1. Lecture No.3

2.  Interest: The Cost of Money  Economic Equivalence  Interest Formulas – Single Cash Flows  Equal-Payment Series  Dealing with Gradient Series  Composite Cash Flows. Power-Ball Lottery

3.  To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different point in time.  To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.

4.  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).  Time value of money is measured in terms of interest rate.  Interest is the cost of money—a cost to the borrower and an earning to the lender

6. Which Repayment Plan? End of Year ReceiptsPayments Plan 1Plan 2 Year 0$20,000.00 $200.00 Year 15,141.850 Year 25,141.850 Year 35,141.850 Year 45,141.850 Year 55,141.8530,772.48 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)

8. Beginning of Interest period End of interest period 0 1 10

9. Methods of Calculating Interest Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

10. Simple Interest P = Principal amount i = Interest rate N = Number of interest periods Example:  P = $1,000  i = 8%  N = 3 years End of Year Beginning Balance Interest earned Ending Balance 0$1,000 1 $80$1,080 2 $80$1,160 3 $80$1,240

12. Compound Interest Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

13. Compound Interest P = Principal amount i = Interest rate N = Number of interest periods Example:  P = $1,000  i = 8%  N = 3 years End of Year Beginning Balance Interest earned Ending Balance 0$1,000 1 $80$1,080 2 $86.40$1,166.40 3 $93.31$1,259.71

14. $1,000 $1,080 $1,166.40 $1,259.71 0 1 2 3

15. 0 $1,000 $1,259.71 1 2 3

17. The Fundamental Law of Engineering Economy

18.  Went public in 1965: $18 per share  Worth today (August 22, 2003): $76,200  Annual compound growth: 24.58%  Current market value: $100.36 Billion  If he lives till 100 (current age: 73 years as of 2003), his company’s total market value will be ?

19.  Assume that the company’s stock will continue to appreciate at an annual rate of 24.58% for the next 27 years.

20. In 1626 the Indians sold Manhattan Island to Peter Minuit of the Dutch West Company for $24. If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now? As of Year 2003, the total US population would be close to 275 millions. If the total sum would be distributed equally among the population, how much would each person receive?

21. =FV(8%,377,0,1) = $3,988,006,142,690

22.  Problem Statement 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?

23. 0 1 2 3 4 5 6 7 8 9 10 $100 $200 F

24.  Problem Statement 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? $1,000 $1,500 $1,210 0 1 23 4 ? $1,000

25. $1,500 $1,210 01 2 3 4 ? $1,000 $1,100 $2,100$2,310 -$1,210 $1,100 $1,210 + $1,500 $2,710 $2,981 $1,000

26. End of Period Beginning balance Deposit made WithdrawEnding balance n = 0 0$1,0000 n = 1 $1,000(1 + 0.10) =$1,100 $1,0000$2,100 n = 2 $2,100(1 + 0.10) =$2,310 0$1,210$1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,5000$2,710 n = 4 $2,710(1 + 0.10) =$2,981 00$2,981