Engineering Economy

The Time Value of Money

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
The objective of Chapter 4 is to
explain time value of money
calculations and to illustrate
economic equivalence.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Money has a time value.
• Capital refers to wealth in the form of
money or property that can be used to
produce more wealth.
• Engineering economy studies involve the
commitment of capital for extended periods
of time.
• A dollar today is worth more than a dollar
one or more years from now (for several
reasons).
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Return to capital in the form of interest and
profit is an essential ingredient of
engineering economy studies.
• Interest and profit pay the providers of capital for
forgoing its use during the time the capital is being
used.
• Interest and profit are payments for the risk the
investor takes in letting another use his or her
capital.
• Any project or venture must provide a sufficient
return to be financially attractive to the suppliers
of money or property.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Simple interest is used
infrequently.
When the total interest earned or charged is linearly
proportional to the initial amount of the loan
(principal), the interest rate, and the number of
interest periods, the interest and interest rate are said
to be simple.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Computation of simple interest
The total interest, I, earned or paid may be computed
using the formula below.

P = principal amount lent or borrowed

N = number of interest periods (e.g., years)
i = interest rate per interest period
The total amount repaid at the end of N interest
periods is P + I.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
If $5,000 were loaned for five years at a
simple interest rate of 7% per year, the
interest earned would be

So, the total amount repaid at the end

of five years would be the original
amount ($5,000) plus the interest
($1,750), or $6,750.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Compound interest reflects both the remaining principal
and any accumulated interest. For $1,000 at 10%…

(1) (2)=(1)x10% (3)=(1)+(2)

Amount owed Interest Amount
at beginning of amount for owed at end
Period period period of period
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331

Compound interest is commonly used in personal and

professional financial transactions.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Simple and Compound Interest
Example: Demonstrate the concept of equivalence using the different loan repayment
plans describe below. Each plan repays a $5000 loan in 5 years at 8% interest per year.

Plan 1: Simple interest, pay all at end. No interest or principal is paid until the end of
year 5. Interest accumulates each year on the principal only.

Plan 2: Compound interest, pay all at end. No interest or principal is paid until the
end of year 5. Interest accumulates each year on the total of principal and all accrued
interest.

Plan 3: Equal payments of compound interest and principal made annually. Equal
payments are made each year with a portion going toward principal repayment and the
remainder covering the accrued interest. Since the loan balance decreases at a rate
slower than “if Compound interest and portion of principal repaid annually”( equal
end-of-year payment), the interest decreases, but at a slower rate.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Solutions (Refer to Table 13 Discrete Cash Flow: Compound Interest @ 8%
End of Year Interest Total Owed at End-of-Year Total Owed
Owed for End of Year Payment after
Year Payment
0 $5,000
1 $400 $5,400 $1,252.28 $4,147.72
2 $331.82 $4,479.54 $1,252.28 $3,227.25
3 $258.18 $3,485.43 $1,252.28 $2,233.15
4 $178.65 $2,411.80 $1,252.28 $1,159.52
5 $92.76 $1,252.28 $1,252.28
Totals $6,261.41
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Economic equivalence allows us to
compare alternatives on a common basis.
• Each alternative can be reduced to an
equivalent basis dependent on
– interest rate,
– amount of money involved, and
– timing of monetary receipts or expenses.
• Using these elements we can “move” cash
flows so that we can compare them at
particular points in time.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We need some tools to find economic
equivalence.
• Notation used in formulas for compound interest
calculations.
– i = effective interest rate per interest period
– N = number of compounding (interest) periods
– P = present sum of money; equivalent value of one or
more cash flows at a reference point in time; the present
– F = future sum of money; equivalent value of one or
more cash flows at a reference point in time; the future
– A = end-of-period cash flows in a uniform series
continuing for a certain number of periods, starting at
the end of the first period and continuing through the
last
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
A cash flow diagram is an indispensable
tool for clarifying and visualizing a
series of cash flows.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Cash flow tables are essential to
modeling engineering economy
problems in a spreadsheet

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We can apply compound interest
formulas to “move” cash flows along the
cash flow diagram.
Using the standard notation, we find that a
present amount, P, can grow into a future
amount, F, in N time periods at interest rate
i according to the formula below.

In a similar way we can find P given F by

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
If an amount P is invested at time t=0, the amount F₁ accumulated 1 year
hence at an interest rate of i percent per year will be;

F₁ = P + Pi
= P (1 + i)
where the interest rate is expressed in decimal form. At the end of the
2nd year, the amount accumulated F₂ is the amount after 1 year plus the
interest from the end of year 1 to the end of year 2 on the entire F₁

F2 = F 1 + F1 i
= P (1 + i) + P (1 + i) i , So the amount of F2 can be expressed as

F2 = P (1+ i + i + i2) = P (1 + 2i + i2)

= P (1 + i)2
Similarly, the amount of money accumulated at the end of year 3 will be: F3 = P (1 + i)3
So therefore the formula can be generalized for n years to F = P (1 + i)N

Insert the Cash flow diagram for single-payment factors: a) find F and b) find P
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
It is common to use standard notation for
interest factors.

This is also known as the single payment

compound amount factor. The term on the
right is read “F given P at i% interest per
period for N interest periods.”

is called the single payment present worth

factor.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We can use these to find economically
equivalent values at different points in time.
$2,500 at time zero is equivalent to how much after six
years if the interest rate is 8% per year?

$3,000 at the end of year seven is equivalent to how

much today (time zero) if the interest rate is 6% per
year?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Example

An engineer received a bonus of $12, 000 that he will invest now. He

wants to calculate the equivalent value after 24 years, when he plans
to use all the resulting money as the down payment on an island
vacation home. Assume a rate of return of 8% per year for each of the
24 years. Find the amount he can pay down.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Example

Hewlett-Packard has completed a study indicating that $50,000 in

reduced maintenance this year (i.e., year zero) on one processing line
resulted from improved wireless monitoring technology.

a) If Hewlett-Packard considers these types of savings worth 20%

per year, find the equivalent value of this result after 5 years.

b) If the $50,000 maintenance savings occurs now, find its equivalent

value 3 years earlier with interest at 20% per year.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Pause and solve
Betty will need $12,000 in five years to pay for a major
overhaul on her tractor engine. She has found an
investment that will provide a 5% return on herinvested
funds. How much does Betty need to invest today so
she will have her overhaul funds in five years?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Uniform Series (P/A,A/P,A/F,F/A)
There are four uniform series formulas that involve A, where A,
means that:
1. The cash flow occurs in consecutive interest periods
2. The cash flow amount is the same in each period

The formulas relate a present worth P or a future worth F to a uniform series

amount A. The two equations that relate P and A are:

P = A (P/A, i, N), Uniform-series present worth and A = P (A/P, i, N), Capital

recovery

The uniform series formulas that relate A and F are:

F = A (F/A, i, N), Uniform-series compound amount and

A = F (A/F, i, N),Sinking fund
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
There are interest factors for a series of
end-of-period cash flows.

How much will you have in 40 years if you

save $3,000 each year and your account
earns 8% interest each year?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Finding the present amount from a series
of end-of-period cash flows.

How much would is needed today to provide

an annual amount of $50,000 each year for 20
years, at 9% interest each year?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding A when given F.

How much would you need to set aside each

year for 25 years, at 10% interest, to have
accumulated $1,000,000 at the end of the 25
years?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding A when given P.

If you had $500,000 today in an account

earning 10% each year, how much could you
withdraw each year for 25 years?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Example

1) How much money should you be willing to pay now for a guaranteed $600 per year
for 9 years starting next year, at a rate of return of 16% per year? P=A(P/A, i, N)

2) Formasa Plastic has major fabrication plants in Texas and Hong Kong. The
presidents wants to know the equivalent future worth of $1million capital investment
each for 8 years starting 1 year from now. Formasa capital earns at a rate of 14% per
year. F = A (F/A, i, N)

3) How much money must an electrical contractor deposit every year in her savings
account starting 1 year from now at 5 ½ % per year in order to accumulate $6,000 seven
years from now? A = F (A/F, i, N)

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Pause and solve
Acme Steamer purchased a new pump for $75,000.
They borrowed the money for the pump from their
bank at an interest rate of 0.5% per month and will
make a total of 24 equal, monthly payments. How
much will Acme’s monthly payments be?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
It can be challenging to solve for N or i.

• We may know P, A, and i and want to find

N.
• We may know P, A, and N and want to find
i.
• These problems present special challenges
that are best handled on a spreadsheet.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding N
Acme borrowed $100,000 from a local bank, which
charges them an interest rate of 7% per year. If Acme
pays the bank $8,000 per year, now many years will it
take to pay off the loan?

So,

This can be solved by using the interest tables and

interpolation, but we generally resort to a computer
solution.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding i
Jill invested $1,000 each year for five years in a local
company and sold her interest after five years for
$8,000. What annual rate of return did Jill earn?

So,

Again, this can be solved using the interest tables

and interpolation, but we generally resort to a
computer solution.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
There are specific spreadsheet functions
to find N and i.
The Excel function used to solve for N is
NPER(rate, pmt, pv), which will compute the
number of payments of magnitude pmt required to
pay off a present amount (pv) at a fixed interest
rate (rate).
One Excel function used to solve for i is
RATE(nper, pmt, pv, fv), which returns a fixed
interest rate for an annuity of pmt that lasts for nper
periods to either its present value (pv) or future value
(fv).
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We need to be able to handle
cash flows that do not occur until
some time in the future.
• Deferred annuities are uniform series that
do not begin until some time in the future.
• If the annuity is deferred J periods then the
first payment (cash flow) begins at the end
of period J+1.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding the value at time 0 of a
deferred annuity is a two-step
process.
1. Use (P/A, i%, N-J) find the value of the
deferred annuity at the end of period J
(where there are N-J cash flows in the
annuity).
2. Use (P/F, i%, J) to find the value of the
deferred annuity at time zero.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Pause and solve
Irene just purchased a new sports car and wants to
also set aside cash for future maintenance expenses.
The car has a bumper-to-bumper warranty for the first
five years. Irene estimates that she will need
approximately $2,000 per year in maintenance
expenses for years 6-10, at which time she will sell the
vehicle. How much money should Irene deposit into
an account today, at 8% per year, so that she will have
sufficient funds in that account to cover her projected
maintenance expenses?

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Sometimes cash flows change by a
constant amount each period.
We can model these situations as a uniform
gradient of cash flows. The table below
shows such a gradient.
End of Period Cash Flows
1 0
2 G
3 2G
: :
N (N-1)G
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
It is easy to find the present value
of a uniform gradient series.
Similar to the other types of cash flows, there is a
formula (albeit quite complicated) we can use to find
the present value, and a set of factors developed for
interest tables.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We can also find A or F
equivalent to a uniform gradient
series.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
The annual equivalent of End of Year Cash Flows ($)
this series of cash flows can 1 2,000
be found by considering an
2 3,000
annuity portion of the cash
flows and a gradient 3 4,000
portion. 4 5,000
End of Year Annuity ($) Gradient ($)
1 2,000 0
2 2,000 1,000
3 2,000 2,000
4 2,000 3,000

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Sometimes cash flows change by
a constant rate, ,each period--this
is a geometric gradient series.

This table presents a End of Year Cash Flows ($)

geometric gradient series. It
1 1,000
begins at the end of year 1
and has a rate of growth, , 2 1,200
of 20%. 3 1,440
4 1,728

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 We can find the present value of a
geometric series by using the appropriate
formula below.

Where is the initial cash flow in the series.

Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Pause and solve
Acme Miracle projects good things for their new weight
loss pill, LoseIt. Revenues this year are expected to
be $1.1 million, and Acme believes they will increase
15% per year for the next 5 years. What are the
present value and equivalent annual amount for the
anticipated revenues? Acme uses an interest rate of
20%.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 When interest rates vary with
time different procedures are
necessary.
• Interest rates often change with time (e.g., a
variable rate mortgage).
• We often must resort to moving cash flows
one period at a time, reflecting the interest
rate for that single period.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
The present equivalent of a cash flow occurring at
the end of period N can be computed with the
equation below, where ik is the interest rate for the
kth period.

If F4 = $2,500 and i1=8%, i2=10%, and i3=11%, then

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Nominal and effective interest rates.
• More often than not, the time between successive
compounding, or the interest period, is less than
one year (e.g., daily, monthly, quarterly).
• The annual rate is known as a nominal rate.
• A nominal rate of 12%, compounded monthly,
means an interest of 1% (12%/12) would accrue
each month, and the annual rate would be
effectively somewhat greater than 12%.
• The more frequent the compounding the greater
the effective interest.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 The effect of more frequent
compounding can be easily
determined.
Let r be the nominal, annual interest rate and M the
number of compounding periods per year. We can
find, i, the effective interest by using the formula
below.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Finding effective interest rates.
For an 18% nominal rate, compounded quarterly, the
effective interest is.

For a 7% nominal rate, compounded monthly, the

effective interest is.

Copyright ©2012 by Pearson Education, Inc.

Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
 Interest can be compounded
continuously.
• Interest is typically compounded at the end
of discrete periods.
• In most companies cash is always flowing,
and should be immediately put to use.
• We can allow compounding to occur
continuously throughout the period.
• The effect of this compared to discrete
compounding is small in most cases.
Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
We can use the effective interest
formula to derive the interest
factors.

As the number of compounding periods gets

larger (M gets larger), we find that

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Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.
Continuous compounding interest
factors.

The other factors can be found from these.

Copyright ©2012 by Pearson Education, Inc.
Engineering Economy, Fifteenth Edition
Upper Saddle River, New Jersey 07458
By William G. Sullivan, Elin M. Wicks, and C. Patrick Koelling
All rights reserved.