Cost Engineering and Analysis
Cost Engineering and Analysis C E 406
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Date Created: 10/05/15
CE406 Chapter 6 100405 Annual Worth Analysis Previoust we used Present Worth analysis to select between alternatives When given alternatives that exist for various lifespans we had several ways to deal with them Least Common Multiple normalize the costs by examining over lifespans that span the same time period repeat all cash flow identically over each lifespan Study Period pick a study period less than the lifespan and only examine costsrevenue over that period include end of life costs salvage value refunds etc Capitalized costs used to determine cost of an item that will last forever Annualize recurring costs over their frequency then sum up all annual costs i to get capitalized worth In all of these methods we determined the Present Worth of a projects and used that to compare Another and more intuitive method is to use the Annual Worth AW turns all cash flows into into annual values This allows direct comparison between costs and revenues on a yearly basis AW PWAPin FW AFin in this case n is the of years for equalservice comparisons either the LCM or the stated study period of the PW or FW analysis However the advantage ofAW analysis is that LCM or study period is NOT needed AW value has to be calculated over only one life cycle However when compared with PW or FW services provided are needed for at least the LCM selected alternatives repeat exactly the same as first life cycle all cash flows have same estimated value in every life cycle Example Consider we wish to lease office space example from last time Page 1 of6 CE406 Chapter 6 100405 First cost 15000 Annual lease cost 3500 Deposit Return 1 000 lease term 6 years Analyze this by PW over 3 life cycles and AW over 1 life cycle Assume i 15 Thus for location A we can write Pwa 15000 15000PF156 15000PF15121000PF156 1000PF1512 1000PF1518 3500PA1518 Pwa 45036 If we examine the AW of the cash flows over 1 life cycle AW 15000AP156 1000AF156 3500 7349 Let39s convert the PW determined previously into an AW AW PW AP1518 45036017102 7349 Thus we can say that the AW over 1 life cycle is equivalent to the PW determine over the LCM In general AW can be used anywhere PW or FW can AW most useful in situations where asset replacement and retention time to minimize overall annual costs are an issue as well as studies dealing with manufacturing costs where a costunit or profitunit is the focus Calculation of Capital Recovery and AW values Each alternative analyzed should have the following cash flow estimates Initial Investment P total first costs of all assets and services required to initiate the alternative If these take place over several years the PW is an equivalent initial investment Salvage value 8 terminal estimated value of assets at the end of their useful life S is 0 if negligible value negative if it will cost money to dispose of If study period is less than the useful life then Sis equal to the market value at end of study period Annual amountA this is the equivalent annual amount cash flows Often this is the annual operating cost Annual worth value for an alternative is comprised of Page 2 of 6 CE406 Chapter 6 100405 capital recovery CR for the initial investment P at stated interest rate equivalent annual amountA AWCRA In practice both CR and A are negative because they are costs A is determined from uniform recurring costs nonrecurring amounts PA and PF are needed to determine present worth amount AP is used to convert this amount to the A value CR the equivalent annual Lst of owning an asset plus the return on the initial investment When capital P is use for an asset there is a time value of money associated with the time it is committed to it AP factor is used to convert P to an equivalent annual cost In addition AF is used to convert any salvage value to an annual value Thus CR PAPin SAFin This is the equivalent annual cost of owning the asset CR is a COST therefore it must be negative However if we write our P39s and 839s with the correct sign and respectively then we do not need the sign in the book Page 3 of 6 CE406 Chapter 6 100405 Example Design a DNA sequencer to analyze monkey DNA faster Estimated that our investment of13 million will be broken up with 8 M now and 5 M at the end of the first year Annual operating costs of our DNA sequencer starting year 1 are 09 M per year We expect the sequencing will take 8 years and at the end of that time the unit will be mostly obsolete and can be salvaged for 05 M What is the AW of the system if the MARR 12 per year diagram 8 now 5 yr 1 09 yr 18 05 yr 8 AW CR A what is CR CR PAPin SAFin what is P equivalent initial investment at time 0 so P 8 5PF121 What is S salvage value at end s 05 CR 8 5PF121AP128 05AF128 CR 247 This tells us that for each and every year for the 8 years the equivalent total revenue from our DNA discoveries MUST be at least 247M just to recover the initial investment plus the required rate of return Just the cost of the machine On top of this we have annual operating costs AOC of 09 Thus the annual worth is AW 247 09 337 M This value is the amount of revenue our discoveries must make each year in order to cover the initial investment ROR as well as the AOC Page 4 of 6 CE406 Chapter 6 100405 Guidelines for evaluating alternatives using AW analysis for mutually exclusive alternatives 1 alternative AW 0 MARR is met or exceeded 2 or more Chose the lowest cost or highest income AW value Note In all of the equations given thus far the signs differ from the book The book prefers to work in absolute values and then assign the sign after the fact I prefer to work in actual values costs revenues Example 65 Chemical engineer considering 2 types of pipes in an process Small pipeline costs 17 M to install and has operating costs of 12000 per month The large pipeline costs 21 M and has a 8000 monthly operating cost Which pipe size is more economical assuming 1 per month interest rate Assume salvage value is 10 the first cost for each pipeline at the end of the 10year project CP PP 1 month n 10 years 12 moyr 120 months i 1 month Small pipeline AW CR A CR PAPin SAFin CR 17AP1120 0117AF120 CR 23656 AW 23656 12000 35655 Big pipeline AW CR A CR PAPin SAFin CR 21AP1120 0121AF120 CR 29222 AW 29222 8000 37221 Based on the annual worth of the problem the smaller pipeline will cost less on a yearly basis than the bigger pipeline Page 5 of 6 CE406 Chapter 6 100405 AW of a permanent investment What is annual worth equivalent of capitalized cost Amount yearly that extends indefinitely CC AWi CC capitalized cost AW annual worth AW CC i Capitalized costs are the present worth of infinite projects AW P i 3 options to increase capacity in a drainage canal this increased capacity should be available indefinitely A Dredging equipment cost 650000 10 year equipment life 17000 salvage value AOC of dredging is 50000 AOC of weeding is 120000 B line the canal lining 4000000 yearly maintenance 5000 repairs every 5 years 30000 C new pipeline along different route initial cost 6000000 annual maintenance for right of way 3000 50 year pipeline lifespan assume i 5 year AW analysis is over 1 lifespan of the project ProjectA 10 years AW 650000AP510 17000AF510 50000 120000 252 824 Project B 5 years AW 4M 5 5000 30000AF55 210429 Project C 50 years AW 6MAP550 3000 331 680 Page 6 of 6