1. Faculty of Applied Engineering and Urban Planning Civil Engineering Department Engineering Economy Lecture 1 Week 1 2 nd Semester 20015/2016 Chapter 3 Interest and Equivalence

2. Chapter 3 :Interest and Equivalence Time Value of Money Simple Interest Compound Examples Interest

3. Time Value of Money Question : Would you prefer $100 today or $100 after 1 year? A dollar today is worth more than a dollar next year  Money makes money -Investments are expected to earn a return  The change in the amount of money (through investment) over a given period of time is called “time value of money.”  Money has a time value because it can earn more money over time (earning power)  Money has a time value because its purchasing power changes over time (inflation)  The “time value of money” is the most important concept in engineering economy  The amount and timing of a project’s cash flows (operating expenses; revenues) are crucial to the value of a project’s worth.

4. Interest  Some people would pay to have money available for use. The charge for its use is called interest rate.  Interest is a compensation for using money and for uncertainties related to future value of money  It also called Rate of Return (RoR) on investment.  Two types of interest:  Simple interest: the practice of charging an interest rate only to an initial sum (principal amount).  Interest on principal only  Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.  Interest earns interest on interest; compounds over time

5. Simple Interest  Simple interest is interest that is computed on the original sum.  Interest = (principal)(number of periods)(interest rate) Example: A bank gave your friend a $1000 for three years at a simple interest rate of 8% per year. How much money the bank should be receiving from your friend at the end of these three years?  Notation P= Principal amount = $1,000 i= Interest rate = 8% n= Number of interest periods = 3 years

6. Simple Interest Formula If you borrow an amount P for n years at a rate of i% a year, then after n years you will have: F = P + n (i P) = P + P in = P (1 + in) where P = Principal amount i= Simple interest rate n = Number of interest periods F = Total amount accumulated at the end of period n Previous Example: F = $1,000 + (3) (0.08) ($1,000) = $1,240 Note: The borrower has used the $80 for 3 years without paying interest on it.

7. Compound Interest Compounded interest is interest that is charged on the original sum and any accumulated, un-paid interest.  Interest = (principal + all accrued interest) (interest rate) Example: A bank gave your friend a $1000 for three years at a compound interest rate of 8% per year. How much money the bank should be receiving from your friend at the end of these three years?

8. Compounding Process

9. Example 3-4: Repaying a Debt You borrowed $5,000 from a bank at 8% interest rate and you have to pay it back in 5 years. There are many ways the debt can be repaid: Plan A: At end of each year pay $1,000 principal plus interest due. Plan B: Pay interest due at end of each year and principal at end of five years. Plan C: Pay in five end-of-year payments. Plan D: Pay principal and interest in one payment at end of five years.

10. Example (cont'd) Plan A: At end of each year pay $1,000 principal plus interest due.

11. Example (cont'd) Plan B: Pay interest due at end of each year and principal at end of five years.

12. Example (cont'd) Plan C: Pay in five end-of-year payments.

13. Example (cont'd) Plan D: Pay principal and interest in one payment at end of five years.

14. Example (cont'd) Summary of Payment Plans Question: What common property do all four plans have? Interest rate = (total interest paid)/(total amount owed) or Total interest paid = (interest rate) * (total amount owed) Since the total amount owed vary for the four plans, but the interest rate does not, the total interest paid also varies. We can simply say that these four plans are EQUIVALENT.

15. Equivalence Different sums of money at different times may be equivalent (equal in economic value) to each other.  Each of the plans on the previous slide is equivalent because each repays $5000 at the same 8% interest rate. $100 now is said to be equivalent to $106 one year from now, if the $100 is invested at the interest rate of 6% per year.  If two or more situations are to be compared, their characteristics must be placed on an equivalent basis.  Equivalence is dependent on interest rate.

16. Compound Interest Formula Math Notation:  i= interest rate (per time period)  n = # of time periods  P = money at present (a.k.a. Principal)  F = money in future after n time periods  Equivalent to P now, at interest rate I  A = payment at end of each time period, e.g., annual Assumptions:  All cash flow occurs at the end of each time period  No inflation, depreciation, or income taxes  Stable prices

17. Single Payment Compound Interest Formula n = 0:P n = 1:F 1 = P +P i= P (1+i) n = 2:F2 = F1(1+i) = P (1+i) 2. n = N:F= P (1+i)N The Fundamental Law of Engineering Economy  If you put Pin the bank now at an interest rate of i% for n years, the future amount F you will have after n years is given by F = P (1+i) n  The term (1+i) n is called the single-payment compound factor.  The factor is used to compute F, given P, and given I and n.

18. Standard Notation to Represent Interest Factors  Consists of two cash flow symbols, the interest rate, and number of compounding periods  General form: (X/Y, i%, n)  X represents what is unknown  Y represents what is known  I and n represent input parameters; can be known or unknown depending upon the problem  Example: (F/P,6%,20) is read as:  To find F, given P when the interest rate is 6% and the number of time periods equals 20.  In problem formulation, the standard notation is often used in place of the closed-form equivalent relations (factor)  Example: The term (F/P, i, n) is equivalent to (1+i) n ; therefore we can write: F = P (1+i)n= P (F/P, i%, n)  Tables at the back of the text provide tabulations of common values for i% and n

19. Single Payment Interest Factors

20. Example 1 A bank gave your friend a $1000 for three years at a compound interest rate of 8% per year. How much money the bank should be receiving from your friend at the end of these three years? F = P (F/P, i, n) = P (1+i) n = $1,000 (1+ 0.08) 3 = $1,259.71

21. Example 2 If you deposit $100 now (n= 0) and $200 two years from now (n= 2) in a savings account that pays 10% interest, how much would you have at the end of year 10? Solution:

22. Example 3 Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn 10% interest, what would be the balance at the end of 4 years?

23. Example 3 -Tabular Solution

24. Example 4 If you want to have $800 in savings at the end of four years, and 5% interest is paid annually, how much do you need to put into the savings account today? Solution: We solve P (1+i) n = F for P with i= 0.05, n = 4, F = $800 P = F/(1+i) n = F(1+i) -n = 800/(1.05)4= 800 (1.05) -4 = 800 (0.8227) = $658.16 Note: Single Payment Present Worth Formula: P = F/(1+i)n= F(1+i) -n P = F (P/F,i,n) The factor (1+i) -n is called the present worth OR discount amount factor

25. Factors in the Book (Appendix B)

26. Tables of Conversion Factors All engineering economics textbooks provide tables of the various conversion factors: Usually in an appendix at the end of the text Refer to the back of your text for those tables The unknown interest rate that you want will typically not be included in those tables: But you can interpolate between two tabulated values to estimate it Linear interpolation is not exact, because: The conversion factors are non-linear! Therefore, interpolation can cause errors: Typically from 2-5% Errors will be small when interpolating between values that are close to each other

27. Example 5 In 3 years, you need $400 to pay a debt. In two more years, you need $600 more to pay a second debt. How much should you put in the bank today to meet these two needs if the bank pays 12% per year?

28. Example 6

29. Solution

31. Remember!

32. Next Class Read Chapter 3 of textbook Try to do some of the problems at end of Chapter 3 Next time we’ll learn more interesting formulas!