1. ENGINEERING ECONOMICS ISE460 SESSION 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

2. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 2 Quiz Results

3. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 3 Quiz Results

4. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 4 Quiz Results

5. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 5 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

6. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 6 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

7. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 7 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)

8. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 8 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?”

9. ENGINEERING ECONOMICS ISE460 SESSION 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.

10. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 10 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?

11. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 11 Contemporary Engineering Economics, 4 th edition, © 2007 Incremental Analysis Lecture No. 13 Chapter 7 Contemporary Engineering Economics Copyright © 2006

12. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 12 Comparing Mutually Exclusive Alternatives Based on IRR Issue: Can we rank the mutually exclusive projects by the magnitude of its IRR? nA1A2 0 1 IRR -$1,000-$5,000 $2,000$7,000 100% > 40% $818<$1,364 PW (10%)

13. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 13 Contemporary Engineering Economics, 4 th edition, © 2007 Who Got More Pay Raise? BillHillary 10%5%

14. ENGINEERING ECONOMICS ISE460 SESSION 15 Annual Equivalent, June 23, 2014 Geza P. Bottlik Page 14 Can’t Compare without Knowing Their Base Salaries BillHillary 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.

15. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 15 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. nProject A1Project A2 Incremental Investment (A2 – A1) 0101 -$1,000 $2,000 -$5,000 $7,000 -$4,000 $5,000 ROR PW(10%) 100% $818 40% $1,364 25% $546

16. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 16 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 IRR A and IRR B are greater than MARR.

17. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 17 Example 7.10 - Incremental Rate of Return nB1B2B2 - B1 01230123 -$3,000 1,350 1,800 1,500 -$12,000 4,200 6,225 6,330 -$9,000 2,850 4,425 4,830 IRR25%17.43%15% Given MARR = 10%, which project is a better choice? Since IRR B2-B1 =15% > 10%, and also IRR B2 > 10%, select B2.

18. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 18 IRR on Increment Investment: Three Alternatives nD1D2D3 0-$2,000-$1,000-$3,000 11,5008001,500 21,0005002,000 38005001,000 IRR34.37%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. IRR D1-D2 =27.61% > 15%, so select D1. D1 becomes the current best. Step 3: Compare D1 and D3. IRR D3-D1 = 8.8% < 15%, so select D1 again. Here, we conclude that D1 is the best alternative.

19. ENGINEERING ECONOMICS ISE460 SESSION 15 Annual Equivalent, June 23, 2014 Geza P. Bottlik Page 19 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 ABCD Initial cost $150$220$300$340 Revenues/ Year $115$125$160$185 Expenses/ Year $70$65$60$80 IRR (%) 27.3224.1331.1128.33

20. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 20 ItemsCMS OptionFMS Option Annual O&M costs: Annual labor cost$1,169,600$707,200 Annual material cost832,320598,400 Annual overhead cost 3,150,0001,950,000 Annual tooling cost470,000300,000 Annual inventory cost141,00031,500 Annual income taxes1,650,0001,917,000 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

21. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 21 nCMS OptionFMS Option Incremental (FMS-CMS) 0-$4,500,000-$12,500,000-$8,000,000 1-7,412,920-5,504,1001,908,820 2-7,412,920-5,504,1001,908,820 3-7,412,920-5,504,1001,908,820 4-7,412,920-5,504,1001,908,820 5-7,412,920-5,504,1001,908,820 6-7,412,920-5,504,100 $2,408,820 Salvage+ $500,000+ $1,000,000 Incremental Cash Flow (FMS – CMS)

22. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 Geza P. Bottlik Page 22 Solution:

23. ENGINEERING ECONOMICS ISE460 SESSION 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*.

24. ENGINEERING ECONOMICS ISE460 SESSION 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.

25. ENGINEERING ECONOMICS ISE460 SESSION 15 Rate of Return, June 22, 2015 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.