# Factors

MER 439
Design of Thermal Fluid Systems
Introduction to
Engineering Economics
2. Factors
Professor Wilk
Winter 2007
Factor Notation
Single Payment Compound Amount Factor
(F/P, i%, n):
F/P = (1 + ieff)n
Single Payment Present Worth Factor
(P/F, i%, n):
P/F = (1 + ieff)-n
n is in years if the ieff is used
Example 5 - Factors
How much inheritance, to be
received 20 years from now, is
equivalent to receiving \$10,000
now? The interest rate is 4% per
year. Use factors.
Factor Tables
Fortunately these factors are
tabulated…
(And Excel has nice built in functions
to calculate them too)
Use of Multiple Factors
Many cash flow situations do not fit the single
factor equations. It is often necessary to
combine factors in multiple steps.
Example 6: What is P for a series of 10
\$100 payments starting 4 years from now?
1 2 3 4 5 6
7 8 9 10 11 12 13
\$100
P =?
years
Use of Multiple Factors
Several Methods:
1. Use P/F of each payment
2. F/P of each and then multiply by P/F
3. Get F =A (F/A,i,10), then P = F (P/F,i,13)
4. Get P3 = A(P/A,i,10) and P0 = P3(P/F,i,3)
1 2 3 4 5 6
7 8 9 10 11 12 13
\$100
P =?
years
Use of Multiple Factors
Example 7:
Someone deposited \$700 per year (at
the end of each year) at an effective
interest rate of 6%. At the beginning of
the tenth year she increased her
deposits to \$1200 per year. How much
money did she have in her account
immediately after she made her last
deposit in the beginning of year 14?
Engineering Economic problems frequently
involve disbursements or receipts that
increase or decrease each year (i.e.
equipment maintenance)
If the increase is the same every year this is
The Uniform
amount of
increase each
period is the
The amount in
the initial year is
called a base
amount,
and it
doesn’t need
to equal the
Present Value @ time zero
To get the Gradient Factors we subtract off the base
amount, and start things in year (period) 2:
PT (total) = PG+PA
PA comes from using the P/A
factor on an annuity
equal to the base amount.
PG = Present worth of
year 2…
This is what is calculated
by the P/G factor.
P/G = factor to convert a gradient series
to a present worth.
A/G = factor to convert a gradient series
to an equivalent uniform annual series.
PG/G and AG/G
(i  1)  in  1
( P / G, i , n ) 
2
n
i (1  i )
n
1
n
( A / G, i, n )  
n
i (1  i )  1
Find the PW of an income series with a
cash flow in Year 1 of \$1200 which
increases by \$300 per year through
year 11. Use i = 12%
Review of Factors
Using the tables..
Single Payment factors (P/F), (F/P)
Uniform Series factors (P/A), (F/A)
Unknown Interest Rates and Years
-Unknown Interest rate:
-i.e. F = \$20K, P = \$10K, n = 9  i = ?
-Or A = \$1770, n = 10, P = \$10K  i =?
-Unknown Years – sometimes want to determine
the number of years it will take for an
investment to pay off (n is unknown)
-i.e. A = \$100, P = \$2000, i = 2%  n = ?
Unknown interest example
Example 9
 If you would like to retire with
\$1million 30 years from now, and
you plan to save \$6000 per year
every year until then, what interest
rate must your savings earn in
order to get you that million?
Doubling Time
The time it will take for an amount
to double, given an annual growth
rate i%. (Rule of 72)
n  72/i%
Alternatively,
i%  72/n

References




Newnan, D.G. and Lavelle, J.P.,
“Essentials of Engineering Economic
Analysis,” Engineering Press, 1998.
Lindeburg, M.R., “Engineer-in-Training
Reference Manual,” 8th ed., Professional
Publications, 1992.
Jaluria, Y., “Design and Optimization of
Thermal Systems,” McGraw-Hill, 1998.
National Council of Examiners for
Engineering and Surveying,
“Fundamentals of Engineering Discipline
Specific Handbook,”NCEES, 1996.