Engineering Economic AnalysisCanadian Edition

Chapter 3:

Interest and Equivalence

3-2

Chapter 3 … Defines the time value of money. Distinguishes between simple and

compound interest in engineering economic analysis.

Explains equivalence of cash flows. Defines nominal, effective, and periodic

interest rates. Applies single payment compound interest

formulas.

3-3

Chapter Assumptions in Solving Economic Analysis Problems End-of-year convention

• simplifies calculations

Viewpoint of the firm Sunk costs

• past has no bearing on current decisions

Owner-provided capital• no debt capital

Stable prices No depreciation No income taxes

3-4

Economic Decision Components Where economic decisions are immediate or

short term, we need to consider:• amount of expenditure (costs) and benefits

(income)• taxes (if after-tax analysis)

Where economic decisions occur over a considerable period of time, we also need to consider interest and inflation.

3-5

Computing Cash Flows Cash flows have:

• Costs (disbursements) > a negative number• Benefits (receipts) > a positive number

Examples 3-1 & 3-2Interest rate: 10.00%

EOY 1 2(NOW) 0 -$29,100 -$5,000

1 $0 -$8,0002 $0 -$6,0003 $0 -$6,0004 $0 -$6,0005 $0 -$6,000

SUM -$29,100 -$37,000NPV -$29,100 -$29,563

Alternative

3-6

Time Value of Money (TVM) Money has value.

• Money can be leased or rented.• The payment is called interest.• If you put $100 in a bank at 9% interest for one

time period, you will receive your original $100 plus $9.

Original amount to be returned = $100Interest to be returned = $100 9% = $9Total amount received = $109

3-7

Importance of Time Time is a critical factor in engineering

economic studies (except for very short term projects). • There is a preference for consuming goods and

services sooner rather than later.• The stronger the preference for current

consumption, the greater the importance of time in investment decisions.

The amount and timing of a project’s cash flows (operating expenses; revenues/ sales) are crucial to the value of a project’s worth.

3-8

TVM Illustration (Projects A & B)

Project Parameters Project A Project B

Initial investment $10,000 $10,000

Salvage value $0 $0

Life 5 years 5 years

End-of-year cash flows

Year 1 $3,000 $5,000

Year 2 $3,000 $4,000

Year 3 $3,000 $3,000

Year 4 $3,000 $2,000

Year 5 $3,000 $1,000

3-9

Projects A and B have identical initial costs, duration, and cumulative cash flows over 5 years ($15,000) but different cash flow patterns.

If the discount rate i = 0%, then A and B are equivalent.

If i ≠ 0%, then A and B are not equivalent.

TVM Illustration (Projects A & B)…

3-10

Interest Rate The importance of time is inherent in the rate

of interest. The rate of interest is

• the rate of return received by a lender for lending money (foregoing or trading off current consumption for future consumption), or

• the rate of return paid by a borrower for the use of a lender’s funds.

3-11

For discrete cash flows, we have:• Simple interest: interest calculated once only and

paid at the end of the term. • Compound interest: finite number (m) of within-

year compounding periods, e.g.12% compounded monthly (m = 12)12% compounded quarterly (m = 4)12% compounded annually (m = 1)

For continuous cash flows, we have continuous compounding:• Infinite number of within-year compounding

periods, i.e. m ∞.

Interest Rate

3-12

Simple Interest Interest that is computed only on the original

sum or principal. The term is usually up to one year. Total interest earned: I = Pin

• whereP – principal i – interest rate (%) per periodn – number of periods (usually years)

For example: P = $100; i = 9% per year; 6 month term

I = $100 9%/period ½ period = $4.50

3-13

Future Value of a Loan with Simple Interest Amount of money due at the end of a loan

• F = P + Pin; or F = P(1 + in), where F = future value.

Would you accept payment with simple interest terms? Would a bank?

F = $100 (1 + 9% ½) = $104.50

3-14

Simple Interest Simple annual interest rate = 14%

End-of-year

Amount Borrowed

Interest for Period

Amount Owed at EOY

0 $1,000

1 $140 $1,140

2 $140 $1,280

3 $140 $1,420

$1,280 = $1,000+2(14%)$1,000

3-15

Compound Interest Interest that is computed on the original

unpaid principal and on the unpaid interest.• Total interest earned: In = P(1 + i)n – P

• whereP = present sum of money i = interest rate per periodn = number of periods

For example: P = $100; i = 9% compounded annually; 5-year term

I5 = $100 (1+9%)5 $100 = $53.86

3-16

Compound annual interest rate = 14%

End-of-year

Amount Borrowed

Interest for Period

Amount Owed at EOY

0 $1,000

1 $140.00 $1,140.00

2 $159.60 $1,299.60

3 $181.94 $1,481.54

$1,299.60 = $1,000(1+14%)2

Compound Interest

3-17

Future Value of a Loan with Compound Interest Amount of money due at the end of a loan F = P(1+i)(1+i)(1+i) = P(1+i)n; where F =

future value Would you be more likely to accept payment

with compound interest terms? Would a bank?

For example: P = $100; i = 9%; 5-year term

F = $100(1 + 9%)5 = $153.86

3-18

Comparison of Simple and Compound Interest Over Time The difference is negligible

over a short period of time, but over a long period of time it may a considerable amount.• Check the table to see the

difference over time.

Principal: 100.00$ Interest rate: 9.00%

PeriodSimple interest

Compound interest

0 $100.00 $100.001 $109.00 $109.002 $118.00 $118.813 $127.00 $129.504 $136.00 $141.165 $145.00 $153.866 $154.00 $167.717 $163.00 $182.808 $172.00 $199.269 $181.00 $217.19

10 $190.00 $236.7411 $199.00 $258.0412 $208.00 $281.2713 $217.00 $306.5814 $226.00 $334.17

Short or long? When is the difference significant? You choose the time period.

3-19

Four Ways to Repay a Debt

PlanRepay

PrincipalRepay

Interest Interest Paid

1 Equal annual amounts

Interest on unpaid balance

Declines

2 End of loan

Interest on unpaid balance

Constant

3 Equal annual amounts Declines at increasing rate

4 End of loan

Compound and pay at end of loan

Compounds at increasing rate until end of loan

3-20

Loan Repayment — Four Options

Complete the calculator in this spreadsheet.

$5,000 Principal10.00% Interest rate

10 YearsPlan 1 Enter 1 through 4

Years

Amount owed

(beginning of year)

Interest owed for

year

Total owed

(end of year)

Principal payment

Total payment

1 $5,000 $500 $5,500 $500 $1,0002 $4,500 $450 $4,950 $500 $9503 $4,000 $400 $4,400 $500 $9004 $3,500 $350 $3,850 $500 $8505 $3,000 $300 $3,300 $500 $8006 $2,500 $250 $2,750 $500 $7507 $2,000 $200 $2,200 $500 $7008 $1,500 $150 $1,650 $500 $6509 $1,000 $100 $1,100 $500 $600

10 $500 $50 $550 $500 $55011 $0 $2,750 $5,000 $7,750

3-21

Equivalence When an organization is indifferent as to

whether it has an amount of money now (present value) or another amount of money in the future (future value), we say that the present amount of money is equivalent to the future amount of money.

Each of the plans on the previous slide is equivalent because each repays an amount of money that has the same value today at the same interest rate.

3-22

Given the choice of these two plans which would you choose?

Year Plan 1 Plan 2

1 $1400 $400

2 $1320 $400

3 $1240 $400

4 $1160 $400

5 $1080 $5400

Total $6200 $7000

To make a choice the cash flows must be altered (discounted) so a comparison can be made.

Equivalence …

3-23

Technique of Equivalence Select a common point in time. Determine a single equivalent value at that

point in time for plan 1. Determine a single equivalent value at that

point in time for plan 2. Judge the relative attractiveness of the two

alternatives from the comparable equivalent values.

3-24

Repayment PlansEstablish the Interest Rate

$5,000 Principal (Present Value)8.00% Interest rate

5 YearsPlan 1 Enter 1 through 4

Years

Amount owed (beginning of

year)

Interest owed for

year

Total owed (end of year)

Principal payment

Total payment

1 $5,000.00 $400.00 $5,400.00 $1,000.00 $1,400.002 $4,000.00 $320.00 $4,320.00 $1,000.00 $1,320.003 $3,000.00 $240.00 $3,240.00 $1,000.00 $1,240.004 $2,000.00 $160.00 $2,160.00 $1,000.00 $1,160.005 $1,000.00 $80.00 $1,080.00 $1,000.00 $1,080.00

Totals $1,200.00 $5,000.00 $6,200.00Interest paid over time $1,200

Total owed over time $15,000

$4,876.63 Principal (Present Value)9.00% New Interest rate

5 YearsPlan 1

Years

Amount owed (beginning of

year)

Interest owed for

year

Total owed (end of year)

Principal payment

Total payment

1 $4,876.63 $438.90 $5,315.52 $975.33 $1,400.002 $3,901.30 $351.12 $4,252.42 $975.33 $1,320.003 $2,925.98 $263.34 $3,189.31 $975.33 $1,240.00

= 8.00%

Principal outstanding over time

Amount repaid over time

For example,• if F = P(1 + i)n,• then i = (F/P)1/n 1

3-25

Application of Equivalence Calculations Pick an alternative. Which would you choose? Change the interest rate to 8%, 15%, 3%, etc.

3-26

Comparing Alternatives … Compare the investment alternatives for

various interest rates. Is there a rate at which you are indifferent among all alternatives?

10.00%

Year A B C D0 -$1,000 -$2,000 -$3,000 -$4,0001 $150 $300 $450 $6002 $150 $300 $450 $6003 $150 $300 $450 $6004 $150 $300 $450 $6005 $150 $300 $450 $6006 $150 $300 $450 $6007 $150 $300 $450 $6008 $150 $300 $450 $6009 $150 $300 $450 $600

10 $150 $300 $450 $600

P -$78.31 -$156.63 -$234.94 -$313.26A -$12.75 -$25.49 -$38.24 -$50.98F -$203.13 -$406.26 -$609.39 -$812.52

AlternativeInterest rate

3-27

Interest Formulas To understand equivalence, the underlying

interest formulas must be analyzed. Notation:

• i = Interest rate per interest period• n = Number of interest periods• P = Present sum of money (Present value)• F = Future sum of money (Future value)

3-28

Single Payment CompoundInterest

YearBeginning balance

Interest for period

Ending balance

1 P iP P(1+i)

2 P(1+i) iP(1+i) P(1+i)2

3 P(1+i)2 iP(1+i)2 P(1+i)3

n P(1+i)n-1 iP(1+i)n-1 P(1+i)n

P at time 0 becomes P(1+i)n at the end of period n. Stated otherwise, Future value = Present value (1+i)n.

3-29

Notation for Calculating Future and Present Values Formula: F = P(1+i)n is the single payment

future value factor. Find the value in five years of $1000 today if

the rate of interest is 7½%. (Ans: $1435.63) Formula: P = F(1/(1+i)n) = F(1+i)n is the single

payment present value factor. Find the value today of $1000 in five years if

the rate of interest is 7½%. (Ans: $696.56) Note: never use financial tables.

3-30

Nominal, Effective and Periodic Interest Rates Interest rate = 8% compounded quarterly

• Nominal rate = 8% (per year) compounded quarterly, i.e. 4 times per year.

• Effective rate = [1+(8%/4)]4 – 1 = 8.2432%.• Periodic rate (per quarter) = 8%/4 = 2%.

Investing $1 at 2% per quarter is equivalent to investing $1 at 8.2432% annually.

3-31

Nominal, Effective and Periodic Interest Rates … Interest rate = 6¼% compounded monthly

• Nominal rate = 6¼% compounded monthly.• Effective rate = [1+(6¼%/12)]12 – 1 = 6.4322%.• Periodic rate (per month) = 6¼%/12 = 0.520833%.

The spreadsheet compares effects of interest specified as nominal and effective rates.

7.50%1210.00 years

i n $1.00 $500.00 $1.00 $500.007.50% 10.00 $2.06 $1,030.52 $0.49 $242.600.63% 120.00 $2.11 $1,056.03 $0.47 $236.74

Comparing the future value and present value generated by the specified interest rate when it is interpreted as effective (annually compounded) and as nominal (compounded the specified number of periods per year).

Future value Present value

Nominal rate:Compounded:

Term:times/year

3-32

Examples If you borrow $5000 today, then $10,000 in

eight months, calculate the two payments you must make in six months and one year, with the second payment being twice the size of the first, to repay the debt if interest is 6¾% per year.• $5129.79 in six months and $10,259.58 in one

year, using one year from now as the focal date• $ 5131.67 in six months and $10,263.34 in one

year, using the sequential payment technique

3-33

Examples … Determine which of the following you would

rather have: A. $10,000 today; B. $5000 today and $7000 in four years; or C. $3000 at the end of each of the next four years. Your opportunity cost of capital (interest rate) is 8½% compounded annually.• B is best with a value of $10,051.02 today.• A is second with a value of $10,000 today.• C is worst with a value $9826.79 today.

3-34

Examples … Find how long it takes for $10,000 invested

today to reach a value of $11,600.78 if the interest rate is 4¼% compounded monthly.• 42 months or 3½ years

You promise to pay $10,000 in 30 months to a lender who gives you $8731.54 today. Find the interest rate the lender is charging. Express it as a nominal rate compounded semiannually and as an effective rate.• 5½% compounded semiannually• 5.575625% effective

3-35

Suggested Problems 3-2, 3-6, 3-9, 3-10, 3-13, 3-15, 3-16, 3-17, 3-

19, 3-20.