Lecture # 4 gradients factors and nominal and effective interest rates

Lecture # 4
Gradient Factors/Shifted gradients
1-1 Dr. A. Alim

Gradients
 Gradients are special cases where a series of
cash flows consists of regular, unequal
amounts that increase or decrease following a
specific pattern
 Gradient factors are the factors used to
calculate equivalent P, A and F for such series
of cash flows.
Slide Sets to accompany Blank & Tarquin, Engineering
Economy, 7th Edition, 2012 1-2 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

ARITHMETIC GRADIENT
An arithmetic gradient is a cash flow series that either increases or decreases by
a constant amount. The cash flow, whether income or disbursement, changes by
the same constant amount each period. The amount of the increase or decrease
is the gradient “G”. For example, if a manufacturing engineer predicts that the
cost of maintaining a machine will increase by $500 per year until the machine is
retired, a gradient series is involved and the amount of the gradient is $500.
GEOMETRIC GRADIENT
It is also common for cash flow series, such as operating costs, construction
costs, and revenues, to increase or decrease from period to period by a constant
percentage, for example, 5% per year. This uniform rate of change defines a
geometric gradient series of cash flows. In addition to the symbols i and n used
thus far, we now need the term g.
g = constant rate of change, in decimal form, by which amounts increase or
decrease from one period to the next

Arithmetic Gradient Factors
(P/G) and (A/G)
 Cash flow profile
0 1 2 3 n-1 n
A1+G
A1+2G
A1+(n-2)G
A1+(n-1)G
Find P, given gradient cash flow G
CFn = A1 ± (n-1)G
Base amount
= A1

Gradient Structure
 As we know, arithmetic gradients are
comprised of two components
1. Gradient component
2. Base amount
 When working with a cash flow containing a
gradient, the (P/G) factor is only for the
gradient component
 Apply the (P/A) factor to work on the base
amount component
 P = PW(gradient) + PW(base amount)

Use of the (A/G) Factor
0 1 2 3 n-1 n
G
2G
(n-2)G
(n-1)G
Find A, given gradient cash flow G
CFn = (n-1)G
Equivalent A of
gradient series
A A A . . . A A
A = G(A/G,i,n)

Use of the (A/G) Factor
A = G(A/G,i,n)
The general equation for PT and AT :
PT = PA ± PG
AT = AA ± AG
Important: First A is end of year 1, while first G is end of year 2. Same n
is used for both equations !

The P/G and A/G factors
 Remember, there are only two ways to
determine these factors:
* From tables
* From formulas
No EXCEL functions for these factors !

1-10Slide Sets to accompany Blank & Tarquin, Engineering
Economy, 7th Edition, 2012
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
The P/G and A/G factors

1-11Slide Sets to accompany Blank & Tarquin, Engineering
The P/G and A/G factors’
formulas

Geometric Gradient Series Factor
 Geometric Gradient
Cash flow series that starts with a base amount A1
Increases or decreases from period to period by a
constant percentage amount
This uniform rate of change defines…
 A GEOMETRIC GRADIENT
 Notation:
g = the constant rate of change, in decimal form, by which
future amounts increase or decrease from one time
period to the next

Typical Geometric Gradient
A1
A1(1+g)
A1(1+g)2
. . . .
0 1 2 3 n-2 n-1 n
A1(1+g)n-1
Required: Find a factor (P/A,g %,i %, n) that will convert future
cash flows to a single present worth value at time t = 0
Given A1, i%, and g%

Two Forms to Consider…
1
(1 )
g
nA
P
i


1
1
1
1
g i
n
g
g
i
P A
i g
  
  
   
 
 
  
Case: g = i Case: g = i
 A1 is the starting cash flow
 There is NO base amount associated with a geometric gradient
 The remaining cash flows are generated from the A1 starting value
 No Excel function or tables available to determine this factor…too many combinations of i% and g%
to support tables
 The easiest way to get equivalent A or F is to use A/P and F/P factors with the P shown above.
To use the (P/A,g%,i%,n) factor

Shifted payment series/gradients
 Shifted uniform series.
 Shifted arithmetic gradients.
 Shifted geometric gradients.
 By definition, a shifted series/gradient means
its present value is NOT at time zero.

1 - Shifted Uniform Series
0 1 2 3 4 5 6 7 8
A = $-500/year
Consider:
1-16

1 - Shifted Uniform Series
0 1 2 3 4 5 6 7 8
A = $-500/year
Consider:
P of this series is at t = 2 (P2)
P2 = - 500 (P/A,i%,4)
P0 = P2 (P/F,i%,2)
P2
P0
1-17

Example of Shifted Series P and F
A = $-500/year
• F for this series is at t = 6; F6 = A(F/A,i%,4)
• P0 for this series at t = 0 is also
P0 = - F6(P/F,i%,6)
0 1 2 3 4 5 6 7 8
P2P0
F6
1-18

2 - Shifted Arithmetic Gradients
 The Present Worth of an arithmetic gradient
(linear gradient) is always located:
One period to the left of the first cash flow
in the series ( “0” gradient cash flow) or,
Two periods to the left of the “1G” cash
flow

Shifted Gradient
 A Conventional Gradient
has its present worth
point at t = 0
 A Shifted Gradient has its
present value point
removed from time t = 0
0 1 2 3 4 5 6 7 8 9 10
P
0 1 2 3 4 5 6 7 8 9 10
P3
P0

Shifted Gradient: Numerical Example
Base Series = $500
G = $+100
0 1 2 3 4 ……….. ……….. 9 10
Cash flows start at t = 3
$500/year increasing by $100/year through
year 10; i = 10%; Find P at t = 0

PW of the base series
Base Series = $500
0 1 2 3 4 ……….. ……….. 9 10
P2 = 500(P/A,10%,8) = 500(5.3349) = $2667.45
nseries = 8 time periods
P0 = 2667.45(P/F,10%,2) = 2667.45(0.8264)
= $2204.38
P2
P0

PW for the gradient component
0 1 2 3 4 ……….. ……….. 9 10
G = +$100
P2 = $100(P/G,10%,8) = $100(16.0287) = $1,602.87
P0 = $1,602.87(P/F,10%,2) = $1,602.87(0.8264)
= $1,324.61
P2
P0

Example: Total Present Worth Value
 For the base series
 P0 = $2204.38
 For the arithmetic gradient
 P0 = $1,324.61
 Total present worth
P = $2204.38 + $1,324.61 = $3528.99

To Find A for a Shifted Gradient
1) Find the present worth at actual time 0
2) Then apply the (A/P,i,n) factor to
convert the present worth to an
equivalent annuity (series)
3) A = P(A/P, i, n)
A = $3528.99 (A/P, 10%,10)
= $547.36

Using Spreadsheet Functions (NPV)
 Excel is a powerful tool to calculate P0 for shifted series or
gradients in one step. We use the NPV function.
 Net Present Value for a shifted series or gradient. Excel
function is:
=NPV(i%,second_cell : last_cell) + first_cell

Spreadsheet Applications – NPV Function
 The NPV function requires that all cells in the defined
time range have an entry
 The entry can be $0…but not blank! A “0” value must
be entered
 Incorrect results can be generated if one or more
cells in the defined range is left blank
 Unlike the PV function, CF values do not need to be
equal. In fact, they can be negative or zero values
 This makes the NPV function particularly suitable for
shifted series and/or gradients.

Base Series = $500
G = $+100
0 1 2 3 4 ……….. ……….. 9 10
Cash flows start at t = 3
$500/year increasing by $100/year through
year 10; i = 10%; Find P at t = 0 using the
NPV function. Then find the equivalent A.

Using NPV Function
YEAR CF $
0 0
1 0
2 0
3 500 present Worth = $3,529.20
4 600
5 700 Equivalent A = $574.36
6 800
7 900
8 1000
9 1100
10 1200

3 - Shifted Geometric Gradient
Conventional Geometric Gradient
0 1 2 3 … … … n
A1
Present worth point is at t = 0 for a conventional
geometric gradient

Shifted Geometric Gradient
Shifted Geometric Gradient
0 1 2 3 … … … n
A1
Present worth point is at t = 2 for this example

Example 3.7, page 83, Blank’s 7th edition

Using NPV Function
YEAR CF
0 $35,000
1 $7,000
2 $7,000
3 $7,000 Present Worth = $83,229.83
4 $7,000
5 $7,000 Equivalent A = $14,907.33
6 $7,840
7 $8,781
8 $9,834
9 $11,015
10 $12,336
11 $13,817
12 $15,475
13 $17,331

CHEE 5369/6369
Homework # 1
Thursday January 23, 2014
The following solved examples from Blank (seventh edition):
Example 1.3 page 11
Example 1.4 page 11
Example 1.5 page 12
Example 1.8 page 14
Example 1.9 page 17
Example 1.11 page 18
Example 2.3 page 44
Example 2.5 page 47
Example 2.8 page 51
Example 3.5 page 80
Example 3.8 page 84

1-37

1-38

Nominal and Effective Interest Rate
Statements
 So far, we have learned:
 Simple interest and compound interest definitions
 Compound Interest –
 Interest computed on interest
 For a given interest period
 The time standard for interest computations – One Year

Two Common Forms of Quotation
 Two types of interest quotation:
 1. Quotation using a Nominal Interest Rate
 2. Quoting using an Effective Interest Rate
 Nominal and Effective interest rates are common in
business, finance, and engineering economy
 Each type must be understood in order to solve
problems where interest is stated in various ways

Examples of Nominal Interest Rates
 18% per year
Same as: 18/12 = 1.5% per month
Same as: 18/4 = 4.5% per quarter
Same as: 18/0.5 = 36% per two years
 1.5% per 6-month period
Same as: (1.5%)(2 six-month periods) = 3% per year
 1% per week
Same as: (1%)(52 weeks) = 52% per year

Effective Interest Rate
Definition:
The effective interest rate is the actual rate
that applies for a stated period of time.
The compounding of interest during the time
period of the corresponding nominal rate is
accounted for by the effective interest rate.
The effective rate is commonly expressed on
an annual basis denoted as “ia”
All interest formulas, factors, tabulated values, and spreadsheet relations must have
the effective interest rate to properly account for the time value of money.

Time Based Units
Time Period – The period over which the interest is
expressed (always stated).
Ex: “8% per year”, or “1% per month”
Compounding Period (CP) – The shortest time unit over
which interest is charged or earned.
Ex: “8% per year, compounded monthly”
Compounding Frequency – The number of times m that
compounding occurs within time period t.
Ex: “10% per year, compounded monthly” has m = 12
Ex: “1% per month, compounded monthly” has m = 1

Effective Interest Rates for Any Time
Period
 The following equation is the effective interest rate per
time period as related to nominal int. rate and
compounding frequency for the same time period.
r
i = (1+ ) 1
m
m
Effective 
where:
r = nominal interest rate per time period
m = number of compounding periods per time period
r/m is the nominal; and also the effective interest rate
per compounding period.

Annual Effective Interest Rates
ONLY in this case of ANNUAL rates, there are excel
functions as follow:
To calculate ia from r use EFFECT(r,m)
To calculate r from ia use NOMINAL(ia,m)
where:
r = Annual nominal interest rate
m = number of compounding periods per year
r/m is the nominal; and also the effective interest rate
per compounding period.

Effective Interest Rates for Any Time
Period
 Note: when the CP is equal to the time period:
Then m = 1, and Effective i = r
meaning nominal interest rate is also the effective interest
rate.
 When we say interest rate of say 8% per year and do not
refer to a compounding period, it usually means that the
compounding is annual and the 8% is also the effective
rate per year.

Example: Calculating the Effective Rate
 Interest is 8% per year, compounded quarterly
 What is the effective int. rate per quarter?
 What is the effective annual interest rate?
Use equation above with r = 0.08, m = 4
Effective quarterly int. rate = nominal quarterly int.
rate = 0.08/4 = 0.02
Effective annual ia = (1 + 0.08/4)4 – 1
= (1.02)4 – 1
= 0.0824 or 8.24%/year
OR: EFFECT(0.08,4) = 8.24%

 In the previous example:
 What is the future value of $10,000 invested for 5
years?
F = 10,000 (1.0824)5 = $ 14,859
OR
F = 10,000 (1.02)20 = $ 14,859

Common Compounding Frequencies
 Interest may be computed (compounded):
 Annually – One time a year (at the end)
 Every 6 months – 2 times a year (semi-annual)
 Every quarter – 4 times a year (quarterly)
 Every Month – 12 times a year (monthly)
 Every Day – 365 times a year (daily)
 …
 Continuous – infinite number of compounding periods in a
year.
One Year is segmented into: 365 days, 52 weeks, 12 months
One Half Year is segmented into: 182 days, 26 weeks, 6 months
One Quarter is segmented into: 91 days, 13 weeks, 3 months

Example: APR and APY
 APR is Annual Percentage Rate ( Nominal)
 APY is Annual Percentage Yield ( Effective)
 Example:
A credit card company charges 18 % APR. The Law
requires that the APY must also be stated. What is the
APY if interest is compounded daily?
APY = (1 + 0.18/365)365 – 1 = 19.716 %

Effective Interest Rate for
Continuous Compounding
 Recall that effective i = (1 + r/m)m – 1
 What happens if the compounding frequency, m,
approaches infinity?
This means an infinite number of compounding
periods within a time period, and
The time between compounding approaches “0”
 A limiting value of i will be approached for a given
value of r

Continuous Compounding Effective Rate
 The effective i for the same time period when interest
is compounded continuously is then:
Effective i = er – 1
 To find the equivalent nominal rate given i when
interest is compounded continuously, apply:
ln(1 )r i 

Examples 4.11, p.150, Blank’s 6th edition.
Example 4.11
Given r = 18% per year, compounded contin., find:
 The effective annual rate
 The effective monthly rate
r/month = 0.18/12 = 1.5%/month
Effective monthly rate is e0.015 – 1 = 1.511%
The effective annual interest rate is e0.18 – 1 = 19.72%
per year.

Example 4.11
Note: The following is critically important.
You obtain the nominal monthly rate by dividing the nominal
annual rate by 12.
You CAN NOT divide the effective annual interest rate by 12
In order to obtain the monthly effective interest rate.
You must follow this sequence:
Nominal annual rate → Nominal monthly rate → Effective
monthly rate.

Example 4.12, page 150, Blank’s 6th edition

Lecture # 4 gradients factors and nominal and effective interest rates

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Lecture # 4 gradients factors and nominal and effective interest rates