1. OUTLINE
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Present Worth Analysis, Annual Equivalent
Internal Rate of Return

2. Example

3. CAPITALIZED EQUIVALENT METHOD
APPLIES TO LONG HORIZONS OR PERPETUAL INVESTMENTS
COMPUTING THE PRESENT WORTH OF AN INFINITE SERIES IS CALLED THE CAPITALIZATION OF THE PROJECT’S COST
APPLYING L’HOPITAL’S RULE TO THE (P/A,i,N), WE GET A/i
SO THE PRESENT WORTH OF A SERIES OF CONSTANT PAYMENT STRETCHING OUT TO INFINITE IS A/i

4. Capitalized Equivalent Method

5. MUTUALLY EXCLUSIVE ALTERNATIVES
IF ONE ALTERNATIVE IS SELECTED, ALL OTHERS ARE REJECTED
“DO NOTHING’ IS AN ALTERNATIVE
REVENUE PROJECT -- INCOME DEPENDS ON THE CHOICE OF THE ALTERNATIVE
SERVICE PROJECT -- REVENUES ARE THE SAME REGARDLESS OF THE CHOICE OF ALTERNATIVE

6. PROJECT LIVES
ANALYSIS PERIOD, STUDY PERIOD OR PLANNING HORIZON:
TIME PERIOD OVER WHICH THE IMPACT OF THE PROJECT WILL BE EVALUATED
WE HAVE TO CONSIDER:
REQUIRED SERVICE PERIOD
USEFUL LIFE OF THE PROJECT
FOR MULTIPLE PROJECTS WE MUST COMPARE THEM OVER AN EQUAL TIME SPAN

7. PROJECT LIVES (CONTINUED)
EXAMPLES
PROJECT LIVES LONGER THAN ANALYSIS PERIOD (5.11)
PROJECT LIVES SHORTER THAN ANALYSIS PERIOD (5.10)
ANALYSIS PERIOD = LONGEST LIFE (5.12)
UNEQUAL LIVES - LOWEST COMMON MULTIPLE METHOD (5.13)

8. Example 5.11 – Project lives longer than the analysis period
THIS ONE IS VERY STRAIGHT FORWARD:
WE ANALYZE FOR THE LENGTH OF TIME THAT WE ARE INTERESTED IN. WHATEVER THE EQUIPMENT IS WORTH AT THE END OF THE ANALYSIS PERIOD, WE CONSIDER AS SALVAGE VALUE

9. Example 5.10 Project lives shorter than analysis period
THIS EXAMPLE HAS TWO OPTIONS, ONE THAT WILL LAST 3 YEARS, THE OTHER 4. UNFORTUNATELY WE HAVE TO PLAN FOR 5 YEARS.
WE HAVE TO CHOOSE AN APPROACH THAT WILL FILL IN THE LAST TWO YEARS OF THE FIRST OPTION, AND THE LAST YEAR OF THE SECOND OPTION

10. Example 5.12 Analysis period = longest life

11. Example 5.13 Unequal lives - lowest common multiple method
Option A has a life of 3 years, Option B 4 years.
We use a lowest common denominator (12) as the analysis period
For each option we calculate NPV and the end of its life and then repeat the investment until we have reached the common denominator

12. A FUN NOTE TO END ON (GARY LARSEN) “Notice all the computation, theoretical scribbling, and lab equipment, Norm … Yes, curiosity killed these cats

13. Annual Equivalent
CALCULATE THE NPW, THEN CONVERT TO THE ANNUAL EQUIVALENT
Examples 6.1 and 6.2

14. Contemporary Engineering Economics
Incremental Analysis
Lecture No. 13
Chapter 7
Contemporary Engineering Economics
Copyright © 2006
Contemporary Engineering Economics, 4th edition, © 2007

15. Comparing Mutually Exclusive Alternatives Based on IRR
Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR?
n A1 A2
-$1,000 -$5,000
$2,000 $7,000
100% > %
$818 < $1,364 1
IRR
PW (10%)

16. Contemporary Engineering Economics, 4th edition, © 2007
Who Got More Pay Raise?
Bill
Hillary
10% 5%
Contemporary Engineering Economics, 4th edition, © 2007

17. 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 on
the magnitude of its IRR. We need to know the size of investment and its timing
of when to occur.

18. Incremental Investment
Project A1
Project A2
Incremental Investment (A2 – A1) 1
-$1,000
$2,000
-$5,000
$7,000
-$4,000
$5,000
ROR
PW(10%)
100%
$818
40%
$1,364
25%
$546
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.

19. 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.

20. Example 7.10 - Incremental Rate of ReturnB1 B2
B2 - B1 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.

21. IRR on Increment Investment: Three Alternatives
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 best
alternative. n D1 D2 D3
-$2,000
-$1,000
-$3,000 1
1,500
800 2
1,000
500
2,000 3
IRR
34.37%
40.76%
24.81%

22. 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

23. Example 7.13 Incremental Analysis for Cost-Only Projects
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

24. Incremental Cash Flow (FMS – CMS)
CMS Option
FMS Option
Incremental
(FMS-CMS)
-$4,500,000
-$12,500,000
-$8,000,000 1
-7,412,920
-5,504,100
1,908,820 2 3 4 5 6
$2,408,820
Salvage
+ $500,000
+ $1,000,000

25. Solution:

26. 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*.

27. 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.

28. For a pure investment, the solving rate of return (i
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.