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

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

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

Contemporary Engineering Economics, 4th edition © 2007 Cash Flow Transactions for Two Types of Loan Repayment End of Year Receipts Payments Plan 1 Plan 2 Year 0 $20,000.00 $200.00 Year 1 5,141.85 1,800 Year 2 Year 3 Year 4 Year 5 21,800.00 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate) Contemporary Engineering Economics, 4th edition © 2007

Cash Flow Diagram for Plan 1 Contemporary Engineering Economics, 4th edition © 2007

End-of-Period Convention In practice, cash flows can occur at the beginning or in the middle of an interest period, or indeed, at practically any point in time. One of the simplifying assumptions we make in engineering economic analysis is the end-of-period convention. End-of-period convention: Unless otherwise mentioned, all cash flow transactions occur at the end of an interest period. Contemporary Engineering Economics, 4th edition © 2007

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

Contemporary Engineering Economics, 4th 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practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th 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 Beginning Balance Interest earned Ending Balance $1,000 1 $100 $1,100 2 $1,200 3 $1,300 Contemporary Engineering Economics, 4th edition © 2007

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

Contemporary Engineering Economics, 4th 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 $1,000 1 $100 $1,100 2 $110 $1,210 3 $121 $1,331 Contemporary Engineering Economics, 4th edition © 2007

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

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

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

Warren Buffett’s Berkshire Hathaway Went public in 1965: $18 per share Worth today (April 05, 2010): $121,700 per share Annual compound growth: 21.65% Current market value: $127.7 Billion If his company continues to grow at the current pace, what will be his company’s total market value when he reaches 100? (80 years as of 2010) Contemporary Engineering Economics, 5th edition © 2010

Contemporary Engineering Economics, 4th edition © 2007 Market Value Assume that the company’s stock will continue to appreciate at an annual rate of 21.65% for the next 20 years. F = $127.7B (1 + 0.2165)20 = $6,433.29B Contemporary Engineering Economics, 4th edition © 2007

Example: Comparing Simple with Compound Interest 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 2010, the total US population would be close to 308 millions. If the total sum would be distributed equally among the population, how much would each person receive? Contemporary Engineering Economics, 5th edition © 2010

Contemporary Engineering Economics, 4th edition © 2007 Excel Solution P = $1 i = 8% N = 384 years F = $1 (1+0.08)384 = $ 6,834,741,711,384.36 ~ $ 6.8 Trillion Excel Formula: F = FV(8%,384,0,1) = $6,834,741,711,384.36 Amount per person = F/308 Million = $22,190.72 Contemporary Engineering Economics, 4th edition © 2007

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

Contemporary Engineering Economics, 4th 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? ? $1,210 1 4 2 3 $1,500 $1,000 $1,000 Contemporary Engineering Economics, 4th edition © 2007

Contemporary Engineering Economics, 4th edition © 2007 Solution End of Period Beginning balance Deposit made Withdraw Ending n = 0 $1,000 n = 1 $1,000(1 + 0.10) =$1,100 $2,100 n = 2 $2,100(1 + 0.10) =$2,310 $1,210 $1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,500 $2,710 n = 4 $2,710(1 + 0.10) =$2,981 $2,981 Contemporary Engineering Economics, 4th edition © 2007