Cost Engineering and Analysis
Cost Engineering and Analysis C E 406
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CE406 Engineering Economics and Administration 083005 Chapter 1 Foundations of Engineering Economy Why study engineering economy As engineers our main task is to design and create Engineers constantly make decisions choosing between various alternatives Many times those decisions involve an investment of funds referred to as capital Available capital is always limited Engineering economy involves formulating estimating and evaluating the economic outcomes when alternatives to accomplish a de ned purpose are available Consider these examples Your contracting firm is going to bid on a bridge rehabilitation project You will need 5 cranes in orderto complete the work Your firm only owns one Should you chose to buy or rent the remaining 4 cranes Again this depends on many factors such as o Rentallease cost of the individual cranes and the time necessary to complete the project 0 Whether you have the current available capital to purchase 4 cranes outright of whether a loan of some sort is needed o If you buy the cranes can you recoup your initial costs by selling the 4 after the project How much will the cranes be worth at the end of the project 0 Expected revenue from project Again after careful analysis of these and other factors the appropriate decision can be made In this case the appropriate decision will be the one that costs the company the least amount of money You wish to purchase a house The house costs 600000 You have a mutual fund with 100000 How do you finance your purchase A number of factors go into this decision some of which are listed below 0 What types of loans are available and how much interest do they charge 0 30 year fixed 0 5 10 year Interest only loans 0 1 loans and more 0 How much does it cost points to buy down the interest rates on these loans 0 What are the tax ramifications of the various mortgage options Interest payments are tax deductible Page 1 of 13 CE406 Engineering Economics and Administration 083005 0 How much interest does the mutual fund earn 0 How long are you going to live in the house and what will the future value of the house be when you sell Ultimately after carefully analyzing the above factors the appropriate amount of cash from your mutual fund will be used as a downpayment and a mortgage of some sort will be taken out Typical Engineering involvement in Economic decisions can be classified as either 0 Equipment or Process Selection which piece of equipment shall we purchase for a given purpose Which choice generates the largest saving or largest return on investment 0 Equipment Replacement decisions involving considering the expenditure necessary to replace wornout or obsolete equipment Eg after several years company switches from metal parts to plastic parts and needs to replace the metal stamping equipment with plastic injection 0 New Product or Product Expansion investments in this category increase company revenues if output is increased Eg lncrease output of existing production or distribution facilities 0 Cost Reduction attempts to lower a firm39s operating costs Buy equipment to perform an operation currently done manually or spend money now in order to save money later 0 Improvement in Service or Quality techniques developed are also applicable to various economic decisions related to improving services or quality of product Fundamental Principles of Engineering Economics Principle 1 A nearby penny is worth a distant dollar Money has a time value associated with it Because we earn interest on money received today it is better to receive money earlier than later Principle 2 All that counts are the differences among alternatives Decision is based on differences among alternatives All that is common is irrelevant to the decision Any economic decision is no better than alternatives considered Each choice involves giving up something Principal 3 Marginal revenue must exceed marginal cost Effective decision making requires comparing the additional costs of alternatives with the additional benefits Each alternative must be justified on its own economic merit before being Page 2 of 13 CE406 Engineering Economics and Administration 083005 compared with other alternatives Any increased economic activity must be justified on the basis of the fundamental economic principle that marginal revenue must exceed marginal cost Marginal revenue means additional revenue made possible by increasing the activity by one unit Marginal cost is similar Principal 4 Additional risk is not taken without expected additional return For delaying consumption investors demand a minimum return that must be greater than the anticipated rate of inflation of any perceived risk If they don39t receive enough to compensate for anticipated inflation and the perceived investment risk investors would purchase whatever goods they desired ahead of time or invest in assets that would provide sufficient return to compensate for any loss from inflation or potential risk The problem solving approach we will emphasize consists of the following steps Understand the problem and define objectives eg a manufacturing process requires extensive annual maintenance 2 Collect relevant information what is the production rate how much does it cost to operate yearly what did it cost when new if we scrap it how much can we get for it how much do new processes cost 3 Define the feasible alternatives keep as it repair and upgrade install an entirely new process 4 Identify the criteria for decision making what time frame do we need to recover our investment how much should we make with this investment do we have to upgrade 5 Evaluate each alternative using sensitivity analysis using previously determined cash flow values compare each alternative to determine which is the most economically sound 6 Select the best alternative 7 Implement the solution and monitor the result over time our new production line will degrade at some point it may be in our best interest to replace it with newer technology Definitions Capital wealth in the form of money or property also referred to as assets Cash Flow actual cash amounts which are receipts and disbursements Time value of money change in amount of money over a given time period All money posses a time value The most important concept in engineering economy Eg a small refrigerator for keeping your beer in costs 100 now You could buy it and you39d be broke If you can put the money in a saving account earning 6 after a year you39d have 106 However inflation tends increases the cost of the ref by 4 Page 3 of 13 CE406 Engineering Economics and Administration 083005 per year After 1 year you39d have 2 left over If inflation were 8 you wouldn39t have enough to buy it Ultimately this means that rate at which interest is earned must be higher than the inflation rate Inflation rate at which the general level of prices and goods and services is rising and subsequently purchasing power is falling Example of time value of money lnterest manifestation of the time value of money Interest is the difference between an ending amount of money and the beginning amount It is always positive If the difference is zero or negative there is no interest Cost of having money available for use Principal initial amount of money in transaction involving debt or investment Interest rate measures the cost or price of money and is expressed as a percentage per period of time interest period determines how frequently interest is calculated Number of interest periods a specified length marking the duration of the transaction Plan for receipts or disbursements a particular cash flow pattern over a specified length of time future amount of money results from cumulative effects of the interest rate over a number of interest periods lnterest can be considered from two perspectivesinterest paid and interest earned If you put money in the bank you earn interest lfyou borrow money you y interest The calculations for determining interest are identical just the point of view is different Borrowed funds Interest owed amount owed now Principal Interest rate interest accrued per unit time I Principal 100 The unit time is called the interest period and is typically 1 year unless specified For example if you borrow 10000 at 7 interest per year how much do you owe after 1 year Interest rate interest accrued per unit time principal 100 007 interest accrued 10000 700 interest accrued lnterest original amount amount owed now 700 10000 10700 Simply put we can say that the future value F P Page 4 of 13 CE406 Engineering Economics and Administration 083005 Conversely if you invest the money you earn interest Invested funds Interest earned total amount now Principal Rate of Return interest accrued per unit time I Principal 100 We use ROR instead of interest rate to signify that your investment is returning money to us For example you have a money market account with a rate of return of 5 per year If the account currently has 1000 in it how much did it have last year This is a bit trickier We can39tjust apply the equations straight because we don39t know the original amount or how much interest we earned Amount now lnterest earned Principal lnterest earned ROR Principal We39ll say X amount of original deposit The interest earned in 1 year is X005 Thus Total amount now original amount interest earned 1000 X X005 1000 105 X X 95238 How much interest was earned lnterest earned 1000 95238 4762 Thus far we have looked at interest accumulating over a single unit time of 1 year What happens if we consider greater times Say 2 years or 5 years or 10 years 5 months We have to specify how the interest accumulates Simple interes calculated using the principal only ignoring any interest accrued in preceding interest periods The total simple interest over several periods Interest interest rateprincipalnumber of periods I i P N Future amount owed therefore is FPP1iN Returning to previous example if you borrow 10000 at 7 per year simple interest how much would you owe after 3 years Interest 100003007 2100 Total owed Principal lnterest 10000 2100 12100 Page 5 of 13 CE406 Engineering Economics and Administration 083005 However not all interest is simple interest In fact many lenders use compound interest Compound interest interest accrued for each interest period is calculated on the principal plus the total amount of interest accumulated in all previous periods Thus if you borrow the same 10000 at 7 per year compound interest after 3 years you would owe 1st year lnterest 10000 007 700 Total amount owed 10000 700 10700 2nd year lnterest 10700 007 749 Total amount owed 10700 749 11449 3rd year lnterest 11449 007 80143 Total amount owed 11449 80143 1225043 which is 15043 greater than the simple interest rate If we consider P as principal and i as our interest rate we could expand it as yr 1 P Pi P1i yr 2 PPi PPii P Pi Pi Pi2 P 2Pi Pi2 P1 2i i2 P1i2 yr 3 PPi PPii PPi PPiii P1i3 We can simplify the calculations by using the general form Future Total due after a number of years principal1 interest ratenumberofyears F P 1 iN Total due 10000 1 0073 Total due 1225043 The vast majority of loans or interest earning accounts are compound That is why people who invest early on have such tremendous gains later in life However occasionally you can find simple interest loans EXAMPLE Compare Compound and Simple interest 1626 Peter Minuit bought Manhattan Island for 24 If he had invested this money in a saving account at 8 interest how much would it be worth in 2007 P 24 i 8 per year N 381 years Find F for simple F P1iN 24 1 008381 75552 compound F P1iN 24 1008381 13021531990901500 Page 6 of 13 CE406 Engineering Economics and Administration 083005 Question You borrow 10000 at 5 compound interest what simple interest rate would be equivalent if you plan to pay off the entire loan at the end of the 5th year F P 1i N F 10000 1055 1276282 For simple interest F P P i N new equation i FPP N 127628210000100005 00553 Thus a 553 simple interest rate for 10000 over 5 years is equivalent to a 5 compound interest rate for the same value and period This is ONLY valid for the 5 year period If we chose a longer or shorter period then the equivalent simple interest rate will be different Terminology amp Symbols P Present value of money The value of money at a time designated as the present or time 0 also called Present Worth PW present value PV capitalized cost CC F Future value of money The value or amount of money at some future value also called Future Value FV or Future Worth FW A Series of consecutive equal endof period amounts of money Also called the Annual Worth AW n number of interest periods days months years i interest rate or rate of return per period of time Commonly refers to compound interest unless specified day month year t time stated in periods days months years Difference between n and t and how it relates to i We carry a 1000 balance on our credit card charging 125 interest per month After a year how much do we owe The time we are concerned with is one year thus t 1 year The period in i is per month i 125 per month n is the number of interest periods In this case there are 12 months in year so n 12 For the first several chapters in the book we will focus on the much simpler case where n t and i have the same units years or per year Cash Flows In order to visualize Engr Econ Problems we use the concept of a cash flow diagram similar to a freebody diagram Page 7 of 13 CE406 Engineering Economics and Administration 083005 Cash flow diagrams are constructed on a horizontal axis with unit interest periods indicated Arrow direction up arrow inflow income receipts down arrow outflow disbursements expenses often cash flows at the same time are combined into a single arrow representing a net cash flow 20000 income with a 10000 expense results in a 10000 net cash flow The arrow direction is always in reference to a particular perspective your wallet vs your bank account would have opposite directions if you withdrew money from the atm We will use cash flow diagrams when dealing with recurring cash flows On the diagram t0 is the present or any time specified as the present and t1 is the end of time period 1 Endof year convention places cash flows at the end of the years period Interest is assumed to accrue at the end of the period and not in between for now at least We will deal with this more complicated case later in the class Thus our cash flow diagram for Manhattan lsland If we deposit 24 in an account earning 8 interest how much is the account worth in 2007 381 years later 0 24 381 130triion i 8 A series of consecutive equal end of period amounts of money yr mo Assume wish to stop working for 5 years I desire to withdraw 10000 a year for the first 3 years and 20000 per year for the final 2 How much money do I need to deposit today to make this happen ifl can earn 8 in a money market A220 000 A A l l l l l l l l l l 4 5 lt4044 H4444 gt N244 ll be gt Page 8 of 13 CE406 Engineering Economics and Administration 083005 P We have to always remember the convention that the cash flow is at the end of period unless speci ed More complicated example Plant chemical plant superintendant arranges to purchase an additive through a 5year contract for 7000 a year starting 1 year from now Afterwards he expects the price to increase 12 a year for the next 8 years Additionally an initial investment of 35000 was made to prepare the site What is the present worth of these expenses assuming a rate of return of 15 Draw cash flow diagram 0 35000 1 7000 2 7000 3 7000 4 7000 5 7000 6 7840 El Series1 7 87808 8 98345 9110146 10123364 11138168 12154748 13173317 Cash flows can occur anytime within a period but by convention are represented at the endof period Equivalence Equivalent terms are used to transfer from one scale to another Eg 1 m 100 cm 1000 mm 394 inches All values are equivalent We can do the same thing with money If the interest rate is 6 then 100 today is worth 106 in 1 year Thus the 2 values are equivalent So whether you invest the 100 for a year or you get a gift of 106 in a year there is no difference Page 9 of 13 CE406 Engineering Economics and Administration 083005 This is a fundamental concept of engineering economy and we will use to to solve all of our problems Essentially what it states is that a cash flow whether a single payment or series of payments can be converted into an equivalent cash flow at any point in time What we will do in this class 0 convert single payments into equivalent values at different times eg the 24 manhattan island problem converts a single value into it39s equivalent value 381 years later by accounting for the interest rate that could be earned or its time value of money Extend this analysis to account for series of payments How much money is an account worth if you made payments of 500 every year for 20 years Ultimately this method will be used to solve engineering economy problems Given two or more alternatives with different cash flows we can determine the equivalent value of each at a common time We can then chose the option which has the greatest value or lowest cost Equivalence Statements Equivalence like other engineering calculations involving units requires a common unit or in our case a common time basis In many cases that time basis can be arbitrarily chosen beginning of project end of project 20 years from now as long as it is equal for all alternatives Equivalence depends on the interest rate A change in rate changes the values and requires recalculation Equivalents typically involves converting multiple cash flows into a single value Following our idea of net cash flow if we convert each cash flow to an individual equivalent value that occurs at the same time we can simply add them all up to get the over all cash flow Regardless of point of view equivalence is maintained Simple problem I offer you either 10000 today or 11000 in one year What interest rate would make these two values equivalent Present interest F 10000 interest 11000 interest 1000 Interest rate interest accrued per unit time original amount 100 interest rate 100010000 10 Page 10 of13 CE406 Engineering Economics and Administration 083005 Thus if we invest our money in an account that earns 10 per year it makes no difference whether we take 10000 now or 11000 one year from now We use the concept of equivalence extensively throughout this class In order to compare different options we need to compare them on an equal footing For example one design might cost 10000 every 10 years to maintain while another design might cost 8000 every 8 years Which should we go with Given the different time frames we must determine an equivalent value for each design that allows for direct comparison Say the current value of all maintenances costs over the next 40 years Minimum attractive rate of return For investments to be profitable you expect to receive more money that the amount invested a rate of return of rate of investment must be realized When making engineering decisions alternatives are evaluated based being able to exceed what is know as the Minimum Attractive Rate of Return MARR which is higher than the rate expected from a bank or other safe investment such as US Treasury Bill Thus in order for an alternative to be selected by must have a ROR greater than the MARR How do we determine the MARR In order to invest in something you must have available capital money also referred to as capital funds or capital investment money There are two ways to develop capital equity financing use of your own funds from cash on hand stock sales or retained earnings or savings investments eg 8 in a money market debt financing borrow money from outside source and repay the principal and interest according to a schedule from bonds loans mortgages venture capital credit cards eg a loan for 4 The cost of capital is either the interest rate on the loan that is taken out or it is the loss of interest incurred by taking money from the savings account Most investments involve some of both For example buy a house involves some equity financing 20 of the house price in cash is typical and some debt financing 80 in the form of a mortgage In this case the Weighted Average Cost of Capital is simply 022 0 8056 488 assume savings account earns 2 and mortgage is 56 Page 11 of13 CE406 Engineering Economics and Administration 083005 Thus in order for an investment to be profitable ROR gt MARR gt cost of capital Estimating Doubling Time and Interest Rate Often we want to estimate the number of years n or rate or return i required for a single cash flow amount to double in size lfl borrow 20 at 5 how many years will it be before I owe 40 As long as either n ori is known the other can be estimated from igtltn72 So that Estimated i 72 n Estimated n 72 i returning to the problem estimated n 725 144 years This is only an approximation for Compound Interest For simple interest an exact solution is found if a rule of 100 is used Homework 111114119121125136148 Page 12 of 13 CE406 Engineering Economics and Administration l A A 01 0 083005 CE 406 I CEM 410 Homework 1 due July 17 2006 A broadband service company borrowed 2 million for new equipment and repaid the principal of the loan plus 275000 interest after 1 year What was the interest rate of the loan A company reported that itjust paid off a loan it received 1 year earlier If the total amount of money the company paid was 16 million and the interest rate on the loan was 10 per year how much money did the company borrow 1 year ago At what interest rate would 100000 now be equivalent to 80000 one year ago A local bank is offering to pay compound interest of 7 per year on new savings accounts An ebank is offering 75 per year simple interest on a 5year certificate of deposit Which offer is more attractive to a company that wants to set aside 1000000 now for a plant expansion 5 years from now Companies frequently borrow money under an arragement that requires them to make yearly payments of only the interest and then pay the principal of the loan all at once A company borrows 400000 for 3 years at 10 per year compound interest under such an arrangement What is the difference in the total amount paid between this arrangement and one in which the company makes no interest payments until the loan is due and then pays it off in one lump sum Five separate projects have calculated rates of return of 8 11 124 14 19 per year The company plans to use a mixture of 25 equity and 75 debt financing to finance the project The cost of capital for company funds is 18 per year and borrowed money currently costs 10 per year If the MARR is established at exactly the weighted average cost of capital which projects should she accept Page 13 of13