Volume 13, Issue 2
57
A Rule-of-Thumb Approximation for Time Value of
Money Calculations
David N. Swingler, Ph.D., Professor, Engineering Division,
Saint Mary’s University, Halifax, Nova Scotia, Canada
A simple rule-of-thumb is presented for the classic financial calculations centered on the
present value of a series of equal future payments. It is demonstrated that it is a useful
addition to the armamentaria of engineering students engaged in engineering economics
and/or finance courses.
Journal of Personal Finance
58
Introduction
Discussion
Engineers have long been associated with approximations, often
codified as “rules-of-thumb” which help them deal with otherwise
complex mathematical situations. These commonly find application not only at the outset of projects where ballpark estimates are
useful, but also during projects as guidance while undertaking a
more complete exposition of the situation, and even at the end for
final “sanity checking”. The sanity-checking mechanism is also at
play while reading third party reports, for instance, just to make
sure that what is written makes on-going sense to the reader. The
most useful, and gratifying, of these rules-of-thumb are perhaps
those that are simple enough to be handled via mental arithmetic
Engineering students are encouraged to use the same philosophy:
nothing is more humiliating than having a nice piece of work
undone by a careless use of a spreadsheet formula or mis-pressing of the buttons on a sophisticated calculator, producing a
significant error which makes all the ensuing results completely
useless. The proverb “spoiling the ship for a ha’porth of tar”
springs to mind here.
,
.
We begin by simply asserting that a useful approximation to b is
given by
, i≤0.2
.
(1A)
This is felt to encompass most situations of
“everyday” practical
, i≤0.2
. use
interest but for completeness for ni > 3 we
, i≤0.2
,
If we compute
the ,percentage
error between
,
i≤0.2 (1B)
and
, i≤0.2
IfEquation
we compute
percentage
errorand
between
A justification
for
(1A) isthe
given
in the appendix
it
might be remarked that the derivation therein requires only the
mathematical skills of a first- or second-year engineering student. In passing,
remark thattoin this
financial
$ b̂ is
the
Thewesolution
is language,
given by
solving
(approximate)
value
of
any
one
of
a
stream
of
n
equal
and
equi, i≤0.2 approximation of (1A) it becomes simply:
spaced future payments with a present value of P = $1.
an
One field where rules-of-thumb play little role is that of finance
The solution to this is given by solving
(with the exception of the ubiquitous rule-of-72). This is almost
If we
compute
the percentage
error between
b̂ and
If
we
compute
the
percentage
error between
and
viab via simply:
approximation
of (1A)
it becomes
certainly due to the notion that monetary amounts must be accu=618 .
rate to the penny, which is true at the time contracts are signed
but may not be true at every point in the process, as outlined
above. Additionally, university courses in finance and economics
=618 .
are relatively new additions to the field of engineering education.
And finally bankers are just not engineers and do not think like
forval-i, with n
The solutionthentothethis
is givenresults
by ofsolving
representative
Figure 1 ensue for various
them.
ues
of
i
from
1
to
20%
and
n
from
1
to
121
periods.
We
see
from
approximation of (1A) it becomes simply: =
The purpose of this brief note is to suggest that there is a useful
Figure 1 that the peak errors are not much more than very roughly
“rule-of-thumb” for the classic and most common time value of
± 6% which is deemed entirely adequate for our rule-of-thumb
money problem which involves dealing with the present value, P, usage, especially as the underlying
= for b seems quite
=618expression
.
of series of n equi-spaced future payments, each of value, A, with subtle over this wide range of n and i. Further we might observe
an interest rate of i, via the formula
that the peak negative errors of about -6% occur in the vicinity of
ni=3 which makes their occurrence easily predictable (and largely
A = positive
= 250,000
correctable if so desired). The peak
errors of about +5% /100 =
occur in the vicinity of ni=1 (roughly), again useful knowledge.
where the factor b is given by
Zero error occurs near ni=2 (roughly). Finally we note that at
= 250,000
/100
A=
ni=3 where the
approxima= .two parts of the piecewise continuous
= 24000/
(2.4/21) =$2
.
tion abut, there
no discontinuity.
This is aexpression
felicitous effect.
Weis start
with the correct
for given by
The first payment is at the end of the first period. As an aside, we
note that Hawawini and Vora [1, p. (v)] have indicated that the
, i≤0.2 and
problem of calculating i from known A, P and n is non-trivial
,
i≤0.2
has a very long history.
,
, i≤0.2
,
We start with the correct expression for
A=
= 250,000
and let
/100 = 250,000
so that the above becomes
and let
so that the above becomes
, i≤0.2We start with the correct expression
for given by
If we compute the percentage error between
compute the percentage error between and via
and
via
©2014, IARFC. All rights of reproduction in any form reserved.
and let
so that the above becomes
given b
/100
Volume 13, Issue 2
59
.
, i≤0.2
,
, i≤0.2
If we compute the percentage error between
.
and
via
Figure 1. Errors in the approximation of Equation (1).
Examples
, i≤0.2
The solution to this is given by solving
approximation of (1A) it becomes simply:
for i, with n=9. With our
We take our second example from a worked example in the
We now present simple examples of the utility of the rule-ofcourse
text Contemporary Engineering Economics by Park et al
thumb of Equation (1A). It might be of interest to note that these
,
, i≤0.2
=618 .
[2, p.78]. Paraphrasing it states: A lottery winner is expecting 21
were literally the first examples that came to the hand of the
annual payments of $24,000 (after tax) which he wants to use
author as he thought about this topic. We start with one ofIfthe
we compute the percentage error between and via
against a bank loan at 10% interest. How much can he borrow
earliest examples of interest calculations, taken from [1, p. 1]
from the bank? Here we have A =$24k, n=21, i=0.1 and we use
A 16th century Italian university was loaned 2814 ducats and
=
= 24000/ (2.4/21) =$210,000
repaid it in nine annual installments of 618 ducats beginning at
the end of the first year. What was the effective interest rate?
The solution to this is given by solving
for i, with n=9. With our
(Recall that interest rate calculations, accurate ones anyway,
are
which
is also
(just about)
manageable by mental arithmetic alone.
approximation
of (1A)
it becomes
simply:
regarded as mathematically non-trivial.)
The correct answer is $207,569 but the book’s authors offer
the interesting caveat that the whether the bank would loan that
=618 . /100 = 250,000
The solution to this is given by solving 2814b = 618 for i, with amount
A = depends
= 250,000
/100
on the applicant’s
creditworthiness, suggestn=9. With our approximation of (1A) it becomes simply:
ing that our $210k figure would nicely suffice for this ballpark
estimate.
1 2 
2814 + i  =618 .
We start with the correct expression for given by
Our last problem is taken from another worked example in the
9 3 
same text=[2, p.76]. Here a company has
borrowed
$250,000
for
= 24000/
(2.4/21)
=$210,000
equipment. The loan carries an interest rate of 8% and is to be
It is convenient to mentally rearrange this as (1 + 6i) = 9 * 618 / 2814
repaid in annual installments over the next six years. What is the
and if we observe that the right-hand side is almost exactly two
amount
annual
installment? Here we have
then an approximate answer immediately falls out as i =16⅔%by
and let
so that of
thethe
above
becomes
mental arithmetic only. (Given that the numbers involved
strongly suggest that the amount repaid was exactly twice that
A=
= 250,000
/100 = 250,000
/100
loaned, one wonders whether there is an ancient typographical
error in the data as given here). With the data as actually given,
or A = 22% of $250,000 = $55,000, again via mental arithmetic
our approximation yields i =16.3% which is the correct answer
We start with the correct expression for given by
only. Note the efficacy of turning b̂ into a percentage figure.
to three significant digits. The fact that the rule-of-thumb is this
The correct answer is $54,079 but requires access to (and correct
accurate lies in the fortuitous positioning of this problem on the
application of) significant computational resources.
left in Figure 1 in the vicinity of n=9, i=16% where the approximation error is small.
and let
so that the above becomes
Journal of Personal Finance
60
Students will of course complain that Equation (1A) is just
another formula to memorize. To this end is useful to refer to it
via the mnemonic “the one ’n two-thirds formula” and to indicate
that, as it deals with loan repayments, it is natural for the quantity
to reduce with n, hence n appears as a denominator, but increase
with i so i appears as a numerator term. It should also be stressed
this is not a replacement for the “correct” results, but simply an
auxiliary tool.
A=
= 250,000
/100 = 250,0
Appendix I : Derivation of the approximation
We start with the correct expression for given by
We start with the correct expression for b given by
and let
so that the above becomes
and let ni = p so that the above becomes
Escalating payment streams
It should be observed that approximation of equation (1A) can be
readily extended to a stream of escalating payments. Rather than
digress here, this material is adumbrated in Appendix II.
We begin
the
by
lettingn be
nbylarge.
beletting
large.
a cla
We approximation
begin
the approximation
nFollowing
be alarge. Follo
We begin
the approximation
by letting
Following
We
begin
the
approximation
by
letting
n
be
large.
Fo
L’hopital’s
rule
for
the
bracketed
term
in
the
denominator,
we
arrive
a
We begin the approximation
by letting
n
be
large.
Following
a
classic
application
of
L’hopital’s
rule
for
the
bracketed
term
in
the
denominator,
classic application of L’hopital’s rule for the bracketed term in the
L’hopital’s
rule
for
the
bracketed
term
in
the
denominato
We begin the approximation
by
letting
n
be
large.
Following
a
classic
application
of
L’hopital’s rule for the bracketed term
in the denominator,
arrive at the approximation
denominator,
we arrive at thewe
approximation
Conclusion
L’hopital’s rule for the bracketed term in the denominator, we arrive at the approximation
.
.
We have presented a useful rule-of-thumb for time value of
.
money problems, of benefit to students in engineering econom.
ics and finance classes and
perhaps
even
to
their
professor,
idly
.
musing, while exam invigilating, what the payments might be
It transpires,It not
altogether
that
is reasonably
linea
transpires,
notintuitively,
altogether intuitively,
that
is reaso
on the $300,000 mortgage she’d need to take out on that new
It transpires,
not altogether
intuitively, that
isthat
reasonIt
transpires,
not
altogether
intuitively,
is
rea
is reasonably linear for
and
transpires,
not altogether
intuitively, that
house she is interested in, It
with,
say, an amortization
of 25 years
ably
for
and
it is this
unitymight
whereuse
p=0.
this use
is reasonably
linear
for
andFor
It transpires,
not altogether
that answer: (100/25
and an interest
rate of 6%. intuitively,
[Her approximate
+ unity
Swingler
paper
it is
where
p=0.
For this
reason
we
≈reaan
itlinear
is unity
where
p=0.
reason
we For
might
2/3 * 6)% of $300,000 = 8%
of
$300,000
=
$24,000pa
or
about
itmight
is unity
where
p=0.
For
this
reason
we
might
use
≈
and
with
a
little
it is unity where p=0. For this reasonsonwewemight
use
use
and with a little numerical
$2000pm].
regression
it is unity
where p=0. For this reason we might use
≈ numerical
and with
a little analysis on the interval 0  p  3 , m is
regression analysis on the interval 0 ≤ p ≤ 3, m is found to be
immediately leads to the final result
about 0.67. This immediately leads to the final result
≈
or
=
=
≈
=
or
or
or
for ni ≤ 3. or
or
for ni ≤ 3.
.
≈
≈=
==
p / n 1  ≈2  1 2 =
=b 
≈ 1  p  =  i = b̂
1  e p n  3  n 3
=
=
for ni ≤ 3.
for ni ≤ 3.
1 2 
3. ni ≤ 3.
bˆ    i  for ni ≤ for
n 3 
Notwithstanding the constraints in this analysis, the main body of
Notwithstanding
thethat
constraints
in this analysis,
the
main body of
this
note demonstrates
the approximation
works well
down
approximation
works
downfortothe
as special
low ascase
n=1offor
i<0.2 (altho
to
as low as n=1 for
i<0.2well
(although
n=1,
. where where
just1+i,
1+i,ofof
course).
completeness,
ni>3,wei<0.2, we
bb isisjust
course).
For For
completeness,
ni>3,. i<0.2,
.
simply. use
bˆˆ=i i.
b
.
Appendix II: Escalating Payment Streams
Escalating payment streams can be readily handled by simple m
In particular, if the payment amounts increase by e% each period
we have
and where
the nth
is
where n(i-e)≤2,
i≤0.25,
e≤i i≤0.25,
Again, then
gives (approximately)
the
n(i-e)≤2,
e≤i Again
gives (approxi
where
i≤0.25,
e≤i
Again
gives
where n(i-e)≤2, i≤0.25, e≤i asymptotically
Again
givesn(i-e)≤2,
(approximately)
the
first
payment.
It
is (appro
correct
as (i-e)→0
has
utility,
for utility,
example,
asymptotically
correct
asand
(i-e)→0
and
has
forinex
1
2
where n(i-e)≤2, i≤0.25, e≤iasymptotically
Again
gives
(approximately)
the
first
payment.
It
is
asymptotically
correct
asin(i-e)→0
and the
has
utility,
for
paymentand
ofpayment
ahas
stream
escalating
retirement
from
a princi
correct as (i-e)→0
utility,
example,
first
ofofafor
stream
escalating
retirement
payouts
fro
bˆ  of 
i estimating
 e  . payouts
n
3
payment
of
a
stream
of
escalating
retirement
payouts
asymptotically correct as (i-e)→0
and
has
utility,
for
example,
in
estimating
the
first
portfolio
yield
is
i
and
the
inflation
rate
is
e.
It
has
a
sweet
spot
nf
payment of a stream of escalating retirement
principal
of Prate
where
portfoliopayouts
yield isfrom
i anda the
inflation
is e. the
It has a sw
portfolio
yield
is
i
and
the
inflation
rate
is
e.
It
has
a
payment of a stream of escalating
payouts
from amakes
principal
ofIt Pfor
where
theforspot
itisideal
retirement
income
streams.
Forstreams.
instanc
portfolio retirement
yield is i and
the which
inflation
rate
e.makes
has
aideal
sweet
n≈30,
(i-e)≤0.6
which
itsuch
suchnear
retirement
income
ˆmakes
gives
(approximately)
the
first
payment
which
is
Here
which
it
ideal
for
such
retirement
income
stream
A
Pn≈30,
bform
portfolio yield is i and thewhich
inflation
rateitisideal
e. Itfor
hasrights
aprincipal
sweet
spot
near
(i-e)≤0.6
ofprincipal
$500,000
and
his
portfolio
has
a
yield
of
i=7%
and
©2014,
IARFC.
All
ofretirement
reproduction
inany
reserved.
makes
such
income
streams.
For
instance
if
retiree
has
a
of $500,000 and his portfolio has a yield
of s
principal
of
$500,000
and
his
portfolio
has
a
yield
which makes it ideal for such
retirement
income streams.
instance
ifaerror
awith
retiree
aover
period.
The
distribution
is of
now
but
there
peak
with
e=3%
over
ahas
period
years
then
theareapprox
principal
of $500,000
and inflation
his For
portfolio
has
yield
of
i=7%
hecomplex,
wants
match
inflation
e=3%
a n=30
period
of to
n=30
years
theno
.
Volume 13, Issue 2
61
It transpires, not altogether≈intuitively, that
=
= is reasonably linear for
Escalating
Payment
Streams
it is unityAppendix
where p=0.II:For
this reason
we might
use
or
Escalating payment streams can
forbenireadily
≤ 3. handled by simple
modification(s) to equation (1A). In particular, if the payment
amounts increase by e% each period such that the first payment is
A and the nth is A*(1 + e) ^ (n – 1), then we have
≈
=
=
.
or
for ni ≤ 3.
Here A = P bˆ gives (approximately) the first payment which is
subsequently indexed by e% per period. The error distribution is
now complex, but there are peak errors of about ±10% for ni≤3,
i≤0.1, e≤i so again it covers a wide range of “everyday” situations. The errors generally are worst as (i-e)→0.
≈
and
and with a Acknowledgement
little
The author would like to acknowledge the helpful comments of
Prof. Moshe Milevsky, York University, Toronto on an early draft
of this document.
References
[1] G. A. Hawawini and A. Vora, The History of Interest Approximations,
Arno Press, USA, 1980, ISBN 0-405-13480-0
[2] C. S. Park, R. Pelot, K. C. Porteous, M. J. Zuo, Contemporary
Engineering Economics, Addison Wesley Longman, Toronto, 2001,
ISBN 0-201-61390-5
.
To avoid this problem perhaps a more useful approximation is for
gives (approximately) the first payment. It is
where
n(i-e)≤2,
i≤0.25,
e≤i
Again
an escalating immediate payment stream similar to the above but
asymptotically
(i-e)→0
and of
hastheutility,
for Inexample, in estimating the first
where the firstcorrect
paymentas
is at
the beginning
first period.
this case
have peak
of aboutretirement
±10% for thispayouts
formula: from a principal of P where the
payment
of we
a stream
oferrors
escalating
portfolio yield is i and the inflation rate is e. It has a sweet spot near n≈30, (i-e)≤0.6
which makes it ideal for such retirement income streams. For instance if a retiree has a
principal of $500,000 and his portfolio has a yield of i=7% and he wants to match
inflation with e=3% over a period of n=30 years then the approximate first annual
payment
isn(i-e)≤2,
abouti≤0.25,
$26,700.
The
true
isgives
about
$27,400k. The difference
is a wholly
where
n(i-e)≤2,
e≤iAgain
Again
(approximately)
the first payment.
It is
where
i≤0.25,
e≤i
(approxiA =value
P bˆ gives
acceptable
mately)-2.6%.
the first
payment.
is asymptotically
correct
as (i-e)→0
asymptotically
correct
as It(i-e)→0
and has
utility,
for example, in estimating the first
and
has
utility,
for
example,
in
estimating
the
first
payment
of from a principal of P where the
payment of a stream of escalating retirement payouts
a stream of escalating retirement payouts from a principal of P
portfolio yield is i and the inflation rate is e. It has a sweet spot near n≈30, (i-e)≤0.6
where the portfolio yield is i and the inflation rate is e. It has a
which
makes
it ideal
such which
retirement
For instance if a retiree has a
sweet
spot near
n≈30,for
(i-e)≤0.6
makes itincome
ideal forstreams.
such
principal
of
$500,000
and
his
portfolio
has
a
yield
of
i=7%
and he wants to match
retirement income streams. For instance if a retiree has a prininflation
e=3%
a period
n=30
years
then the approximate first annual
cipal ofwith
$500,000
and over
his portfolio
has a of
yield
of i=7%
and he
wants is
to match
withThe
e=3%
overvalue
a period
of n=30$27,400k.
years
payment
aboutinflation
$26,700.
true
is about
The difference is a wholly
then
the
approximate
first
annual
payment
is
about
$26,700.
The
acceptable -2.6%.
true value is about $27,400k. The difference is a wholly acceptable -2.6%.