© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-1

Slightly Modified Lecture

Slides to Accompany

Engineering Economy

7th edition

By: Prof. Ramzi Taha

(Most Credit Goes to Leland

Blank & Anthony Tarquin)

Chapter 2

Factors: How Time

and Interest Affect

Money

© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-2

LEARNING OUTCOMES

1. F/P and P/F Factors

2. P/A and A/P Factors

3. F/A and A/F Factors

4. Factor Values

5. Arithmetic Gradient

6. Geometric Gradient

7. Find i or n

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Single Payment Factors (F/P and P/F)

Single payment factors involve only P and F. Cash flow diagrams are as follows:

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

Formulas are as follows:

Terms in parentheses or brackets are called factors. Values are in tables for i and n values

Factors are represented in standard factor notation such as (F/P,i,n) ,

where letter to left of slash is what is sought; letter to right represents what is given

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F/P and P/F for Spreadsheets

Future value F is calculated using FV function:

= FV(i%,n,,P)

Present value P is calculated using PV function:

= PV(i%,n,,F)

Note the use of double commas in each function

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Example: Finding F uture V alue

A person deposits $ 5000 into an account which pays interest at a rate of 8%

per year. The amount in the account after 10 years is closest to:

(A) $ 2,792 (B) $ 9,000 (C) $ 10 ,795 (D) $ 12 ,165

The cash flow diagram is:

Solution:

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

= 5000 (F/P, 8%, 10 )

= $10 ,794.50

Answer is (C)

= 5000 (2.1589 )

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Example: Finding Present Value

A small company wants to make a single deposit now so it will have enough money to

purchase a backhoe costing $ 50 ,000 five years from now. If the account will earn

interest of 10 % per year, the amount that must be deposited now is nearest to:

(A) $ 10 ,000 (B) $ 31 ,050 (C) $ 33 ,250 (D) $ 319 ,160

The cash flow diagram is: Solution:

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

= 50 ,000 (P/F, 10 %, 5 )

= 50 ,000 (0.6209 )

= $31 ,045

Answer is (B)

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U niform Series Involving P/A and A/P

0 1 2 3 4 5

A = ?

P = Given

The cash flow diagrams are:

Standard Factor Notation P = A(P/A,i,n) A = P(A/P,i,n)

Note: P is one period Ahead of first A value

(1) Cash flow occurs in consecutive interest periods

The uniform series factors that involve P and A are derived as follows:

(2) Cash flow amount is same in each interest period

0 1 2 3 4 5

A = Given

P = ?

Example: Uniform Series Involving P/A

© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-8

A chemical engineer believes that by modifying the structure of a certain water

treatment polymer, his company would earn an extra $ 5000 per year. At an interest

rate of 10 % per year, how much could the company afford to spend now to just

break even over a 5 year project period?

(A) $ 11 ,170 (B) 13 ,640 (C) $ 15 ,300 (D) $ 18 ,950

The cash flow diagram is as follows:

P = 5000 (P/A, 10 %, 5 )

= 5000 (3.7908 )

= $ 18 ,954

Answer is (D)

0 1 2 3 4 5

A = $ 5000

P = ?

i = 10 %

Solution:

Uniform Series Involving F/A and A/F

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(1) Cash flow occurs in consecutive interest periods

The uniform series factors that involve F and A are derived as follows:

(2) Last cash flow occurs in same period as F

0 1 2 3 4 5

F = ?

A = Given

0 1 2 3 4 5

F = Given

A = ?

Note: F takes place in the same period as last A

Cash flow diagrams are:

Standard Factor Notation F = A(F/A,i,n) A = F(A/F,i,n)

© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-10

Example: Uniform Series Involving F/A

An industrial engineer made a modification to a chip manufacturing

process that will save her company $ 10 ,000 per year. At an interest rate

of 8% per year, how much will the savings amount to in 7 years?

(A) $ 45 ,300 (B) $ 68 ,500 (C) $ 89 ,228 (D) $ 151 ,500

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

Answer is (C)

Group Exercise # 1

? Please Work on Problem # 2.5 , Page # 64

? Please Work on Problem # 2.7 , Page # 64

? Please Work on Problem # 2.8 , Page # 65

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Factor Values for Untabulated i or n

3 ways to find factor values for untabulated i or n values

Use formula

Use spreadsheet function with corresponding P, F, or A value set to 1

Linearly interpolate in interest tables

Formula or spreadsheet function is fast and accurate

Interpolation is only approximate

© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-13

Example: Untabulated i

Determine the value for (F/P, 8.3 %, 10 )

Formula: F = ( 1 + 0.083 )10 = 2.2197

Spreadsheet: = FV( 8.3 %, 10 ,,1) = 2.2197

Interpolation: 8% —— 2.1589

8.3 % —— x

9% —— 2.3674

x = 2.1589 + ( 8.3 – 8.0 )/( 9.0 – 8.0 ) 2.3674 – 2.1589

= 2.2215

Absolute Error = 2.2215 – 2.2197 = 0.0018

OK

OK

(Too high)

Group Exercise # 2

? Please Work on Problem # 2.21 , Page # 66

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Arithmetic Gradients

Arithmetic gradients change by the same amount e ach period

The cash flow diagram for the P G

of an arithmetic gradient is:

0

1 2 3 n

G

2G

4

3G

(n -1)G

P G = ?

G starts between periods 1 and 2

(not between 0 and 1)

This is because cash flow in year 1 is

usually not equal to G and is handled

separately as a base amount

(shown on next slide)

Note that P G is located Two Periods

Ahead of the first change that is equal

to G

Standard factor notation is

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

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Typical Arithmetic Gradient Cash Flow

PT = ?

i = 10 %

0 1 2 3 4 5

400

450

500

550

600

PA = ?

i = 10 %

0 1 2 3 4 5

400 400 400 400 400

PG = ?

i = 10 %

0 1 2 3 4 5

50

100

150

200

+

This diagram = this base amount plus this gradient

PA = 400 (P/A, 10 %, 5) PG = 50 (P/G, 10 %, 5)

P T = P A + P G = 400 (P/A, 10 %, 5) + 50 (P/G, 10 %, 5)

Amount

in year 1

is base

amount

Amount in year 1

is base amount

Converting Arithmetic Gradient to A

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i = 10 %

0 1 2 3 4 5

G

2G

3G

4G

i = 10 %

0 1 2 3 4 5

A = ?

Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n)

General equation when base amount is involved is

A = base amount + G(A/G,i,n)

0 1 2 3 4 5

G

2G

3G

4G

For decreasing gradients,

change plus sign to minus

A = base amount – G(A/G,i,n)

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Example: Arithmetic Gradient

The present worth of $ 400 in year 1 and amounts increasing by $ 30 per year

through year 5 at an interest rate of 12 % per year is closest to:

(A) $ 1532 (B) $ 1,634 (C) $ 1,744 (D) $ 1,829

0

1 2 3 Year

430

460

4

490

520

P T = ?

5

400

i = 12 %

G = $ 30

= 400 (3.6048 ) + 30 (6.3970 )

= $ 1,633.83

Answer is (B)

PT = 400 (P/A, 12 %, 5) + 30 (P/G, 12 %, 5)

The cash flow could also be converted

into an A value as follows:

A = 400 + 30 (A/G, 12 %, 5)

= 400 + 30 (1.7746 )

= $ 453.24

Solution:

Group Exercise # 3

? Please Work on Problem # 2.25 , Page # 66

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Geometric Gradients

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Geometric gradients change by the same percentage each period

0

1 2 3 n

A1

A 1(1+g) 1

4

A 1(1+g) 2

A 1(1+g) n-1

P g = ?

There are no tables for geometric factors

Use following equation for g ? i:

P g = A 1{1- ( 1+g)/( 1+i) n}/(i -g)

where: A 1 = cash flow in period 1

g = rate of increase

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

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

Cash flow diagram for present worth

of geometric gradient

Note : g starts between

periods 1 and 2

Example: Geometric Gradient

© 2012 by McGraw -Hill, New York, N.Y All Rights Reserved 2-21

Find the present worth of $ 1 ,000 in year 1 and amounts increasing

by 7 % per year through year 10 . Use an interest rate of 12 % per year.

(a) $ 5,670 (b) $ 7,333 (c) $ 12 ,670 (d) $ 13 ,550

0

1 2 3 10

1000 1070

4

1145

1838

P g = ? Solution:

Pg = 1000 1-(1+0.07 /1+0.12 )10/( 0.12 -0.07 )

= $ 7,333

Answer is (b)

g = 7%

i = 12 %

To find A, multiply P g by (A/P, 12 %, 10 )

Group Exercise # 4

? Please Work on Problem # 2.33 , Page # 67

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

2-23

Unknown Interest Rate i

Unknown interest rate problems involve solving for i,

given n and 2 other values (P, F, or A )

(Usually requires a trial and error solution or interpolation in interest tables)

A contractor purchased equipment for $ 60 ,000 which provided income of $ 16 ,000

per year for 10 years. The annual rate of return of the investment was closest to:

(a) 15 % (b) 18 % (c) 20 % (d) 23 %

Can use either the P/A or A/P factor. Using A/P: Solution:

60 ,000 (A/P,i%, 10 ) = 16 ,000

(A/P,i%, 10 ) = 0.26667

From A/P column at n = 10 in the interest tables, i is between 22 % and 24 % Answer is (d)

Procedure: Set up equation with all symbols involved and solve for i

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Unknown Recovery Period n

Unknown recovery period problems involve solving for n,

given i and 2 other values (P, F, or A)

(Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables)

Procedure: Set up equation with all symbols involved and solve for n

A contractor purchased equipment for $ 60 ,000 that provided income of $ 8,000

per year. At an interest rate of 10 % per year, the length of time required to recover

the investment was closest to:

(a) 10 years (b) 12 years (c) 15 years (d) 18 years

Can use either the P/A or A/P factor. Using A/P: Solution:

60 ,000 (A/P, 10 %,n) = 8,000

(A/P, 10 %,n) = 0.13333

From A/P column in i = 10 % interest tables, n is between 14 and 15 years Answer is (c)

Group Exercise # 5

? Please Work on Problem # 2.38 , Page # 67

? Please Work on Problem # 2.45 , Page # 68

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Summary of Important Points

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In P/A and A/P factors, P is one period ahead of first A

In F/A and A/F factors, F is in same period as last A

To find untabulated factor values, best way is to use formula or spreadsheet

For arithmetic gradients, gradient G starts between periods 1 and 2

Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount

For geometric gradients, gradient g starts been periods 1 and 2

In geometric gradient formula, A 1 is amount in period 1

To find unknown i or n, set up equation involving all terms and solve for i or n