chapter 2a

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MS-291: Engineering Economy(3 Credit Hours)

MS291: Engineering Economy

Chapter 2Factors: How Time and Interest

Affect Money

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Content of the Chapter

Single-Payment Compound Amount Factor (SPCAF) Single-Payment Present Worth Factor (SPPWF)

Uniform Series Present Worth Factor (USPWF) Capital Recovery Factor (CRF)

Uniform Series Compound Amount Factor Sinking Fund Factor (SFF)

Arithmetic Gradient Factor Geometric Gradient Series Factor

Simple InterestHereP=$100,000n= 3i= 10%Simple interest = P X i x nInterest = 100,0000(0.10) (3)

= $30,000Total due = 100,000 +

30,000= $130,000

Compound Interest Interest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: F1 = 100,000 + 10,000

=$110,000

Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: F2 = 110,000 + 11,000

= $121,000

Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: F3 = 121,000 + 12,100

= $133,100

Simple and Compound Interest:Comparison

Example: $100,000 lent for 3 years at interest rate i= 10% per year. What is repayment after 3 years ?

Simple: $130,000: Compounded: $133,100

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Single Payment CompoundAmount Factor (SPCAF)

If an amount P is invested at time t=0 the amount accumulated after a yearis given as

F1 = P + Pi= P(1 + i) . (1)

At the end of second year, the accumulated amount F2 is given as;

F2 = F1 + F1 i= P(1+i) + P(1+i)i (from Eq. 1)= P + Pi + Pi+ Pi2= P(1+i)2 (2)

Similarly; F3 = F2 + F2 i= P(1+i)3 ..(3)

to generalize the process for period n we can write as;

F = P(1+i)n

P = 1000@5%After one year ?F1 = 1000 + 50

After two year ?F2 = 1050 + 52.5

Single Payment CompoundAmount Factor (SPCAF)

The term (1+i)n is known as Single PaymentCompound Amount Factor (SPCAF)

It is also refer as F/P factor

This is a converting factor, when multiplied by Pyields the future amount F of initial amount Pafter n years at interest rate i

F = P(1+i)n

Do not forget i .. Refers tocompoundinterest rate

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Simple InterestHereP=$100,000n= 3i= 10%Simple interest = P X i x nInterest = 100,000(0.103)(3)

= $30,000Total due = 100,000 +

30,000= $130,000

Compound Interest

Now we have F = P(1+i)nP = $100,000n=3i=10%

So F = 100,000 (1+0.10)3

F= 100,000 (1.331)F = 133100

Simple and Compound InterestExample: $100,000 lent for 3 years at interest rate i= 10% per year. What is repayment after 3 years ?

Simple: $130,000: Compounded: $133,100

From SPCAF to SPPWF

Now we have the formula how to convert singlepresent amounts into future amount at a giveninterest rate i.e. F = P(1+i)n

What if we are given a future amount (F) and weare asked to calculate present worth (P) ?

F = P(1+i)n

=> P = F [1/(1+i)n]or P = F(1+i)-n

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Single Payment PresentWorth Factor (SPPWF)

The term (1+i)n is known as Single PaymentPresent Worth Factor (SPPWF)

It is also refer as P/F factor

This is a converting factor, when multiplied by Fyields the present amount P of initial amount Fafter n years at interest rate i

Compounding andDiscounting

When we convert a P value into a F using somerate we call this process . COMPOUNDINGand the rate use is called Interest rate

When we convert F into P using some rate wecall the process Discounting and the rate weuse is called Discount rate

Compounding increase your amount (as itscompounded).discounting decrease your amountas its (discounted)

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Example Find the present value of $10,000 to be

received 10 years from now at a discount rateof 10%F = $10,000i or r = 10%n = 10

P = F (1+i)-n=> P = 10,000 (1+0.1)-10

= 10,000 x 0.385= $3850

Class Practice:Allowed time 5 minutes

Sandy, a manufacturing engineer, justreceived a year-end bonus of $10,000 thatwill be invested immediately. With theexpectation of earning at the rate of 8% peryear, Sandy hopes to take the entire amountout in exactly 20 years to pay for a familyvacation when the oldest daughter is due tograduate from college. Find the amount offunds that will be available in 20 years?

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Class AssignmentAllowed time 5 minutes

Take a paper sheet Write down your name and Registration number on

top of the paper sheet Check for your registration numbers If it is Odd

(end with number 1, 3, 5, 7, 9) then attemptquestion one

For Even registration numbers attempt question2

Those who talk will get Zero When I announce time overstop writing

Class PracticeAllowed Time 5 Minutes

For Odd registration numbers: Question 11. What is present worth of $50,000 money that a

person will get, 8 years from now, given that rateof discount is 7% per year.

For Even registration numbers: Question 22. A family that won a $100,000 prize in a lottery and

decided to put one-half of the money in a collegefund for their child. If the fund earned interest at6% per year, how much was in the account 14years after it was started?

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A Standard Notation Instead of writing the full formulas of SPCAF and SPPWF for

simplicity there is a standard notation

This notations includes two cash flows symbols, interest rate andnumber of periods

General form is: (X/Y, i, n) which means X represents what issought, Y is given, i is interest rate and n is number of periods

Examples:

Name Equation withfactor formula

Notation Standard NotationEquation

Find/Given

Single-payment compoundamountSingle-payment presentworth

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

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

Using Standard NotationExample

What will be the future value of Rs. 100,000compounded for 17 years at rate of interest10% ?

F= (1+ i)n or F = P(1+0.1)n now writing thatin standard notation we have

F = P(F/P, i, n) F = 100,000(F/P, 10%, 17) F = 100,000 (5.054) F= 505400

That value youget from Table

(F/P, i, n)

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From Single Payments toAnnuity

Normally, in real world we do not face Singlepayments mostly instead faces cash flowssuch as home mortgage payments andmonthly insurance payments etc

An annuity is an equal annual series of cashflows. It may be equal annual deposits,equal annual withdrawals, equal annualpayments, or equal annual receipts. The keyis equal, annual cash flows

Uniform Series Present WorthFactor (P/A factor)

P = ?t = given

A = given

n2

t=0

1 n13

A A A A A1(1 + ) 1(1 + ) ( ) . 1(1 + ) 1(1 + )= [( )+ ( ) +( ) + .+ ( ) +( ) ] . (1)( ) = ( ) [( )+ ( ) +( ) + .+ ( ) +( )Multiply Eq(1) by (P/F, i, n) factor and subtract the equation(1) from Eq (2)( ) = [( ) + ( ) +( ) + .+ ( ) +( ) ..(2)

Can we use SinglePayment Present WorthFactor (P/F, i%, n), to getP for this cash flow ?

= + + + + +

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= [( )+ ( ) +( ) + .+ ( ) +( ) ] . (1)( ) = [( ) + ( ) +( ) + .+ ( ) +( ) ] ..(2)Uniform Series Present Worth

Factor (P/A factor)Subtracting Eq(1)from Eq(2)

(1 + ) = [ 11 + 1(1 + )]( ) = [( ) ]( ) = [( ) ]( ) = [( ) 1 ]

= [( ) 1 ]= [( ) 1 ]= (1 + ) 1(1 + )

Uniform Series Present Worth Factor(USPWF)

P = ?t = given

A = given

n2

t=0

1 n13

To find P of such series of A we cannot use P=F(P/F, i, n) because wedo not have a single amount but a uniform series in which cash flowsoccurs in equal amounts (in each period) and in consecutive interestperiods.

Yes P/F can be use for each A separatelybut thats a lengthy process

Uniform Series Present Worth Factor (USPWF) represented as P/A isused to calculate the equivalent P value in year 0 for uniform end-of-periodseries of A values beginning at the end of period 1 and extend

Mathematically: = (1 + ) 1(1 + )

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Capital Recovery Factor (CRF)

Name Equation with factorformula

Notation Standard NotationEquation

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

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

Uniform SeriesPresent Worth

Capital Recovery

P = givent = given

A = ? n2

t=0

1 n-13

= (1 + ) 1(1 + )= (1 + )(1 + ) 1

The term in bracket is called Capital Recovery Factor (CRF)or A/P factor and it calculates the equivalent uniform annualworth A over n years for a given P in year 0, when theinterest rate is i

= (1 + )(1 + ) 1= (1 + ) 1(1 + )

Example 1: Uniform SeriesPresent Worth (P/A)

A chemical engineer believes that by modifying the structure of acertain water treatment polymer, his company would earn an extra$5000 per year. At an interest rate of 10% per year, how muchcould the company afford to spend now to just break even over a 5year project period?The cash flow diagram is as follows:

P = 5000(P/A,10%,5)= 5000(3.7908)= $18,954

0 1 2 3 4 5

A = $5000

P = ?i =10%

Solution:A = 5000 i = 10% n = 5

P = A(P/A, i, n)Which factor should be used ?

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Example 2.3How much money should you be willing topay now for a guaranteed $600 per year for9 years starting next year, at a rate of returnof 16% per year?

Solution:P = ?i = 16%n = 9

Which factor should be used ?P = ?

A = $600

2t=0 1 3 4 9

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

A = $600(P/A,16%, 9) 600(4.6065) = $2763.90

Example 1: Uniform SeriesPresent Worth (P/A)

A chemical engineer believes that by modifying the structure of acertain water treatment polymer, his company would earn an extra$5000 per year. At an interest rate of 10% per year, how muchcould the company afford to spend now to just break even over a 5year project period?The cash flow diagram is as follows:

P = 5000(P/A,10%,5)= 5000(3.7908)= $18,954

0 1 2 3 4 5

A = $5000

P = ?i =10%

Solution:A = 5000 i = 10% n = 5

P = A(P/A, i, n)Which factor should be used ?

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Class Practice: Allowed Time 7A chemical product company is considering investment in cost

saving equipment. If the new equipment will cost $220,000 topurchase and install, how much must the company save eachyear for 3 years in order to justify the investment, if the interestrate is 10% per year?

Class Practice: Allowed Time 7

A chemical productcompany is consideringinvestment in cost savingequipment. If the newequipment will cost$220,000 to purchase andinstall, how much must thecompany save each yearfor 3 years in order tojustify the investment, if theinterest rate is 10% peryear?

0 1 2 3

A = ? i =10%

P = $220,000

The cash flow Diagram

A = 220,000(A/P,10%,3)= 220,000(0.40211)= $88,464

Solution:P = 220,000I = 10%n = 3

A = P(A/P, i, n)Which factor should be used ?

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Practice: 5 minutesHow much money should you be willing to pay now fora guaranteed $600 per year for 9 years starting nextyear, at a rate of return of 16% per year?

Practice: 3 minutesHow much money should you be willing to pay now fora guaranteed $600 per year for 9 years starting nextyear, at a rate of return of 16% per year?Solution:P = ?i = 16%n = 9

Which factor should be used ? P = ?

A = $600

2

t=0

1 3 4 9

P = A(P/A, i, n)A = $600(P/A,16%, 9)600(4.6065)= $2763.90

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Spread Sheet Functions

Our course book (Blank and Tarquin) from timeto time refers to Spread sheet functions

You will be getting some formulas as well in textwhile reading the book

We are going to Ignore that parts because youcannot use it in exams. Yes you will need it whiledoing projects in real life, but applying that fromEXCEL is not difficult if you know what it mean.

Uniform Series Compound AmountFactor (USCAF)

F = ?t = given

A = given

n20 1 n-13

Similar to Uniform Series Present Worth Factor we haveUniform Series Compound Amount Factor which is givenas follows (only the term in the parenthesis)= (1 + ) 1

The term in the parenthesis is called Uniform SeriesCompound Amount Factor (USCAF) represented as F/A, tomultiply with given uniform amount gives future worth of auniform series

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Sinking Fund Factor (SFF)

F = given

t = given

A = ?

n20 1 n-13

Sinking Fund Factor can be obtained from USCA and given as:= (1 + ) 1= (1 + ) 1

The term in the brackets is Sinking Fund Factor and is used to determinesthe uniform annual series A that is equivalent to a given future amount F

Example 3: Uniform Series Involving F/A

An industrial engineer made a modification to a chip manufacturingprocess that will save her company $10,000 per year. At an interest rateof 8% per year, how much will the savings amount to in 7 years?

The cash flow diagram is:

A = $10,000

F = ?

i = 8%0 1 2 3 4 5 6 7

Solution:

F = 10,000(F/A,8%,7)= 10,000(8.9228)= $89,228

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

A =10,000i =8%n =7

PracticeExample 2.5

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What is meant by Sinking Fund ?

Sinking Fund Factor (SFF)F = givent = given

A = ?n

20 1 n-13

Sinking Fund Factor can be obtained from USCA and given as:= (1 + ) 1= (1 + ) 1

The term in the brackets is Sinking Fund Factor and is used to determinesthe uniform annual series A that is equivalent to a given future amount F

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PracticeAn industrial engineer made a modification to a chip manufacturing process thatwill save her company $10,000 per year. At an interest rate of 8% per year, howmuch will the savings amount to in 7 years?

The cash flow diagram is:

A = $10,000F = ?i = 8%

0 1 2 3 4 5 6 7

Solution:

F = 10,000(F/A,8%,7)= 10,000(8.9228)= $89,228

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

A =10,000i =8%n =7

Class Practice:8 Minutes

Question No. 1The president of Ford MotorCompany wants to know theequivalent future worth ofa $1 million capitalinvestment each year for 8years, starting 1 year fromnow. Ford capital earns at arate of 14% per year.

Question No. 2A company that makes self-clinching fasteners expectsto purchase newproduction-line equipmentin 3 years. If the new unitswill cost $350,000, howmuch should the companyset aside each year, if theaccount earns 10% peryear?

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Class Practice

i= 14% n = 8 years A= 10,000

Which factor should be used ?

F = A(F/A, i, n) F = 1000( F/A, 14%,8)

= 1000(13.2328)$13,232.80

Question No. 2 i= 10% n = 3 years A= ? F = 350, 000

A = 350,000 (A/F, i, n)A = 350,000(A/F,10%,3)

= 350,000(0.30211)= $105,739

A = $10,000

F = ?i = 14%

1 2 3 4 5 6 7 8

Question No. 1

A = ?

F = 350,000i = 10%1 2 3

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From Uniform Series toGradient

We started with single amount payments

We moved to uniform series (Annuities) its presentor future values

Real world dont have uniform series benefits/costsnormally

We cannot exclude the possibilities of Arithmetic orGeometric Gradient Series cash flows

Example You bought a used car expected costs during first year will

be fuel and insurance that is $2500 let assume that cost of repair is

increasing by $200 every year what will be the amount in Second Year ?

n20 1 n-13

$2500

$2700

$2900

$2500+(n-2)200$2500+(n-1)200

Base amount

Gradient (G) = $200

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Arithmetic Gradient Factors(P/G, A/G)

Cash flows that increase or decrease by a constant amountare considered arithmetic gradient cash flows.

The amount of increase (or decrease) is called the gradient

CFn = base amount + (n-1)GCash Flow Formula

G = $25Base = $100

$100$125

$150$175

0 1 2 3 4

$500

G = -$500Base = $2000

$2000$1500

$1000

0 1 2 3 4

When we have a Gradient Series wecannot apply Single Amount PresentWorth/Future Worth factors or UniformSeries factors

We have to use a different methodology toaddress problems related to gradient cashflows.


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Solving Arithmetic Gradientrelated problems

Present value of the Arithmetic Gradient series can becalculated as follows:

1. Find the gradient and base2. Cash flow diagram maybe helpful if u draw it3. Break the gradient series into a Uniform series

and a Gradient Series as shown on next slide4. The formula for calculating present value of the

Arithmetic Gradient series is as follows;

5. Calculate PA and PG and use the aboveformula to get the present value of the ArithmeticGradient

PT = PA + PG





A A A A A

AA+G

A+2GA+3G

A+(n-1) G

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

0

G2G

3G

(n-1)G

0 1 2 3 4 n

=+

PT = PA + PG

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PA = A(P/A, i, n) or Uniform Series Present worthFactor

PG = G(P/G, i, n) or Arithmetic Gradient PresentWorth Factor you can use table for it too.

Alternatively, PG can also be calculated byfollowing formula

PT = PA + PG

nn

n

G in

iii

iGP )1()1(

1)1(


$100$125

$150$175

G = $25Base = $100

0 1 2 3 4 0 1 2 3 4

$100

0 1 2 3 4

$25$50

$75

$P = $100(P/A,i,4) + $25(P/G,i,4)

PT = PA + PG

Where PA = Present worth uniform series (P/A, i,n) and PG = present worth of the gradient series (P/G,i, n)

nn

n

G in

iii

iGP )1()1(

1)1(

Equivalent cash flows:

=> +

Note: the gradientseries by conventionstarts in year 2.

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Example (Problem 2.25)

Profits from recycling paper, cardboard, aluminium,and glass at a liberal arts college have increased ata constant rate of $1100 in each of the last 3 years.If this years profit (end of year 1) is expected to be$6000 and the profit trend continues through year5,(a) what will the profit be at the end of year 5 and(b) what is the present worth of the profit at an

interest rate of 8% per year?

G = $1100, Base = $6000

Example (Problem 2.25)(a) what will the profit be at the end of year 5 &(b) what is the present worth of the profit at an interest rate of 8% per

year?

G = $1100 Base = $6000

$6000$7100

$8200$9300

0 1 2 3 4 5Find the cash flows as follows:CF = Base + G(n-1)GCF1 = 6000 + 1100(1-1)= 6000CF2 = 6000 + 1100(2-1)= 7100CF3 = 6000 + 1100(3-1)= 8200CF4 = 6000 + 1100(4-1)= 9300CF5 = 6000 + 1100(5-1)= 10400

$10400

=>0 1 2 3 4 5

$6000

+0 1 2 3 4 5

$1100$2200

$3300$4400

P = A(P/A, i, n) G(P/G, i, n)+P = 6000(P/A, 8%, 5) + 1100(P/G, 8%, 5)P = 6000(3.9927) + 1100(7.3724)

P = 326066

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CFn = base amount + (n-1)GCash FlowFormula

A A A A A

AA+G

A+2GA+3G

A+(n-1) G

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

0

G2G

3G

(n-1)G

0 1 2 3 4 n

=+

PT = PA + PG



PG = G(P/G, i, n) or Arithmetic Gradient PresentWorth Factor


PT = PA + PG

nn

n

G in

iii

iGP )1()1(

1)1(

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1. Find the gradient and base2. Cash flow diagram maybe helpful if u draw it3. Break the gradient series into a Uniform series and a

Gradient Series as shown on previous slide4. The formula for calculating present value of the


5. Calculate PA and PG and use the above formula toget the present value of the Arithmetic Gradient

PT = PA + PG

From Uniform Series toGradient

We started with single amount payments

We moved to uniform series (Annuities) its presentor future values

Real world dont have uniform series benefits/costsnormally

We cannot exclude the possibilities of Arithmetic orGeometric Gradient Series cash flows

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Example You bought a used car with one year warranty

expected costs during first year willbe fuel and insurance that is $2500

let assume that cost of repair isincreasing by $200 every year

what will be the amount in Second Year ?

n20 1 n-13

$2500

$2700

$2900

$2500+(n-2)200$2500+(n-1)200

Base amount

Gradient (G) = $200





G = $25Base = $100

$100$125

$150$175

0 1 2 3 4

$500

G = -$500Base = $2000

$2000$1500

$1000

0 1 2 3 4

Gradientseriescould beboth: cashinflow (asgivenhere) orOutflows

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When we have a Gradient Series wecannot apply Single Amount PresentWorth/Future Worth factors or UniformSeries factors

We have to use a different methodology toaddress problems related to gradient cashflows.




1. Find the gradient and base2. Cash flow diagram maybe helpful if you draw it3. Break the gradient series into a Uniform series and a

Gradient Series as shown on next slide4. The formula for calculating present value of the



PT = PA + PG

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The base amount is A and the Gradient is G in thefollowing graph


A A A A AAA+G

A+2GA+3G

A+(n-1) G

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

0 G2G

3G(n-1)G

0 1 2 3 4 n

= +

PT = PA + PG



PG = G(P/G, i, n) or Arithmetic Gradient PresentWorth Factor you can use table for it too.


PT = PA + PG

nn

n

G in

iii

iGP )1()1(

1)1(

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$100$125

$150$175

G = $25Base = $100

0 1 2 3 4 0 1 2 3 4

$100

0 1 2 3 4

$25$50

$75

$P = $100(P/A,i,4) + $25(P/G,i,4)

PT = PA + PG

Where PA = Present worth uniform series (P/A, i,n) and PG = present worth of the gradient series (P/G,i, n)

nn

n

G in

iii

iGP )1()1(

1)1(

Equivalent cash flows:

=> +

Note: the gradientseries by conventionstarts in year 2.

Example (Problem 2.25)

Profits from recycling paper, cardboard, aluminium,and glass at a liberal arts college have increased at aconstant rate of $1100 in each of the last 3 years.If this years profit (end of year 1) is expected to be$6000 and the profit trend continues through year 5,(a) what will the profit be at the end of year 5 and(b) what is the present worth of the profit at an interest

rate of 8% per year?

G = $1100, Base = $6000

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Example (Problem 2.25)(a) what will the profit be at the end of year 5 &(b) what is the present worth of the profit at an interest rate of 8% per

year?

G = $1100 Base = $6000

$6000$7100

$8200$9300

0 1 2 3 4 5Find the cash flows as follows:CF = Base + G(n-1)GCF1 = 6000 + 1100(1-1)= 6000CF2 = 6000 + 1100(2-1)= 7100CF3 = 6000 + 1100(3-1)= 8200CF4 = 6000 + 1100(4-1)= 9300CF5 = 6000 + 1100(5-1)= 10400

$10400

=>0 1 2 3 4 5

$6000

+0 1 2 3 4 5

$1100$2200

$3300$4400

P = A(P/A, i, n) G(P/G, i, n)+P = 6000(P/A, 8%, 5) + 1100(P/G, 8%, 5)P = 6000(3.9927) + 1100(7.3724)

P = 32066

PT = PA + PG

Practice QuestionNeighboring parishes in Louisiana have agreed to poolroad tax resources already designated for bridgerefurbishment. At a recent meeting, the engineersestimated that a total of $500,000 will be deposited atthe end of next year into an account for the repair ofold and safety-questionable bridges throughout the area.Further, they estimate that the deposits will increaseby $100,000 per year for only 9 year thereafter, thencease. Determine the equivalent: present worth, if publicfunds earn at a rate of 5% per year.

5% Uniform Series Factors Athematic Gradientn Sinking

Fund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

GradientPresent Worth(P/G)

GradientUniformSeries (A/G)

9 0.09069 11.0266 0.14069 7.1078 26.1268 3.675810 0.07950 12.5779. 0.12950 7.7217 31.6520 4.0991

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Solution

Base = 500,000 Gradient = 100,000 Taking units in 1000 Base = 500 Gradient =100 i= 5% n=9+1 = 10

0 1 2 3 4 5 6 7 8 9 10

$500$600

$700 $800$900

$1000 $1100 $1200$1300

$1400

PT = 500(P/A,5%,10) + 100(P/G,5%,10)= 500(7.7217) + 100(31.6520)=$7026.05 or .. ($7,026,050)

PT = PA + PG

P = ?



1. Find the gradient and base2. Cash flow diagram maybe helpful if you draw it3. Break the gradient series into a Uniform series and a

Gradient Series as shown on next slide4. The formula for calculating present value of the



PT = PA + PG

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. What about A/G or F/G?

Arithmetic Gradient UniformSeries Factor (A/G)

Similar procedure as done for Arithmetic Gradient Presentworth Factor

Following formula: AT = AA + AG

= 1 (1 + ) 1AG and AA can be getfrom factor tablesAA = P(A/P, i, n) and AG = G(A/G, i, n)

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Arithmetic Gradient FutureWorth Factor (F/G)

Another factor in Gradient family is Future worthof an Arithmetic Gradient series (F/G)

It can be obtained by multiply (P/G) and (F/P) factors

=

Note: To get future value of Arithmetic Gradient . We do not need todivide the gradient into two separate cash flows like Present worth ofArithmetic gradient series. This F/G will be the future value of entiregradient series.

/ = 1 (1 + ) 1 No factor table values is available so only formula can be use for calculating F/G factor

Final Words about ArithmeticGradient

Present Worth(PW) or Annual Worth(AW) of Arithmetic Gradient (P/Gor A/G). Base and Gradient considered separately for both P/G and A/G. Get Two series a PA/AA series and one PG/AG series.use the factor tables to get values for PA/AA & PG/AG Add both to get PT/AT.

Future worth of Arithmetic Gradient (F/G) Base and Gradient are not considered separately. No factor values are available so have to relay on formula.formula directly calculate the future worth of Arithmetic Gradient/ = 1 (1 + ) 1

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Geometric Gradient Factors(Pg /A)

A Geometric gradient is when the periodic payment isincreasing (decreasing) by a constant percentage:

the rate at which the cash flow is increasing is g The initial amount of Geometric Gradient is A1 Pg is the present worth of entire Gradient Series including A1

It is important to note that Initial amount is not consideredseparately while working with Geometric Gradient

Note: If g is negative, change signs in front of both g values

= 1 1 +1 +1 for g i: for g = i:= 1 +

Geometric Gradient Factors(Pg /A)

A Geometric gradient is when the periodic payment isincreasing (decreasing) by a constant percentage:

A1 = $100, g = 10% or 0.1A2 = $100(1+g)A3 = $100(1+g)2

An = $100(1+g)n-1 0 1 2 3 4 nwhere: A1 = cash flow in period 1 and g = rate of increase

Note: If g is negative, change signs in front of both g values

= 1 1 +1 +1 for g i:

for g = i:= 1 +

A1A1 (1+g) A1 (1+g)

2

A1 (1+g)n-1

It maybe noted that A1is not consideredseparately in geometricgradients

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Future Worth (F/G) and Annuity(A/G) from Geometric Gradient

Series

We just learned how to get Present Value/Worth ofa Geometric Gradient

We can first derive the Present Worth of the GeometricGradient and then can use F/P factor forcalculating future value of a geometric gradient

Similarly, A/P factor can be applied to P/G factor tocalculate the Annual worth/Annuity series fromGeometric Gradient

Class Practice: 4 Minutes

Determine the present worth of a geometricgradient series with a cash flow of $50,000 inyear 1 and increases of 6% each yearthrough year 8. The interest rate is 10% peryear.

gi

ig

APg

n

111

06.010.0

10.016.011

50000

8

04.0

257.0000,50 )425.6(000,50250,321$

04.0743.01000,50

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THANK YOU

chapter 2a

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