Time Value of Money Lecture No.4 Chapter 3 Contemporary Engineering Economics Copyright © 2006 Contemporary Engineering Economics, 4th 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? Contemporary Engineering Economics, 4th edition © 2007 Year 0 1 2 3 Option A (Lump Sum) $2.912M Option B (Installment Plan) $280,000 $280,000 $280,000 $280,000 19 $280,000 Contemporary Engineering Economics, 4th 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. 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 The Interest Rate Contemporary Engineering Economics, 4th edition © 2007 Cash Flow Transactions for Two Types of Loan Repayment End of Year Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Receipts $20,000.00 Payments Plan 1 $200.00 5,141.85 5,141.85 5,141.85 5,141.85 5,141.85 Plan 2 $200.00 0 0 0 0 30,772.48 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 2 Contemporary Engineering Economics, 4th edition © 2007 End-of-Period Convention 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 Simple Interest P = Principal amount End of i = Interest rate Year N = Number of 0 interest periods 1 Example: P = $1,000 i = 10% N = 3 years Beginning Balance Interest earned Ending Balance $1,000 $1,000 $100 $1,100 2 $1,100 $100 $1,200 3 $1,200 $100 $1,300 Contemporary Engineering Economics, 4th edition © 2007 Simple Interest Formula F P ( iP ) N w here P = P rincipal am ount i = sim ple interest rate N = num ber of interest periods F = total am ount accum ulated at the end o f period N F $1, 000 (0.10)($1, 000)(3) $1, 300 Contemporary Engineering Economics, 4th edition © 2007 Compound Interest P = Principal amount End i = Interest rate of N = Number of Year interest periods 0 Example: P = $1,000 i = 10% N = 3 years Beginning Balance Interest earned Ending Balance $1,000 1 $1,000 $100 $1,100 2 $1,100 $110 $1,210 3 $1,210 $121 $1,331 Contemporary Engineering Economics, 4th edition © 2007 Compounding Process $1,100 $1,210 0 $1,331 1 $1,000 2 3 $1,100 $1,210 Contemporary Engineering Economics, 4th edition © 2007 Cash Flow Diagram $1,331 0 1 2 3 F $1, 000(1 0.10) $1,000 $1, 331 Contemporary Engineering Economics, 4th edition © 2007 3 Relationship Between Simple Interest and Compound Interest Contemporary Engineering Economics, 4th edition © 2007 Compound Interest Formula n 0:P n 1 : F1 P (1 i ) n 2 : F2 F1 (1 i ) P (1 i ) n N : F P (1 i ) N Contemporary Engineering Economics, 4th edition © 2007 2 Some Fundamental Laws F V E m i R m a c 2 The Fundamental Law of Engineering Economy F P (1 i ) Contemporary Engineering Economics, 4th edition © 2007 N Compound Interest “The greatest mathematical discovery of all time,” Albert Einstein Contemporary Engineering Economics, 4th 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) Contemporary Engineering Economics, 4th 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. F $115.802 M (1 0.2315) $17.145 trillions Contemporary Engineering Economics, 4th edition © 2007 24 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? Contemporary Engineering Economics, 4th edition © 2007 Excel Solution P $1 i 8% N 3 8 0 years F $ 1(1 0 .0 8) 380 $ 5, 0 2 3, 7 3 9,1 9 4, 0 2 0 =FV(8%,380,0,1) = $5,023,739,194,020 Am ount per person $5, 023, 739,194, 020 300, 000, 000 $16, 746 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 Solution F 0 1 2 3 4 5 6 7 8 9 10 $100(1 0.10) $100(2.59) $259 10 $200(1 0.10) $200(2.14) $429 8 $100 $200 F $259 $429 $688 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 0 1 4 2 $1,000 $1,000 ? 3 $1,500 Contemporary Engineering Economics, 4th edition © 2007 ? $1,210 0 1 3 2 $1,000 $1,000 4 $1,500 $1,100 $1,000 $2,100 $2,310 $1,210 -$1,210 + $1,500 $1,100 $2,710 Contemporary Engineering Economics, 4th edition © 2007 $2,981 Solution End of Period Beginning balance Deposit made Withdraw Ending balance n=0 0 $1,000 0 $1,000 n=1 $1,000(1 + 0.10) =$1,100 $1,000 0 $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 $1,500 0 $2,710 n=4 $2,710(1 + 0.10) =$2,981 0 0 $2,981 Contemporary Engineering Economics, 4th edition © 2007