1. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Quiz results Go over quiz Homework due tonight Next Homework due 6/17 Recommendation Chapter 5

2. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 2 Quiz Results The good news: –Substantial improvement over quiz 1 –Did not change class GPA (very good – it usually drops) –No scores below 70%, also a good sign Cautions –Some of you are still somewhat unsure about the use of the formulas –You need to fix this to do well in the rest of the class!

3. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 3 Quiz Results

4. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 4

5. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 5 Recommendation British Library (on Euston, next to St. Pancras) –Treasures exhibit »Gutenberg Bible »Mozart manuscript »Leonardo Da Vinci sketches »Beatles’ notes »And much more And of course free! About a 15 minute walk from Accent

6. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 6 CHAPTER 5 PRESENT WORTH ANALYSIS PROJECT CASH FLOWS PAYBACK PERIOD NET PRESENT WORTH CAPITALIZED EQUIVALENT METHOD MUTUALLY EXCLUSIVE ALTERNATIVES PROJECT LIVES

7. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 7 PROJECT CASH FLOWS INVESTMENTS ARE MADE EARLY IN THE PROJECT BENEFITS ARE EXPECTED OVER A PERIOD OF YEARS IN THE FUTURE LOAN CASH FLOW - PAYMENTS RECEIVED BY THE LENDER PROJECT CASH FLOW - INCOME LESS EXPENSES PAYBACK PERIOD - THE TIME THAT IT TAKES TO RECOVER THE ORIGINAL COST, e.g., A WAREHOUSE COSTS $4M TO BUILD AND YIELDS A PROFIT OF $500,000/YEAR. THE PAYBACK IS 8 YEARS. THIS IS THE SIMPLEST ANALYSIS DISCOUNTED PAYBACK PERIOD – REDUCE THE VALUE OF THE INVESTMENT BY THE DISCOUNT RATE

8. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 8 PROJECT CASH FLOWS (CONTINUED) FOR UNEQUAL CASH FLOWS, DETERMINE THE TIME WHEN CASH FLOW EQUALS THE INITIAL INVESTMENT SALVAGE VALUE - WHAT YOU CAN GET FOR USED EQUIPMENT THE SALVAGE VALUE OF THE OLD EQUIPMENT IS SUBTRACTED FROM THE INITIAL INVESTMENT DO NOT USE PAYBACK PERIOD FOR COMPARING PROJECTS, -- ONLY FOR ROUGH SCREENING

9. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 9 NET PRESENT WORTH DISCOUNTED CASH FLOW TECHNIQUES (DCF) ONE SUCH METHOD IS NET PRESENT WORTH (NPV) BRING ALL INVESTMENTS AND BENEFITS BACK TO TIME ZERO COMPARE THEM COMPARE TO OTHER ALTERNATIVES SELECT LARGEST PRESENT WORTH PROJECT

10. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 10 NET PRESENT WORTH WHAT INTEREST RATE DO WE USE? COMPANIES DETERMINE WHAT THEY CALL: –MINIMUM ATTRACTIVE RATE OF RETURN (MARR) –MANY COMPANIES INSIST ON A HIGH RATE OF RETURN BECAUSE THEY HAVE LIMITED RESOURCES AND WANT TO INVEST IN THE BEST PROJECTS NET PRESENT WORTH SHOULD BE POSITIVE TO ACCEPT A PROJECT INVESTMENT POOL - ALL AVAILABLE FUNDS FOR INVESTMENT IF THERE ARE NO FUNDS, THEY MUST BE BORROWED

11. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 11 NET PRESENT WORTH (CONTINUED) EXAMPLE - ALL PAYMENTS AT THE END OF THE PERIOD

12. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 12 Example

13. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 13 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

14. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 14 Capitalized Equivalent Method

15. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 15 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

16. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 16 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

17. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 17 PROJECT LIVES (CONTINUED) EXAMPLES –PROJECT LIVES LONGER THAN ANALYSIS PERIOD (5.11)(5.11) –PROJECT LIVES SHORTER THAN ANALYSIS PERIOD (5.10)(5.10) –ANALYSIS PERIOD = LONGEST LIFE (5.12)(5.12) –UNEQUAL LIVES - LOWEST COMMON MULTIPLE METHOD (5.13) (5.13)

18. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 18 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

19. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 19 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

20. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 20 Example 5.12 Analysis period = longest life

21. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 21 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

22. ENGINEERING ECONOMICS ISE460 SESSION 10 CHAPTER 5, June 15, 2015 Geza P. Bottlik Page 22 A FUN NOTE TO END ON (GARY LARSEN) “Notice all the computation, theoretical scribbling, and lab equipment, Norm … Yes, curiosity killed these cats