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## Cost Engineering and Analysis

by: Charley Wintheiser DVM

18

0

7

# Cost Engineering and Analysis C E 406

Charley Wintheiser DVM

GPA 3.96

Jeremy Redman

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COURSE
PROF.
Jeremy Redman
TYPE
Class Notes
PAGES
7
WORDS
KARMA
25 ?

## Popular in Civil Engineering

This 7 page Class Notes was uploaded by Charley Wintheiser DVM on Monday October 5, 2015. The Class Notes belongs to C E 406 at California State University - Long Beach taught by Jeremy Redman in Fall. Since its upload, it has received 18 views. For similar materials see /class/218764/c-e-406-california-state-university-long-beach in Civil Engineering at California State University - Long Beach.

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Date Created: 10/05/15
CE406 Chapter 8 102005 ROR Multiple Alternatives Previoust we looked at determining ROR for a project Given a series of cash flows the ROR was the value of i that exactly balanced them We further expanded this by recognizing that positive cash flows from an investment could be reinvested at a new rate the external ROR and by combining this with the internal ROR we determined the composite ROR Now we take this 1 further step and look at how we compare alternatives We have done this already with PW FW and AW analysis here we do it for ROR analysis We have 90000 to invest and a MARR of 16 year AltA requires 50000 and has irr of i 35 year Alt B requires 85000 and has irr of i 29 year Which do we chose Depends on what happens with left over money As we saw last time with CRR any positive cash flows were invested at MARR Assume that is true in this case Composite rate of return for each alternative Overall ROR a 5000035 4000016 90000 266 Overall ROR b 8500029 500016 90000 283 In this case B would be the better investment ROR analysis may not provide the same ranking of alternatives as PWFWAW analysis As it merely tells us the rate of return on a project but does not consider the magnitude of return A 20 rate of return on a 100 investment is fine But a 10 rate of return on a 1000 investment yields more money in one year One of the first rules we mentioned for engineering economics was that it was only the differences between project that really matter Incremental Cash Flow Analysis makes use of this rule by comparing the differences in the cash flows between investments and how those differences affect the money not invested in the project but invested in an outside project To conduct an incremental cash flow analysis we begin with a tabulation In order to compare differences between projects we need to make sure the cash flows all line up To do so we may have to use the LCM If the projects have the same life then table goes to n end of service life If different life span we must use LCM and table Page 1 of7 CE406 Chapter 8 102005 goes to n LCM The ordering of the columns is important The investments are always ranked from lowest to most expensive initial cost Such at the cheapest to implement will be option A and the most expensive to implement is option B Alt Byr0 LstgtAltAyr0Lst Example Machine 1 cost 40K annual 10 salvage 12 3 year life ALTA Machine 2 cost 65K annual 8 salvage 25 6 year life ALT B Cash Flow year Alternative A Alternative B Incremental Cash Flow 0 40000 65000 25000 extra investment 1 10000 8000 2000 extra savings 2 10000 8000 2000 extra savings 3 1000012000 8000 30000 extra savings 40000 38000 4 10000 8000 2000 extra savings 5 10000 8000 2000 extra savings 6 1000012000 8000 25000 15000 extra savings 2000 15000 incremental cash flows represent extra investment or cost required if the alternative with the larger first cost is selected value in year 0 represents the extra initial cost of the most expensive alternative This savings if we go with alternative A can be reinvested in another project We compare the equivalent worth of the cash flows in yr 1n against the equivalent worth of the extra investment at the MARR t0 to determine which alternative should be chosen lfthe ROR available through incremental cash flow equals or exceeds the MARR alternative B should be selected Otherwise chose alternative A Incremental ROR method Essentially a value At A is calculated which represents the internal ROR of the incremental series We may get several sign changes Thus we need to be vigilant and check for the of real roots Steps Page 2 of 7 CE406 Ch apter 8 order alternatives by initial cost with B being greatest cost Develop cash flow and incremental cash flow series using LCM 1 2 3 Draw incremental cash flow diagram 4 count sign changes in incremental cash flow to determine if multiple rates exist 1 If necessary use Norstrom39s criterion to determine if a single positive root exists 701 Select the alternative as follows 1 if Ai lt MARR selectA 2 if Ai3 MARR select B Let39s solve our example assume MARR i Descartes 3 syn changes s15 Set up PW equation for the incremental cash flow and solve 102005 Incremental Cash Flow Cumulative 25000 extra investment 25000 2000 extra savings 23000 2000 extra savings 21000 30000 extra savings 9000 2000 extra savings 11000 2000 extra savings 13000 15000 extra savings 28000 Norstrom39s says 80 lt0 and only 1 sign c hange so we have only 1 di value PW 25 2PA Ai 6 28PF Ai 3 13PFAi 6 28 is the biggest lump there PW 25 26 28 13PFAi3 25 53 047 PFAi3 Ai 30 solve at 30 PW 25 2PA 30 6 28PF 30 3 13PF30 6 428 solve at 25 PW 25 2PA 25 6 28PF 25 3 13PF25 6 135 iterate further until we converge on right answer Ai 23 Page 3 of 7 CE406 Chapter 8 102005 25 MARR Ai 2o 7 15 7 4 10 7 E 5 7 o 5 710 7 15 l l l l l l l l l 0 5 10 15 20 25 30 35 40 45 50 We can also see graphically that Ai is about 23 So what does this mean It means that the extra savings we get from the more expensive alternative B earns us 23 The MARR on this project is 15 The savings are greater than this meaning the extra cost is justified and we go with alternative B What do we do if we have more than 1 possible root Use composite approach with the incremental cash flow series Solve it exactly the same as we did previously However we don39t know exactly what the reinvestment rate will be 2 options set reinvestment rate at Ai or specify a value of c perhaps the MARR Correctly using ROR method gives same value as PW FW and AW analysis However for AW we have options use incremental cash flow over LCM just as we have done so far annualize all single cash flow values find AW for each alternative39s actual cash flow and solve for 0 AWb AWa Page 4 of 7 CE406 Chapter 8 102005 To this point what have we learned Determine the equivalent worth of cash flows at any points in time Determine Present Future and Annual Worth of an investment Single investments we calculate PW at the MARR if 0 we accept it For alternatives calculate PW at the MARR and select the largest value compare investments with different lifespans using LCM Annual Worth Analysis avoids LCM We have determined PW and AW for investments with infinite lifespans Determine the ROR of an investment given a cash flow Internal Rate of Return on the investment value for i giving a PW 0 select projects with ROR gt MARR External Rate of Return on positive cash flows not needed revenues initial funds not fully invested Composite Rate of Return weighted average of lRR and ERR Incremental ROR analysis to select between alternatives compare the savings by going with 1 alternative over another If the difference in ROR between the 2 options is less than MARR invest in the cheaper option and reinvest the remaining funds in another investment if the difference is greater than MARR you pickthe more expensive option Rules Requires LCM for alternative comparison Multiple roots must be determined External funds eg revenues should get reinvested at the External ROR Mutually Exclusive Alternatives Determine ROR for all alternatives if only 1 exceeds MARR select it If gt1 exceed MARR gt Incremental Cash Flow analysis Order all alternatives from smallest to largest initial investment calculate i for the first alternative if i lt MARR eliminate it and go to next investment first alternative with i gt MARR is the DEFENDER Next alternative on list is CHALLENGER Determine incremental cash flow between CHALLENGER and DEFENDER Calculate AP If AP 3 MARR DEFENDER is eliminated and CHALLENGER becomes DEFENDER If A1 lt MARR eliminate CHALLENGER Repeat until 1 alternative remains standing Page 5 of 7 CE406 Chapter 8 102005 Let39s consider you got an inheritance from grandma You want to invest the money for 5 years until you graduate college and grad school The safe option is a CD earning 3 per year Some friends have connections and can promise you the following yr A B C D 0 2500 2700 3000 3200 15 615 660 710 735 Which do you go with Lets order by lowest to highest initial cost A B C D If the alternatives were independent we simply check each against the MARR Thus PWA 2500 615PAi5 PAi5 4065 which if we look up in table is i 732 A is acceptable PW B 2700 660PAi5 i 708 acceptable PW C 3000 710PAi5 i 589 acceptable PW D 3200 735PAi5 i 48 acceptable So based on comparison with the safe investment we can invest in aA B C D However which should we go with Incremental ROR analysis A is defender B is challenger BA CB DB 200 300 500 45 5o 75 45 5o 75 45 5o 75 45 5o 75 45 5o 75 0 579 888 4064 Page 6 of 7 CE406 Chapter 8 102005 PW ba 200 45PAde i 5 del i 406 The extra initial cost with going for the more expensive option yields a greater incremental return than the safe option so we will pick it A is eliminated NOTICE del i ba ltgt ib ia 406 ltgt 708 732 del i refers to the yearly difference in revenues between alternatives Thus we eliminate A Now C is CHALLENGER B is DEFENDER PW CB 300 50PA del i 5 del i 58 A negative del i means that the additional 300 we spend initially 30002700 buying C does not compensate for the 50 extra we get per year In fact compared to B C is a losing proposition This is not surprising if we look at IRR for B C we see 708 and 589 7 is much greater than 6 Repeating this for PW DB 500 75PAdel i5 del i 888 Again the extra 500 initially spent on D yields a negative ROR We see some obvious and not obvious answers here That C and D were eliminated is somewhat obvious as they definitely were underperforming compared to A and B However B beats A is a little surprising and is simply due to the fact that the extra 45 earned yearly by buying B offset the initial 200 cost WHEN COMPARED TO THE SAFE OPTION Page 7 of 7

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