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

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
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

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

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
= 5000 (2.1589 )

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

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
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
0 1 2 3 4 5
A = \$ 5000
P = ?
i = 10 %
Solution:

Uniform Series Involving F/A and A/F
(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)

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

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

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

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

Arithmetic gradients change by the same amount e ach period
The cash flow diagram for the P G
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)

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

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
change plus sign to minus
A = base amount – G(A/G,i,n)

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
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

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
Note : g starts between
periods 1 and 2

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
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

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

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