1 EGGC4214 Systems Engineering & Economy Lecture 4 Interest Formulas

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EGGC4214EGGC4214Systems Engineering & EconomySystems Engineering & Economy

Lecture 4Lecture 4Interest FormulasInterest Formulas

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INTEREST FORMULASINTEREST FORMULAS

Example Example Jane deposits $500 in a credit union at the end of Jane deposits $500 in a credit union at the end of each year for five years. The CU pays 5% interest, each year for five years. The CU pays 5% interest, compounded annually. At the end of five years, compounded annually. At the end of five years, immediately following her fifth deposit, how much will Jane immediately following her fifth deposit, how much will Jane have in her account?have in her account?

AA

00 11 22 33 44 55

AA AA AA AA

FF

00 11 22 33 44 55

AA

FF

AA AA AA AAJane’s point of view:Jane’s point of view:

CU’s point of view:CU’s point of view:

33

UNIFORM SERIES FORMULA UNIFORM SERIES FORMULA DERIVATIONDERIVATION

n = 5 periodsn = 5 periods

F = A (1+i)F = A (1+i)44 + A (1+i) + A (1+i)33 + A (1+i) + A (1+i)22 + A (1+i) + A + A (1+i) + A

F = A [(1+i)F = A [(1+i)44 + (1+i) + (1+i)33 + (1+i) + (1+i)22 + (1+i) + 1 ] + (1+i) + 1 ]

(1+i) F = A [(1+i)(1+i) F = A [(1+i)55 + (1+i) + (1+i)44 + (1+i) + (1+i)33 + (1+i) + (1+i)22 + + (1+i)] (1+i)]

(1+i) F – F = A [(1+i)(1+i) F – F = A [(1+i)55 – 1] – 1]

F = A [(1+i)F = A [(1+i)55 – 1]/i – 1]/i

A = F i/[(1+i)A = F i/[(1+i)55 – 1] – 1] F = A [(1+i) F = A [(1+i)55 – – 1]/i1]/iUniform Series Factors: (A/F,i,n)=i/[(1+i)Uniform Series Factors: (A/F,i,n)=i/[(1+i)nn – 1] – 1] (F/A,i,n)=[(1+i) (F/A,i,n)=[(1+i)nn– 1]/i– 1]/i

0 1 2 3 4 5

AA

FF

AA AA AA AA

44

UNIFORM SERIESUNIFORM SERIES

Uniform series compound amount factorUniform series compound amount factor : :

(F/A,i,n) = [(1+i)(F/A,i,n) = [(1+i)nn – 1]/i, i > 0 – 1]/i, i > 0

Uniform series fundUniform series fund

(A/F,i,n) = i/[(1+i)(A/F,i,n) = i/[(1+i)nn – 1] – 1]

Example 4-1.Example 4-1. Jane deposits $500 in a credit union at the end of Jane deposits $500 in a credit union at the end of each year for five years. The CU pays 5% interest, each year for five years. The CU pays 5% interest, compounded annually. At the end of five years, immediately compounded annually. At the end of five years, immediately following her fifth deposit, how much will Jane have in her following her fifth deposit, how much will Jane have in her account?account?

F = A (F/A,i,n) = A [(1+i)F = A (F/A,i,n) = A [(1+i)nn – 1]/i = $500[(1.05) – 1]/i = $500[(1.05)55 – 1]/(0.05) – 1]/(0.05) = $500 (5.5256) = $2,762.82 = $500 (5.5256) = $2,762.82 $2,763. $2,763.

55


ExampleExample A lot of land is available for $1000. Jane will put the A lot of land is available for $1000. Jane will put the same amount in the bank each month to have $1,000 at same amount in the bank each month to have $1,000 at the end of the year. The bank pays 6% annual interest, the end of the year. The bank pays 6% annual interest, compounded monthly. How much does Jim put in the bank compounded monthly. How much does Jim put in the bank each month?each month?

Solution Solution A = 1000 (A/F, ½%,12) = 1000 (0.0811) = $81.10/monthA = 1000 (A/F, ½%,12) = 1000 (0.0811) = $81.10/month

Example Example Suppose on January 1 Jane deposits $5000 in a Suppose on January 1 Jane deposits $5000 in a credit union paying 8% interest, compounded annually. She credit union paying 8% interest, compounded annually. She wants to withdraw all the money in five equal end-of year wants to withdraw all the money in five equal end-of year sums, beginning December 31sums, beginning December 31stst of the first year. of the first year.

1 2 3 4 5

AAAAA

P

66


We know We know

A = F i/[(1+i)A = F i/[(1+i)nn – 1] – 1]

F = P (1 + i)F = P (1 + i)nn. .

Hence Hence

A = P{[i (1 + i)A = P{[i (1 + i)nn]/[(1+i)]/[(1+i)nn – 1]} =P(A/P,i,n) – 1]} =P(A/P,i,n)

This is a formula to determine the value of a series of end-This is a formula to determine the value of a series of end-of-period payments, A, when the present sum P is known. of-period payments, A, when the present sum P is known.

(A/P,i,n) is called the (A/P,i,n) is called the uniform series capital recovery factoruniform series capital recovery factor . .

77


Example Example Suppose on January 1 Jim deposits $5000 in a credit union Suppose on January 1 Jim deposits $5000 in a credit union paying 8% interest, compounded annually. He wants to withdraw paying 8% interest, compounded annually. He wants to withdraw all the money in five equal end-of year sums, beginning all the money in five equal end-of year sums, beginning December 31December 31stst of the first year. of the first year.

P = $5000 n = 5 i = 8% A = unknownP = $5000 n = 5 i = 8% A = unknown

A = P (A/P,8%,5) = P{[i (1 + i)A = P (A/P,8%,5) = P{[i (1 + i)nn]/[(1+i)]/[(1+i)nn – 1]} – 1]}

= 5000 (0.2504564545) = $1252.28= 5000 (0.2504564545) = $1252.28

The withdrawal amount is about $1252The withdrawal amount is about $1252

1 2 3 4 5

AAAAA

P

88


ExampleExample

An investor holds a time payment purchase contract on An investor holds a time payment purchase contract on some machine tools. The contract calls for the payment of some machine tools. The contract calls for the payment of $140 at the end of each month for a five-year period. The $140 at the end of each month for a five-year period. The first payment is due in one month. He offers to sell you the first payment is due in one month. He offers to sell you the contract for $6800 cash today. If you otherwise can make contract for $6800 cash today. If you otherwise can make 1% per month on your money, would you accept or reject 1% per month on your money, would you accept or reject the investor’s offer?the investor’s offer?

A =140, n=60, i=1%A =140, n=60, i=1%

……………………..

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

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

= A (P/A,i,n) = 140 (P/A,1%,60) = 140 (44.955) = $6293.70= A (P/A,i,n) = 140 (P/A,1%,60) = 140 (44.955) = $6293.70

Imagine Imagine 60 +ve 60 +ve arrowsarrows

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The Reverse Point of View.The Reverse Point of View. We can solve We can solve

A = P{[i (1 + i)A = P{[i (1 + i)nn]/[(1+i)]/[(1+i)nn – 1]} – 1]}

for P to obtain: for P to obtain:

P = A {[(1+i)P = A {[(1+i)nn – 1]/[i (1 + i) – 1]/[i (1 + i)nn]} = A (P/A,i%,n)]} = A (P/A,i%,n)

In this case In this case

(P/A,i%,n) = {[(1+i)(P/A,i%,n) = {[(1+i)nn – 1]/ [i (1 + i) – 1]/ [i (1 + i)nn]}]}

is called the is called the uniform series present worth factoruniform series present worth factor..

1010


Uniform series compound amount factorUniform series compound amount factor : :

(F/A,i,n) = [(1+i)(F/A,i,n) = [(1+i)nn – 1]/i, i > 0 – 1]/i, i > 0

Uniform series sinking fundUniform series sinking fund : :

(A/F,i,n) = i/[(1+i)(A/F,i,n) = i/[(1+i)nn – 1] – 1]

Uniform series capital recoveryUniform series capital recovery : :

(A/P,i,n) = [i (1 + i)(A/P,i,n) = [i (1 + i)nn]/[(1+i)]/[(1+i)nn – 1] – 1]

Uniform series present worthUniform series present worth : :

(P/A,i,n) = [(1+i)(P/A,i,n) = [(1+i)nn – 1]/[i (1 + i) – 1]/[i (1 + i)nn]]

1111


Example Example

Compute the value of the following cash flows at the first of Compute the value of the following cash flows at the first of year 5.year 5.

Year Cash flowYear Cash flow

1 + 1001 + 100

2 + 1002 + 100

3 + 1003 + 100

4 04 0

5 - F5 - F

The Sinking Fund Factor diagram is based on the The Sinking Fund Factor diagram is based on the assumption that the withdrawal coincides with the last assumption that the withdrawal coincides with the last deposit. This does not happen in this example. deposit. This does not happen in this example.

Note the zero

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The standard approach is:A A A

-F1 2 3 1 2 3 4 5

A A AWhat we have is:

-F

First Approach:

A A A

F1 F

Use the “standard” approach to compute F1.

Then compute the future value of F1 to get F.

F1 = 100 (F/A,15%,3) = 100 (3.472) = $347.20

F = F1 (F/P,15%,2) = 347.20 (1.322) = $459.00

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Second Approach:Second Approach: Compute the future values of Compute the future values of each each deposit and add them.deposit and add them.

FF = F= F11 + F + F22 + F + F33

= 100(F/P,15%,4) + 100(F/P,15%,3) + 100(F/P,15%,2)= 100(F/P,15%,4) + 100(F/P,15%,3) + 100(F/P,15%,2)

= 100 (1.749) + 100 (1.521) + 100 (1.322)= 100 (1.749) + 100 (1.521) + 100 (1.322)

= $459.20= $459.20

=

A A A

F

A

F1

A

F2

A

F3

+ +

1414


ExampleExample i = 15% i = 15%

This diagram is This diagram is not in a standard formnot in a standard form. .

Approach 1.Approach 1. Compute the present value of each flow and add Compute the present value of each flow and add them.them.

P = PP = P11 + P + P22 + P + P33

= 20 (P/F,15%,2) + 30 (P/F,15%,3) + 20 (P/F,15%,4)= 20 (P/F,15%,2) + 30 (P/F,15%,3) + 20 (P/F,15%,4)

= 20 (0.7561) + 30 (0.6575) + 20 (0.5718) = $46.28= 20 (0.7561) + 30 (0.6575) + 20 (0.5718) = $46.28

20 2030

0

P

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Approach 2.Approach 2. Compute the future value, F, of the flows at Compute the future value, F, of the flows at the the end of year 4. Then compute the end of year 4. Then compute the present valuepresent value

of the future value, F.of the future value, F.

Approach 3.Approach 3. Compute the present values of the flows at Compute the present values of the flows at the the end of year 1, Pend of year 1, P11. Then compute P, . Then compute P,

the present the present value of Pvalue of P11..

Other approaches will also work. Other approaches will also work.

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Relationships Between Compound Interest Relationships Between Compound Interest FactorsFactors

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

P = A[(1+i)P = A[(1+i)nn–1]/[i(1 + i)–1]/[i(1 + i)nn] =A(P/A,i,n) ] =A(P/A,i,n) A = P[i(1 + i)A = P[i(1 + i)nn]/[(1 + i)]/[(1 + i)n n – 1]– 1] = P(A/P,i,n)= P(A/P,i,n)

F = A{[(1 + i)F = A{[(1 + i)nn – 1]/i} = A(F/A,i,n) A = F{i/[(1 + i) – 1]/i} = A(F/A,i,n) A = F{i/[(1 + i)n n – 1]} = – 1]} = F(A/F,i,n)F(A/F,i,n)

Present Worth factorCompound Amount factor

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

(A/P,i,n) = 1/(P,A,i,n)(A/P,i,n) = 1/(P,A,i,n)

Uniform Series Capital Recovery Factor

(A/F,i,n) = 1/ (F/A,i,n)(A/F,i,n) = 1/ (F/A,i,n)Uniform Series

CompoundAmount Factor

Uniform Series Sinking Fund Factor

Uniform SeriesPresent Worth Factor

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P P = A(P/A,i,n) = A(P/A,i,n) = A[ (1+i)= A[ (1+i)-1 -1 + (1+i)+ (1+i)-2 -2 + … + (1+i)+ … + (1+i)-n -n ] ] = A[(P/F,i,1)+(P/F,i,2)+...+(P/F,i,n)] = A[(P/F,i,1)+(P/F,i,2)+...+(P/F,i,n)]

F F = A (F/A,i,n) = A[ 1 + (1+i) + (1+i)= A (F/A,i,n) = A[ 1 + (1+i) + (1+i)22 + ... + (1+i) + ... + (1+i)n-1 n-1 ]]= A[ 1 + (F/P,i,1) + (F/P,i,2) + ... + (F/P,i,n-1) ]= A[ 1 + (F/P,i,1) + (F/P,i,2) + ... + (F/P,i,n-1) ]

(P/A,i,n) = [(P/F,i,1) + (P/F,i,2) + ... + (P/F,i,n)]

(F/A,i,n) = 1 + (F/P,i,1) + (F/P,i,2) + ... + (F/P,i,n-n-11)

0 1 2 3 n

A

F

A A A…..

0 1 2 3 n

A A A A…..

P


1818

(A/P,i,n) = [i(1 + i)(A/P,i,n) = [i(1 + i)nn]/[(1+i)]/[(1+i)nn–1]–1]

(A/F,i,n) = {i/[(1+i)(A/F,i,n) = {i/[(1+i)nn–1]}–1]}

We start with an identity:We start with an identity:

i (1+i)i (1+i)nn = i + i (1+i) = i + i (1+i)nn – i = i + i [(1+i) – i = i + i [(1+i)nn – 1] – 1]

Now divide by {(1+i)Now divide by {(1+i)nn – 1} to get: – 1} to get:

[i (1+i)[i (1+i)nn]/[(1+i)]/[(1+i)nn – 1] = i/[ (1+i) – 1] = i/[ (1+i)nn – 1] + i – 1] + i

This gives:This gives:

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


1919

Arithmetic GradientArithmetic Gradient

Suppose you buy a car. You wish to set up enough money in a Suppose you buy a car. You wish to set up enough money in a bank account to pay for standard maintenance on the car for bank account to pay for standard maintenance on the car for the first five years. You estimate the maintenance cost the first five years. You estimate the maintenance cost increases by G = increases by G = $30$30 each year. The maintenance cost for each year. The maintenance cost for year 1 is estimated as year 1 is estimated as $120$120. Thus, estimated costs by year are . Thus, estimated costs by year are $120$120, , $150$150, , $180$180, , $210$210, , $240$240..

0 1 2 3 4 5

$120$150

$180$210

$240

2020

We break up the cash flows into two components:We break up the cash flows into two components:

andand1 2 3 4 5

G = 30

030

60

90

120

A = 120

P1 P2

P1 = A (P/A,5%,5) = 120 (P/A,5%,5) = 120 (4.329) = $519.48

P2 = G (P/G,5%,5) = 30 (P/G,5%,5) = 30 (8.237) = $247.11

P = P1 + P2 = $766.59Note: 5 and not 4Note: 5 and not 4

Standard form diagram for Standard form diagram for an Arithmetic Gradient: (n) an Arithmetic Gradient: (n) periods and (n-1) non-zero periods and (n-1) non-zero flows in an increasing flows in an increasing orderorder


1 2 3 4 5

2121

F = G(1+i)F = G(1+i)n-2 n-2 + 2G(1+i)+ 2G(1+i)n-3n-3 + … + (n-2)G(1+i) + … + (n-2)G(1+i)11 + (n-1)G(1+i) + (n-1)G(1+i)00 F = G [(1+i)F = G [(1+i)n-2 n-2 + 2(1+i)+ 2(1+i)n-3n-3 + … + (n-2)(1+i) + … + (n-2)(1+i)11 + n-1] + n-1]

(1+i) F (1+i) F = G [(1+i)= G [(1+i)n-1 n-1 + 2(1+i)+ 2(1+i)n-2n-2 + 3(1+i) + 3(1+i)n-3n-3 + … + (n-1)(1+i) + … + (n-1)(1+i)11] ] iFiF = G [(1+i)= G [(1+i)n-1 n-1 + (1+i)+ (1+i)n-2n-2 + (1+i) + (1+i)n-3n-3 + … + (1+i) + … + (1+i)11 – n + 1] – n + 1]

= G [(1+i)= G [(1+i)n-1 n-1 + (1+i)+ (1+i)n-2n-2 + (1+i) + (1+i)n-3n-3 + … + (1+i) + … + (1+i)11 + 1] – + 1] – nGnG

= G (F/A, i, n) - nG = G [(1+i)= G (F/A, i, n) - nG = G [(1+i)nn-1]/i – nG-1]/i – nG FF = G (F/G, i, n) = G [(1+i)= G (F/G, i, n) = G [(1+i)nn-in-1]/i-in-1]/i22

PP = G (P/G, i, n) = G [(1+i)= G (P/G, i, n) = G [(1+i)nn-in-1]/[i-in-1]/[i22(1+i)(1+i)nn]] AA = F (A/F, i, n)= F (A/F, i, n)

= G [(1+i)= G [(1+i)nn-in-1]/i-in-1]/i2 2 × i/[(1+i)× i/[(1+i)nn-1]-1] AA = = G (A/G, i, n)G (A/G, i, n)

= G [(1+i)= G [(1+i)nn-in-1]/-in-1]/[i(1+i)[i(1+i)nn-i] -i] 0 1 2 3 …. n

G2G

……

(n-1)G

0

F


2222

Arithmetic Gradient Uniform SeriesArithmetic Gradient Uniform Series

Arithmetic Gradient Present WorthArithmetic Gradient Present Worth

(P/G,5%,5) (P/G,5%,5) = {[(1+i)= {[(1+i)nn – i n – 1]/[i – i n – 1]/[i22 (1+i) (1+i)nn]}]} = {[(1.05)= {[(1.05)55 – 0.25 – 1]/[0.05 – 0.25 – 1]/[0.0522 (1.05) (1.05)55]}]} = 0.026281562/0.003190703 = = 0.026281562/0.003190703 =

8.236916768.23691676

(P/G,i,n) = { [(1+i)n – i n – 1] / [i2 (1+i)n] }

(A/G,i,n) = { (1/i ) – n/ [(1+i)n –1] }

(F/G,i,n) = G [(1+i)n – in – 1]/i2

=1/(G/P,i,n)

=1/(G/A,i,n)

=1/(G/F,i,n)


2323

Example 4-6.Example 4-6. Maintenance costs of a machine Maintenance costs of a machine start at $100 and go up by start at $100 and go up by $100$100 each year for each year for 4 years4 years. What is the equivalent uniform . What is the equivalent uniform annual maintenance cost for the machinery if annual maintenance cost for the machinery if i i = 6%= 6%..

100

200

300

400

A A A A

This is not in the standard form for using the gradient equation, because the first year cash flow is not zero. We reformulate the problem as follows:


0 1 2 3 4

2424

= +A1=100

G =100

100

200

300

0

The second diagram is in the form of a $100 uniform series. The last diagram is now in the standard form for the gradient equation with n = 4, G = 100.

A = A1+ G (A/G,6%,4) =100 + 100 (1.427) = $242.70

100200

300

400

0 1 2 3 4 0 1 2 3 4 0 1 2 3 4


2525

ExampleExample

With i = With i = 10%,10%, n = n = 44, find an equivalent uniform payment for:, find an equivalent uniform payment for:

This is a problem with This is a problem with decreasingdecreasing costs instead of increasing costs instead of increasing costs. The cash flow can be rewritten as the DIFFERENCE of costs. The cash flow can be rewritten as the DIFFERENCE of the following two diagrams, the second of which is in the the following two diagrams, the second of which is in the standard form we need, the first of which is a series of standard form we need, the first of which is a series of uniform payments.uniform payments.

2400018000

120006000

1 2 3 40


2626

= _18000

24000

120006000

0 1 2 3 4

A1 = 24000

0 1 2 3 4

A1 A1 A1 A13G

2G

0 1 2 3 4

G

A = A1 – G(A/G,10%,4)

= 24000 – 6000 (A/G,10%,4)

= 24000 – 6000(1.381) = $15714.00

G = 6000


2727

ExampleExample Find P for the following diagram with i = Find P for the following diagram with i = 10%10%

P

1 2 3 4 5 6

50100

150

This is not in the standard form for the arithmetic gradient. However, if we insert a “present value” J at the end of year 2, the diagram from that point on is in a standard form.

J

Thus:J = 50 (P/G,10%,4) = 50 (4.378) = $218.90P = J (P/F,10%,2) = 218.90 (0.8264) = $180.90


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Geometric GradientGeometric Gradient

Example Example Suppose you have a vehicle. The first year maintenance Suppose you have a vehicle. The first year maintenance cost is estimated to be cost is estimated to be $100$100. The rate of increase in each . The rate of increase in each subsequent year is subsequent year is 10%10%. You want to know the present worth of . You want to know the present worth of the cost of the first five years of maintenance, given i = the cost of the first five years of maintenance, given i = 8%8%..

Repeated Present-Worth (Step-by-Step) Approach:Repeated Present-Worth (Step-by-Step) Approach:

Year n Mnt. Cost (P/F,8%,n) PW of mnt.

1 100 0.9259 $92.59 2 110 0.8573 $94.30 3 121 0.7938 $96.05 4 133.1 0.7350 $97.83 5 146.41 0.6806 $99.65

$480.42

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Geometric Gradient.Geometric Gradient. At the end of year j, j = 1, ..., n, we incur a cost A At the end of year j, j = 1, ..., n, we incur a cost A jj = A(1+g) = A(1+g)j-1j-1..

P = A(1+i)P = A(1+i)-1-1+A(1+g)+A(1+g)11(1+i)(1+i)-2-2+A(1+g)+A(1+g)22(1+i)(1+i)-3-3+…+… … …+A(1+g)+A(1+g)n-2n-2(1+i)(1+i)-n+1-n+1+A(1+g)+A(1+g)n-1n-1(1+i)(1+i)-n-n

P(1+g)P(1+g)11(1+i)(1+i)-1-1=A(1+g)=A(1+g)11(1+i)(1+i)-2-2+A(1+g)+A(1+g)22(1+i)(1+i)-3-3++

…………+A(1+g)+A(1+g)n-1n-1(1+i)(1+i)-n-n+A(1+g)+A(1+g)nn(1+i)(1+i)-n-1-n-1

P - P(1+g)P - P(1+g)11(1+i)(1+i)-1-1= A(1+i)= A(1+i)-1-1 -- A(1+g) A(1+g)nn(1+i)(1+i)-n-1-n-1

P (1+i-1-g) = A (1 P (1+i-1-g) = A (1 -- (1+g) (1+g)nn(1+i)(1+i)-n-n))

i ≠ g: P = A (1 - (1+g)n(1+i)-n)/(i-g)

i = g: P = A n(1+i)-1

0 1 2 3 …. n

A A(1+g)1A(1+g)2

A(1+g)n-1

(P/A,g,i,n)(P/A,g,i,n)

3030


Example Example n = 5, A n = 5, A11 = 100, g = 10%, i = 8%. = 100, g = 10%, i = 8%.

Year n Mnt. Cost (P/F,8%,n) PW of Mnt. 1 100 0.9259 = $92.59 2 110 0.8573 $94.30 3 121 0.7938 $96.05 4 133.1 0.7350 $97.83 5 146.41 0.6806 $99.65

$480.42

(P/A,g,i,n) = (P/A,10%,8%,5) = 4.8042P = A(P/A,g,i,n) = 100 (4.8042) = $480.42

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Ten thousand dollarsTen thousand dollars is borrowed for is borrowed for two yearstwo years at an interest at an interest rate of rate of 24% per year24% per year compounded compounded quarterlyquarterly..If this same sum of money could be borrowed for the same period If this same sum of money could be borrowed for the same period at the same interest rate of at the same interest rate of 24% per year24% per year compounded compounded annuallyannually, how much could be saved in interest charges?, how much could be saved in interest charges?

interest chargesinterest charges for for quarterlyquarterly compounding: compounding: $10,000(1+24%/4)$10,000(1+24%/4)2*42*4-$10,000 = $5938.48-$10,000 = $5938.48

interest charges for interest charges for annuallyannually compounding: compounding: $10,000(1+24%)$10,000(1+24%)22 - $10,000 = $5376.00 - $10,000 = $5376.00

Savings: Savings: $5938.48 - $5376 = $562.48$5938.48 - $5376 = $562.48

Compounding is not less Compounding is not less important than interest. You important than interest. You have to know all the have to know all the information to make a good information to make a good decisiondecision

Nominal and Effective Interest Nominal and Effective Interest RateRate

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Sometimes one interest rate is quoted, sometimes another is quoted. If you confuse the two you can make a bad decision.

A bank pays 5% compounded semi-annually. If you deposit $1000, how much will it grow to by the end of the year?

The bank pays 2.5% each six months.

You get 2.5% interest per period for two periods.

1000 1000(1.025) = 1,025 1025(1.025) = $1,050.625

With i = 0.05/2, r = 0.05,

P (1 + i) P (1+r/2)2 P = (1+ 0.05/2)2 P = (1.050625) P


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Terms the example illustrates:

r = 5% is called the nominal interest rate per one yeari = 2.5 % is called the effective interest rate per interest period

ia = 5.0625% is called the effective interest rate per one year

In the example: m = 2 is the number of compounding sub-periods per time period.

ia = (1.050625) – 1 = (1.025)2 – 1 = (1+0.05/2)2 – 1

r = nominal interest rate per year m = number of compounding sub-periods per yeari = r/m = effective interest rate per compounding sub-period.

ia = (1 + r/m)m – 1 = (1 + i)m – 1

The term i we have used up to now The term i we have used up to now is more precisely defined as the is more precisely defined as the

effective interest rate effective interest rate per interest periodper interest period. .

If the interest period is one year If the interest period is one year (m = 1) then i = r. (m = 1) then i = r.


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ExampleExample A bank pays A bank pays 1.5%1.5% interest every interest every three monthsthree months. . What are the nominal and effective interest rates per year?What are the nominal and effective interest rates per year?

Solution:Solution:

Nominal interest rate per year is Nominal interest rate per year is

r = 4 r = 4 1.5% = 6% a year 1.5% = 6% a year

Effective interest rate per year:Effective interest rate per year:

iiaa = (1 + r/m) = (1 + r/m)mm – 1 = (1.015) – 1 = (1.015)44 – 1 = 0.06136 – 1 = 0.06136 6.14% per year. 6.14% per year.


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Table Nominal & Effective Interest expressed in percent

Effective rates, ia = (1 + r/m)m - 1

Nominal rate Yearly Semi-ann.

Monthly Daily Continuously

r m = 1 m = 2 m = 12 m = 365 m = ∞ 1 1.0000 1.0025 1.0046 1.0050 1.0050 2 2.0000 2.0100 2.0184 2.0201 2.0201 3 3.0000 3.0225 3.0416 3.0453 3.0455 4 4.0000 4.0400 4.0742 4.0808 4.0811 5 5.0000 5.0625 5.1162 5.1267 5.1271 6 6.0000 6.0900 6.1678 6.1831 6.1837 8 8.0000 8.1600 8.3000 8.3278 8.3287

10 10.0000 10.2500 10.4713 10.5156 10.5171 15 15.0000 15.5625 16.0755 16.1798 16.1834 25 25.0000 26.5625 28.0732 28.3916 28.4025

Remark. The formula used for the continuous rate is er - 1, expressed in percent. It is the limit of (1 + r/m)m as m goes to infinity.


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ExampleExample Joe lends money on the following terms: “ If I give Joe lends money on the following terms: “ If I give you 50 dollars on Monday, then you owe me 60 dollars you 50 dollars on Monday, then you owe me 60 dollars the following Monday.”the following Monday.”

1.What is the 1.What is the nominal ratenominal rate, r?, r?

We first note Joe charges i = 20% a week, We first note Joe charges i = 20% a week, since 60 = (1+i)50 since 60 = (1+i)50 i = 0.2. Note we have solved F = 50(F/P,i,1) i = 0.2. Note we have solved F = 50(F/P,i,1)

for i. for i. We know m = 52, so r = 52 We know m = 52, so r = 52 i = 10.4, or 1040% a year. i = 10.4, or 1040% a year.

2.2. What is the What is the effective rateeffective rate, i, iaa ? ?

From iFrom iaa = (1 + r/m) = (1 + r/m)mm – 1 we have i – 1 we have iaa = (1+10.4/52) = (1+10.4/52)5252 – 1 – 1 13104 13104

This means exactly 1310363.1% per year.This means exactly 1310363.1% per year.


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Suppose Joe can keep the $50, as well as all the Suppose Joe can keep the $50, as well as all the money he receives in payments, out in loans at money he receives in payments, out in loans at all times? How much would Joe have at the end of all times? How much would Joe have at the end of the year?the year?

We use F = P(1+i)We use F = P(1+i)nn to get F = 50(1.2) to get F = 50(1.2)5252 $655,232 $655,232

Words of Warning.Words of Warning. When the various compounding periods in When the various compounding periods in a problem all match, it makes calculations much simpler. a problem all match, it makes calculations much simpler.

When they do not match, life is more complicatedWhen they do not match, life is more complicated..

Recall Recall ExampleExample. We put $5000 in an account paying 8% . We put $5000 in an account paying 8% interest, compounded annually. We want to find the five interest, compounded annually. We want to find the five equal EOY withdrawals.equal EOY withdrawals.

We used A = P(A/P,8%,5) = 5000 We used A = P(A/P,8%,5) = 5000 (0.2505) (0.2505) $1252 $1252 Suppose the various periods are not the same in this problem.Suppose the various periods are not the same in this problem.


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Example:Example: Sally deposits Sally deposits $5000$5000 in a CU paying in a CU paying 8%8% nominalnominal interest, interest, compounded quarterlycompounded quarterly. She wants to withdraw the money in five . She wants to withdraw the money in five equal yearly sums, beginning equal yearly sums, beginning Dec. 31Dec. 31 of the first year. How much of the first year. How much should she withdraw each year? Note: effective interest is i = 2% = should she withdraw each year? Note: effective interest is i = 2% = r/4 = 8%/4 quarterly, and there are 20 periods.r/4 = 8%/4 quarterly, and there are 20 periods.

Solution: Solution:

$5000

i = 2%, n = 20

WW W W W


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The withdrawal periods and the compounding periods are not the The withdrawal periods and the compounding periods are not the same. If we want to use the formulasame. If we want to use the formula

A = P (A/P,i,n)A = P (A/P,i,n)then we must find a way to put the problem into an equivalent then we must find a way to put the problem into an equivalent form where all the periods are the same.form where all the periods are the same.

Solution 1.Solution 1. Suppose we withdraw an amount A quarterly. We Suppose we withdraw an amount A quarterly. We compute:compute:

A = P (A/P,i,n) = 5000 (A/P,2%,20) = 5000 (0.0612) = A = P (A/P,i,n) = 5000 (A/P,2%,20) = 5000 (0.0612) = $306.00$306.00

These withdrawals are equivalent to P = These withdrawals are equivalent to P = $5000$5000

$5000i = 2%, n = 20

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Now consider the following:Now consider the following:

Consider a one-year period:Consider a one-year period:

This is now in a standard form that repeats every year.This is now in a standard form that repeats every year.W = A(F/A,i,n) = 306 (F/A,2%,4) = 306 (4.122) = $1261.33W = A(F/A,i,n) = 306 (F/A,2%,4) = 306 (4.122) = $1261.33Sally should withdraw about $1261 at the end of each year.Sally should withdraw about $1261 at the end of each year.

A

WW WW

Wi = 2%, n = 4


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Solution 2Solution 2

iiaa = (1 + r/m)= (1 + r/m)mm – 1 = (1 + i) – 1 = (1 + i)mm – 1 – 1= (1.02)= (1.02)44 – 1 = 0.0824 – 1 = 0.0824 8.24% 8.24%

Now use:Now use:WW = P(A/P,8.24%,5) = P {[i (1+i)= P(A/P,8.24%,5) = P {[i (1+i)nn]/[(1+i)]/[(1+i)nn – 1]} – 1]}

= 5000(0.252043) = $1260.21 per year= 5000(0.252043) = $1260.21 per year

ia = 8.24%, n = 5

W

$5000

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Continuous CompoundingContinuous Compounding

Table Nominal & Effective Interest expressed in percent

Effective rates, ia = (1 + r/m)m - 1

Nominal rate Yearly Semi-ann.

Monthly Daily Continuously

r m = 1 m = 2 m = 12 m = 365 m = ∞

1 1.0000 1.0025 1.0046 1.0050 1.0050 2 2.0000 2.0100 2.0184 2.0201 2.0201 3 3.0000 3.0225 3.0416 3.0453 3.0455 4 4.0000 4.0400 4.0742 4.0808 4.0811 5 5.0000 5.0625 5.1162 5.1267 5.1271 6 6.0000 6.0900 6.1678 6.1831 6.1837 8 8.0000 8.1600 8.3000 8.3278 8.3287

10 10.0000 10.2500 10.4713 10.5156 10.5171 15 15.0000 15.5625 16.0755 16.1798 16.1834 25 25.0000 26.5625 28.0732 28.3916 28.4025

Remark. The formula used for the continuous rate is er - 1, expressed in percent. It is the limit of (1 + r/m)m as m goes to infinity.

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rr = nominal interest rate per year = nominal interest rate per year mm = number of compounding sub-periods per year= number of compounding sub-periods per yearii = r/m = effective interest rate per compounding sub-period.= r/m = effective interest rate per compounding sub-period.

Continuous compounding can sometimes be used to simplify computations, Continuous compounding can sometimes be used to simplify computations, and for theoretical purposes. The table illustrates that eand for theoretical purposes. The table illustrates that e rr - 1 is a good - 1 is a good approximation of (1 + r/m)approximation of (1 + r/m)mm for large values of m. This means there are for large values of m. This means there are continuous compounding versions of the formulas we have seen earlier.continuous compounding versions of the formulas we have seen earlier.For example, For example,

F = P eF = P ernrn is analogous to F = P (F/P,r,n):is analogous to F = P (F/P,r,n): (F/P,r,n) (F/P,r,n)infinf= = eernrn

P = F eP = F e-rn-rn is analogous to P = F (P/F,r,n):is analogous to P = F (P/F,r,n): (P/F,r,n) (P/F,r,n)infinf= e= e--

rnrn

Continuous CompoundingContinuous Compounding

ia = (1 + i)m – 1 = (1 + r/m)m – 1

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i: i: effective effective iinterest rate per interest period (stated as a nterest rate per interest period (stated as a

decimal) decimal)

n: n: nnumber of interest periods umber of interest periods

P: P: ppresent sum of money resent sum of money

F: F: ffuture sum of money: an amount, n interest periods from uture sum of money: an amount, n interest periods from

the present, that is equivalent to P with interest rate i the present, that is equivalent to P with interest rate i

A: A: end-of-period cash receipt or disbursement end-of-period cash receipt or disbursement aamount in a mount in a

uniform series, continuing for n periods, the entire series uniform series, continuing for n periods, the entire series

equivalent to P or F at interest rate i.equivalent to P or F at interest rate i.

Summary NotationsSummary Notations

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G: G: arithmetic arithmetic ggradient: uniform period-by-period radient: uniform period-by-period

increase or decrease in cash receipts or disbursements increase or decrease in cash receipts or disbursements

g: g: ggeometric gradient: uniform rate of cash flow increase or eometric gradient: uniform rate of cash flow increase or

decrease from period to perioddecrease from period to period

r: r: nominal interest nominal interest rrate per interest period (usually one year)ate per interest period (usually one year)

iiaa: : effective effective iinterest rate per year (nterest rate per year (aannum) nnum)

m: num: nummber of compounding sub-periods per periodber of compounding sub-periods per period

Summary NotationsSummary Notations

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Single Payment formulasSingle Payment formulas::

Compound amount:Compound amount: F = P (1+i)F = P (1+i)nn = P (F/P,i,n) = P (F/P,i,n)

Present worth:Present worth: P = F (1+i)P = F (1+i)-n-n = F (P/F,i,n) = F (P/F,i,n)

Uniform Series FormulasUniform Series Formulas

Compound Amount:Compound Amount: F = A{[(1+i)F = A{[(1+i)nn –1]/i} = A (F/A,i,n) –1]/i} = A (F/A,i,n)

Sinking Fund:Sinking Fund: A = F {i/[(1+i)A = F {i/[(1+i)nn –1]} = F (A/F,i,n) –1]} = F (A/F,i,n)

Capital Recovery:Capital Recovery: A = P {[i(1+i)A = P {[i(1+i)nn]/[(1+i)]/[(1+i)nn – 1] = P (A/P,i,n) – 1] = P (A/P,i,n)

Present Worth:Present Worth: P = A{[(1+i)P = A{[(1+i)nn – 1]/[i(1+i) – 1]/[i(1+i)nn]} = A (P/A,i,n)]} = A (P/A,i,n)

Summary FormulasSummary Formulas

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Arithmetic Gradient FormulasArithmetic Gradient Formulas

Present WorthPresent Worth ,, P = G {[(1+i)P = G {[(1+i)nn – i n – 1]/[i – i n – 1]/[i22 (1+i) (1+i)nn]} = G ]} = G

(P/G,i,n)(P/G,i,n)

Uniform Series,Uniform Series, A = G {[(1+i)A = G {[(1+i)nn – i n –1]/[i (1+i) – i n –1]/[i (1+i)nn – i]} = G (A/G,i,n) – i]} = G (A/G,i,n)

Geometric Gradient FormulasGeometric Gradient Formulas

If i If i g, g, P = A {[1 – (1+g)P = A {[1 – (1+g)nn(1+i)(1+i)-n-n]/(i-g)} = A (P/A,g,i,n)]/(i-g)} = A (P/A,g,i,n)

If i = g,If i = g, P = A [n (1+i)P = A [n (1+i)-1-1] = A (P/A,g,i,n)] = A (P/A,g,i,n)


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Nominal interest rate per yearNominal interest rate per year, r: the annual interest rate , r: the annual interest rate

without considering the effect of any compounding without considering the effect of any compounding

Effective interest rate per yearEffective interest rate per year, i, iaa::

iiaa = (1 + r/m) = (1 + r/m)mm – 1 = (1+i) – 1 = (1+i)mm – 1 with i = r/m – 1 with i = r/m

Continuous compoundingContinuous compounding, :, :

r – one-period interest rate, n – number of periodsr – one-period interest rate, n – number of periods

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

(F/P,r,n)(F/P,r,n)infinf= e= ernrn

Documents

1 EGGC4214 Systems Engineering & Economy Lecture 4 Interest Formulas