1 / 44

IEN 10143

IEN 10143. Engineering Economy Winter 2012 Leland T.Blank & Anthony J. Tarquin 5 th Edition. Syllabus Info. T/Th: 5:00 – 9:00 pm, Room 311 Instructor: Carolus Kaswandi Office: 3 rd floor, 311;phone: 081311470126 E-mail address: ckaswandi @gmail.com, carolus_kaswandi@yahoo.com

robertbgonzalez
Download Presentation

IEN 10143

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. IEN 10143 Engineering Economy Winter 2012 Leland T.Blank & Anthony J. Tarquin 5th Edition

  2. Syllabus Info T/Th: 5:00 – 9:00 pm, Room 311 Instructor: Carolus Kaswandi Office: 3rd floor, 311;phone: 081311470126 E-mail address: ckaswandi@gmail.com, carolus_kaswandi@yahoo.com Office Hours: 10.00 – 21.00 (Tues, Wednesday, Thursday)

  3. Economic News http://www.investors.com/learn/b.asp

  4. Textbook: Engineering Economy, Sixth Edition, Leland T. Blank and Anthony J. Tarquin, McGraw-Hill, ISBN 0-07-320382-3 http://www.csun.edu/~bavarian/mse_304.htm

  5. Course Text Overview • Level 1 This is How It All Starts Chapter 1: Foundations of Engineering Economy Chapter 2: Factors: How Time and Interest Affect Money Chapter 3: Combining Factors Chapter 4: Nominal and Effective Interest Rates • Level 2 Tools for Evaluating Alternatives Chapter 5: Present Worth Analysis Chapter 6: Annual Worth Analysis Chapter 7: Rate of Return Analysis: Single Alternative Chapter 8: Rate of Return Analysis: Multiple Alternatives Chapter 9: Benefit/Cost Analysis and Public Sector Economics • Level 3 Making Decisions on Real-World Projects Chapter 11: Replacement and Retentions Decisions • Level 4 Rounding Out the Study Chapter 14: Effects of Inflation Chapter 17: After-Tax Economic Analysis Chapter 18: Formalized Sensitivity Analysis and Expected Value Decisions

  6. Tentative Schedule

  7. Grade Determination • 40% - Final Exam: open formula sheet (2 pages of A4),calculator, no neighbors. • 30% - Midterm Exams: open formula sheet (1 page of A4), calculator, no neighbors. • 15% - Quiz (2 times), open book, calculator, no neighbors. • 10% - Home Work, 5% Attendance

  8. Chapter 1 Foundations of Engineering Economy

  9. Why Engineering Economy is Important to Engineers • Decisions made by engineers, managers, corporation presidents, and individuals are commonly the result of choosing one alternative over another. • Decisions often reflect a person’s educated choice of how to best invest funds (capital). • The amount of capital is usually restricted, just as the cash available to an individual is usually limited. The decision of how to invest capital will invariably change the future, hopefully for the better; that is, it will be value adding. • Engineers play a major role in capital investment decisions based on their analysis, synthesis, and design efforts. • The factors considered in making the decision are a combination of economic and noneconomic factors. • Fundamentally, engineering economy involves formulating, estimating, and evaluating the economic outcomes when alternatives to accomplish a defined purpose are available.

  10. Problem Solving Approach • Understand the Problem and define the objective. • Collect all relevant data/information • Define the feasible alternative solutions and make realistic estimates. • Evaluate each alternative • Select the “best” alternative • Implement and monitor

  11. Time Value of Money • An important concept in engineering economy • Money can “make” money if invested. • The change in the amount of money over a given time period is called the time value of money.

  12. The Big Picture • Engineering economy is at the heart of making decisions. • These decisions involve the fundamental elements of cash flows of money, time and interest rates. • Chapter 1 introduces the basic concepts and terminology necessary for an engineer to combine these three essential elements in organized, mathematically correct ways to solve problems that will lead to better decisions.

  13. Parameters and Cash Flows • Parameters • First cost (investment amounts) • Estimates of useful or project life • Estimated future cash flows (revenues and expenses and salvage values) • Interest rate • Cash Flows • Estimate flows of money coming into the firm – revenues, salvage values, etc. – positive cash flows--cash inflows • Estimates of investment costs, operating costs, taxes paid – negative cash flows -- cash outflows

  14. The Cash Flow Diagram: CFD

  15. Net Cash Flows • A NET CASH FLOW is • Cash Inflows – Cash Outflows • (for a given time period) • We normally assume that all cash flows occur: • At the END of a given time period • End-of-Period Assumption

  16. Interest Rate • INTEREST - THE AMOUNT PAID TO USE MONEY. • INVESTMENT • INTEREST = VALUE NOW - ORIGINAL AMOUNT • LOAN • INTEREST = TOTAL OWED NOW - ORIGINAL AMOUNT • INTEREST RATE- INTEREST PER TIME UNIT

  17. Interest – Lending Example • Example 1.3 • You borrow $10,000 for one full year • Must pay back $10,700 at the end of one year • Interest Amount (I) = $10,700 - $10,000 • Interest Amount = $700 for the year • Interest rate (i) = 700/$10,000 = 7%/Yr

  18. Interest Rate - Notation • Notation • I = the interest amount is $ • i = the interest rate (%/interest period) • N = No. of interest periods (1 for this problem) • Interest – Borrowing • The interest rate (i) is 7% per year • The interest amount is $700 over one year • The $700 represents the return to the lender for the use of funds for one year • 7% is the interest rate charged to the borrower

  19. Interest – Example • Borrow $20,000 for 1 year at 9% interest per year • i = 0.09 per year and N = 1 Year • Pay $20,000 + (0.09)($20,000) at end of 1 year • Interest (I) = (0.09)($20,000) = $1,800 • Total Amt Paid in one year: • $20,000 + $1,800 = $21,800

  20. T=0 t = 1 Yr Economic Equivalence • Two sums of money at different points in time can be made economically equivalent if: • We consider an interest rate and, • number of Time periods between the two sums $20,000 is received here $21,800 paid back here $20,000 now is economically equivalent to $21,800 one year from now IF the interest rate is set to equal 9%/year

  21. Equivalence Illustrated • $20,000 now is not equal in magnitude to $21,800 1 year from now • But, $20,000 now is economically equivalent to $21,800 one year from now if the interest rate in 9% per year. • To have economic equivalence you must specify: • timing of the cash flows • interest rate (i% per interest period) • Number of interest periods (N)

  22. Simple and Compound Interest • Two “types” of interest calculations • Simple Interest • Compound Interest • Compound Interest is more common worldwide and applies to most analysis situations

  23. Simple and Compound Interest • Simple Interest is calculated on the principal amount only • Easy (simple) to calculate • Simple Interest is: • (principal)(interest rate)(time); $I = (P)(i)(n) • Borrow $1000 for 3 years at 5% per year • Let “P” = the principal sum • i = the interest rate (5%/year) • Let N = number of years (3) • Total Interest over 3 Years...

  24. P=$1,000 1 2 3 I1=$50.00 For One Year • $50.00 interest accrues but not paid • “Accrued” means “owed but not yet paid” • First Year:

  25. P=$1,000 1 2 3 Pay back $1000 + $150 of interest I2=$50.00 I1=$50.00 I3=$50.00 End of 3 Years • $150 of interest has accrued The unpaid interest did not earn interest over the 3-year period

  26. Compound Interest • Compound Interest is different • In this application, compounding means to calculate the interest owed at the end of the period and then add it to the unpaid balance of the loan • Interest “earns interest”

  27. P=$1,000 1 2 3 I1=$50.00 I2=$52.50 I3=$55.13 Compound Interest Cash Flow • For compound interest, 3 years, we have: Owe at t = 3 years: $1,000 + 50.00 + 52.50 + 55.13 = $1157.63

  28. Compound Interest: Calculated • For the example: • P0 = +$1,000 • I1 = $1,000(0.05) = $50.00 • Owe P1 = $1,000 + 50 = $1,050 (but, we don’t pay yet) • New Principal sum at end of t = 1: = $1,050.00

  29. Compound Interest: t = 2 • Principal and end of year 1: $1,050.00 • I1 = $1,050(0.05) = $52.50 (owed but not paid) • Add to the current unpaid balance yields: • $1050 + 52.50 = $1102.50 • New unpaid balance or New Principal Amount • Now, go to year 3…….

  30. Compound Interest: t = 3 • New Principal sum: $1,102.50 • I3 = $1102.50 (0.05) = $55.125 = $55.13 • Add to the beginning of year principal yields: • $1102.50 + 55.13 = $1157.63 • This is the loan payoff at the end of 3 years • Note how the interest amounts were added to form a new principal sum with interest calculated on that new amount

  31. Terminology and Symbols P = value or amount of money at a time designated as the present or time 0. F= value or amount of money at some future time. A= series of consecutive, equal, end-of-period amounts of money. n = number of interest periods; years i= interest rate or rate of return per time period; percent per year, percent per month t = time, stated in periods; years, months, days, etc

  32. 0 1 2 … … n-1 n $P P and F • The symbols P and F represent one-time occurrences: • It should be clear that a present value Prepresents a single sum of money at some time prior to a future value F $F

  33. $A $A $A $A $A 0 1 2 3 .. N-1 n Annual Amounts • It is important to note that the symbol A always represents a uniform amount (i.e., the same amount each period) that extends through consecutive interest periods. • Cash Flow diagram for annual amounts might look like the following: ………… A = equal, end of period cash flow amounts

  34. Spreadsheets • Excel supports (among many others) six built-in functions to assist in time value of money analysis • Master each on your own and set up a variety of the homework problems (on your own)

  35. Excel’s Financial Functions • To find the: • present value P: PV (i%,n,A,F) • future value F: FV (i%,n,A,P) • equal, periodic value A: PMT (i%,n,P,F) • number of periods n: NPER (i%,A,P,F) • compound interest rate i: RATE (n,A,P,F) • These built-in Excel functions support a wide variety of spreadsheet models that are useful in engineering economy analysis.

  36. The MARR • Firms will set a minimum interest rate that the financial managers of the firm require that all accepted projects must meet or exceed. • The rate, once established by the firm is termed the Minimum Attractive Rate of Return (MARR) • The MARR is expressed as a per cent per year • In some circles, the MARR is termed the Hurdle Rate

  37. Example 1.17 • A father wants to deposit anunknownlump‑sum amount into an investment opportunity 2 years from now that is large enough to withdraw $4000 per year for state university tuition for 5 years starting 3 years from now. • If the rate of return is estimated to be 15.5% per year, construct the cash flow diagram.

  38. Rule of 72’s for Interest • A common question most often asked by investors is: • How long will it take for my investment to double in value? • Must have a known or assumed compound interest rate in advance • Assume a rate of 13%/year to illustrate….

  39. Rule of 72’s for Interest • The Rule of 72 states: • The approximate time for an investment to double in value given the compound interest rate is: • Estimated time (n) = 72/i • For i = 13%: 72/13 = 5.54 years

  40. Rule of 72’s for Interest • Can also estimate the required interest rate for an investment to double in value over time as: • i approximate = 72/n • Assume we want an investment to double in say 3 years. • Estimate i – rate would be: 72/3 = 24%

More Related