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Cost Engineering and Analysis

by: Charley Wintheiser DVM

Cost Engineering and Analysis C E 406

Charley Wintheiser DVM

GPA 3.96

Jeremy Redman

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Jeremy Redman
Class Notes
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This 8 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 32 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 7 101005 Rate of Return Analysis We can consider ROR from two different perspectives The lender and the borrower If you borrow the money the interest rate is applied to the unpaid balance so that the total load amount and interest are paid in full exactly with the last loan payment If you lend the money there is an unrecovered balance at each time period The interest rate is the return on this unrecovered balance so that the total amount lent and interest are recovered exactly with the last receipt Rate of Return is o the rate paid on the unpaid balance of borrowed money or o the rate earned on the unrecovered balance of an investment so that the final payment or receipt brings the balance exactly to zero with interest considered Previously we have specified interest rates on money In this chapter we will calculate ROR on alternatives and then evaluate whether these values are sufficient to meet the MARR Calculating ROR To determine ROR of a cash flow series set up ROR equation using either PW or AW relations The present worth of costs or disbursements PWd is equated to the present worth of incomes or receipts PWr PWd PWr Using proper sign convention we could say that 0PWRPWD recognizing that PWD is going to have a negative sign We can write similar expressions for AW AWd AWr or 0 AWr AWd The Rate of Return is the value i that satisfies this equation We refer to this value as i When comparing alternatives against the MARR If iz MARR the alternative is economically viable if i lt MARR the alternative is not economically viable Page 1 of 8 CE406 Chapter 7 101005 It is important to remember that the basis for all of our calculations is EQUIVALENCE All terms must be written in terms of equivalent present worths future worths or annual worths We solve these terms for i such that they become equivalent You borrow 10000 from the bank and are required to repay 4021 each year for three years What is the interest that the bank is charging 10000 4021 PAi3 PAi3 24869 We look in the interest tables at the PA column for 3 year row for this value We find it to be 10 For example if you deposit 1000 now and are promised 500 in 3 years and an additional 1500 in 5 years what is the rate of return Using PW factors PW 1000 500PFi3 1500PFi5 We then must solve for i We have 2 options trial and error and computer using spreadsheets Tips for trial and error By trial and error we can just pick an i value and see what the PW is If PW gt0 that indicates our chosen interest rate is not fully capturing all of the money and we must chose the higher value if PW lt 0 that indicates our chose interest rate is accounting for more money than available and we must chose a lower value Trial and error will take a LONG time Obviously the closer we get to i on our initial guess the easier it will be What we will do is lump all the factor values into a single factor and then solve that for our initial guess convert all disbursements into a single amount P F or A by neglecting the time value of money We typically lump the values into the higher of the values if we convert As into an F simply multiplyA by n if Fs into As then divide by n Minimize error of time value of money by shifting to the larger of the values convert all receipts similarly use table to find approximate i which gives a close value JON Page 2 of 8 CE406 Chapter 7 101005 0 1000 500PFi3 1500PFi5 this value isjust a guess Initial Estimate As a guess let39s lump them all at year 5 2000PFi5 1000 PFi5 05 looking through tables we see that 14 05194 and 15 04972 pick one i Evaluation iteration i 15 1000 500 PF153 1500PF155 0 74 A positive value indicates our choice this indicates that our value of i is too small not accounting all of the money i 16 0 3550 40 i 17 35 a c 365 30 7 Interpolation 25 ab C d 20 350 365 1716 350 0i16 E 15 a c 10 5 if we solve 5 7 d 0 l l Linear lnterploation 5 15 1H 18 1 16 3500355 365 b 1 169 10 Example with Trial and Error We spend 500000 now on a new system This will save 10000 each year for 10 years and 700000 at the end of 10 years in equipment refurbishment costs Find ROR 0 500000 10000PAi10 700000PFi10 Comparing the factors 10000 for 10 years is going to be 100000 wo accounting for interest vs 700000 700000 is much greater so we will lump everything with it Page 3 of 8 CE406 Chapter 7 101005 ESTIMATE 0 500000 800000 PFi10 PFi10 58 0625 if we look in tables PF 10 years i 4 06756 i 5 06139 We start with i 5 because our approximate PF value is going to be lower than the true value when we incorporate the true time value of money i 5 0 500000 10000PA510 700000PF510 0 6946 Positive result meaning that the rate is greater than 5 i 6 0 500000 10000PA610 700000PF610 0 35519 Result negative meaning rate of return is less than 6 lnterpolate i 5 6946 0 6946 355196 5 i 516 Now tre SPREADSHEET method year 0 1 2 3 4 5 6 7 8 9 10 cash 500 10 10 10 10 10 10 10 10 10 710 flow lRRrange of cash flows i guess 516 Problem 75 What is rate of return over 25 year period if you invest 150000 to produce a portable air compressor and estimate monthly costs of 27000 with income of 33000 use AW analysis pp monthly therefore n 2512 30 months disbursements 150000APi30 27000 receipts 33000 0 Awd Awr 0 150000APi30 27000 33000 Page 4 of 8 CE406 Chapter 7 101005 0 150000APi30 6000 APi30 6000 150000 004 look up 004 in tables i 125 NP 004018 i 1 AP 003875 125 0 27 1 0 1875 i A 187501875 271251 122 per month ROR Method is used to select from alternative as we will see next week ROR method used to select among alternatives will give the same decision as the PW or AW analysis Multiple ROR values One of the problems with ROR analysis is sign change in the cash flows When there is only 1 sign change we have only 1 unique i value However if we have more than 1 sign change it is possible to have many i values If we have only 1 sign change we refer to this as a conventional simple cash flow senes If we have 2 or more sign changes it is a nonconventional nonsimple cash flow series 2 Rules Descartes39 rule of signs total of realnumber roots is always less than or equal to the number of sign changes in the series Norstrom39s criterion if there is only 1 sign change in the series of cumulative cash flows that begins negatively there will be only 1 real i value the cumulative cash flow sign test Determine cumulative cash for at each point Sum of Cash Flow from t0 gt t cash flow If 80 lt 0 if only 1 sign change in 8 only 1 real i value Page 5 of 8 CE406 Chapter 7 EX year cash flow Cumulative 0 1000 1000 1 500 1500 2 0 1500 3 1000 500 4 1000 500 5 500 1000 6 500 500 7 1000 1500 101005 through year 5 we have a conventional series after year 5 it becomes a nonconventional series Descartes rule say that there may be 3 real positive roots Nostrom39s says that Slt0 and there is only 1 sign change so there must only be 1 root consider the following cash flow year cash flow Sequence cumulative 0 2000 SO 2000 1 500 S1 1 500 2 8100 S2 6600 3 6800 S3 200 descartes rule says there are 2 sign changes so there may be at most 2 real roots Nostroms rule since s0gt0 we may have up to 2 roots if SOlt0 we still may have 2 roots because 8 changes sign twice To solve PW 2000 500PFi1 8100PFi2 6800PFi3 solve this for different values of i result is parabola crossing i axis at 8 and 41 So which is the right value Both fit however some values are too large to be ridiculous 41 ROR is too high A more reasonable value is 8 But there is an even better way to determine the real ROR Page 6 of 8 CE406 Chapter 7 101005 Composite Rate of Return i values determined in this method exactly balance the cash flows this i is known as the internal rate of return the IRR is the ROR on the unrecovered balance These are funds that are still tied up in the inevestment IRR refers to funds earned until the project breaks even Funds beyond that once a net positive cash flow is realized are released as external funds to the project These funds are not included in true ROR analysis i refers to i rate for break even over the cash flow When the project becomes profitable the money might be invested by the company in another project at a rate MARR What happens to any positive cash flows during the break even period obviously the company has already made the initial investment Perhaps they invest this positive cash flow at a different rate maybe the MARR or anther rate c reinvestment rate or external rate of return this is commonly the MARR We now have a new interest rate Composite Rate of Return CRR i39 unique rate of return for a project that assumes the net positive cash flows not immediately needed by the project are reinvested at reinvestment rate c How do we get i39 NetInvestment procedure Find future worth of net investment 1 year in future If we are net positive we invest that amount at c if net negative it is at i39 find the future worth of the projects net investment value FF 1z C t0 12 n Ct net cash flow in yeart iCith1gt0 ii39ifF1lt0 Fn is set at 0 and the equation is solved for i39 Essentially if the previous year used available funds then money is invested at i39 if the previous year generated extra funds then the funds are invested at c Lets look at earlier data and assume MARR is 15 Page 7 of 8 CE406 Chapter 7 101005 0 2000 1 500 2 8100 3 6800 F0 2000 F1 20001015 500 1800 F2 1800115 8100 6030 F3 60301i396800 F3 Fn o o 60301i396800 1i39 1128 i39 128 What happens if project goes longer than 3 years 4 1000 5 500 6 50 F3 60301i396800 F4 60301i3968001i39 1000 F5 F41i39 500 F6 F51i39 50 0 F51i39 50 0 60301i3968001i39 10001i39 5001i39 50 solve this expression for i39 195 Importantly i39 195 is greater than MARR 15 Thus the project is viable Homework 7 11 26 29 31 32 Page 8 of 8


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