This chapter discusses various factors used in engineering economy to analyze cash flows over time under conditions of interest. It introduces the single-payment compound amount factor (SPCAF) and single-payment present worth factor (SPPWF) for analyzing single payments compounded over time. It also discusses the uniform series present worth factor (USPWF) and capital recovery factor (CRF) for analyzing uniform series of cash flows. Examples are provided to illustrate the calculation of future and present values using these factors under simple and compound interest conditions. Arithmetic and geometric gradient factors are also introduced for cash flows that regularly increase or decrease by a constant amount or percentage.

1. Chapter 2
Factors: How Time and
Interest Affect Money
MS291: Engineering Economy

2. 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 (Optional Topic)

3. Simple and Compound Interest
Example: $100,000 lent for 3 years at interest rate i = 10%
per year. What is repayment after 3 years ?
Simple Interest
Compound Interest
Here
Interest, year 1: I1 = 100,000(0.10) = $10,000
 Total due, year 1: F1 = 100,000 + 10,000
=$110,000
P=$100,000
n= 3
i= 10%
Simple interest = P X n x i
Interest = 100,000(3)(0.10)
= $30,000
Total due = 100,000 + 30,000
= $130,000
 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: $130,000: Compounded: $133,100

4. Single Payment Compound Amount
Factor (SPCAF)
If an amount “P” is invested at time “t=0”
the amount accumulated after a year is
given as
to generalize the process for period “n” we
can write as;
F1 = P + Pi
= P(1 + i) ………. (1)
F = P(1+i)n
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)
The term “(1+i)n” is known as Single
Payment Compound Amount Factor
(SPCAF)

It is also refer as F/P factor

This is a converting factor, when
multiplied by “P” yields the future
amount “F” of initial amount “P”
after “n” years at interest rate “i”

5. Simple and Compound Interest
Example: $100,000 lent for 3 years at interest rate i = 10%
per year. What is repayment after 3 years ?
Simple Interest
Compound Interest
Here
Now we have F = P(1+i)n
P=$100,000
n= 3
i= 10%
Simple interest = P X n x i
Interest = 100,000(3)(0.10)
= $30,000
Total due = 100,000 + 30,000
= $130,000
P = $100,000
n=3
i=10%
So F = 100,000 (1+0.10)3
F= 100,000 (1.331)
F = 133100
Simple: $130,000: Compounded: $133,100

6. From SPCAF to SPPWF
 Now we have the formula how to “convert” present
amounts into future amount at a given interest rate i.e.
F = P(1+i)n
 What if we are given a future amount (F) and we are
asked to calculate present worth (P) ?
F = P(1+i)n
=> P = F [1/(1+i)n]
or P = F(1+i)-n

7. Single Payment Present Worth
Factor (SPPWF)

The term “(1+i)-n” is known as Single Payment Present
Worth Factor (SPPWF)

It is also refer as P/F factor

This is a converting factor, when multiplied by “F”
yields the present amount “P” of initial amount “F”
after “n” years at interest rate “i”

8. Example
 Find the present value of $10,000 to be
received 10 years from now at a discount rate
of 10%
F = $10,000
r = 10%
n = 10
P = F (1+i)-n
=> P = 10,000 (1+0.1)-10
= 10,000 x 0.385
= $385

9. 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 and
number of periods
 General form is: (X/Y, i, n) which means “X” represents what is sought, Y
is given, i is interest rate and n is number of periods
 Examples:
Name
Single-payment compound
amount
Single-payment present
worth
Equation with
factor formula
Notation
Standard Notation
Equation
Find/
Given
F = P(F/P, i, n)
F/P
A = P(1+i)n
(F/P, i, n)
P = F(1+i)-n
(P/F, i, n) P = F(P/F, i, n)
P/F

10. Using Standard Notation
Example
 What will be the future value of Rs. 100,000
compounded for 17 years at rate of interest 10% ?
 F= (1+ i)n or F = P(1+0.1)n now writing that in
standard notation we have
(F/P, i, n)
 F = P(F/P, i, n)
 F = 100,000(F/P, 10%, 17)
That value you
 F = 100,000 (5.054)
get from “Table”
 F= 505400

11. Single Payments to Annuity
 Normally, in real world we do not face Single
payments mostly instead faces cash flows such as
home mortgage payments and monthly insurance
payments etc
 An annuity is an equal annual series of cash flows.
It may be equal annual deposits, equal annual
withdrawals, equal annual payments, or equal
annual receipts. The key is equal, annual cash
flows

12. Uniform Series Present Worth Factor
(USPWF)
P=?
t = given
1
2
3
n-1
t=0
A = given
n

13. Capital Recovery Factor (CRF)
P = given
t = given
1
2
3
A=?
n-1
n
t=0
Name
Uniform Series
Present Worth
Capital Recovery
Equation with factor
formula
Notation
Standard Notation Equation
(P/A, i, n)
P = A(F/P, i, n)
(A/P, i, n)
A = P(A/P, i, n)

14. Example 1: Uniform Series Present
Worth (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?
The cash flow diagram is as follows:
Solution:
A = 5000
A = $5000
I = 10%
P = A(P/A, i, n)
0
1
2
i =10%
P=?
3
4
5
P = 5000(P/A,10%,5)
= 5000(3.7908)
= $18,954
n=5

15. Example 2: Uniform Series
Capital Recovery (A/P)
A chemical product company is considering investment in cost
saving equipment. If the new equipment will cost $220,000 to
purchase and install, how much must the company save each year
for 3 years in order to justify the investment, if the interest rate is
10% per year?
Solution:
The cash flow diagram is as follows:
A=?
P = 220,000
I = 10%
n=3
A = P(A/P, i, n)
0
1
P = $220,000
2
i =10%
3
A = 220,000(A/P,10%,3)
= 220,000(0.40211)
= $88,464

16. Uniform Series Compound Amount Factor
(USCAF)
F=?
t = given
0
1
2
n-1
3
A = given
n

17. Sinking Fund Factor (SFF)
t = given
0
1
2
n-1
3
n
A=?
F = given

18. Example 3: 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?
Solution:
The cash flow diagram is:
A =10,000
i =8%
n =7
F=?
i = 8%
0
1
2
3
4
A = $10,000
5
6
7
F = A(F/A, i, n)
F = 10,000(F/A,8%,7)
= 10,000(8.9228)
= $89,228
Practice
Example 2.5

19. Uniform Series Compound Amount Factor
(USCAF)
F=?
t = given
0
1
2
n-1
3
A = given
n

20. Sinking Fund Factor (SFF)
F = given
t = given
0
1
2
n-1
3
A=?
n

21. Example 3: 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?
Solution:
The cash flow diagram is:
A =10,000
i =8%
n =7
F=?
A = $10,000
0
1
2
i = 8%
3
4
5
6
7
F = A(F/A, i, n)
F = 10,000(F/A,8%,7)
= 10,000(8.9228)
= $89,228
Practice
Example 2.5

22. Arithmetic Gradient Factors
(P/G, A/G)
Cash flows that increase or decrease by a constant amount
are considered arithmetic gradient cash flows.
The amount of increase (or decrease) is called the gradient
CFn = base amount + (n-1)G
Cash Flow Formula
(n-1)G
A+(n-1) G
A
A+3G
A+2G
A+G
A
A
A
A
3G
2G
A
G
=
+
0
1
2
3
4
PT =
0
n
1
2
3
4
0
n
0
PA
+
PG
1
2
3
4
n

23. Arithmetic Gradient Factors
(P/G, A/G)
PT =
PA
+
PG
 PA = A(P/A, i, n) or Uniform Series Present worth Factor
 PG = G(P/G, i, n) or Arithmetic Gradient Present Worth Factor
 Alternatively, PG can also be calculated by following
formula
PG
G

i
 (1  i ) n  1
n


 i (1  i ) n
(1  i ) n






24. Solving Arithmetic Gradient related
problems
Present value of the Arithmetic Gradient series can be calculated as
follows:
1. Find the gradient and base
2. Cash flow diagram maybe helpful if u draw it
3. Break the gradient series into a Uniform series and a
Gradient Series as shown on next slide
4. The formula for calculating present value of the Arithmetic
Gradient series is as follows;
PT =
PA
+
PG
5. Calculate PA and PG and use the above formula to get
the present value of the Arithmetic Gradient

25. Example (Problem 2.25)
Profits from recycling paper, cardboard, aluminium,
and glass at a liberal arts college have increased at a
constant rate of $1100 in each of the last 3 years.
If this year’s 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

26. 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
$10400
$9300
$8200
1
2
$6000
$1100
+
=>
$7100
$6000
0
0
3
$4400
$3300
$2200
4
5
Find the cash flows as follows:
CF = Base + G(n-1)G
CF1 = 6000 + 1100(1-1)= 6000
CF2 = 6000 + 1100(2-1)= 7100
CF3 = 6000 + 1100(3-1)= 8200
CF4 = 6000 + 1100(4-1)= 9300
CF5 = 6000 + 1100(5-1)= 10400
1
2
3
4
0
5
P = A(P/A, i, n)
P = 6000(P/A, 8%, 5)
P = 6000(3.9927)
P = 326066
+
+
+
1
2
3
4 5
G(P/G, i, n)
1100(P/G, 8%, 5)
1100(7.3724)

27. Arithmetic Gradient Future Worth
The following formula is used for calculating future value of arithmetic
gradient given the values of gradient (the term in bracket is called
Arithmetic gradient future worth factor)
FG
  1   (1  i ) n  1 

  n 
 G   

  i 

i




(n-1)G
3G
2G
G
0
Converting Arithmetic Gradient (G) into
Uniform Series (A) (to convert G into (A/G, i, n)
Formula for conversation from G to (A/G, i, n)
1

n
AG  G  

n
 i
(1  i )  1 


Also AT = AA + AG
Where AA = P(A/P, i, n) and AG = G(A/G, i, n)
F?

28. Geometric Gradient Factors
(Pg /A)
A Geometric gradient is when the periodic payment is
increasing (decreasing) by a constant percentage:
A1 is the initial cash in Year 1
i = given
g = given
A1 (1+g)n-1
A1 (1+g)2
Pg is the sum of whole series including A1
A1 (1+g)
A1
It maybe noted that A1 is not considered
separately in geometric gradients
 1  g 
1 

  1 i 
Pg  A 
ig



nA
Pg 
1 i
n


 for g  i




0
Pg
f or g  i
Note: If g is negative, change signs in front of both g values
where: A1 = cash flow in period 1 and g = rate of increase
1
2
3 ----- n

29. Example: Geometric Gradient
 Determine the present worth of a geometric gradient
series with a cash flow of $50,000 in year 1 and
increases of 6% each year through year 8. The interest
rate is 10% per year.
n

1  g 
1 


 1 i 
Pg  A
ig










8

 1  0 .6  
1 
 

 1  0 .10    50 ,000  1  0 .743 
 50000 


0 .10  0 .06 
 0 .04 






 0 .257 
 50 ,000 

 0 .04 
 50 ,000 ( 6 .425 )
 $ 321 ,250

30. Thank You

31. Using Factor Tables

32. Using Factor Tables