1. Contemporary Engineering Economics, 4 th edition ©2007 Time Value of Money Lecture No.4 Chapter 3 Contemporary Engineering Economics Copyright © 2006

2. Contemporary Engineering Economics, 4 th edition © 2007 Chapter Opening Story —Take a Lump Sum or Annual Installments  Mrs. Louise Outing won a lottery worth $5.6 million.  Before playing the lottery, she was offered to choose between a single lump sum $2.912 million, or $5.6 million paid out over 20 years (or $280,000 per year).  She ended up taking the annual installment option, as she forgot to mark the “Cash Value box”, by default.  What basis do we compare these two options?

3. Contemporary Engineering Economics, 4 th edition © 2007 YearOption A (Lump Sum) Option B (Installment Plan) 0 1 2 3 19 $2.912M$283,770 $280,000

4. Contemporary Engineering Economics, 4 th edition © 2007 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. Contemporary Engineering Economics, 4 th edition © 2007 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.  Interest is the cost of money—a cost to the borrower and an earning to the lender This a two-edged sword whereby earning grows, but purchasing power decreases (due to inflation), as time goes by.

6. Contemporary Engineering Economics, 4 th edition © 2007 The Interest Rate

7. Contemporary Engineering Economics, 4 th edition © 2007 Cash Flow Transactions for Two Types of Loan Repayment End of YearReceiptsPayments 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. Contemporary Engineering Economics, 4 th edition © 2007 Cash Flow Diagram for Plan 2

9. Contemporary Engineering Economics, 4 th edition © 2007 End-of-Period Convention

10. Contemporary Engineering Economics, 4 th edition © 2007 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.

11. Contemporary Engineering Economics, 4 th edition © 2007 Simple Interest P = Principal amount i = Interest rate N = Number of interest periods Example:  P = $1,000  i = 10%  N = 3 years End of Year Beginnin g Balance Interest earned Ending Balance 0$1,000 1 $100$1,100 2 $100$1,200 3 $100$1,300

12. Contemporary Engineering Economics, 4 th edition © 2007 Simple Interest Formula

13. Contemporary Engineering Economics, 4 th edition © 2007 Compound Interest P = Principal amount i = Interest rate N = Number of interest periods Example:  P = $1,000  i = 10%  N = 3 years End of Year Beginning Balance Interest earned Ending Balance 0$1,000 1 $100$1,100 2 $110$1,210 3 $121$1,331

14. Contemporary Engineering Economics, 4 th edition © 2007 Compounding Process $1,000 $1,100 $1,210 $1,331 0 1 2 3

15. Contemporary Engineering Economics, 4 th edition © 2007 0 $1,000 $1,331 1 2 3 Cash Flow Diagram

16. Contemporary Engineering Economics, 4 th edition © 2007 Relationship Between Simple Interest and Compound Interest

17. Contemporary Engineering Economics, 4 th edition © 2007 Compound Interest Formula

18. Contemporary Engineering Economics, 4 th edition © 2007 Some Fundamental Laws The Fundamental Law of Engineering Economy

19. Contemporary Engineering Economics, 4 th edition ©2007 Compound Interest “The greatest mathematical discovery of all time,” Albert Einstein

20. Contemporary Engineering Economics, 4 th edition © 2007 Practice Problem: Warren Buffett’s Berkshire Hathaway Went public in 1965: $18 per share Worth today (June 22, 2006): $91,980 Annual compound growth: 23.15% Current market value: $115.802 Billion If his company continues to grow at the current pace, what will be his company’s total market value when reaches 100? ( lives till 100 (76 years as of 2006)

21. Contemporary Engineering Economics, 4 th edition © 2007 Market Value Assume that the company’s stock will continue to appreciate at an annual rate of 23.15% for the next 24 years.

22. Contemporary Engineering Economics, 4 th edition © 2007 EXCEL Template 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 2006, 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?

23. Contemporary Engineering Economics, 4 th edition © 2007 Excel Solution =FV(8%,380,0,1) = $5,023,739,194,020

24. Contemporary Engineering Economics, 4 th edition © 2007 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?

25. Contemporary Engineering Economics, 4 th edition © 2007 Solution 0 1 2 3 4 5 6 7 8 9 10 $100 $200 F

26. Contemporary Engineering Economics, 4 th edition © 2007 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

27. Contemporary Engineering Economics, 4 th edition © 2007 $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

28. Contemporary Engineering Economics, 4 th edition © 2007 Solution 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