1. Fundamentals of Engineering Economics, ©2008 Time Value of Money Lecture No.2 Chapter 2 Fundamentals of Engineering Economics Copyright © 2008

2. Fundamentals of Engineering Economics, 2 nd edition © 2008 Take a Lump Sum or Annual Installments  Mr. Robert Harris and his wife Tonya won a lottery worth $276 million on February 25, 2008.  The couple weighted a difficult question: whether to take $167 million now or $275 million over 26 years (0r $10.57 million a year).  What basis should the couple compare these two options?

3. Fundamentals of Engineering Economics, 2 nd edition © 2008 YearOption A (Lump Sum) Option B (Installment Plan) 0 1 2 3 25 $167M$10.57M

4. Fundamentals of Engineering Economics, 2 nd edition © 2008 What Do We Need to Know?  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.

5. Fundamentals of Engineering Economics, 2 nd edition © 2008 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).  Time value of money is measured in terms of interest rate.

6. Fundamentals of Engineering Economics, 2 nd edition © 2008 The Interest Rate  Interest is the cost of money—a cost to the borrower and an earning to the lender

7. Fundamentals of Engineering Economics, 2 nd edition © 2008 Cash Flow Transactions for a Loan Repayment Series End of YearReceiptsPayments Year 0$30,000.00 $300.00 Year 17,712.77 Year 27,712.77 Year 37,712.77 Year 47,712.77 Year 57,712.77 The amount of loan = $30,000, origination fee = $300, interest rate = 9% APR (annual percentage rate)

8. Fundamentals of Engineering Economics, 2 nd edition © 2008 Cash Flow Diagram for Plan 1

9. Fundamentals of Engineering Economics, 2 nd edition © 2008  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practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. Methods of Calculating Interest

10. Fundamentals of Engineering Economics, 2 nd edition © 2008 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 Simple Interest

11. Fundamentals of Engineering Economics, 2 nd edition © 2008 Simple Interest Formula

12. Fundamentals of Engineering Economics, 2 nd edition © 2008 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 $80.00$1,080.00 2$1,080$86.40$1,166.40 3 $93.31$1,259.71 Compound Interest

13. Fundamentals of Engineering Economics, 2 nd edition © 2008 Compounding Process $1,000 $1,080 $1,116.40 $1,259.71 0 1 2 3

14. Fundamentals of Engineering Economics, 2 nd edition © 2008 0 $1,000 $1,259.71 1 2 3 Cash Flow Diagram

15. Fundamentals of Engineering Economics, 2 nd edition © 2008 Compound Interest Formula

16. Fundamentals of Engineering Economics, 2 nd edition © 2008 The Fundamental Law of Engineering Economy

17. Fundamentals of Engineering Economics, ©2008 Compound Interest “The greatest mathematical discovery of all time,” Albert Einstein

18. Fundamentals of Engineering Economics, 2 nd edition © 2008 Practice Problem: Warren Buffett’s Berkshire Hathaway (BRK.A) Went public in 1965: $18 per share Worth today (January 9, 2009): $94,750 Annual compound growth: 21.50% Current market value: $104.721 Billion If his company continues to grow at the current pace, what will be his company’s total market value when he reaches 100? (78 years old as of 2009)

19. Fundamentals of Engineering Economics, 2 nd edition © 2008 Estimated Market Value in 2031 Assume that the company’s stock will continue to appreciate at an annual rate of 21.50% for the next 22 years.

20. Fundamentals of Engineering Economics, 2 nd edition © 2008 With EXCEL 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 2009, the total US population would be close to 300 millions. If the total sum would be distributed equally among the population, how much would each person receive?

21. Fundamentals of Engineering Economics, 2 nd edition © 2008 Excel Solution =FV(8%,383,0,1) = $6,328,464,547,578

22. Fundamentals of Engineering Economics, 2 nd edition © 2008 Practice Problem 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. Fundamentals of Engineering Economics, 2 nd edition © 2008 Solution 0 1 2 3 4 5 6 7 8 9 10 $100 $200 F

24. Fundamentals of Engineering Economics, 2 nd edition © 2008 Practice Problem Problem Statement Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn a 10% interest, what would be the balance at the end of 4 years? $1,000 $1,500 $1,210 0 1 2 3 4 ? $1,000

25. Fundamentals of Engineering Economics, 2 nd edition © 2008 $1,000 $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 Solution: Graphical Approach

26. Fundamentals of Engineering Economics, 2 nd edition © 2008 $1,000 $1,500 $1,210 01 2 3 4 ? $1,000 $2,981 Solution: Analytical Approach

27. Fundamentals of Engineering Economics, 2 nd edition © 2008 Solution: Tabular Approach 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