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Design of Thermal System

by: Rowan Spinka DVM

Design of Thermal System MAE 412

Rowan Spinka DVM
GPA 3.62

Stephen Terry

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Stephen Terry
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This 25 page Class Notes was uploaded by Rowan Spinka DVM on Thursday October 15, 2015. The Class Notes belongs to MAE 412 at North Carolina State University taught by Stephen Terry in Fall. Since its upload, it has received 10 views. For similar materials see /class/224017/mae-412-north-carolina-state-university in Aerospace Engineering (AE) at North Carolina State University.

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Date Created: 10/15/15
MAE 412 Design of Thermal Systems Course Notes Engineering Economic Topics Topics 1 2 3 0 9 0 Introduction De nitions Cost Estimates Visualizations a Fixed vs Variable Cost b Pro tLoss Diagram 0 Cash Flow Diagram d Cumulative Cash Flow Diagram Time Value of Money Present Worth and Future Worth a Formulae b Analysis Annual Cash Flow Analysis Payback Period Annual Rate of Return Bene tCost Ratio Analysis In ation and Price Change 1 Introduction One difference between science and engineering is that scientists are only concerned with the physics of a problem Engineers must always be concerned about the cost of a problem even to the point of deciding whether it is cheaper to let a problem continue or to x it A personal example is periodic maintenance of your automobile Do you spend the money to change the oil and oil lter periodically or do you just top up the oil level occasionally and leave the car alone If you are going to get rid of the car in a few years you may decide not to spend the money on preventive maintenance This will minimize your expenditures on the car although it will take its toll on the next owner An engineering rm is always faced with making a choice between alternatives in a project and the ultimate goal is to select an alternative that will minimize costs over the life of the project The project life does not just include purchase and installation of equipment or manufacture and sales of products but includes the entire projected life of the equipment or product line In this section of MAE 412 we will examine a set oftools that can be used to analyze alternatives for minimum project cost One of the most important aspects of any project is the schedule of expenditures and anticipated income or savings due to the project For example if you are responsible for developing the manufacturing facilities for a new product line is it better to spend more money initially to purchase longlived equipment or to purchase cheap equipment that reduces your initial outlay but will have to be replaced after the production line has run for a few years The advantage of the latter approach is that sales of the product will pay for the equipment more quickly and you will not have as much money at risk at any given time Before we begin to look at the analytical tools available to us we need to de ne a number of terms that we will use The next section will present those de nitions Equot P 5quot 0 a 5 Lquot F 39quot F De nitions Fixed Cost cost unrelated to the level of production Example the mortgage on Broughton Hall which is independent of the number of students using the building The building might be closed for asbestos removal but the mortgage must still be paid Variable cost cost related to the level of production Example enrollment in MAE courses rises by 200 students We must hire additional faculty until they graduate and there will be additional expenditures for project courses like MAE 416 and more xeroxing cost electricity use in the building etc Marginal cost this is the additional cost for one more unit of production Total Cost Sum of total xed costs plus total variable costs This must be de ned for a speci c time period such as the length of a project or another period of interest Average Cost per unit Total Cost divided by total production Sunk Cost Money spent as the result of past decisions These costs are ignored in engineering analyses because they can t be changed therefore they can t impact choices between alternatives The only time they are of interest is when the engineer is deciding whether a project should be abandoned and must report the total cost to date A good example is the abandonment of three of the planned generating units at the Shearon Harris Nuclear Plant A tremendous amount of equipment had been purchased and much construction had been completed before the units were abandoned The decision to stop building the units was based on the lack of need for more generation capacity not the cost to date However the sunk cost had to be tabulated in order to decide how to recover those expenses Opponunity Cost This is the cost of using project resources in one activity instead of another For example if the MAE Dept is given 200000 more in the budget because of increasing enrollments should it be spent on more faculty or more classrooms Recurring costs expenses that occur at regular intervals such as your telephone bill Non recurring costs oneofakind costs occurring at irregular intervals such as replacing a broken windshield on a car Incremental Cost this is the difference in total cost between two alternatives This term is often used also as a synonym for Marginal Cost Life cycle Costing summing up all of the costs over the entire life of a project There is an interesting rule of thumb during the design phase of a project a the later L in a project that design changes are made the higher the costs and the corollary b decisions made early in the life cycle tend to lock in costs that are incurred later Interest this is the amount of money paid to borrow a sum of money It is expressed as the ratio of the extra amount paid diVided by the sum borrowed The payment is made or credited to the owner at the end of a speci ed period of time In simple language if you lend money to a bank ie put it in a certi cate of deposit they will pay you an additional sum at the end of a speci ed period of time such as a year They will tell you ahead of time the nominal rate of interest they will pay which is calculated once at the end of the year This is called simple interest If they tell you they are compounding the interest daily this means they take the nominal rate and diVide it by 365 and add that amount to your account each day Thus you get interest on interest as well as on your deposit This is called compound interest It may be compounded daily or monthly or quarterly Example 1000 deposited for one year at 6 nominal interest At the end of the year your account will have 1000 06 X 1000 1060 Ifthe interest is compounded monthly you will have 1000 X 1 0644 106136 Ifthe interest is compounded daily you will have 1000 X 1 06365365 106183 Every little bit helps 3 Cost Estimation Cost estimates are timeconsuming to develop and various phases of a project require different levels of effort For example if you wanted to decide how to heat your house your initial estimate will be very crude in order to decide whether to use an electric hot water heater a solar collector or a gas furnace All of these options require attached heating systems and controls You might go on the internet and look for crude estimates of cost for the three options Once you settle on a system you will re ne your numbers by looking at the cost of major pieces of equipment so that you can estimate a total installed system cost That can be used for budgeting purposes so that you can get a bank loan You always need to include a sum for contingencies because no project proceeds smoothly Once you have obtained the loan you then have to plan in detail the quantities of each material you need such as sizing ductwork and piping and determining total length required This would be out of place in the initial planning stage because you don t need that much detail for the alternatives that were not chosen It would be a waste of your time to develop all of the drawings and schedules of material for options that were not chosen Thus the planning effort for estimates is a function of the detail needed to make a decision at that stage of the project The three types of project cost estimates I have mentioned above are a Rough estimates initial crude estimates used for highlevel planning Semi detailed estimates used for project budgeting at the beginning Detailed estimates developed once a specific design is chosen These are the most accurate possible and contain all of the detail required to complete the project 09 At each stage the effort invested in estimating cost should be optimized with regard to the decisions that will be made at that point in the project Do not spend time on unnecessary details One quick way of developing estimates is to estimate by analogy what did it cost in previous projects 4 Visualizations ere are a number of tools that can be used to demonstrate your cost analysis of a project Some of them are shown be 0W a Graph of Fixed vs Variable Costs 5194 605 rumif To ml gt7 av 05an as 11th Vartle 425139 HA cos b ProfitLoss Chart 12 Exemal or In comc ProJLuoHou c Cash Flow Diagram The cash ow diagram is used to show the speci c points in time When income and expenditures occur It is a useful tool for recognizing the exposure and risk that a project has at any given time It is very important to create a cash ow diagram before other analysis is begun because the diagram contains all of the nancial information in visual form In come Income T r I V Z 3 4 I J nm 1 1 l c 39 year airEMAquot Ewen In Phil Exrend BT anvmiw ml a m came an expemlfhwu Miresz me summed aj 441 and n d Net Cash Flow Diagram in some l 2 3 A E MAI Trim Iaav as 5 Time Value of Money Would you be willing to lend an acquaintance 1000 for the next ve years And then receive just the 1000 back at the end Even if you knew the acquaintance would de nitely pay you back Probably not If you put the money in the bank you would have more money at the end of the ve years You d probably offer to lend the money if the acquaintance agreed to pay you interest You have just made a project decision between alternatives You could earn interest on the money in the bank or you could earn interest on the money from the loan or you could lend the money at no interest If you have assets such as money you want to earn the greatest return possible on them contingent on intangible factors such as the value of friendship How do we compare project alternatives We analyze the interest we can earn on the money we risk I ll introduce some basic terms here so that we have a language we can use for the analysis Principal this is the money we will invest or will borrow Simple interest interest calculated only on the principal Compound interest interest calculated on principal plus previous interest Variables P principal current amount of money i interest rate per period oftime decimal 5 005 n number of interest periods often in years months or days F amount of principal at some point in the future A annual amount Example of simple interest If we start with P and the interest rate per year is i and our money earns interest for n years then the total interest we receive is Total interest Pin Thus at the end of n years we have F P Pin P 1 in Example of Compound Interest If we start with P and the interest rate per year is I and our money earns interest for n years then the total sum we accumulate is Year0 F0 P Yearl F1 PPi P1i Year2 F2 F1F1iP1i2 Yearn Fn P 1 iquot If I wanted to have a certain sum F in the future and I wanted to set aside enough money now to accumulate it by earning interest at a rate of i per year I could invert the above equation to nd out how much I need now P F 1 iquotn This leads us to a whole series of expressions that can be used for analyzing the time value of money always assuming that interest is compounded Future Sum of Current Principal Invest a sum P at i for n years F P 1 iquot Example Buy a Cert of Deposit at the bank 10000 4 for 5 years F 10000 1045 12167 Current Principal to Get Future Sum Invest unknown P now at i for n years to receive F P F 1 i 1 Example Your parents want to invest P at 5 for 10 years to have 5000 to send your little brother to college How much do they need now P 5000 10510 3070 Future Worth of a Uniform Series Future sum if you invest A per year for n years at i F A 1in 1 i Example You invest 1000 each year for 10 years in a mutual fund that earns 7 per year How much will you have in 10 years F 1000 10710 107 13816 Present W01th of a Uniform Series If you receive uniform payments of A for n years and the available interest rate is I what is the current value of that stream of payments P A 1iquot 1 i1iquot Example you win a lottery for 25000000 and can be paid l000000year for 25 years If the available interest rate is 5 what is the equivalent present value P 1000000 105n 1 0510510 14090000 Uniform Payments to Accumulate Future Sum This is often called a Sinking Fund calculation You make periodic payments into an account that will be a specified size in the future in order to pay off a debt at that time What should those payments be A no 1 i 11 Example Your parents want to accumulate 50000 in 10 years for your little brother to go to college If available interest is 5 what should they set aside each year A 5000005 10510 1 3975year Uniform Payments to Pay Off an existing principal This is known as Capital Recovery because the lender to whom you make the payments is recovering hisher capital investment as well as receiving interest This is the expression you use to calculate mortgage payments A P 00 i 1 1 i 11 Example you apply for a 20 year mortgage of 100000 at 6 per year What are the monthly payments I 0612 per month 005 n 2012 240 months A 100000 00510524 1005Z4 1 71643 per month Present Worth of an Arithmetic Gradient Series This means the present worth P of a series of increasing payments each payment increasing by an increment G from the previous one Thus A is the base payment and A1 A A2 A G A3 A 2G A4 A 3G Define G increment increase from payment to payment PG G 1in i n 1 i2 1iquot for the periodic increases P A A 1in 1 i1in for the series of base payments P G 139 PA Example To control future costs you agree to a contract to increase annual maintenance charges by 4000 per year each year after the rst year s 100000 You can earn 5 on your money What equivalent current principal is required Break the series of payments into two series the rst of 100000 per year and the second is the increase of 4000 per year after the rst year A 100000 G 4000 i 05 n 10 PG 4000 10510 0510 105210510 126600 PA 10000 1051 1051051 772200 P PGPA 126600772200 898800 Present Worth of a Geometric Gradient Series This is the present worth of a series that increases geometrically with time For example you agree to pay a base amount for your rst payment then successive payments increase by a certain percentage calculated on each previous payment Define g rate of cash ow increase per year decimal percent Let A1 cash ow at end of year 1 P A1 1 1gquot 1i quot i g for i not equal to g P A1n 1 i for i g Example Power plant coal pulverizers currently cost 100000 to maintain It is expected that those costs will increase by 4 per year due to in ation You can earn 5 on your money What equivalent principal do you need now to represent 10 years of maintenance A1 100000 i 005 g 004 n 10 P 100000 1 1041 105391 05 04 912600 6 Present Worth and Future Worth There are a number of ways of comparing project income and costs I ll enumerate several here 1 Cash ow diagrams Compare income and expense without consideration of the time value of money Present worth of all income and expenses bring all sums to the present and sum Future worth of all income and expenses take all sums to a common future point in time 5 Uniform payments method turn all sums into an equivalent uniform series of annual payments or receipts over the project life 59 You have seen the use of cash ow diagrams We ll comment on the limitations of making comparisons of project alternatives without considering the time value of money when we get to the section on the Payback Method I want now to lay out the ground rules for making calculations of the Present Worth also called Present Value of the cash ow in a project We ll make some assumptions to simplify the analysis 1 All expenditures and income during a year are totaled at the end of the year for analysis purposes 2 Sunk costs are past history and do not affect our decisions Ignore them We want differences between alternatives 3 In ation is a factor and will be included 4 Income taxes are very complex We will ignore the topic because the engineer rarely is entrusted to make those calculations 5 Depreciation is another topic that is very complex and will be ignored Again the engineer generally has someone else do those calculations if they are needed during a project There is another problem that surfaces quite often in projects It is the question of comparing alternatives that have different useful lives For example in the current design project if we can select tubing with different expected lifetimes we need a method of accounting for the financial impact on the project If a heat exchanger is expected to last 50 years but one tubing alternative has an estimated life of 20 years and another has a life of 30 years what do we do a Least common multiple method perform the analysis over a period of time that is the least common multiple of their useful lives Thus the evaluation would extend over 60 years in the example b Evaluate over the longest life of the alternatives or some obvious lifetime In the above example we could do our evaluation over 30 years the life of the longerlived tubing However it makes more sense to select the 50year life of the heat exchanger as the evaluation period in the example To use that period we need to repurchase the tubing when the original tubing reaches the end of its lifetime Since the second set of tubing still has life left at the end of 50 years we will include its salvage value in our analysis If we present our analysis for the life of the heat exchanger our management will have the most comprehensive view of the future The method I have found most useful is to select the longest life option and do all calculations on that basis In that way I avoid projecting costs and income too far into the future I39d like to show you some simple examples of the use of the Present Value Method so that you can see how future sums are brought back into the present using the expressions I de ned earlier Example 01 Perpetual Motion Toys You have an opportunity to start up a new garage factory with a friend to make perpetual motion toys for engineers You have two choices as you consider what to do Choice A Initial expenditure of 7000 lst year pro t 5000 2quotd year pro t 4000 3rd year pro t 2000 43911 amp 5Lh years no Lpro t Salvage at end of 5 year 6000 Choice B Initial expenditure of 9000 1St amp 2n year no pro t 3rd year pro t 2000 43911 year pro t 4000 5Lh year pro t 5000 Salvage at end of 5Lh year 8000 Interest rate is 5yr Choice A Choice B Cash Flow Present Cash Flow Present Value Value Year End 0 7000 7000 9000 9000 1 5000 1051 4762 0 0 2 4000 1052 3628 0 0 3 2000 1053 1728 2000 1053 1728 4 0 0 4000 1054 3291 5 6000 1055 4701 13000 1055 10186 Total 10000 7819 10000 6205 Example 02 Bar Scanning Machine Your store is trying to decide which of two bar code scanners it should purchase and install The ideal bar code scanner would never make a mistake and require no maintenance If both scanners have lives equal to the 6 year analysis period which one should be selected Assume an 8 interest rate Uniform End of Life Alternatives Cost Annual Bene t Salvage Value Trueread Scanner 20000 4500 1000 Perfect Scanner 30000 6000 7000 True read Scanner PW ofbenefits PW of costs P4500 uniform series P1000 6 yr 20000 P4500 4500 1086 1081086 20804 P1000 10001086 630 PW net 20804 630 20000 1434 Perfect Scanner PW ofbenefits PW of costs P6000 uniform series P7000 6 yr 30000 P6000 6000 1086 1081086 27737 P7000 70001086 4411 PW net 27737 4411 30000 2148 Note that the salvage value is treated as a benefit of each alternative Since we re looking at net benefits the Trueread Scanner is our choice Example 03 Arithmetic Series You are considering two different gadgets to install on your home heating system to save money Both gadgets cost 1000 and have useful lives of 5 years and no salvage value Gadget A will save 300 each year Gadget B will save 400 the first year but due to maintenance will decline in savings by 50 each year making the second year savings 350 the third year savings 300 etc Interest is 7 Note that the total of the savings for both options is 1500 over the 5 years if we ignore the time value of money 61174 A A 300 1 ryeau FW 043 w 5 beneth hemRf GadgetA PW ofbene ts 300 1075 1071075 1230 Gadget B PW of bene ts P400 uniform series P50 arithmetic series 400 1075 1071075 50 1075 0751 075 1075 1258 Gadget B is our choice for the heating system Example 04 Different Equipment Lifetimes You work for a manufacturer of spark plugs You are considering two alternative production machines with the following data Alt 1 Alt 2 Initial Cost 50000 75000 Estimated salvage value at end oflife 10000 12000 Useful life of equipment years 7 13 Your manager tells you to use an interest rate of 8 and to use the PW Method to compare the alternatives over an analysis period of 10 years Estimated market value end of 10 yrs 20000 15000 based on remaining life of equipment 41 th 1 TWM Value In 1w um Salmi Iof 2 3 4 r 5 7 5 p ya u m r 394 Tl 77m HP J15 7er HP 4 mix epsf lm szulm Allma Z Tm mu 9 My Year In I z 3 5 A I 8 9 I lo M n 3 ra year we I H 7 t loymP Ala1 1 yenJ lt a Inil ial 5139 51100 Note that we Will have to replace Alt 1 machine after 7 years and we39ll assume it Will have the same price at that time Alternative 1 PW initial purchase trans 7th year 10th year market value PW 50000 10000 salv 50000 buy 1087 20000 10810 PW 50000 23340 9264 5364 076 Alternative 2 PW initial purchase 10Lh year market value PW 75000 15000 10810 68052 It turns out to be less expensive to purchase the cheaper shortlife equipment and replace it partway through the project life 7 Annual Cash Flow Analysis a For this we want to convert all expenses to equivalent uniform cash ows over some period of time Example 05 Student furniture A student buys 1000 of furniture It is expected to last 10 years and the interest rate 7 1111111111 A aguivalenf win Hay F can The equivalent annual uniform cash ow is AP 1000 071071 10710 1 14238 equiv annual cost Suppose he can sell it for 200 at the end ofthe 10 years Asamgae 200 07 10710 1 1448 Therefore the equivalent annual cash ow is And 14238 1448 12790 b If you have costs at different points at time they must rst be brought back to a Present Worth then annualized over the required period of time c If the analysis period does not match the useful life of the equipment you must establish a salvage value for the equipment at the end of the analysis period then treat it as a bene t at that point in time Example 06 Annualization of Example 04 You work for a manufacturer of spark plugs You are considering two alternative production machines with the following data Alt 1 Alt 2 Initial Cost 50000 75000 Estimated salvage value at end of1ife 10000 12000 Useful life of equipment years 7 13 Your manager tells you to use an interest rate of 8 and to use the PW Method to compare the alternatives over an analysis period of 10 years Estimated market value end of 10 yrs 20000 15000 based on remaining life of equipment Al anmlz 1 Twmlml valuaw yeu WM 1015 In fu l 2 J 4 f 7 E B A 1 IL I M t T hayme L 7W 11h 9 ml xi 1 plan z m lim quot Alina2139 Z Tammi Va 3 1m Year 15 I 2 3 r L 39I 5 9 If 1 IL 3 r3 Year We I l lt laymr nvmlyn yeaJ 7 7 9i Inil ial us l 59 Note that we Will have to replace Alt 1 machine after 7 years and we39ll assume it Will have the same price at that time Alternate 1 A0 50000081081 10810 1 7451 P7 50000 100001087 23340 brought back to present A7 23340 081081 1081 1 3478 A10 20000 081081 1 1381 salvage value Equivalent annual cost 7451 3478 1381 9548 per year Alternate2 A0 75000081081 10810 1 11177 A10 15000 081081 1 1035 salvage value Equivalent annual cost 11177 1035 10142 per year Example 07 Assembly line alternative investigation You are investigating assembly line improvement alternatives Interest 8 10 year life for all equipment Which alternative is the best choice A B C D Installed cost of equipment 15000 25000 33000 0 Material amp labor savings per year 14000 9000 14000 0 Annual operating expenses 8000 6000 6000 0 Salvage value 1500 2500 3300 0 Evaluation annualized values Equivalent annual uniform bene t Material amp labor savings 14000 9000 14000 0 Salvage value 104 172 228 0 14104 9172 14228 0 Equivalent uniform annual cost Installed cost 2235 3725 4917 0 Operating Expenses 8 000 6 000 6 000 9 10235 9725 10917 0 Net bene t annualized 3869 553 3311 0 Therefore the best alternative is A 8 Payback Period In its simplest form the payback period is the time required for the bene ts of an investment to equal the cost Sometimes people include depreciation interest income taxes or other factors However it generally is calculated without even an accounting for the time value of money This is a very crude way to represent the cash ow in a project but it is immediately understood by managers and other nontechnical types The intersection of the bene t and cost curves is the breakeven point and the time at that point is called the payback period Example 08 Payback Period We are given two production machines as follows Find the payback periods Installed Cost 30000 35000 Net annual bene t 12000 15L yr 100015 yr Declining 3000yr increasing 3000yr Useful life 4 years 8 years 41 4o 1 2 G LT Bmkavea u m 30 n emf 1 13 Zn in Q Q 7 quot 9mm quot ya 5 PMs ox 4 7r 1 I 9 Annual Rate of Return De nition Internal rate of return is the interest rate at which the bene ts equal the costs Search for the interest rate that satis es the following equation PW of bene ts PW of costs Calculating Rate of Return 1 Convert all income and expenses into present worth expressions but leave the interest rate unknown 2 Find the interest rate by trial and error that satis es the above equation 3 That interest rate is the Internal Rate of Return Example 09 Internal Rate of Return An 8200 investment returns 2000yar over 5 years What is the rate of return PW of costs PW of bene ts 8200 2000 1 i5 1i1i5 By successive iteration we nd the interest rate to be 70 Example 10 IRR with Arithmetic Gradient Year Cash Flow 700 investment 1 100 2 175 3 250 4 325 PW of costs PW ofbene ts 700 1001i41i1i4751i4i41i21i4 Solve iteratively for i approx 7 There is an associated expression that is used Minimum Attractive Rate of Return MARR If you set an MARR for your project then i must be at least that great or the project is not worth investing in The MARR is the minimum rate of interest you want to earn on your money if you risk it in the project 10 Bene tCost Ratio Analysis If a project is to be justi ed on economic grounds then PW ofbene ts 3 PW of costs If the two equal each other there must be additional grounds to implement a project Otherwise we haven t gained anything One way of evaluating the pro tability of a project is to calculate the IRR Another way is to calculate the ratio of bene ts to costs ie how much do we earn on each dollar spent PW ofbene ts PW of costs returned spent Example 11 Bene tCost Ratio Two machines are being considered for purchase Assume i 10 Which machine should we buy Initial cost 200 Uniform annual bene t 95 Endoflife salvage value 50 Useful 1ife years 6 700 120 150 12 Use a 12year evaluation period The cash ow for the analysis is the following Year Machine X Machine Y 0 200 700 l 5 95 120 6 200 2 d machine 120 6 95 6 50 salvage 1St machine 7 l 1 95 120 12 95 120 12 50 2quotd salvage 150 salvage Ppurchase 20000 70000 Pumform series 647 30 81764 11289 2 d machine 28 22 salvage P6 0050 Pfinal salvage 1593 Sum PWbenems 69145 Sum PWmsts 31289 Ratiobene tcosts 2 21 4779 86543 70000 1236 On this basis Machine X is the better purchase because it gives a better return on each dollar 11 In ation and Price Change In ation means that our money won t buy as much as it used to quotloss of purchasing powerquot Future dollars are worth less than today39s dollars An excellent example is the recent rise in gasoline price When I was teaching at the Univ of Connecticut in the 197039s gas sold for 33 cents gallon My salary was 18th of what it would be today That39s the effect of in ation The relative purchasing power of the dollar is the same in both periods the numbers just look bigger Some countries have had in ation rates as high as 100 a year We need to be able to make comparisons on an equivalent basis when we analyze projects One way to handle the problem is to de ne a 39real interest rate which is the difference between the interest rate we can get on our money and the in ation rate If f in ation rate i market interest rate the quoted rate we can get such as bank interest auto loan interest mortgage rate etc i real interest rate the real rate of growth of our money after excluding in ation We define i by considering interest for a year P 1i P 1i 1f Thus i is what39s left over after we remove the effect of in ation Solving for i i if1f Example 12 Effect of In ation Your bank is paying 15 annual interest on savings accounts and in ation is 25 i 015 025 1025 0098 This says we are losing 98 buying power per year on our savings Example 13 Analysis in Real Dollars Price in ation is 35 per year A project has two choices for producing a product over 5 years Case A production cost 150000 the 1St year will increase 5 per year Case B production cost 150000 all 5 years in today39s dollars Year 0 Case A Year Cost Constant 1 150000 1500001035 144928 2 150000x105 157500 15750010352 147028 3 150000Xl052 165375 16537510353 149159 4 150000Xl053 173644 17364410354 151321 5 150000Xl054 183 326 18332610355 153 514 829845 745950 Case B 5 years X 150000 today39s dollars 750000 This looks like the two cases are very similar How much money do we need to have on hand at present to nance either alternative if we can get 4 interest on our money Case A Year Present Value 1 150000104 144230 2 150000x1051042 145617 3 150000x10521043 147017 4 150000x10531044 148458 5 150000x10541045 149 913 735235 Case B Year Present Value 1 150000104 144230 150000x10351042 143538 150000x103521043 142849 150000x103531044 142163 150000x103541045 141262 UIAUJN Therefore Case B will save us 20973 in today s dollars


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