chapter 1 - equivalence and time value of money 5 · types of cash flow single payment r100 012 354...

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Chapter 1 - Equivalence and Time Value of Money 1

Introduction

RECOGNIZE A DECISION PROBLEM

DEFINE THE GOALS AND

OBJECTIVES

COLLECT DATA AND

INFORMATION

IDENTIFY A SET OF DECISION

ALTERNATIVES

SELECT DECISION CRITERION

TO USE

SELECT THE BEST

ALTERNATIVE

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Introduction

BUY A CAR

ALTERNATIVES DECISION

GOALS & OBJECTIVES

- Lower monthly payment- Enough room- Oil usage- Suitable for site visit, etc.

SURVEYING

- Type of car- Price and salvage value- Convenience- Specification, etc

- Sedan cars- Pick-up cars- Family cars

CRITERION TO USE

- Less RM 1000/month- Enough for 7 persons- Suitable for far distances- Diesel, 4WD, etc.

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Time Value of Money

$100 CASHTODAY$100 A YEAR FROM

NOW, AFTER SAVEDINTO A BANK

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Time Value of Money

$7,000$5,000$2,000$5,400$5,000$5,400$400$5,0005

$400$0$5,400$400$5,0004

$400$0$5,400$400$5,0003

$400$0$5,400$400$5,0002

$400$0$5,400$400$5,0001

PAY INTEREST DUE AT END OF EACH YEAR AND PRINCIPAL AT END OF FIVE YEARSPlan 2

$6,200$5,000$1,200

$1,080$1,000$1,080$80$1,0005

$1,160$1,000$2,160$160$2,0004

$1,240$1,000$3,240$240$3,0003

$1,320$1,000$4,340$320$4,0002

$1,400$1,000$5,400$400$5,0001

AT THE END OF EACH YEAR PAY $ 1000 PRINCIPAL + INTEREST DUEPlan 1

(f)(e)(d) = (b) + (c)(c) = 8% x (b)(b)(a)

TOTAL END-OF YEARPAYMENT

PRINCIPALPAYMENT

TOTAL OWED AT END OF YEAR

INTEREST OWED FOR THAT YEAR

AMOUNT OWED ATTHE BEGINNING OF YEAR

YEAR

Table 1.1 Four Plans for Repayment of $5000 in Five Years with Interest at 8%

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Time Value of Money

TOTAL OWED AT END OF YEAR

$7,347$5,000$2,347$7,347$5,000$7,347$544$6,8035

$0$0$6,803$504$6,2994

$0$0$6,299$467$5,8323

$0$0$5,832$432$5,4002

$0$0$5,400$400$5,0001

PAY PRINCIPAL AND INTEREST IN ONE PAYMENT AT END OF FIVE YEARSPlan 4

$6,260$5,000$1,260$1,252$1,159$1,252$93$1,1595

$1,252$1,074$2,411$178$2,2334

$1,252$994$3,485$258$3,2273

$1,252$921$4,479$331$4,1482

$1,252$852$5,400$400$5,0001PAY IN FIVE EQUAL END-OF-YEAR PAYMENTSPlan 3

(f)(e)(d) = (b) + (c)(c) = 8% x (b)(b)(a)

TOTAL END-OF YEARPAYMENT

PRINCIPALPAYMENT

INTEREST OWED FOR THAT YEAR

AMOUNT OWED ATTHE BEGINNING OF YEAR

YEAR

Table 1.1 Four Plans for Repayment of $5000 in Five Years with Interest at 8% (cont.)

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EquivalenceWhen we are indifferent as to whether we have a quantity of money now or the assurance of some other sum of money in the future, or series of future sums of money, we say that the present sum of money is equivalent to the future sum or series of future sums

Table 1.2 Alternative Plans for Repayment of $5000 in Five Years with Interest at 8%

$7,000$6,200

$5,400$1,0805

$400$1,1604

$400$1,2403

$400$1,3202

$400$1,4001

PLAN 2PLAN 1YEAR

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Types of Cash Flow

SINGLE PAYMENT

R100

0 1 2 43 5

IRREGULAR SERIES

R70 R50

R100

R80R90

0 1 2 43 5

GEOMETRIC GRADIENT

R50R50(1+g)

R50(1+g)2R50(1+g)3

R50(1+g)4

0 1 2 43 5

LINEAR GRADIENT

R50R50+G

R50+2GR50+3G

R50+4G

0 1 2 43 5

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NotesP : A present sum of moneyF : A future sum of moneyi : Interest rate per interest period (decimal)n : Number of interest periods

Single Payment Formulas

P(1+i)n=iP(1+i)n-1+P(1+i)n-1nth year

P(1+i)3=iP(1+i)2+P(1+i)2Third Year

P(1+i)2=iP(1+i)+P(1+i)Second Year

P(1+i)=iP+PFirst Year

AMOUNT AT THE END OF INTEREST PERIOD

=INTEREST FOR PERIOD

+AMOUNT AT BEGINNING OF INTEREST PERIOD

Table 1.3 The Progression of Single payment Formulas

- Compound Amount F = P (1+i)n, orF = P (F/P, i , n)

- Present Worth P = F (1+i) -n, orP = F (P/F, i , n)

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Single Payment FormulasE X A M P L E 1. 1

If you had R 2,000 now and invested it at 10%, how much would it be worth in 8 years?

Cash Flow Diagram

Based on Formulas

F = P (1+i)n

= 2,000 (1+0.1)8

= R 4,287.18

Based on Compound Interest Table

F = P (F/P, i, n)= 2,000 (F/P, 10%, 8)= 2,000 (2.1436)= R 4,287.200 1 2 4 5 6 7 83

+

-

?

R 2,000

i = 10%

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Suppose that R 1,000 is to be received in 5 years. At an annual interest rate of 12%, what is the present worth of this amount?

Cash Flow Diagram

Based on Formulas

P = F (1+i) -n

= 1,000 (1+0.12) -5

= R 567.40


P = F (P/F, i, n)= 1000 (P/F, 12%, 5)= 1,000 (0.5674)= R 567.40

0 1 2 3 4 5

+

-

R 1,000

?

i = 12%

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Suppose you buy a share for R10 and sell it for R20, your profit is R10. If that happens within a year, your rate of return is an impressive 100%. If that takes 5 years, what would be the average annual rate of return on your investment?

Cash Flow Diagram


F = P(F/P, i, n)20 = 10 (F/P, i, 5)

If i = 15%

20 ? 10 (2.0114)20 ? 20.114

If i = 14%

20 ? 10 (1.9254)20 ? 19.254

With iteration,i = 14.87%0 1 2 3 4 5

+

-

R 20

R 10

i = ? %

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You have just purchased 100 shares of General Electric stock at R60 per share. You will sell the stock when its market price has doubled. If you expected the stock price to increase 20% per year, how long do you expect to wait before selling the stock?

Cash Flow Diagram

Based on Formula

F = P (1+i) n

12,000 = 6,000 (1+0. 20) n

2 = (1+0. 20) n

log 2 = n log 1.2n = log 2

log 1.2n = 3.8 4 years

0 n = ?

+

-

R 12,000

R 6,000

i = 20 %

≈

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Uniform Series Formulas

CAPITAL RECOVERY FACTOR

SERIES PRESENT WORTH FACTOR

SERIES COMPOUND AMOUNT FACTOR

SINKING FUND FACTOR

( )( )

n) i,(A/P, PA

1i1i1i PA n

n

=

⎥⎦

⎤⎢⎣

⎡−+

+= ( )

n)i,(F/A, A F

i1i1 A F

n

=

⎥⎦

⎤⎢⎣

⎡ −+=

( )( )

n)i,(P/A,A P

i1i1i1A P n

n

=

⎥⎦

⎤⎢⎣

⎡

+−+

= ( )

n)i,(A/F, F A

1i1i F A n

=

⎥⎦

⎤⎢⎣

⎡−+

=

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Uniform Series FormulasE X A M P L E 1. 5

Suppose you make an annual contribution of R 3,000 to your savings account at the end of each year for 10 years. If your savings account earns 7% interest annually, how much can be withdrawn at the end of 10 years?

Cash Flow Diagram

Based on Formulas


F = A (F/A, i, n)= 3,000 (P/F, 7%, 10)= 3,000 (13.8164)= R 41,449.20

0 n = ?

+

-

R 12,000

R 6,000

i = 20 %

( )

( )

41,449.20R 0.07

10.071 3000

i1i1 A F

10

n

=

⎥⎦

⎤⎢⎣

⎡ −+=

⎥⎦

⎤⎢⎣

⎡ −+=

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Uniform Series FormulasE X A M P L E 1. 6

Biogen company has borrowed R250,000 to purchase an equipment for gene splicing. The loan carries an interest rate of 8% per year and is to be repaid in equal installments over the next 6 years. Compute the amount of this annual installment.

Cash Flow Diagram

Based on Formulas


A = P (A/P, i, n)= 250,000 (A/P, 8%, 6)= 250,000 (0.2163)= R 54,075

( )( )

( )( )

54,075R 10.081

0.0810.08 250,000

1i1i1i PA

6

6

n

n

=

⎥⎦

⎤⎢⎣

⎡−+

+=

⎥⎦

⎤⎢⎣

⎡−+

+=

0 1 2 3 4 6

+

-

R 250,000

?/year

i = 8 %

5

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Combination of P, F and AE X A M P L E 1. 7

To help you reach a R5,000 goal for 5 years from now, your father offers to give you R500 now. You plan to get a part time job and make five additional deposits at the end of each year (the first deposit is made at the end of the first year). If all your money is deposited in a bank that pays 7% interest, how large must your annual deposit be?

Cash Flow Diagram


A = 5,000 (F/P,7%,5)(A/P,7,5) -500(A/P,7%,5)

= 5,000 (1.4026)(0.2439) –500(0.2439)

= R 747.55

0 1 2 3 4 5

+

-

R 5,000

R 500

i = 7 %

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Arithmetic Gradient

NotesP : a present sum of moneyF : a future sum of moneyA : an end-of-period cash receipt or disbursement in a uniform seriesG : uniform period-by-period increase or decrease in cash receipts

(arithmetic gradient)g : uniform rate of cash flow increase or decrease from period to

period (geometric gradient)i : interest rate per interest period (decimal)n : number of interest periods

ARTIHMETIC GRADIENT UNIFORM SERIES

( )( )

n)i,(A/G,GA

ii)1i1ini)1 GA

n

n

=

⎥⎦

⎤⎢⎣

⎡−+−−+

=

( )( )

n)i,(P/G,GP

i1i1ini1 GP n2

n

=

⎥⎦

⎤⎢⎣

⎡+

−−+=

0 1 2 43 5

G 2G3G 4G 5G

P A

ARTIHMETIC GRADIENT PRESENT WORTH

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Arithmetic GradientE X A M P L E 1. 8

A textile mill has just purchased a lift truck that has a useful life of 5 years. The engineer estimates that the maintenance costs for the truck during the first year will be R1,000. Maintenance cost are expected to increase as the truck ages at a rate of R250 per year over the remaining life. Assume the maintenance account that earns 12% annual interest. All future maintenance expenses will be paid out of this account. How much does the firm have to deposit in the account now?

Cash Flow Diagram

0 1 2 3 4 5+

-

P i = 12 %

R 1,000R 1,250

R 1,500R 1,750

R 2,000

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Cash Flow Diagram

P = P1 + P2

= A1 (P/A,12%,5) + G (P/G,12%,5)= 1000 (3.6048) + 250 (6.397)= R 5,204

Arithmetic Gradient

0 1 2 3 4 5

P i = 12 %

R 1,000 per year

0 1 2 3 4 5

P i = 12 %

R 250R 500

R 750R 1,000

+

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Arithmetic GradientE X A M P L E 1. 9

Suppose that you make a series of annual deposits into a bank account that pays 10% interest. The initial deposits at the end of the first year is R1,200. The deposit amounts decline by R200 in each of the next 4 years. How much would you have immediately after the 5th deposit?

Cash Flow Diagram

0 1 2 3 4 5+

-

Fi = 10 %

R 400R 600R 800

R1,000R 1,200

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Cash Flow Diagram

F = F1 - F2

= A1 (F/A,10%,5) - 200 (P/G,10%,5)(F/P,10%,5)= 1200 (6.105) - 200 (6.862)(1.611)= R 5,115

Arithmetic Gradient

0 1 2 3 4 5

Fi = 10 %

R 1,200/year

0 1 2 3 4 5

Fi = 10 %

R 200 R 400R 600

R 800

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Geometric Gradient

NotesP : a present sum of moneyF : a future sum of moneyA : an end-of-period cash receipt or disbursement in a uniform seriesG : uniform period-by-period increase or decrease in cash receipts

(arithmetic gradient)g : uniform rate of cash flow increase or decrease from period to

period (geometric gradient)i : interest rate per interest period (decimal)n : number of interest periods

GEOMETRIC SERIESPRESENT WORTH (when i = g)

GEOMETRIC SERIESPRESENT WORTH (when i g)

( )[ ]

n)i,g,,(P/AAP

i1nAP

11

1

1

=

+= −

≠( ) ( )

gii1g11AP

nn

1 ⎥⎦

⎤⎢⎣

⎡−

++−=

−

n)i,g,,(P/AAP 11=

0 1 2 43 5

A1 A2 A3 A4A5

P Aj=A1(1+g)j-1

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Geometric GradientE X A M P L E 1. 10

A self-employed individual, Jimmy Carpenter, is opening a retirement account at a bank. His goal is to accumulate R1,000,000 in the account by the time he retires from work in 20 years time. A local bank is willing to open a retirement account that pays 8% interest, compounded annually, throughout the 20 years. Jimmy expects his annual income will increase at a 6% annual rate during his working career. He wishes to start with deposit at the end of year 1 (A1) and increase deposit at a rate of 6% rate each year thereafter. What should be the size of his first deposit (A1)? The first deposit will occur at the end of year 1, and subsequent deposit will be made at the end of each year. The last deposit will be made at the end of year 20.

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Cash Flow Diagram

F = A1 (F/A,g,I,n)1,000,000 = A1 (F/A,6%,8%,20)1,000,000 = A1 (72.6911)A1 = R 13,757

Geometric Gradient

0 1 2 3 4 5 6 7 8 9 10 11 2012 13 14 15 16 17 18 19

A1= R13,757

A20= A1 (1+0.06)19

= R 41,623

R 1,000,000+

-

i = 8 %

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Faedah Mudah

JUMLAH FAEDAH = PRINSIPAL (P) X JUMLAH TAHUN (n) X KADAR FAEDAH (i%)

R1,150R503

R1,100R502

R1,050R501

R1,000R1,0000

5% setahun

JUMLAH HUTANGFAEDAH MUDAHJUMLAH PINJAMANAKHIR TAHUN

Table 1.4 Faedah Mudah dan Jumlah Hutang

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Faedah Kompaun

JUMLAH FAEDAH = [PRINSIPAL (P)+ FAEDAH TERKUMPUL ] X KADAR FAEDAH (i%)

R1,157.63R55.133

R1,102.50R52.502

R1,050.00R50.001

R1,000R1,0000

5% setahun

JUMLAH HUTANGFAEDAH MUDAHJUMLAH PINJAMANAKHIR TAHUN

Table 1.5 Faedah Kompaun dan Jumlah Hutang

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Kaedah Pengiraan Kadar Faedah Nominal dan Efektif

Faedah Kompaun

Jenis Faedah KompaunKadar faedah nominalKadar faedah efektif

Kadar faedah nominal & efektif dinyatakan bagi satu jangka masa kurang daripada 1 tahun

Kaedah pengiraan kadar faedah nominal dan efektif adalah sebagai berikut :

Tidak dinyatakan sama ada kadar faedah tersebut nominal atau efektif

Tidak dinyatakan tempoh pengkompaunan

Dinyatakan sebagai kadar faedah efektif

Kaedah Pengiraan Kadar Faedah Nominal dan Efektif KAEDAH PENGIRAAN KADAR FAEDAH NOMINAL DAN EFEKTIF

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Faedah Kompaun

Tidak dinyatakan tempoh pengkompaunan1 = 10% setahun: kadar efektif 10% setahun dikompaun tiap tahun1 = 6% setengah tahun : kadar efektif 6% setengah tahun dikompaun setengah tahun

Tidak dinyatakan sama ada kadar faedah terusebut nominal atau efektifI = 12% setahun, dikompaun setiap suku tahundapat dinyatakan sebagai kadar efektif atau nominal, 12% setahun, dikompaun setiap suku tahun

Dinyatakan sebagai kadar faedah efektifI = kadar efektif 3% sesuku tahun, dikompaun tiap bulankadar efektif 3% sesuku tahun, dikompaun tiap bulan

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I L L U S T R A T I O N

Suppose you deposit R10,000 in a savings account that pays you at an interest rate of 9% compounded quarterly. The following is an example of how interest is compounded when it is paid quarterly.

Nominal & Effective Interest

The nominal interest rate (r) : 9%

The interest rate per quarter (i) : 2.25%

In fact, you are earning

= R10,930.98 – R10,000

R10,000

= 9.3083%

Effective interest rate (ieff) :9.3083%

R10,930.83= Value after one year

R240.53+ interest (2.25%)

R10,690.30= New base amount4th Quarter

R235.24+ interest (2.25%)

R10,455.06= New base amount3rd Quarter

R230.06+ interest (2.25%)

R10,225.00= New base amount2nd Quarter

R225.00+ interest (2.25%)

R10,000.00Base amount1st Quarter

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F O R M U L A

or

r : Nominal interest rate per year

i : Effective interest rate per compounding sub period

m : Number of compounding sub periods per year

Nominal & Effective InterestNOMINAL INTEREST RATE PER YEAR (r)

the annual interest rate without considering the effect of any compounding

EFFECTIVE INTEREST RATE PER YEAR (ieff)

The annual interest rate taking into account the effect of any compounding during the year

1mr1i

m

eff −⎟⎠⎞

⎜⎝⎛ +=

( ) 1i1i meff −+=

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Nominal & Effective InterestE X A M P L E 1.11

If a savings bank pays 1.5% interest every three months, what are the nominal and effective interest rate per year?

NOMINAL INTEREST RATE PER YEAR

r = 4 x 1.5% = 6%

EFFECTIVE INTEREST RATE PER YEAR

ALTERNATIVELY,

r : Nominal interest rate per yeari : Effective interest rate per compounding sub period

m : Number of compounding sub periods per year

( )( )

6.1%0.06110.0151

1i1i4

meff

==−+=

−+=

6.1%0.061

14

0.061

1mr1i

4

m

eff

==

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

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Nominal & Effective InterestE X A M P L E 1.12

On January 1st, a woman deposits R5,000 in a credit union that pays 8% nominal annual interest, compounded quarterly. She wishes to withdraw all the money in five equal yearly sum, beginning December 31st of the first year. How much should she withdraw each year?


ANNUAL WITHDRAWAL (W)

0 1 2 3 4 5

+

-R5,000

i = 8.24%n = 5 periods

W WWWW

yearper 8.24%0.0824

14

0.081

1mr1i

4

m

eff

==

−⎟⎠⎞

⎜⎝⎛ +=

−⎟⎠⎞

⎜⎝⎛ +=

( )( )

year per 1,260 R10.08241

0.082410.0824 5000

,5)(A/P,8.24% 5000n)i,P(A/P,W

5

5

=

⎥⎦

⎤⎢⎣

⎡

−++

=

==

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SINGLE PAYMENT INTEREST FACTORS – CONTINUOUS COMPOUNDING

COMPOUND AMOUNT

PRESENT WORTH


Continuous Compounding

F = P (er n) = P (F/P, r, n)

P = F (e-r n) = F (P/F, r, n)

ieff = er - 1

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Continuous CompoundingE X A M P L E 1.13

If you were to deposit R2,000 in a bank that pays 5% nominal interest, compounded continuously, how much would be in the account at the end of two years?

SINGLE PAYMENT COMPOUND AMOUNT –CONTINUOUS COMPOUNDING

F = P er n

= 2000 e(0.05 x 2)

= R 2210.40 at the end of 2 years

E X A M P L E 1.14

How long will it take for money to double

at 10% nominal interest, compounded

continuously?

SINGLE PAYMENT COMPOUND AMOUNT –CONTINUOUS COMPOUNDING

F = P er n

2 = 1 e(0.10)n

e(0.10)n = 20.10 n = ln 2n = 6.93 years

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CC – SERIES COMPOUND AMOUNT

CC – SERIES PRESENT WORTH

UNIFORM PAYMENT SERIES – CONTINUOUS COMPOUNDING

CC - SINKING FUND

CC – CAPITAL RECOVERY

Continuous Compounding

( )11,,/−−

= nr

r

eenrFA

( ) ( )11,,/

−−

= nr

rnr

eeenrPA

( )11,,/

−−

= r

nr

eenrAF

( ) ( )11,,/−−

= rnr

nr

eeenrAP

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A man deposited R500 per year into a credit union that paid 5% interest, compounded

annually. At the end of five years, he had R2,763 in the credit union. How much would he

have if they paid 5% nominal interest, compounded continuously?

SERIES COMPOUND AMOUNT - CONTINUOUS COMPOUNDING

( )

2769.84 R1e1e500

1e1eA

n)r,A(F/A,F

0.05

50.05

r

nr

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

=

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Jim wished to save a uniform amount each month so he would save R1,000 at the end of

one year. Based on 6% nominal interest, compounded monthly, he had to deposit R81.10

per month. How much would he have to deposit if his credit union paid 6% nominal interest,

compounded continuously?

SINKING FUND – CONTINUOUS COMPOUNDING

( )

81.07 R1e

1e10001e1eF

n)r,F(A/F,A

120.005

0.005

nr

r

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

=

chapter 1 - equivalence and time value of money 5 · types of cash flow single payment r100 012 354...

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