Cost Engineering and Analysis
Cost Engineering and Analysis C E 406
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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 71 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
Replacement Study Most assets have finite lifespans Determining when to replace an asset is based on the expected lifespan as well at the projected salvage values and annual operating costs Reduced performance physical deterioration reduced reliability or productivity ofan asset Altered requirements new requirements of accuracy speed or other speci cations cannot be met with the existing equipment Obsolescence rapidly changing technology makes currently used asset perform acceptably but much less productively than equipment coming available AW value annual worth value is often called the EUAC equivalent uniform annual cost Economic service life number ofyears at which the lowest AW of cost occurs When conducting a replacement study take the point ofview ofan outsider evaluating both the existing and alternative asset as if neither is currently owned with the services provided by each available at a cost equal to the market value A replacement study is based on AW calculations and as such the following assumptions are included Services provided for inde nite future 2 The chosen alternative isthe best available now and in the future to replace the existing asset and when it replaces the existing asset it will be repeated for succeeding life cycles 3 Cost estimates for every life cycle ofthe new asset will be the same Economic Service Life The ESL is the number of years n at which the equivalent uniform annual worth AW of costs is the minimum considering the most current cost estimates over all possible years that the asset may provide a needed service 2 types of cost Capital Recovery initial investment less any salvage value Annual Operating Costs yearly cost to operate the asset which generally changes over time Total AW CR AW ofAOC AOC 800 year 0 1 2 3 4 5 Market value 10000 8000 5000 3000 2000 1500 i 10 Annual cost of ownership for 2 years AW of CR 10000AP102 5000AF102 338095 AW ofAOC 800 Total AW2 338095 800 418095 Annual cost of ownership for 5 years AW of CR 10000AP105 1500AF105 23923 AW ofAOC 800 Total AW5 239230 800 31923 What this tells us is that if we only own the car for 2 years before selling it costs us 4200 year lfwe hold onto it for longer say 5 years it only costs us 3200 year Thus on an annual basis it is cheaper for us to hold onto it for a longer period oftime We could even plot the total AW as a function of how long we hold onto our asset AW of costs What the graph shows us is that capital recovery costs increase in the 2quotd year because of the relatively high drop off in the resale value between year 1 and 2 and then decrease over time as our initial expense is spread out over a longer period oftime However in most cases the annual operating costs are not fixed values Our car example over time the factors that go into AOC gasoline costs increase insurance decrease maintenance increasel Thus to properly account for this effect we have to examine each year39s AOC and annualize it over the life of ownership This becomes a bit more complicated It Toa1AW PAPikSAFik ZA0CPFij APz39k 11 The value P represents the initial investment or the current market value Consider example where you buy a car Capital Recovery yearly PV ofAOC yr Value Current Market AOC Cum AOC AW ofAOC Total AW 159 41 781 Current market value ofyour car is 15000 The blue book values after years 13 are listed The annual operating cost is also presented This includes cost of gasoline oil changes major and minor maintenance Assume an interest rate of 5 per year When is the best time to trade this car in for a new car Determine Economic Service Life This is AW 80 ALL TERMS MUST BE IN AW The capital recovery is equal to the PAP 8 NF So for year 1 15000AP51 11 000AF51 15000105 11 0001 4750 year 2 15000AP52 8000AF52 1500005378 800004878 41646 This represents the annual cost of purchasing and selling the asset over a given lifespan Next we consider the annual operating costs Typically these increase with the life ofthe asset Our car has a big jump in year 3 because of major repair work that is expected These calculations are a bit tricker While these are annual costs we need to account for the fact that they change yearly What we do is determine the PV of each year39s AOC then nd the cumulative AOC for a given year and annualize that year 1 AOC 1000 PV 1000PF51 100009524 9524 Cum AOC for year 1 is 9524 AW ofAOC through year 1 9524AP51 9524105 1000 The total AW ofthe vehicle in year 1 is 1000 4750 5750 Year 2 AOC in year 2 is 1200 In order to determine the total AW ofthe AOCs we must nd equivalent values and then annualize them So we determine the PW of this term 1200PF52 120009070 10884 This is the AOC for year 2 We add the AOC for year 1 to get total AOC over 2 years 10884 9524 20408 This value is then annualized over the 2 year period 20408AP52 2040805378 10976 Total cost then is 10976 4164 5262 We repeat the same calculations for year 3 We can continue this for as many years as we estimate we will keep the asset 7000 00 5000 00 5000 00 4000 00 3000 00 2000 00 1000 00 0 00 Ifwe plot the data for each ofthe 3 years we see that there exists a minimum in year 2 Thus the Economic Service Life of this asset is 2 years from now And any replacement asset will have to beat this AW cost value of 5265yr What this means The economic service life of an asset isthe n value we have been using all along Equipment is retained until it becomes too expensive The shape of the curve drops initially as our investment is spread out over the time but then increases as the cost of maintenance increases with age The AW at the ESL is the actual cost of owning the equipment for the lifespan n ESL Let39s work problem 1127 2 years ago a GPS tracker was purchased for 1500000 Estimated salvage value was 50000 after 9 years Currently remaining life is 7 years with AOC of 75000 per year A new unit costs 400000 with a 12 year life and an AOC of 50000 and a salvage value of 35000 12 MARR What is the minimum tradein value necessary NOW to make the challenger economically advantageous First let39s look at the annual cost of ownership Assuming we keep both products for their full lives Current AW 15MAP129 75000 50000AF129 3531344 New AW 400000AP1212 50000 35000AF1212 1131244 Obviously the new system is cheaper to operate Given the choice initially we would have bought it However we didn39t have the choice so we make do The question is what value should we sell the current system at NOW so that the future cost of ownership between the two options is equal What we calculate is the REPLACEMENT VALUE We set the AW challenger AW defender and solve for the current market value NOW that makes the two equal for the remainder of their service lives The current system was n 9 We have owned it for 2 years so it really is n 7 Our equation for total AW CR AOC Let39s say that RV is the replacement value NOW that would make the 2 options equally attractive RVAP127 50000AF127 75000 400000AP1212 35000AF1212 50000 RVAP127 43080 RV021912 43080 RV 196606 Thus if we can sell it NOW for 196606 we are better off going with the new system OthenNise we keep it and see what we can sell it for next year How do we know this 196606 is the right value Let39s examine the cost of ownership for 7 years ifthe current market value is 196606 AW 196606AP127 75000 50000AF127 113124 orthe same annual cost of owning the new system Essentially the replacement value tells us If we bought the defender today at the replacement value it would cost us the same as if we bought the challenger today Annual Cost of ownership and replacement value are fonNard looking tools they don39t look at the history of ownership or how much it originally cost us To quote Rafiki from Lion King It doesn39t matter it39s in the past What we spent on an asset is irrelevant All that is relevant is how much is it worth today how much it will be worth when we sell it and how much it costs to operate on a yearly basis We can perform the same analysis for a car Let39s make it simple Your current car has a blue book value of10000 and you39re paying 1000 a year to maintain it You figure it has 5 more years of life in it before you will get rid of it at that time you probably can sell it for 2000 However you see this sweet ride which you can lease for 250 a month for 3 years with AOC from similar models being around 300 The interest rate is 5 per year compounded yearly this is how much your money is earning in your money market Based on the total AW of costs will it be cheaper for you to lease the new vehicle AW defender 10000AP55 2000AF55 1000 294780 yr AW challenger 25012 300 3300 yr 80 this tells us you are better off sticking with your current car If however RVAP55 2000AF55 1000 3300 RV 3300 1000 2000AF55AP55 11525 So if you really wanted the new car and you could sell your old one for 11525 then it is a worthwhile deal Replacement Study Rules For a new replacement study Conduct AW analysis of both existing and alternative assets Replace immediately ifAW alternative cheaper than AW existing and keep for it39s ESL Otherwise keep existing asset and plant to use for it39s ESL After 1 year check all estimates and repeat analysis if values have changed and reanalyze lfa replacement study is conducted over a speci ed study period we have to make special considerations as it violates our assumptions we made earlier In particular what occurs ifthe existing asset39s service life is shorter than the study period We must then include the cost from the end of ESL to end of study period ofour new asset For example if we have a 5 year study period and our existing asset only has a 3 year service life remaining Option Existing Alternative W 3 years 2 years X 2 3 Y 1 4 Z 0 5 Inflation Inflation is an increase in the amount of money necessary to obtain the same amount of product or service before the in atted price was present Inflation occurs because the value of currency has changed it has gone down in value in t1 in t2 in ation rate between t1 and t2 in t1 constantvalue dollars or today39s dollars in t2 future dollars or thencurrent dollars iffrepresents the inflation rate per period and n is the number of time periods between t1 and 3amp1 35 t2 1fquot or t2 t1 1 0quot These equations are used to produce cost estimates and the Consumer Price Index From this you could predict how much something would cost in the future if you knew the inflation rate i real or in ationfree interest rate what we39ve been using all along f inflation rate rate of change in value of currency i in ationadjusted interest rate the interest rate that is adjusted to take in ation into account it is the market interest rate orthe in ated interest rate Present worth calculations PF 1i L 1f 1z39 7 1z39fz39f ififif 1 FPFz39fN f PF Thus we now have an expression that takes into account the effects of in ation fthe interest rate were 10 and the in ation rate were 4 the inflated interest rate would be if 01 004 01004 0144 or 144 Example A former student wishes to donate to the department39s scholarship fund He has three options A 60000 now B 15000 per year for 8 years beginning 1 year from now C 50000 three years from now and another 80000 ve years from now Which is the best deal from the point ofview of the department to maximize the buying power ofthe dollars received assuming a real 10 interest rate per year and in ation is predicted to be 3 per year We have 2 options to solving this problem 1 easiest way solve for present worth using the in ated interest rate if 2 convert all dollar values to present value dollars using f and then determine present worth using i Both are correct f 01 003 01 003 0133 Pwa 60000 PWb 50000 PA 133 8 71262 ch 50000 PF 133 3 80000 PF133 5 77227 Future Worth Calculations a future worth F can one of4 interpretations 1 The actual amount of money that will be accumulated by time n Actual amount accumulated if we are quoted a market rate of 10 that rate includes in ation 2 A F P FP if n The purchasing power of the actual amount accumulated at time n but stated in today39s dollars Constantvalue with purchasing power n 1 PFPz fn FP1zf n n 1f 1f This represents recognized that the value in the future PFP is worth less than now 11fquot Thus while we account for in ation we compare it to today39s purchasing power The number of future dollars required at time n to maintain the same purchasing power as today39s dollars that is in ation is considered but interest is not Future Amount required no interest This is the case if someone asks how much will something cost in n years Since we39re just looking at it39s cost we don39t include interest only in ation F P1fquot PFPfn The number of dollars required at time n to maintain purchasing power and earn a stated real interest rate accounts for both price increase inflation and time value of money interest lfthe growth of capital is to keep up funds must grow at a rate equal to or above the real interest rate i plus a rate equal to the in ation rate lfwe are talking about a business MARRf i i if F P FP MARRf n
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