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Chapter 3:Interest and Equivalence

BSEN 206. Engineering Economy Jeffrey C. Woldstad, Ph.D., P.E.

Time Value of Money - The most important concept in this class is that the value

of money changes over timeand that you must taketime into account when comparing economic alternatives.

- Would you rather have $1000 today or $1000 ten yearsfrom now?

- Money must be thought of as an asset. If you have money,other people are willing to pay for the use (or rent) of it.If you do not have any money, you must be will to pay forthe use of other peoples money.

- The amount paid for the use of money is called interest.

- The interest required to borrow money depends on boththe market value and how likely other people judge youare to return the money you borrowed.

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Problem 2If we borrowed $20,000 (to buy a car) at 6%simple annual interest for 4 years, how much

would we need to pay back at the end on this

period?(A) $20,000(B) $24,200(C) $24,800(D) $28,400

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Problem 3If we borrowed $200,000 (to buy a house) at 4%simple annual interest for 30 years, how much

would we need to pay back at the end on this

period?(A) $200,000(B) $240,000(C) $320,000

(D) $440,000

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Compound Interest- Most people who lend money think about

both the interest to be paid and therepayment of the loan in specific timeintervals notice that even for simpleinterest, the rate is specified as an annualrate.

- If we associate loan interest and repaymentwith a particular interval, then if makessense to calculate the interest owed at theend of each period.

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EXAMPLE- $10,000 borrowed for 10 years at an annual

interest rate of 8%

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YEAR+ PRINCIPLE (P) INEREST OWED AMOUNT DUE1 $10,000 $10,000*0.08*1= $800 $10,8002 $10,800 $10,800*0.08*1= $864 $11,6643 $11,664 $11,664*0.08*1= $933 $12,5974 $12,597 $12,597*0.08*1= $1008 $13,6055 $13,605 $13,605*0.08*1= $1088 $14,6936 $14,693 $14,693*0.08*1= $1175 $15,8697 $15,869 $15,869*0.08*1= $1269 $17,1388 $17,138 $17,138*0.08*1= $1371 $18,5099 $18,509 $18,509*0.08*1= $1481 $19,99010 $19,990 $19,990*0.08*1= $1599 $21,589

Compared to a simple interest loan - $10,000*(1+(0.08*10)) = $18,000

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EXAMPLE: Repay of a loan of $5000 in 5 yrs at

interest rate of 8%

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Plan #1: Constant principal payment plus interest due

Yr

Balance at

the

Beginning

of year Interest

Balance at

the end of

yearInterest

PaymentPrincipal

PaymentTotal

Payment

1 $5,000.00 $400.00 $5,400.00 $400.00 $1,000.00 $1,400.00

2 $4,000.00 $320.00 $4,320.00 $320.00 $1,000.00 $1,320.00

3 $3,000.00 $240.00 $3,240.00 $240.00 $1,000.00 $1,240.00

4 $2,000.00 $160.00 $2,160.00 $160.00 $1,000.00 $1,160.00

5 $1,000.00 $80.00 $1,080.00 $80.00 $1,000.00 $1,080.00

Subtotal $1,200.00 $5,000.00 $6,200.00


interest rate of 8%

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Plan #2: Annual interest payment and principal payment at end of 5 yrs

Yr

Balance at

the

Beginning

of year Interest

Balance at

the end of

yearInterest

PaymentPrincipal

PaymentTotal

Payment

1 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00

2 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00

3 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00

4 $5,000.00 $400.00 $5,400.00 $400.00 $0.00 $400.00

5 $5,000.00 $400.00 $5,400.00 $400.00 $5,000.00 $5,400.00

Subtotal $2,000.00 $5,000.00 $7,000.00

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interest rate of 8%

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Plan #3: Constant annual payments

Yr

Balance atthe

Beginning

of year Interest

Balance at

the end of

yearInterest

PaymentPrincipal

PaymentTotal

Payment

1 $5,000.00 $400.00 $5,400.00 $400.00 $852.28 $1,252.28

2 $4,147.72 $331.82 $4,479.54 $331.82 $920.46 $1,252.28

3 $3,227.25 $258.18 $3,485.43 $258.18 $994.10 $1,252.28

4 $2,233.15 $178.65 $2,411.80 $178.65 $1,073.63 $1,252.28

5 $1,159.52 $92.76 $1,252.28 $92.76 $1,159.52 $1,252.28Subtotal $1,261.41 $5,000.00 $6,261.41


interest rate of 8%

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Plan #4: All payment at end of 5 years

Yr

Balance at

the

Beginning

of year Interest

Balance at

the end of

yearInterest

PaymentPrincipal

PaymentTotal

Payment

1 $5,000.00 $400.00 $5,400.00 $0.00 $0.00 $0.00

2 $5,400.00 $432.00 $5,832.00 $0.00 $0.00 $0.00

3 $5,832.00 $466.56 $6,298.56 $0.00 $0.00 $0.00

4 $6,298.56 $503.88 $6,802.44 $0.00 $0.00 $0.00

5 $6,802.44 $544.20 $7,346.64 $2,346.64 $5,000.00 $7,346.64Subtotal $2,346.64 $5,000.00 $7,346.64

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Equivalence- Note that if you believe an 8% annual interest rate

is correct, you would not care which of theseplans was used to repay the loan.

- Using the concept of equivalence, one canconvert different types of cash flows at differentpoints of time to an equivalent value at a

common reference point. - Equivalence is dependent on interest rate

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Single Payment Compound Interest Formula

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YearAmount at the

Beginning of

Interest Period + Interest forPeriod = Amount at End of InterestPeriod1 P + iP = P+ iP = P(1+i)

2 P(1+i) + i P(1+i) = P(1+i) + i P(1+i) = P(1+i)(1+i)= P(1+i)2

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

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

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Single Payment Compound Interest Formula

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Notation:i = interest rate per compounding period n = number of compounding periods P = a present sum of moneyF = a future sum of money

n)i,P(F/P,F =

Find F, given P, at i, over n

ni)P(1F +=

Single Payment Compound Amount Formula

Compound Interest Table 6.0%

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Problems 4 & 5

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$500 (P) is deposited in a saving account that pays 6%compounded annually for 3 years what is the future

amount F?

0 1 2 3

P=500

F=?

i=6%

Is this the same as if we borrowed $500 for 3 years at 6%

interest? (A) YES; (B) NO

(A) $590 (B) $595 (C) $600 (D) $610

Problem 6

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P=?

F=800

i=4% 0 1 2 3 4

You wish to have $800 (F) at the end of 4 years, how muchshould be deposited (P) in an account that pays 4% annually?

(A) $684(B) $712(C) $756(D) $800

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Single Payment Present Worth Formula

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Notation:i = interest rate per compounding period n = number of compounding periods P = a present sum of moneyF = a future sum of money

P = F(P/F, i, n)Find F, given P, at i, over n

P = F(1+ i)!n

Single Payment Present Worth Formula


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Problem 7

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$500 (P) is deposited in a saving account that pays6%, compounded quarterly for 3 years, what is thefuture amount F?

i = 6%/4 = 1.5%

n = 3 x 4 = 12 quarters

F = P(1+i)n= P(F/P, i, n)

= 500(1+0.015)12= 500(F/P,1.5%,12)

= 500(1.196) = $598.00

P=500

F=?

i=1.5%

0 1 2 1211

(A) $598 (B) $604 (C) $610 (D) $648

Nominal and Effective Interest Rate - Because compounding charges interest on the

interest, it should be clear that the morecompounding periods used, the more interest that

can be collected.- Consider $1000 borrowed for 1years at a 6%

nominal annual interest.

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Number ofcompounding periods Amount owed after10 years Effective interest rate1 (annually) F= (1000)(1+0.06)1 = $1060.00 6.000 % 4 (quarterly) F= (1000)(1+0.015)4 = $1061.36 6.136 % 12 (monthly) F= (1000)(1+0.005)12 = $1061.67 6.167%24 (bi-monthy) F= (1000)(1+0.0025)24 = $1061.75 6.175 % 365 (daily) F= (1000)(1+0.00016)365 = $1061.83 6.183 %

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Nominal and Effective Interest Rate - The Nominal Interest Rateper year (r) is

the annual interest rate without consideringcompounding this is also called the Annual

Percentage Rate (APR).- The Effective Interest Rate per year (ia) is

the annual interest rate considering compounding.

It increases as the number of compoundingperiods mincreases.

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ia = 1+

r

m

!

"#

$

%&m

'1

Problem 8

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You borrow money for a car at a nominal (APR) interestrate of 8%. This interest in compounded monthly whatis the effective interest rate.

ia = 1+

r

m

!

"#

$

%&

m

'1= 1+0.08

12

!

"#

$

%&

12

'1= 0.0830 = 8.3%

(A) 8.0%(B) 8.3%(C) 8.4%(D) 8.8%

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EXAMPLE

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If I give you $50 today, you owe me $60 on the a weekfrom today.

%13104001%)201(1)i1(i 52ma =!+=!+=

a) Weekly interest rate = ($60-50)/50 = 20% Nominal annual rate = 20% * 52 = 1040%

c) End-of-the-year balance

200,655$%)201(50)i1(PF 52n =+=+=

b) Effective annual rate

Continuous Compounding- Notice that as the number of compounding periods

increases, the effective interest rate increases, but ata slower and slower rate.

- By using calculus you can show that as the number ofperiods goes to infinity, the effective annual interestrate converges to:

-

The single payment compound amount formula for nyears at a nominal annual interest rate r then is:

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ia = e

r!1

F = P ern( ) = P[F/ P, r,n]

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Continuous Compounding

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Nominal Effective Annual Rate when compounded

Rate Yearly Semiannually Quarterly Monthly Daily Continuously1% 1% 1.0025% 1.0038% 1.0046% 1.0050% 1.0050%

2% 2% 2.0100% 2.0151% 2.0184% 2.0201% 2.0201%

3% 3% 3.0225% 3.0339% 3.0416% 3.0453% 3.0455%

4% 4% 4.0400% 4.0604% 4.0742% 4.0808% 4.0811%

5% 5% 5.0625% 5.0945% 5.1162% 5.1267% 5.1271%

6% 6% 6.0900% 6.1364% 6.1678% 6.1831% 6.1837%

8% 8% 8.1600% 8.2432% 8.3000% 8.3278% 8.3287%

10% 10% 10.2500% 10.3813% 10.4713% 10.5156% 10.5171%

15% 15% 15.5625% 15.8650% 16.0755% 16.1798% 16.1834%25% 25% 26.5625% 27.4429% 28.0732% 28.3916% 28.4025%

Problems 9-11 (3.6 &3.10) 3-6. How long will it take for aninvestment to double at 4% simpleinterest per year? 3-10. How long will it take for aninvestment to double at 4%compounded annually?How long will it take for an investment

to double at 4% compoundedcontinuously?

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(A) 15.0 years(B) 17.3 years(C) 17.7 years(D) 25.0 years

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Problem 12 (3-27) 3-27. In 1995 an anonymous private collector

purchased a painting by Picasso entitledAngel Frenandez

de Soto for $29,152,000. The painting was done in 1903

and was valued then at

$600. If the painting was

owned by the same familyuntil its sale in 1995, what

rate of return did they

receive on the $600investment.

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(A) 8.3%(B) 12.4%(C) 22.2%(D) 110%

F=P*(1+i)^(n) => 29152000 = 600*(1+i)^92 => 48586.67 = (1+i) 92 => LOG(48586.67) =92*LOG(1+i) => LOG(1+i) = 0.05094 => 1+i = 10^0.05094 = 1.124 => i = 12.4%

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EXAMPLE- 3-32. Sally is buying a car that costs $12,000. Shewill pay $2000 immediately and the remaining

$10,000 in four annual end-of-year principlepayments of $2500 each. In addition to the$2500, she must pay 15% interest on the unpaidbalance of the loan each year. Prepare a cash flowdiagram to represent this situation.

33$2000

$12,000i=15%

0 1 2 3 4

$2500$1500$4000 $2500

$1125$3625 $2500

$750$3250 $2500

$375$2875

TOTAL PAYMENTS - $15,750

Problem 13 (3-61) 3-61. A friend was left $50,000 by his uncle. He hasdecided to put it into a savings account for the nextyear or so. He finds there are various interest rates atsavings institutions:

(A) 4.375% compounded annually,(B) 4.25% compounded quarterly, and(C) 4.125% compounded continuously.

Which interest rate should he choose to get the highestreturn?

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Case 1 4.375% compounded annually is the nominal rateCase 2 4.25% compounded quarterly - Ia= (1+(r/m))^m -1 = (1+(0.0425/4))^4 -1 =

(1.010625)^4 1 = 1.043182 -1 = 0.043182 = 4.318% Case 3 4.125% compounded continuously - Ia= e^r -1 = e^(0.04125) 1 = 1.0142113 1 =

0.042113 = 4.211%

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