Chapter 2 Time Value of Money Interest: The Cost of Money Economic Equivalence Development of Interest Formulas Unconventional Equivalence Calculations

Decision Dilemma—Take a Lump Sum or Annual Installments A suburban Chicago couple won the Power-ball. They had to choose between a single lump sum $104 million, or $198 million paid out over 25 years (or $7.92 million per year). The winning couple opted for the lump sum. Did they make the right choice? What basis do we make such an economic comparison?

Option A (Lump Sum) Option B (Installment Plan) 1 2 3 25 $104 M $7.92 M

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 points 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.

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

Delaying Consumption

Delaying Consumption

Notation for interest calculations An: A discrete payment or receipt occurring at the end of some interest period i: Interest rate per period N: Total number of interest periods P: A sum of money at a time chosen for purposes of analysis as time zero – present value/worth F: A future sum of money at the end of the analysis period

Example: Paying back a loan You get a loan of $20000 from a bank at a 9% annual interest rate. You also pay a $200 loan origination fee when the loan commences (begins). The bank offers two repayment plans. Plan 1: Equal payments at the end of every year for the next 5 years Plan 2: Single payment at the end of the loan period (5 years)

Repayment Plans End of Year Receipts Payments Plan 1 Plan 2 Year 0 $20,000.00 $200.00 Year 1 A=? Year 2 Year 3 Year 4 Year 5 F=?

Repayment Plans End of Year Receipts Payments Plan 1 Plan 2 Year 0 $20,000.00 $200.00 Year 1 5,141.85 Year 2 Year 3 Year 4 Year 5 30,772.48 P = $20,000, A = $5,141.85, F = $30,772.48

Cash Flow Diagram

End-of-Period Convention Beginning of Interest period End of interest period 1 Interest Period Important simplifying assumption: All cash flows are placed at the end of an interest period.

Methods of Calculating Interest Simple interest: the practice of charging an interest rate only to an initial sum (principal amount) – even though you do not withdraw it Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

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 $1,000 1 $80 $1,080 2 $1,160 3 $1,240

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 $1,000 1 $80 $1,080 2 $86.40 $1,166.40 3 $93.31 $1,259.71

Compounding Process $1,080 $1,166.40 $1,259.71 1 $1,000 2 3 $1,080 $1,259.71 1 $1,000 2 3 $1,080 $1,166.40

Compound Interest The Fundamental Law of Engineering Economy End of 1st period: End of 2nd period: End of 3rd period: At the end of N periods: The Fundamental Law of Engineering Economy

Comparing Simple to Compound Interest

Economic Equivalence Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal. Economic equivalence refers to the fact that a cash flow (which can either be a single payment or a series of payments) can be converted to an equivalent cash at any point in time.

Economic Equivalence The compound interest formula expresses the equivalence between some present amount P and future amount F for given i and N. Equivalent cash flows are equivalent at any common point in time.

Economic Equivalence N F P If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N. F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i.

Typical Repayment Plans for a Bank Loan of $20,000 Repayments Plan 1 Plan 2 Plan 3 Year 1 $5,141.85 $1,800.00 Year 2 5,141.85 1,800.00 Year 3 Year 4 Year 5 $30,772.48 21,800.00 Total of payments $25,709.25 $29,000.00 Total interest paid $5,709.25 $10,772.48 $9,000.00

Equivalence Between Two Cash Flows You are offered the alternative of receiving either $3000 at the end of 5 years or P dollars today. You have no current need for the money, you would deposit the P dollars in an account paying 8% interest. What value of P would make you indifferent to your choice between P dollars today and the promise of $3000 at the end of 5 years?

Equivalence Between Two Cash Flows Step 1: Determine the base period, say, year 5. Step 2: Identify the interest rate to use. Step 3: Calculate equivalence value. $3,000 $2,042 5

Equivalent Cash Flows are Equivalent at Any Common Point In Time

Equivalence Calculations with Multiple Payments

Practice problem 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? ? $1,210 1 4 2 3 $1,500 $1,000 $1,000

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

Solution n = 0 n = 1 n = 2 n = 3 n = 4 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

Practice Problem 2P How many years would it take an investment to double at 10% annual interest? N = ? P

Solution P 2P N = ?

Practice Problem $500 $1,000 0 1 2 3 0 1 2 3 $502 $502 $502 A B At what interest rate would you be indifferent between the two cash flows?

Approach $1,000 $500 Step 1: Select the base period to compute the equivalent value (say, n = 3) Step 2: Find the net worth of each at n = 3. A 0 1 2 3 $502 $502 $502 B 0 1 2 3

Establish Equivalence at n = 3 Find the solution by trial and error, say i = 8%

Types of Cash Flows Single cash flow Equal (uniform) payment series Linear gradient series Geometric gradient series Irregular payment series

Single Cash Flow Formula Single payment present worth factor (discount factor) Given: Find: N P

Uneven Payment Series Wilson Technology wishes to set aside money now to invest over the next 4 years. The company can earn 10% on a lump sum deposited now. The money will be withdrawn in the following increments: Year 1: $25000 to purchase computer hardware and software Year 2: $3000 for additional hardware Year 3: no expenses Year 4: $5000 for software upgrades How much must be deposited now?

Uneven Payment Series

Equal Payment Series A 0 1 2 3 4 5 N-1 N F P

Equal Payment Series Compound Amount Factor 0 1 2 3 N A Example 4.13: Given: A = $3,000, N = 10 years, and i = 7% Find: F Solution: F = $3,000(F/A,7%,10) = $41,449.20

Given: F = $5,000, N = 5 years, and i = 7% Find: A Sinking Fund Factor F 0 1 2 3 N A Example 4.15: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F,7%,5) = $869.50

Handling Time Shifts in a Uniform Series 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000 i = 6% First deposit occurs at n = 0

Capital Recovery Factor 1 2 3 N A Example 4.16: Given: P = $250,000, N = 6 years, and i = 8% Find: A Solution: A = $250,000(A/P,8%,6) = $54,075

Equal Payment Series Present Worth Factor 1 2 3 N A Example 2.14:Powerball Lottery Given: A = $7.92M, N = 25 years, and i = 8% Find: P Solution: P = $7.92M(P/A,8%,25) = $84.54M

Example 2.13 Deferred Loan Repayment Plan You borrowed $21061.82 to finance educational expenses. The loan will be paid with five payments. You want to defer the first payment until the end of year 2, but still desire to make five annual equal installments. With 6% interest, what are the annual payments?

Example 2.13 Deferred Loan Repayment Plan 0 1 2 3 4 5 6 A A A A A i = 6% A’ A’ A’ A’ A’ P’ = $21,061.82(F/P, 6%, 1) Grace period

Two-Step Procedure

Example 2.15 Early Savings Plan – 8% interest 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? ? Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000

Option 1 – Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? Age 31 65

Option 2: Deferred Savings Plan 0 11 12 44 Option 2: Deferred Savings Plan $2,000 ?

Option 1: Early Savings Plan $396,644 Option 1: Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 $2,000 $317,253 Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000

Present Value of Perpetuities Perpetuity: stream of cash flows that continues forever

Present Value of Perpetuities  1 2 3 4 5 6 7 8 N  ∞ i = 10 % P=10000

Linear Gradient Series P

Linear Gradient Series

Linear Gradient Series Arithmetic-Geometric Series Let S be the sum

Linear Gradient Series Geometric Series

Linear Gradient Series Letting x=1/(1+i) Gradient Series Present Worth Factor

Gradient Series as a Composite Series

Example $2,000 $1,750 $1,500 $1,250 $1,000 1 2 3 4 5 How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? P =?

Method 1: $2,000 $1,750 $1,500 $1,250 $1,000 1 2 3 4 5 $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03 P =?

Method 2:

Geometric Gradient Series =(P|A1,g,i,N)

Geometric Gradient Series

Geometric Gradient Series

Geometric Gradient: Find P, Given A1,g,i,N N = 5 years A1 = $54,440 Find: P

Geometric Gradient: Find A1, Given F,g,i,N Planning for retirement: accumulate $1000000 in 20 years time Local bank account that pays 8% interest per year Expects that annual income will increase at 6% annually. Start with a deposit of A1 at the end of year 1.

Unconventional Equivalence Calculations If you make 4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount can be withdrawn over 4 subsequent years?

Method 3: A=100(F|P,10%,4)=146.41

Composite Cash Flows $200 $150 $150 $150 $150 $100 $100 $100 $50 $150 $150 $150 $150 $100 $100 $100 $50 1 2 3 4 5 6 7 8 9

Multiple Interest Rates F = ? Find the balance at the end of year 5. 6% 4% 4% 6% 5% 2 4 5 1 3 $400 $300 $500

Solution

Cash Flows with Missing Payments 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Missing payment

Solution P = ? i = 10% Add this cash flow to offset the change $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 Pretend that we have the 10th payment i = 10%

Approach P = ? $100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $100 i = 10% Equivalent Cash Inflow = Equivalent Cash Outflow

Equivalence Relationship

Unconventional Regularity in Cash Flow Pattern $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C Payment is made every other year

Approach 1: Modify the Original Cash Flows $10,000 i = 10% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A A A A A A A A A A A A A A

Relationship Between A and C $10,000 i = 10% 2 3 4 5 6 7 8 9 10 11 12 13 14 1 C C C C C C C $10,000 i = 10% 2 3 4 5 6 7 8 9 10 11 12 13 14 1 A A A A A A A A A A A A A A

Solution i = 10% C A A A =$1,357.46

Approach 2: Modify the Interest Rate Idea: Since cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period. How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%. (1+0.10)(1+0.10) = 1.21

Solution $10,000 i = 21% 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 C C C C C C C

Summary Money has a time value because of its earning power and inflation rate. Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same value. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal. The purpose of developing various interest formulas was to facilitate the economic equivalence computation.