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Civil Engineering 202 with Tanzler at Lehigh University
About this note
By: Katie Brinkhoff
Textbook:
Engineering Economic Analysis
Engineering Economic Analysis
Created: 2012-02-27
File Size: 30 page(s)
Views: 0
Textbook:
Engineering Economic Analysis
Engineering Economic AnalysisCreated: 2012-02-27
File Size: 30 page(s)
Views: 0
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Copyright Oxford University Press 2009 Chapter 4 More Interest Formulas Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Uniform Series Compound Interest Formulas Arithmetic Gradient Geometric Gradient Using Spreadsheets for Economic Analysis Chapter Outline Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Uniform Series Compound Interest Formulas Notation: A = an end-of-period cash flow in a uniform series, continuing for n periods Examples: Automobile loans, mortgage payments, insurance premium, rents, and other periodic payments Estimated future costs and benefits A A A A n-1 0 1 2 n Copyright Oxford University Press 2009 Solar Example Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Derive Equations Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Uniform Series Compound Interest Formulas Uniform Series Compound Amount Factor (4-5) (4-6) Uniform Series Sinking Fund Factor Copyright Oxford University Press 2009 Equivalence These equations convert Uniform Cash Flows into a Single ?equivalent? amount I.E. Present or Future lump sum amount Or Vis-Versa converting a Present or Future Lump Sum into ?equivalent? uniform Annual Payments or Receipts Of course you can also solve for interest rate i, essentially giving you the ?average rate of return? for given cash flow. Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-1 Uniform Series Compound Interest Formulas $500 were deposited in a credit union (pays 5% compounded annually) at the end of each year for 5 years, how much do you have after the 5th deposit? 4 0 1 2 3 5 F=? 500 500 500 500 500 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-2 Uniform Series Compound Interest Formulas Jim wants to save money at the end of each month to pay for some equipment of $1000 at the end of the year. If bank pays 6% interest, compounded monthly, how much Jim has to deposit every month? 1000 4 0 1 2 3 5 11 9 10 12 A A A A A A A A A=? Copyright Oxford University Press 2009 Important Reminder ? Not Inflation!!! Inflation has nothing to do with the selection of interest rate or the calculation of Present or Future $ amount! Inflation (covered later) affects how much you can buy with the money that you have, not the time value of how much money you have which is due to opportunity costs, investment possibilities, desired return on investment, etc. Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Uniform Series Compound Interest Formulas Uniform Series Capital Recovery Factor (4-7) (4-8) Uniform Series Present Worth Factor Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-3 Uniform Series Compound Interest Formulas A machine costs $5000 and has a life of 5 years. If interest rate is 8%, how much must be saved every year to recover the investment? 4 0 1 2 3 5 5000 A A A A A Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-4 Uniform Series Compound Interest Formulas An investor holds a time payment purchase contract on some machine tools that will give him $140/mo. for 5yrs. First payment is due in one month, he offers to sell you contract for $6,800 is it a good deal if you can make 1% on your money elsewhere? P A=140 n=60 i=1% The Bond is only worth $6293.70, so reject the offer. Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-5 Uniform Series Compound Interest Formulas If $6800 is paid for a Contract that pays $140 at the end of each month for 5 years, what is the monthly rate of return? P=6800 A=140 n=60 i=1% From the compound interest tables, (P/A, i, 60) = 51.726 when i = 0.5% (P/A, i, 60) = 48.174 when i = 0.75% By linear interpolation, rate of return would be 0.72% Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-6 Uniform Series Compound Interest Formulas A student is borrowing $100 per year for 3 years. The loan will be repaid 2 years later at a 15% interest rate. Year Cash Flow 1 +100 2 +100 3 +100 4 0 5 -F 4 0 1 2 3 5 F=? 100 100 100 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-6 Uniform Series Compound Interest Formulas Solution #1: = F3 4 0 1 2 3 5 A + + F2 4 0 1 2 3 5 A F1 4 0 1 2 3 5 A F 4 0 1 2 3 5 A A A Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-6 Uniform Series Compound Interest Formulas Solution #2: = 0 4 1 2 3 5 F3 4 0 1 2 3 5 A A A=100 = F 4 0 1 2 3 5 F3 4 0 1 2 3 5 A A A=100 F Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Relationships Between Compound Interest Factors Single Payment (4-9) (4-10) Uniform Series (4-11) Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Relationships Between Compound Interest Factors (4-12) Uniform Series (4-14) (4-13) Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Arithmetic Gradient Examples: Operating and maintenance costs Salary packages = + 4 0 1 2 3 5 G 2G 3G 4G 4 0 1 2 3 5 A A A A A 4 0 1 2 3 5 A A+G A+3G A+4G A+2G Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Arithmetic Gradient Notation: G = a fixed amount increment or decrement per time period G 2G (n-2)G n-1 0 1 2 n 3 0 (n-1)G Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Arithmetic Gradient Compound Interest Formulas Arithmetic Gradient Present Worth Factor (4-20) (4-21) Arithmetic Gradient Uniform Series Factor Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-8 Application of Arithmetic Gradient Interest Factors Year Cash Flow 1 120 2 150 3 180 4 210 5 240 = 4 0 1 2 3 5 150 120 180 210 240 + 4 0 1 2 3 5 30 60 90 120 4 0 1 2 3 5 A=120 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Geometric Gradient Notation: g = a constant growth rate (+ or -) per period A1 = cash flow at period 1 A1(1+g) n-1 0 1 2 n 3 A1 A1(1+g)2 A1(1+g)n-2 A1(1+g)n-1 Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Geometric Gradient Compound Interest Formulas Geometric Gradient Present Worth Factor (4-29) (4-31) Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Example 4-12 Application of Geometric Gradient Interest Factors Using 8% interest rate, calculate the Present Worth of maintenance cost of the first 5 years where the first year?s cost is estimated at $100, and it increases at a constant rate of 10% per year. Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Spreadsheets for Economic Analysis Constructing tables of cash flows Using annuity functions to calculate P, F, A, n, or i Using block functions to find the present worth or internal rate of return for a table of cash flows Making graphs for analysis and presentations Calculating ?what-if? for different scenarios Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Spreadsheet Annuity Functions Excel Functions Purpose -PV (i, n, A, [F], [Type]) To find P given i, n, and A (F is optional) -PMT (i, n, P, [F], [Type]) To find A given i, n, and P (F is optional) -FV (i, n, A, [P], [Type]) To find F given i, n, and A (P is optional) NPER (i, A, P, [F], [Type]) To find n given i, A, and P (F, optional) RATE (n, A, P, [F], [Type], [guess]) To find i given n, A, and P (F, optional) Note: Type is the number 0 or 1 and indicates when payments are due (0 for end-of-the-period, and 1 for beginning-of-the-period). If type is omitted, it is assumed to be 0. Microsoft Excel solves for one financial argument in terms of the others. P(F/P, i, n) + A(F/A, i, n) + F = 0 Therefore, negative signs are added to find the equivalent values in P, A, and F. Copyright Oxford University Press 2009 Copyright Oxford University Press 2009 Spreadsheet Block Functions Excel Functions Purpose NPV (i, range) To find net present value of a range of cash flows (from period 1 to n) at a given interest rate IRR (range, [guess]) To find internal rate of return from a range of cash flows (from period 0 to n) Copyright Oxford University Press 2009 Assignment Chapter 4 - Read and do problems: 4-9, 4-13, 4-19, 4-22, 4-56, 4-67, 4-68 and 4-83 Due Tuesday beginning of class.
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About this note
By: Katie Brinkhoff
Textbook:
Engineering Economic Analysis
Engineering Economic Analysis
Created: 2012-02-27
File Size: 30 page(s)
Views: 0
Textbook:
Engineering Economic Analysis
Engineering Economic AnalysisCreated: 2012-02-27
File Size: 30 page(s)
Views: 0
About StudyBlue
STUDYBLUE makes things that make you better at school.
Things like online flashcards with photos and audio.
Things like personalized quizzes and friendly reminders about when (and what) to study next.
Think of it as a digital backpack™: access to all of your study materials online and on your phone.
STUDYBLUE exists to make studying efficient and effective for every student, for free. Join us.
“Simply amazing. The flash cards are smooth, there are many different types of studying tools, and there is a great search engine. I praise you on the awesomeness.”
Dennis
Dennis