JIA 109 (1982) 435-445
UNIQUENESS
OF NON-NEGATIVE
RATE OF RETURN
INTERNAL
BY ALLFN M. RUSSELL* AND JOHN A. RICKARD†
*Department of Mathematics,
University of Melbourne, Parkville 3052, Victoria,
Australia and †School of Social and Industrial Administration, Griffith University,
Nathan 4111, Queensland, Australia
INTRODUCTION
WE will be considering a productive investment project or financial security
which yields a sequence of cash flows, positive or negative, over time. Let a,
(dollars) be the cash flow from the project at time t, where t takes the values 0, 1,
2, ..., n. Given the known cash flows a, from the project, and a known market
rate of interest, i per period, at which money may be borrowed or invested, a
common procedure is to accept the project if its present value P is greater than
zero, where
An internal rate of return of the project is defined to be a solution of the equation
P(r) = 0
(1)
in (– I, ),
if, of course, one exists.
Considerable
effort has recently been directed towards obtaining sufficient
conditions for determining a unique non-negative internal rate of return. One of
the early contributions
in this area was given by Soper(10) who showed that if any
one value of r can be found satisfying P(r) = 0, then a sufficient condition for that
value of r to be unique can easily be stated. The difficulty with this procedure is of
course that it is often not easy to determine an exact solution to P(r) = 0. Soper’s
work also introduced
the important
concept of unrecovered
investment.
Bernhard(4)(5) for example, has used this concept to obtain sufficient conditions
for a unique non-negative internal rate of return.
In paper(5), Bernhard generalizes sufficient conditions obtained previously by
himself, de Faro and Norstrøm,
for example. The principal result of (5) is an
extension of what is called the basic Bernhard-de
Faro condition which deals
with a single sign change in the sequence B0, B1, . . ., Bn, where
However,
in (5) Bernhard
presents a simple example which does not comply with
435
436
Uniqueness of Non-negative
Internal
Rate of Return
his more general sufficient conditions, thus highlighting the need for still more
general results.
In this paper we generalize the main results of Bernhard’s paper (5), present an
easier and more straightforward
procedure
for the calculation
of certain
coefficients, and give a condition which is satisfied by Bernhard’s counter-example.
PRELIMINARY RESULTS
Without loss of generality we will assume that a0<0 which means that the
project requires an initial outlay of money. If the project produces an initial
receipt of money, then all we have to do is change the signs of all coefficients in
equation (1).
In the subsequent analysis, instead of (1), it will be more convenient to consider
the equation
(1 + r)nP(r)
This equation
can alternatively
= 0.
be written as a polynomial
in r, namely
where B0, B1, . . ., Bn,have been previously defined.
It is our objective to obtain sufficient conditions to determine
to the equation
f(r)
a unique solution
= 0
in the interval [0, ). That is, we shall be seeking conditions that will guarantee
the existence of a unique non-negative internal rate of return.
An easy procedure for calculating the Bt‘s from the ak’s is by means of the
following Pascal-type
triangle, similar to that used for evaluating binomial
coefficients.
Row 0
a0
Row1
a0
Row2
a0
Row3
Row4
a0
a0
3a0+a1
a1
a2
a0 +a1
a0+a1 +a2
2a0 + a1
3a0 +2a1+ a2
a3
a0 + a1 + a2 +a3
a4
Uniqueness of Non-negative
Internal
Rate of Return
437
We observe that B0, B1, . . ., Bn are the first n + 1 entries in row n + 1, n 1; note
that B0=a0. Like Bernhards’ Ajk table, this table does not require any explicit
evaluation of binomial coefficients, and has the added advantage of being more
direct and straightforward
to construct.
Definition:
Let Sm =
m = 0, 1, 2, ..., n;
and note that
S0 = B0 = a0 < 0.
The following
theorem
(2)
enables us to transform
into a form that we shall use extensively. Furthermore,
it will be shown that
repeated
applications
of this transformation
lead to additional
sufficient
conditions.
Theorem I: The polynomial f(r)
can be written in the form
(3)
Proof
If we define S 1 = 0, then it follows from equation
Bj = Sj–Sj-1,
(2) that
j = 0, 1, . . ., n.
Consequently,
as required.
Corollary:
If Si 0,
i= 0, 1, 2, . . ., n – 1, and if
Sn
min {S0, S1, . . ., Sn–1},
then there is no solution to f(r) = 0
in [0, ).
If Sn = Sn–1, that is, Bn=0, then r=0 is the unique solution
Proof: Since Sn min
in [0, ).
{S0, S1, . . ., Sn–1}, Sn < 0, and therefore f(l) 0.
That is,
438
Uniqueness of Non-negative
r = 1 is not a root of the equation f(r)
equation
Internal
Rate of Return
= 0. Hence, assume r
1, and consider the
(4)
If 0 < r < 1, so that
rn converges,
then
Both the left and right sides of equation (4) are decreasing functions of r, with
–S,/(r – 1) having an asymptote at r = 1. In view of (5), (4) cannot have a
solution in [0, 1). Since for r > 1, the left side of (4) is negative, whereas the right
side is positive, equation (4) cannot have a solution in [0, ).
If Sn = Sn–1, (4) can be written
from which it follows that r = 0 or
(6)
However, if r > 0, a repetition of the argument above shows that equation
does not have a solution. Hence r = 0 is the only solution of (4) in [0, ).
(6)
EXTENSIONS OF RERNHARDS' RESULTS
Theorem
2: Let max{Sn–t, . . ., Sn–1) Sn min
{S0, .. ., Sn–t–1),
where
1 tn–1,
Si 0,
i= 1, 2,. . ., n – 1, and let Bn 0. Then f(r)=0
has a unique
solution in [0, ); more specifically, that solution lies in the interval (0, 1).
Proof: Since Sn < 0, r = 1 is not a solution of f(r) = 0, so we assume in what follows
that r 1.
If r > 1, the functions
Uniqueness of Non-negative
Internal
439
Rate of Return
and
are always negative and positive respectively, so f(r) = 0 will not be satisifed for
such values of r. Similarly we assume r 0 since Bn 0.
Assume now that 0 < r < 1, in which case
converges.
Then,
In (0,1) the function above in square brackets is a strictly increasing function of r,
ranging from –
to + . Hence it can be zero for only one value of r in (0,1).
Remark: Theorem 2 generalizes Proposition
II(b) in paper (5). Note that in
and
S0 S1 S2
. . . Sn–t–1 n,
Bernhards’
Proposition
II(b)
Sn>Sn–1
. . . Sn–t.
In Theorem 1, Corollary, and Theorem 2 there were no sign changes in the
sequence S0, S1, . . ., Sn; the next result admits one sign change in that sequence.
Theorem 3: Let Bn 0, Sn>0 and max{S0, S1, . . ., St – 1} 0 min{St,
where 2 t n – 1. If in addition
Sn
then f (r) = 0 has a unique solution
in (1, ).
. . ., Sn},
max{Sn–1, . . ., St},
in [0, );
more specifically,
that solution
lies
440
Uniqueness of Non-negative
Proof: As in the previous
If 0 < r < 1 then
converges,
theorem
Internal
Rate of Return
we may assume that r
0 and r
1.
and consequently
because Sn max{Sn–1r
equation
. . ., St} and SO, S1, . . ., St–1 are non-positive.
Thus the
(7)
does not have a solution in (0, 1). However, when r > 1 the left side is a strictly
increasing function from –
to 0, whereas the right side is a strictly decreasing
function from S0 + S1 + . . . + Sn – 1(corresponding
to r = 1) to – . Thus there is
a unique solution to f(r)=0
in (1, ).
Remark: Theorem 3 generalizes Proposition II(a) in paper (5). Observe that in
Bernhards’ Proposition II(a), St St+1
. . . Sn.
The previous two theorems have been established under conditions which
require
In Proposition II(c) of paper (5), a sufficient condition is given for r = 1 to be the
unique solution of f(r)=0.
In that result S0, S1, . . ., Sn – 1 are all non-positive.
The next result, a negative one, deals with Sn=0 and one change of sign in the
sequence S0, S1, . . ., Sn– 1.
Theorem 4: If Sn=0, S0 + S1 + . . . + Sn – 1 0, and there is one sign change in the
sequence S0, S1, . . ., Sn–1, then f(r) = 0
has two solutions in [0, ).
Proof: It follows from (2) that
Uniqueness of Non-negative
If S0+S1 +
. . . + Sn–1
Internal
441
Rate of Return
0, then
when r = 1. Since there is one sign change in the sequence
follows from the Bernhard-de
Faro condition in (5) that
S0, S1, . . ., Sn–1 it
has a unique solution not equal to 1, in (0, ). This solution, together with r = 1
gives two solutions of f(r)=0
in [0, ).
All of the results so far have been relevant to either no sign change or just one
sign change in the sequence S0, S1, . . ., Sn. We now obtain some results that
permit two or more sign changes in the sequence S0, S1, . . ., Sn.
The proof of the following lemma is straightforward,
and will be omitted.
Lemma:
Let 0, 1, . . ., n be any sequence
of real numbers.
Let
and
Then the number of sign changes in the sequence
exceed the number of sign changes in the sequence
more, there exists a positive integer m such that
same sign, the sign being that of 0.
Definition:
cannot
Furtherall have the
Define
where S0, S1, . . ., Sn are defined in equation
Theorem 5: The polynomial f(r)
(2); note that T0 = S0<0.
can be written in the form
(8)
Proof: Apply Theorem
1 to the summation
term in equation
Corollary: If Sn = 0 = Tn – 1, and T0, T1, . . ., Tn–2 are of constant
has a unique solution r = 1.
(3).
sign, then f (r)=0
442
Uniqueness of Non-negative
Proof: This follows immediately
Internal
Rate of Return
from equation
(8).
Theorem 6: If Sn = 0, max{T0, T1, . . ., Tn–2} 0,
and
Tn–1
then f(r)
min {T0,
= 0 has a unique solution
T1,
. .,. Tn–2),
r = 1.
Proof: Write
Then, an application
cannot
of Theorem
be zero when r
1, Corollary,
shows that
[0, ).
Example: We consider the example on p. 205 of paper (5) which Bernhard
correctly asserts does not satisfy either the basic Bernhard-de
Faro condition or
the extensions in Proposition II of that same paper. We show that the example
does satisfy the conditions of Theorem 6. The example referred to consists of a
sequence of cash flows given by a0 = – 1, a1 = 5, a2 = – 9, a3 = 6. Then SO= – 1,
S1 = 1, S2 = – 1, S3 = 0 and consequently
T0 =
– 1, T1 = 0, T2 =
– 1, T3 =
– 1.
Hence, the conditions of Theorem 6 are satisfied, and so r = 1 is the unique
solution in [0, ). This of course can be verified directly.
FURTHER GENERALIZATIONS
successive sequences
If from the sequence S0, S1, . . ., Sn we construct
according to the Lemma, then ultimately we obtain a sequence whose elements
are of constant sign, that sign necessarily being negative since we have assumed
S0<0. For example, in the previous theorem the relevant sequence is T0,
T1, . . ., Tn–2 and it did not have a sign change. Unique solutions to f (r) = 0 can
still be obtained, however, when there are sign changes in the sequence TO,
T1, . . ., Tn–2. The next theorem allows for up to two sign changes in this
sequence.
Theorem 7: Let
Uniqueness of Non-negative
and max{T0, T1, . . .,Tn–3} 0. Then f(r)
solution actually lying in (0, 1).
Internal
Rate of Return
= 0 has a unique solution
Proof: Since Sn < 0, r = 1 is not a solution off(r)
consider the equation
443
in [0, ),
= 0. Hence we assume r
the
1, and
(9)
Let
from which it is easy to show that g’(r) is non-negative in (0, 1) if 2Sn Tn–1 0.
Thus g is an increasing function of r in (0, 1), ranging from Tn – 1– Sn to + .
When r > 1, g(r) is a decreasing function of r, ranging from +
to 0. The
function
is a strictly decreasing function of r, ranging from Tn–2 to – . Since Sn–1 < Sn
implies that Tn–1 – Sn < Tn–2, it now follows that equation (9) has only one
solution in [0, ), that solution lying in (0, 1).
Corollary 1: If Sn–1 = Sn, and all other conditions
then r = 0 is the unique solution of f(r) in [0, ).
of the theorem
are satisfied,
Corollary 2: If Sn–1 > Sn, and all other conditions
then f(r) = 0 does not have a solution in [0, ).
of the theorem
are satisfied,
Examples:
The following
examples
satisfy the conditions
(i) a0 =
B0 =
S0 =
T0 =
– 1,
– 1,
– 1,
– 1,
a1 = 6,
B1 =3,
S1 = 2,
T1 = 1,
a2 = – 14,
B2 = –5,
S2 = – 3,
T2 = – 2,
a3 = 19/2
B3=1/2
S3 = – 5/2
T3 = –9/2.
(ii) a0
B0
S0
T0
–1,
– 1,
– 1,
– 1,
a1
B1
S1
T1
a2
B2
S2
T2
a3
B3
S3
T3
=
=
=
=
=
=
=
=
5,
2,
1,
0,
=
=
=
=
– 10,
–3,
– 2,
– 2,
=
=
=
=
of Theorem
7
1
–1
– 3.
The next result deals with one sign change in the sequence
Theorem 8: Let
Bn
7:
0, Sn Tn–1
0,
T0, T1, . . ., Tn–2.
444
Uniqueness of Non-negative
Internal
Rate of Return
and
If there exists one sign change in the sequence T0, T1, . . ., Tn–2, then f (r)
unique solution in [0, ),
or more precisely, in (1, ).
Proof: Consider
= 0 has a
the equation
(10)
Since Bn 0, r = 0 is not a solution of f(r) = 0. We therefore assume r
(10) in the form
0 and write
(11)
where Tn– –2<0,
Tn– –1>0.
Since Tn–2 0, the right hand side of (11) is a
strictly decreasing function of r in (0, co), ranging from +
to – .
We now consider the behaviour of
as a function of r. As r 1
– 0, g(r) – since Sn 0; and as r 0 +, g(r) –
since Sn Tn–1. Furthermore,
since Sn Tn–1 and Tn–1 0, g(r) 0
for < r <
1.
For r > 1, g(r) is a strictly increasing function ranging from –
to 0. Since the
right hand side of (11) is a strictly decreasing function of r and takes the value
T0+T1+ . . . + Tn–2 0 when r = 1, it is clear that (11) has a unique solution in
(1, ).
Example:
The following
a0
B0
S0
T0
example
=
=
=
=
–1,
–1,
–1,
–1,
a1
B1
S1
T1
satisfies the conditions
=
=
=
=
7,
4,
3,
2,
a2
B2
S2
T2
= – 15,
= –4,
= –1,
= 1,
a3 =
B3 =
S3 =
T3 =
of Theorem
8:
14
5
4
5.
REMARKS
1. It is possible to obtain further sufficient
Theorems 7 and 8. For example, the restriction
conditions
similar
to those in
Uniqueness of Non-negative
in Theorem
8 can be weakened
Internal
Rate of Return
445
to
where m is the maximum value of g(r) in (0, 1).
2. The essential feature of the last theorem was the single sign change in the
sequence T0, T1, . . ., Tn–2. If the sequence T0, T1, . . ., Tn–2 has more than one
sign change, it is possible, according to the Lemma, to construct
another
sequence which has less sign changes. If the summation term in (8) is repeatedly
transformed
by the procedure
of Theorem
1 until a sequence is obtained
containing no more than one sign change, then appropriate
conditions can be
obtained to guarantee a unique solution to f (r) = 0 in [0, ). The accompanying
analysis, however, does increase in complexity, and in the interests of simplicity
we do not present any further results.
REFERENCES
D. C. & ECKARDT
JR, W. L. (1976) A sufficientcondition for a unique non-negative
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R. H. (1967)On the inconsistencyof the Soper and Sturm–Kaplan conditions for
(2) BERNHARD,
uniqueness of the internal rate of return. J. Indust. Eng. 18:8 (August), 498–500.
R. H. (1971) A comprehensive comparison and critique of discounting indices
(3) BERNHARD,
proposed for capital investment evaluation. Eng. Econ. 16:3 (Spring), 157–86.
R. H. (1979)A more general sufficientcondition for a unique non-negativeinternal
(4) BERNHARD,
rate of return. J. Fin. Quant. Anal. 14:2 (June), 337–41.
R. H. (1980)A simplificationand an extension of the Bernhard–de Faro sufficient
(5) BERNHARD,
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(6) DEFARO,C. (1973) A sufficientcondition for a unique non-negative internal rate of return. J.
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(9) NORSTRØM,
C. J. (1972)A sufficientcondition for a unique non-negative internal rate of return.
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(10) SOPER,
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