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# Note for INDE 3333 with Professor Schulze at UH foundation of engineering

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## Reviews for Note for INDE 3333 with Professor Schulze at UH foundation of engineering

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Date Created: 02/06/15

CHAPTER 1 FOUNDATIONS OF ENGINEERING Section 13 Taxes oTaxes represent a significant negative cash flow to the forprofit firm 0A realistic economic analysis must assess the impact of taxes Called and AFTERTAX cash flow analysis oNot considering taxes is called a BEFORE TAX Cash Flow analysis Taxes 0 A BeforeTax cash flow analysis while not as accurate is often performed as a preliminary analysis 0A final more complete analysis should be performed using an AfterTax analysis 0 Both are valuable analysis approached Fact 50 of the US working population does not pay taxes oThe top 5 of wage earners pay 90 of the taxes 250000 S YOU ARE RICH ll CUl HOESWOT 00 V SV GHSSHHdX3quot39A3NON SES39H HNOHNOS lO BSD 3H1 80d CIIVd 33d 39IVLNEH LNnOINV IVNISIHO 39 VON GEIVO 39VLOL LSEIHELLNI NVO39I LNnOV 39IVNISIHO 39 VON Eln IVA iSEIHELLNI lNEIWlSEIANI 39AEINOW 38 Cl GIVd LNnOWV EHL 39AEINOW IO Eln39IVA EIVL EHL IO NOLVLSEIIINVV 39 iSHHHiNI 3193 153me Interest Lending Interest can be viewed from two perspectives Lending situation 0 Investing situation 0 You borrow money renting someone else39s money 0 The lender expects a return on the money lent oThe return is measured by application of an interest rate Interest Lending Example 0 You borrow 10000 for one full year 0 Must pay back 10700 at the end of one year 0 Interest Amount I 10700 10000 0 Interest Amount 700 for the year 0 Interest rate i 70010000 7Yr Interest Rate Notation oFor 13 the interest rate is Expressed as a per cent per year oNotation I the interest amount is oi the interest rate Oointerest period N No of interest periods 1 for this problem luthored m Don Smith TX i Ji Interest Borrowing oThe interest rate i is 7 per year c The interest amount is 700 over one year The 700 represents the return to the lender for this use of hisher funds for one year c 7 is the interest rate charged to the borrower 7 is the return earned by the lender Interest Example oBorrow 20000 for 1 year at 9 interest per year oi 009 per year and N 1 Year oPay 20000 00920000 at end of 1 year oInterest I 00920000 1800 oTotaI Amt Paid one year hence 20000 1800 M r r m lid 9amp1 gbl 3 Interest Example Note the following oTotaI Amount Due one year hence is 20000 00920000 20000109 21800 oThe 109 factor accounts for the repayment of the 20000 and the interest amount oThis will be one of the important interest factors to be seen later i J 14 Interest Investing Perspective oAssume you invest 20000 for one year in a venture that will return to you 9 per year At the end of one year you will have oOriginal 20000 back oPlus oThe 9 return on 20000 1800 We say that you earned 9year on the investment This is your RATE of RETURN on the investment iii19595314 M Inflation Effects 0 A socialeconomic occurrence in which there is more currency competing for constrained goods and services 0 Where a country s currency becomes worth less over time thus requiring more of the currency to purchase the same amount of goods or services in a time period 0 Fact The market sets the price If there is a surplus prices go down if there is a scarcity prices go up Inflation Rates oIanation impacts gtPurchasing Power reduces gtOperating Costs increases gtRate of Returns on Investments reduces gtSpeci cally covered in Chapter 14 ECONOMIC EQUIVALENCE oEconomic Equivalence oTwo sums of money at two different points in time can be made economically equivalent if oWe consider an interest rate and 0N0 of Time periods between the two sums Equality in terms of Economic Value 14 Equivalence Illustrated oDiagram the loan Cash Flow Diagram oThe company s perspective is shown 20000 is received here T0 t1Yr 21800 paid back here 20000 now is economically equivalent to 21800 one year from now IF the interest rate is set to equal 9lyear 15 Equivalence Illustrated Another wa20000 now is not equal in magnitude to 21800 1 year from now 0 But 20000 now is economically equivalent to 21800 one year from now if the interest rate in 9 per year c If you were told that the interest rate is 9 Which is worth more o20000 now or 21800 one year from now 0 The two sums are economically equivalent but not numerically equal 16 Equivalence To have economic equivalence you must specify 0 Timing of the cash flows oAn interest rate We per interest period c Number of interest periods N Simple and Compound Interest 0 Two types of interest calculations 0 Simple Interest Car Loans Loan Sharks Compound Interest Mortgage Compound Interest is more common worldwide and applies to most analysis situations Simple Interest 0 Simple Interest Calculated on the principal amount only oEasy simple to calculate oSimple Interest is principalinterest ratetime 1 Pi1 Simple and Compound Interest Borrow 1000 for 3 years at 5 per year 0 Let P the principal sum oi the interest rate 5year Let n number of years 3 I 10000053 5m 0 Total Interest over 3 Years Accrued Simple Interest 3 Years 150 of interest has accrued P1000 115ooo 125ooo I35000 Pay back 1000 150 of interest The unpaid interest did not earn interest over the 3year pe od Simple Interest Summary In a multiperiod situation with simple interest r r m lid 9amp1 Jul 3 oThe accrued interest does not earn interest during the succeeding time period Normally the total sum borrowed lent is paid back at the end of the agreed time period PLUS the accrued owed but not paid interest l J Compound Interest Compounding means to compute the interest owed at the end of the period and then add it to the unpaid balance of the loan 0 Interest then earns interestquot 0 To COMPOUND stop and compute the associated interest and add it to the unpaid balance 0 When interest is compounded the interest that is accrued at the end of a given time period is added in to form a NEW principal balance 0 That new balance then earns or is charged interest in the succeeding time period A 4 e4 Compound Interest Example 0 Assume P 1000 0 i 5 per year compounded annually CA 0N 3 years c For compound interest 3 years we have P1000 Owe at t 3 years 1000 5000 5250 5513 12525 115763 115000 I35513 24 Example Determining best Loan Payoff Five plans are shown that will pay off a loan of 5000 over 5 years with interest at 8 per year Plan1 Simple Interest pay all at the end Plan 2 Compound Interest pay all at the end Plan 3 Simple interest pay interest only at the end of each year Pay the principal at the end of N 5 Plan 4 Compound Interest and part of the principal each year pay 20 of the Principal Amount each year Plan 5 Equal Payments of the compound interest and principal reduction over 5 years with end of year wmpayments Plan 1 Simple Interest Pay all at end on 5000 Loan 1139 2 33 M1 E511 End 9F Intarast DWEd Tami Dated at Endvnf I ear Tata Dwed quotfear far Year End of Year Payment 1 after Faginrent Plan 1quot Simple Ingram Fay AH 111 End 1 U 55mm I 400100 5430 54W E 5300313 5309 3 400130 EEWDU l 4 660000 5 4100100 700000 57115000 Tnmls Sir00000 10192014 25 Plan 2 Compound Interest Pay all at the End of 5 Years 10192014 11 2 l3 Eml all luterest Dwed Tartal Unwed at EndnfTear Hear far quotfear End nf 1fear 0 1 5mm 5 D 2 431013 53311110 3 466515 l g j 4 58133 39244 5 5442 T346 l4 l5 Tntal famed Payment alter Payment 551mm 5mm d 5332M 529355 6510144 573mm 57364 27 Plan 3 Simple Interest Interest Paid Annually Principal Paid at the End Balloon Note ll l El El l4 lSl Encl of interest Dried Total Wed at EndnufTear Total Dwed V l39ear for quotfear End of Year Payment after Payment ll 500003 1 5mm 554430110 490M 5mm 2 mm mm SWIM 3 mm 5mm 4mm 5mm 4 3mm 4mm 5mm Authored by Don Smith TX ASAM University 10192014 Plan 4 Compound Interest Interest and 20 of Principal Paid back annually ll 21 3139 l4 5 Encl nf lnterest Dwed Tatal Dunrad at Endenfm ear Tntal Wadi quotfear r Tear End at Year Pia ent after Pa 39ent l 354061141 HillHill 324mm 2 l W l39 39 39 10192014 29 Plan 5 Equal Repayment Plan Annual Payments Part Principal and Part Interest l ll El 13139 4 El End 91 lmterest Dwed Tatal Owed at End I39ear Tutal Owed fur Year End of quotfer Payment alter Payment l mama 39 549030 125223 4147 447254 12522 2122225 242542 125222 223215 24 122 125222 1 15252 l25222 52 222451 10192014 30 Comparisons 5 Plans Plan 1 Simple interest original principal008 Plan 2 Compound interest total owed previous year008 Plan 3 Simple interest original principal008 Plan 4 Compound interest total owed previous year008 Plan 5 Compound interest total owed previous year008 Analysis 0 Note that the amounts of the annual payments are different for each repayment schedule and that the total amounts repaid for most plans are different even though each repayment plan requires exactly 5 years c The difference in the total amounts repaid can be explained 1 by the time value of money 2 by simple or compound interest and 3 by the partial repayment of principal prior to year 5 lCllgg EGi l 32 Terminology and Symbols P value or amount of money at a time designated as the present or time 0 Also P is referred to as Present Worth PW Present Value PV Net Present Value NPV Discounted Cash ow DCF and capitalized cost CC dollars Terminology and Symbols 0 F value or amount of money at some future time oAlso F is called Future Worth FW and Future Value FV dollars PandF The symbols P and F represent onetime occurrences It should be clear that a present value P represents a single sum of money at some time prior to a future value F P Terminology and Symbols 0A series of consecutive equal endof period amounts of money 0 Also A is called the Annual Worth AW Annual Annuity AA and Equivalent Uniform Annual Worth EUAW dollars per year dollars per month 0 n number of interest periods years months days Uniform Amounts It is important to note that the symbol A always represents a uniform amount ie the same amount each period that extends through consecutive interest periods 0 Cash Flow diagram for annual amounts might look like the following A A A A A 0 1 2 3 N1 n l A equal end of period cash flow amounts I Terminology and Symbols oi interest rate or rate of return per time period percent per year percent per month oThe interest rate i is assumed to be a compound rate unless specifically stated as simple interest oThe rate i is expressed in percent per interest period for example 12 per year t time stated in periods years months days etc Terminology and Symbols 0 many engineering economy problems Involve the dimension of time oAt least 4 of the symbols P F A i and n 0 At least 3 of 4 are either estimated or assumed to be know with certainty Minimum Attractive Rate of Return MARR Firms will set a minimum interest rate that the financial managers of the firm require that all accepted projects must meet or exceed Project Return gt Project Cost by MARR 00 c The rate once established by the firm is termed the Minimum Attractive Rate of Return MARR The MARR is expressed as a per cent 00 per year 0 Numerous models exist to aid the financial managers is estimating what this rate should be in a given time pe od MARR Hurdle Rate In some circles the MARR is termed the Hurdle Rate 0 Capital investment funds is not free 0 It costs the firm money to raise capital debt or to use the owners of the firm s capital retained earnings cash on hand equity 0 This cost is often expressed as a per year interest rate Setting the MARR Safe Investment 0 First start with a safe investment possibility A firm could always invest in a short term CD paying around 45 But investors will expect more that that they want to make a profit ROI The firm should compute it s current weighted average cost of capital This cost will almost always exceed a safe external investment rate 0 This is often indicated as a buffer to account for risk and uncertainty Graphical Presentation MARR ROR o 000 10192014 11m warn a 3 eat Two Alternatives Example 0 Assume a firm s MARR 12 0 Two projects A and B o A costs 400000 and presents an estimated 13 per year c B cost 100000 with an estimated return of 145 i J 14 Opportunity Forgone oThe firm has a budget of say 150000 0 A cannot be funded insufficient funds 0 B is funded and earns 145 return or more 0 A is not funded hence the firm looses the opportunity to earn 13 oThis often happens Cash Flow Diagramming Engineering Economy has developed a graphical technique for presenting a problem dealing with cash flows and their timing the Cash Flow Diagram Cash Inflows Money flows in from the outside and point up Revenues Savings Salvage Values 0 Cash Outflows Money flows out from the inside and point down Disbursements First costs of assets labor salaries taxes paid utilities rents interest Cash Flows Assumed known with some degree of certainty Estimated A range of possible realistic values provided Generated from an assumed distribution and simulated 0 Net Cash Flow Cash Inflows Cash Outflows for a given time period 0 Normally assume that all cash flows occur at the end of a given time period n w J A a lid5 egv39Jli ll ii Sample gash flow Diagram 10192014 48 c 39r m lid 9f gth The Cash Flow Diagram CFD Extremely valuable analysis tool First step in the solution process Graphical Representation on a time scale 0 Do not have to be drawn to exact scale 0 Should be neat and properly labeled l J 2 Problem Perspectives Before solving one must decide upon the perspective of the problem Most problems will present two perspectives Assume a borrowing situation for example Perspective 1 From the lender s view Perspective 2 From the borrower s view 0 Impact upon the sign convention c m lid93 bll Lending Perspective Borrowing Assume 5000 is borrowed and payments are 1100 per year Draw the cash flow diagram for this First whose perspective will be used Lender s or the Borrower s Problem will infer or you must decide lulzhored by Don Smith u xi l l l J l 39 lvaersrty Lending Borrowing From the Lender s Perspective A 1100yr iiiii 0 1 2 3 4 5 5000 Lending Borrowing From the Borrower s Perspective P 5000 0112114151 A 1100yr Example 0 A father wants to deposit an unknown lumpsum amount into an investment opportunity 2 years from now that is large enough to withdraw 4000 per year for tin of his child s state university tuition for 5 years starting 3 years from now If the rate of return is estimated to be 155 per year construct the cash flow diagram CF Diagram 19192514 Rule of 72 Estimating doubling Time and Interest Rate 0 A common question most often asked by investors is How long will it take for my investment to double in value 0 Must have a known or assumed compound interest rate in advance 0 The approximate time for an investment to double in value given the compound interest rate is n 72i 0 For i 13 7213 554 years Rule of 72 s for Interest The formula to estimate the required interest rate for an investment to double in value over time is i approximate 72n Assume you want an investment to double in 3 years 0 Estimate i rate would be 723 24 this is your minimum attractive rate of return MARR Chapter 1 Summary Engineering Economy is The application of economic factors and criteria to evaluate alternatives considering the time value of money interest and time 10192514 Engineering Economy Study 0 Involves modeling the cash flows 0 Computing specific measures of economic worth 0 Using an interest rates 0 Over a specified period of time Equivalence helps in the understanding of how different sums of money at different times are equal in economic terms Simple and Compound Interest Simple interest involves the calculation of interest once for the entire lending borrowing period Compound interest involves the notion of computing interest on interest The power of compounding is very noticeable especially over long periods of time The Minimum Attractive Rate of Return MARR o The MARR is a reasonable rate of return established as a hurdle rate to determine if an alternative is economically viable o The MARR is always higher than a return from a safe investment Attributes of Cash Flows o Difficulties with their estimation 0 Difference between estimated and actual value 0 End of year convention for cash flow location 0 Net cash flow computation 0 Different perspectives in determining the cash flow sign convention 0 Construction of a cash flow diagram Homework 118 120 122 126 129 134 139 142 152 and159

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