NPW or other equivalence measures.
Mathematically we can determine the rate of return for a given project cash flow series by locating an interest rate that equates the net present worth of its cash flows to zero. This break-even interest rate is denoted by the symbol i*.
ENGINEERING ECONOMICS ISE460SESSION 15
Rate of Return, June 22, 2015
Geza P. Bottlik Page 24
Internal rate of return (IRR) is another term for ROR that stresses the fact that we are concerned with the interest earned on the portion of the project that is internally invested, not those portions that are released by (borrowed from) the project.
To apply rate of return analysis correctly, we need to classify an investment into either a simple or a nonsimple investment.
A simple investment is defined as one in which the initial cash flows are negative and only one sign change occurs in the net cash flow, whereas a nonsimple investment is one for which more than one sign change occurs in the net cash flow series.
Multiple i*s occur only in nonsimple investments. However, not all nonsimple investments will have multiple i*s either.
ENGINEERING ECONOMICS ISE460SESSION 15
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Geza P. Bottlik Page 25
• For a pure investment, the solving rate of return (i*) is the rate of return internal to the project; so the decision rule is:
If IRR > MARR, accept the project.
If IRR = MARR, remain indifferent.
If IRR < MARR, reject the project.
IRR analysis yields results consistent with NPW and other equivalence methods.
• For a mixed investment, we need to calculate the true IRR, or known as the “return on invested capital.” However, if your objective is simply to make an accept or reject decision, it is recommended that either the NPW or AE analysis be used to make an accept/reject decision.
• To compare mutually exclusive alternatives by the IRR analysis, the incremental analysis must be adopted.