Cost Engineering and Analysis
Cost Engineering and Analysis C E 406
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This 4 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 20 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
Chapter 3 Shifted Series The problem with our factors and equations is that they specify certain conditions that must be met For example A must being 1 period after P F must coincide with the final A The gradient G starts on period 2 For example given a uniform series of value X from 4 to 8 what is present value How can we solve this Options 0 Do PF for all 5 values P XPFi4 XPFi5 XPFi8 0 Use FP for all 5 values then PF on the sum F XFPi4 XFPi3 XFPi0 P F PFi8 0 Use FA to find a future end value then determine the present value F XFAi5 P FPFi8 XFAi5PFi8 OR use PA to determine present worth at year 3 and then transform that to year 0 P39 XPAi5 P P39PFi3 APAi5PFi3 All of these methods give equivalent nal values All are acceptable methods to solve the problem The latter 2 are the easiest as they involve factors which can be combined To solve problems involving time shifting do the following 0 Draw a diagram Locate the present or future worth of each series Determine n for each series by renumbering the diagram Set up and solve the equations Example 1 A developer is building an apartment complex The complex will be completed in 2 years Expected annual income from the property is 12 million starting end of year 3 lfthe developer is using a 10 year planning horizon and a MARR of15 what is the present worth ofthe project We have a cash ow series that is offset from what we need by 1 year So we have a 2step problem Step 1 renumber CF diagram according to rules Step 2 determine P39 APA15 8 1244873 53847600 Step 3 timeshift this value to the real time 0 P P39PF152 5384760007561 407142 Done Personally I nd it easier to solve these problems and crunch numbers only 1 time so I would write P39 APA158 P APA158PF152 P 12 M 4487308696 407141704 We can also solve it using future worth rst F A FA158 12M 137268 1647216 P F PF1510 164721602472 407192 4071920 which is within round off of the previous term Example 2 Given the following cash ow series yrs 13 20000 yr 4 8000 year 5 20000 Determine the present value ifi 10 We can approach this from several different angles A uniform series from 13 1 and two individual payments in yrs 4 2 and 5 3 Pt P1 P2 P3 P1 APAin 20000PA103 2000024869 49738 P2 FPFin 8000PF104 800006830 5464 P3 FPFin 20000PF105 2000006209 12418 Pt 67620 However this is a bit of work We have to consider three sets of cash ows and look up three different values in the tables Perhaps there39s an easier way Given that we can add and subtract cash ows occurring at a given time can we manipulate this set of cash ows to make only 2 sets of CFs Yes A single CF series of 20000 from years 1 through 5 and then in year 4 we have a CF of12000 How do we solve P APAin FPFin P 20000PA105 12000PF104 P 2000037908 1200006830 67620 Example 3 Let39s say we have he following cash low series 1 Yr 0 1 2 l 3 l 4 2000 2000 2000 2000 2000 Find the future value at i 5 We can evaluate this in a couple ofdifferent methods 2 separate series of cash flows 0 a single amount at yr 0 and a series from yr 14 0 a series from yr 04 As long as we properly apply our rules solving by either method is acceptable Method 1 two series F F1 F2 P 2000FP54 2000FA54 3511 051 Method 2 one series in a uniform series Aalways starts at yr 1 so we have to renumberthe CF diagram F 2000PA55 200055256 11051 Example 4 The trust fund left to you by a relative will pay 10000 for 3 years thereafter the amount paid will decrease by 2000 a year until it has been depleted What is the present value of the trust fund assuming it earns 10 year Recognize immediately that we have two cash ows a uniform series of 1 0000 and a gradient of2000 Based on drawing the CF diagram we see that the account will be empty at the end ofthe 8th year To calculate the present value ofthis we break it into two separate series Pt P1 P2 P1 uniform series APAin 10000PA108 1000053349 53349 P239 gradient GPGin 2000PG106 200096842 193684 however P239 occurs in yr 2 so we cannot simply add P1 P239 We must translate it to year 0 P2 P239PFin 193684PF102 19368408264 1600605 Finally Ptotal 53349 16006 37343 Example 5 Let39s work a complex example lwish to establish a college fund for my newborn son In 18 years he will start a 4 year college College will cost 30000 per year for 4 years yr 18 through 21 lfthe rate of return ofthe college fund is 8 how much money should I deposit in yrs1 through 18 to meet his college needs We actually have 2 uniform series to deal with 18 years of deposits and then 4 years of withdrawals We will use equivalence to solve this problem We start by determining the equivalent amount of the four withdrawals gure We do this by determining the equivalent present value for the uniformseries present worth series Remember that P is at t0 and A begins at the end ofthe end of the first period Thus at year 17 P39APAin30000PA84300003312199363 So we need 99363 in the account by year 17 Next step is to determine the equivalent annual amount needed to give us this future amount However forAF we need to have F on the nal year So rst we have to determine the equivalent value of 99363 at year 17 for year 18 To do that we simply use FP39 99363FP81 99363 108 107312 Finally we determine the amount we need to deposit every year so that we have this value in the future AF 107312AF818 107312002670 286523 80 in order for my daughter to afford college I need to deposit 2865 per year for 18 years How does this calculation change if instead ofa constant cost of college it increases by 2000 a year At year 17 Pt39 Pa39 Pg39 APAin GPGin Pt39 99363 2000PG84 Pt39 99363 200046501 108663 Let39s bring this to time 0 P Pt39 PFin Pt39 PF817 108663 02703 29372 The equivalent 18 year annual series A P APin 29372AP818 29372010670 3134 Thus we need to add in an additional 300 per year in order to make up the final value Now instead of putting money into his college fund on his 5th birthday I buy him a horse instead How does this affect the calculations Instead of breaking the problem into two series that somehow must add up to a total present worth we can just recognize that we still have 1 uniform series but we also have a single negative value equal to A occurring at year 5 Pt Pa P5 29372 APAin FPFin 29372 APA818 FPF818 now F A 29372 APA818 PF818 29372 A93719 02502 A 3220 Thus we need to increase yearly payments by an additional 86 in order to account for the missing year39s worth lfl wanted to minimize the increase in yearly payments should I give him the pony on his 5th birthday or 16th birthday Why 16 Money placed into the account at earliertimes compounds interest every year which is why it is to your bene t to start a retirement account ASAP Think ofthe pony in terms of negative money On his 5th birthday she39ll have 13 years of negative interest compounded On his 16th birthday she39ll only have 2 years of negative interest compounded
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