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 A T :

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 !



1-10 Slide 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

1-11 Slide Sets to accompany Blank & Tarquin, Engineering


© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

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

nAP

i

1

11

1 g i

n

g

g

iP 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:



© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 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




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

P2 P0

F6




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



Shifted Gradient: Numerical Example 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 $17,331Slide Sets to accompany Blank & Tarquin, Engineering




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








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

mEffective

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% Slide Sets to accompany Blank & Tarquin, Engineering


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 Slide Sets to accompany Blank & Tarquin, Engineering




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

Business

Lecture # 4 gradients factors and nominal and effective interest rates