chapter 2a
TRANSCRIPT
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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.
Arithmetic Gradient Factors(P/G, A/G)
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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
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
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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Arithmetic Gradient Factors(P/G, A/G)
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(
Arithmetic Gradient Factors(P/G, A/G)
$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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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 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
Arithmetic Gradient Factors(P/G, A/G)
PA = A(P/A, i, n) or Uniform Series Present worthFactor
PG = G(P/G, i, n) or Arithmetic Gradient PresentWorth Factor
Alternatively, PG can also be calculated byfollowing formula
PT = PA + PG
nn
n
G in
iii
iGP )1()1(
1)1(
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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 previous slide4. The formula for calculating present value of the
Arithmetic Gradient series is as follows;
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
-
3/31/2015
27
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
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
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.
Arithmetic Gradient Factors(P/G, A/G)
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 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
Arithmetic Gradient series is as follows;
5. Calculate PA and PG and use the above formula toget the present value of the Arithmetic Gradient
PT = PA + PG
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Arithmetic Gradient Factors(P/G, A/G)
The base amount is A and the Gradient is G in thefollowing graph
CFn = base amount + (n-1)GCash Flow Formula
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
Arithmetic Gradient Factors(P/G, A/G)
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(
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Arithmetic Gradient Factors(P/G, A/G)
$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 = ?
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 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
Arithmetic Gradient series is as follows;
5. Calculate PA and PG and use the above formula toget the present value of the Arithmetic Gradient
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