Cost Engineering and Analysis
Cost Engineering and Analysis C E 406
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This 12 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 26 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 Engineering Economy Chapter 2 090605 Chapter 2 How Time and Interest Affect Money Single Payment Factors Fundamental factor in Engineering Economy is the one that determinesthe amount of money F accumulated after n years or periods from a single present worth P with interest compounded one time per year Compound interest interest paid on top ofthe previous period39s interest F1PPi FP1z39 F2F1F1i F2P1iP1ii F2P1iii2 F2P1i2 ltCash flow diagram from bank account POVgt F3F2F2i F3P1i2P1i2i F3P1z 21i F3P1z 3 FP1z39 Ifwe wanted to we could look at this equation in a slightly different manner F P x 1iquot could be rewritten as F P x factor where in this case the factor 1iquot is referred to as the singlepayment compound amount factor SPCAF or the FP factor What the FP factor is is a conversion that when multiplied by P yields the future amount F for a given interest rate i after n years In general we can notate these factors as XYin where XY refers to the fact that we seek a value X given a known Y for i compound interest over n years We abbreviate this factor as FPin This is read as Find F given P at i interest rate for n periods In this notation the rst letter refers to what is being sought The second refers to what is known The third refers to the interest rate and the fourth is the number of years So if we wanted to know how much money will be in an account after 10 years if 1000 is deposited today earning 5 we could write the equation as F P 1iquot 1 000 1 00510 162889 or we could write F P FPin 1000 FP510 And both would be correct Let39s say we need to save up for a big purchase in 5 years How much money should we Page 1 of 12 CE406 Engineering Economy Chapter 2 090605 deposit today We have an idea how much the item will cost and know that we need to save up F dollars but how much should we invest today Beginning with the equation we have just derived F P 1 iquot we can reverse the calculations and look at what present value is needed for a given future value 1 1 We can use this expression to calculate how much money we have to deposit now in order to have a given amount at some point in the future The bracketed value 11iquot isthe single payment present worth factor SPPWF or the PF factor Likewise it is abbreviated as PFin To nd Find Notation Equation Factor Future value FPFPin FP1iquot Singlepayment compound amount Present Value PFFPin PF11iquot Singlepayment present worth Application of these equations is fairly straightfonNard Given one value and the interest rate and time period they can simply be solved for the other However it gets even easier An appendix in the book hastabulated values for i ranging from 025 to 50 and n from 1 to really large 80 ultimately there39s no work involved Example How much money do l have to deposit in my investment account today if I want to be a millionaire by the time I39m 60 I39m now 33 Assume get 6 rate ofreturn FP or PF We solve for present value given a future value F 1000000 i 6 n 60 33 27 ltCash ow diagram with withdrawal at F and deposit at P Let39s solve this the long way P F 1 000000100627 20736795 or the short way P FPFin FPF627 see table 11 in back of book The present worth factor for 27 years is 02074 Thus P 1000000 02074 207400 which is essentially the same as the above answer This is easy let39s do another example Page 2 of 12 CE406 Engineering Economy Chapter 2 090605 A company moves it39s telephone support to the Philippines lt expects to save 50000 this year from reduced costs ie wages lfthis savings is worth 20 per year what is the equivalent value after 5 years say with this years savings they can invest in another higher yielding project We have Present value and want to know the future value ltCash ow diagram P up F down F PFPin 50000FP205 5000024883 124415 Cash ow diagram rules 0 P always occurs at time 0 o F always occurs at end of year speci ed eg after 5 years F is on yr 5 o the interest rate compounds at the end of each tickmark period lfyou are using EXCELto solve these problem AND I DON39T RECOMMEND lT lfyou do be careful because EXCEL has very speci c convention regarding signs returning the opposite sign for the future and present value calculations Let39s solve a homework problem Assume you borrow money and are given 2 options to pay back the loan over a three year period The rst option is to repay only the interest earned at the end of each year and the principal at the end of the third year The second option is to pay nothing until the end ofthe third year What is the total amount paid at the end ofthe three years assuming compound interest Consider option 1 interest only loan End of year Interest owed Total owed Endofyear Total owed for year end of year payment after payment 0 P 1 Pi P Pi Pi P 2 Pi P Pi Pi P 3 Pi PPi Pi P 0 Thus for option 1 the total we have repaid is P 3 P i or P 13i Page 3 of 12 CE406 Engineering Economy Chapter 2 090605 Consider Option 2 End of year Interest owed Total owed Endofyear Total owed for year end of year payment after payment 0 P 1 Pi P Pi 0 P1i 2 P1ii P1iquot2 0 P1iquot2 3 P1iquot2 i P1iquot3 P1iquot3 0 Thus for option 2 the total we have repaid is P1iquot3 Algebraic manipulation of factors The factors act in a algebraic manner If we want F given P we must multiply by the FP factor F P FP In the above shortened equation we notice that if we consider the HP factor as a fraction then on the right hand side we have P F P which is equal to F which is the value we are solving for on the left hand side We can demonstrate this algebraic nature in a slightly more complicated example Determine the value of PFPinPFin Let us first solve this mathematically using n 1 the value of the factors P1 1 which reduces to simply P We can also solve this by associating each factor with a fraction P F P P F which also reduces to P This is has important implications in that we can write new factors as combinations of other factors UniformSeries Factors Present Values Thus far we have looked at time value of money for single payments either in the present or future This is identical to the equivalent value of money at 2 different times However often we deal with uniform series of money eg paychecks rent payments etc Let39s consider the present value P for a series of F values which we will refer to as A A series of consecutive equal end of period amounts of money yr mo 1 1 1 111 1z392 1z39 Each term here represents the individual PF values for years 1 through n Factor outA PA 1z39 1 Page 4 of 12 CE406 Engineering Economy Chapter 2 090605 4 1 1 111 112 11 1 11 PA To simplify this we will manipulate this equation a bit by multipling by 11 i which is the PF factor for 1 year This shifts each term back 1 year P 1 I 1 I I 1 I 1 1H 11 113quotquot3911 3911 1 Ifwe subtract the former from the latter P 1 I 1 I I 1 I 1 1 112 113 11 11 1 PA 1 121 III II 11 11 11 11 iPA 1n 1 1 1 11 11 11 134 1 1 1 11 PA 11 1 111 We now have a new factor the Uniformseries present worth factor USPWF This factor can be used to calculate the equivalent value P in year 0 for a uniform endof periods series ofA values beginning at the end ofthe period 1 and extending for n periods Likewise we can determine the equivalent annual worth A over n years for a given P in year 0 111 AP 11 1 These formulas are derived with P and the first uniform annual amount A one period apart P must always be one period prior to the first A ltFiguregt As with the FP and PF factors these are also tabulated Page 5 of 12 CE406 Engineering Economy Chapter 2 090605 Let39s say I need 60000 a year for the next 50 years to live a comfortable live of luxury without working How big does my lump sum lottery check have to be ifl invest my winnings at a 15 rate ofreturn We need to determine present value given annual amounts P APAin 60000 PA15 50 60000 66605 399640 If it were in the bank 2 rate of return P 60000 PA2 50 60000 314236 1885416 At the end of the 50th year and my nal 60000 withdrawal the account will have a balance of 0 Future Values Let39s say instead of present payment we want a future value from some annual values We can derive this expression by substituting the FP equation in to the NP z391z39 1 1 1 i1i 1i 1z39 1 z39 1i 1 AP AF This is the NF or Sinking Fund factor It determines the uniform annual series that is equivalent to a given future worth F The uniform series A begins at the end of period 1 and continues through period of the given F We can rearrange this expression to determine F for a stated A 1i 1 l FA This is the uniform series compound amount factor USCAF or FA factor This tells us the future amount of an annual series of values Remember A is constant and end of every period even final period F occurs during the nal period both an A and F value Page 6 of 12 CE406 Engineering Economy Chapter 2 090605 Examples lam looking forward to when I need to buy a new car in 6 years lfl am willing to spend 20000 on a new car If my money earns 3 interest in the bank how much should I deposit at the end of each year in order to have the full amount by the end of the 6th year Find Agiven F use sinking fund A F AFin 20000 AF36 20000 015460 3128 How much money could RTT environmental services borrow to nance a site reclamation project if it expects revenues of 280000 per year over a 5year cleanup period Expenses associated with the project are expected to be 90000 per year Assume the interest rate is 10 per year Net cash flow per year 280000 90000 190000 Maximum amount they can borrow before exceeding net cashflow P A PA105 190000 37908 720252 Page 7 of 12 CE406 Engineering Economy Chapter 2 090605 Ifthe Tables don39t have the value for i or n that you are interested in USE THE EQUATIONS Ifyou are really lazy and don39t want to do real math you can interpolate but these results are going to be slightly wrong depending on how far i or n are from the tabulated values Arithmetic Gradient Factors Let39s say that instead of a constant annual value we actually have one that varies per year Ifthe period to period changes are constant ie they go up or down the same amount each year we refer to this as an arithmetic gradient The gradient is the actual amount ofthe yearly change For example ifyou buy a car it might cost you 1500 for gas insurance for the rst year This value stays relatively constant through the life of the car but future years incur more cost as it needs repair Thus in year 2 you might be paying 1550 and year 3 1600 etc Thus we can say that the total cost per year is 1500 n150 Base amount 1500 Gradient 50 G constant arithmetic change in the magnitude of receipts or disbursements from one time period to the next G may be positive or negative The cash ow in year n CF is calculated as CF base amount n1G Notes about gradients o Gradients are broken down into two separate cash ows the base amount and the gradient proper first payment begins end of rst period and is equal to the base amount ifthere is no base amount this is 0 The first increase in amount paid occurs end of second period base amount G lfthere is no base amount then the series starts with the gradient at end of period 2 Yearly increase is a xed amount not a percentage Begin by assuming simply a conventional gradient with no base amount Gradient beings year 2 We solve this problem by taking present value of each year P GPFi2 2GPFi3 3GPFi4 n2GPFin1 n1GPFin Page 8 of 12 CE406 Engineering Economy Chapter 2 090605 1 2 3 n 2 71 1 6 112 113 114 39 11 1 1i 11111 111 I 112 113 I 1z39 1 1i iPG First term on right side is the equation for PA Final result is G 1iquot 1 n i z391z39 1i This expression converts an arithmetic gradient into present worth This does NOT include the base amount which must be handled separately PA calculationl The arithmetic gradient present worth factor PG factor PIG z39 701 lt1quotn 1 quot 139 z391z39 1i PG j nlt1 iinn 1 1 11 The gradient starts in year 2 The expression we use to calculate the nal value PGPGz39 n This tells us the present worth of a gradient lfwe want the future worth of the gradient the arithmetic gradient future worth factor can be derived by simply multiplying the PG factor by the FP factor to get e lt11111 Last but not least we can convert a gradient into a uniform annual series of payments FG Let39s say we want a series of annual payments that is equivalent to this gradient A PAPin A GPGinAPin AGi L I 1i 1 The bracketed term is the arithmeticgradient uniform sen39es factor AGin Page 9 of 12 CE406 Engineering Economy Chapter 2 090605 Important PG AG do NOT include the base amount so they must be calculated separately De ne PT total present worth of a gradient series PA present worth of base amount that begins in year 1 PG present worth ofgradient amount that begins in year 2 PTPAiPG depending on whether gradient is increasing or decreasing for equivalent annual series ATAA1AG Example A disaster relief fund is being established The long term budget allows for 500000 to be allocated next year with an increase of 50000 per year forthe following 7 years lfthe ROR on the account is 9 what is the equivalent present value lt guregt Write it all in terms of 1000 We have 2 series that we are concerned with the present value ofthe annual series of 500 starting year 1 the present value ofthe gradient series G50 starting year 2 Pt Pa Pg Pt APAin G PGin Pt APA98 GPG98 Pt 50059952 50168877 384199 3841990 example 230 A company has budgeted 300000 per year to pay for certain parts over the next 5 years If the company expects the cost of the parts to increase uniformly according to an arithmetic gradient of10000 per year what is it expecting the cost to be in year 1 if the interest rate is 10 per year 2 step problem Determine present value ofthe budgeted money PAPA105 30000037908 1137240 Compare this to the present value ofthe gradient to determine the present value ofthe base cost Page 10 of12 CE406 Engineering Economy Chapter 2 090605 Pt Pa Pg Pt 1137240 Pg GPG105 10000 68618 68618 Thus Pa Pt Pg 1137240 68618 1068622 Now this is the present value ofan annual cost thus we have to distribute this over the 5 year period A P AP105 1068622 026380 28190248 Geometric Gradient Series Not all series are arithmetic constant value but rather increase by a constant percentage such as 5 per year We define a new variable g constant rate of change in decimal form that amounts adjust from one period to the next Pg the present worth for the entire cash flow series 2 A1 IA1lt1g I A1lt1g2 A1lt1gn1 3 Hi 1i2 1i3 n1quot Which can be solved to yield 2 different solutions n 1g 1i g ltgti PgA1 lg or m41 P g Hi forg I Thus P9 A1PAgin Unfortunately there are no tables for this particular case Example A carnival ride was modified to be more exciting The modification cost 8000 and it expected to last for 6 years with a 1300 salvage value of the mechanism Maintenance costs are expected to be 1700 for the first year increasing by 11 each year thereafter Determine the equivalent present worth of the modification and maintenance cost Assume 8 interest rate per year Pt 8000 Pg F PF86 Page 11 of 12 CE406 Engineering Economy Chapter 2 090605 Pt 8000 17001111108quot6008011 130006302 Pt 8000 1700 59559 81926 1730585 Part 2 The boost to park attendance from a new ride typically drops off by 10 a year How much of a first year boost does the park need to break even on the ride 1 1g 1i PgA1 ig APg quotg n 11g 1i in this case i 8 but g 10 If we solve for this case we get as the A1 PgAPgin PgAP1086 Pg 02706 173058502706 468357 Determination of Unknown ior n If a problem asks for a time or interest rate rather than a monetary value you cannot use the tables directly For example Wasteco estimates they will need 1000000 for costs incurred during the closure of their landfill If they can afford to deposit 30000 at the bank and earn 6 interest how many years prior to closure do they need to start the deposits FAFAin 2 ways 1i 1 139 n logiFA1 log1i substituting in known values i 006 F 1000000 A 30000 we calculate 1885 or 19 years Or we can interpolate from the tables FA 1000000 30000 3333 6 table looking for this value in the HA column we see that for yr 18 HA 309057 and yr 19 HA 337600 Thus we know our value must be greater than 18 years If need be we can interpolate to determine the exact n FA Page 12 of 12
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