Why Compare About Seven Elements

5. Why Compare About Seven Elements

There are two explanations which one can give to justify the use of not many more than seven elements in a comparison scheme.

Consistency Explanation In making pairwise comparisons, errors arising out of inconsistency in judgments affect the final answer. We distribute inconsistency proportionately among the alternatives because in computing the eigenvector, transitivities of all order are considered and the average is taken over all these transitivities. If the number of elements is small their relative priorities would be large. These priorities would be less affected by inconsistency adjustment. Thus for example if there are 10 homogeneous elements, each would have a relative priority of .10 and is unaffected by a 1% inconsistency distributed among the 10 so that its value would now be .10 .01. If the number of elements is large, the relative priority of each would be small and would be more affected by inconsistency. Still the number should be large enough to enable one to make redundant judgments to improve the validity of the outcome. For this reason seven elements is found to be a reasonable choice for high average priority and high validity.

Neural Explanation This explanation has to do with the brain limit on the identification of simultaneous events. The perception or simultaneity span is the ratio of the buffer-delay time to the attentional integration time. Some psychologists have found that the more intense the stimulus the greater the perception or simultaneity span. The reason is that with increased intensity the time of rapid integration is reduced. The most common duration time estimate for the short term memory (buffer-delay) is 750 milliseconds and that for item-integration time is 100 milliseconds. Their ratio is about seven.

Both forgoing explanations have the following mathematical justification.

Stability of the Eigenvector Requires a Small Number of Homogeneous Elements

The question often arises, how sensitive the priorities give by the eigenvector components are to slight changes in the judgment values. Clearly, it is desirable that the priorities do not fluctuate widely with small changes in judgment. There are essentially three ways to test this sensitivity: (1) by finding a mathematical estimate of the fluctuation; (2) by deriving answers based on a large number of computer runs appropriately designed to test the sensitivity; (3) by a combination of the two, particularly when it is not possible to carry out the full demonstration analytically.

We have already pointed out, in the case of consistency, that λ max is equal to the trace of the matrix which consists of unit entries. In this case one would expect the eigenvector corresponding to the perturbed matrix to undergo an overall change by an amount inversely proportional to the size of the matrix.

In general, the eigenvalues of a matrix lie between its largest and smallest row sums. Changing the value of an entry in the matrix changes the correspondence row sum and has a tendency to change

λ max by an equal amount. However, since a change in the eigenvector should also be influenced by the size of the matrix, we expect that the larger the matrix, the smaller the change in each component.

We begin the analytical treatment of the question by considering a matrix A with the characteristic equation

det (A - λ I) = n +

Following standard procedures, let A + ε B be the matrix obtained by introducing a small perturbation in

A . The corresponding characteristic equation is

a 1 ε ) λ + K + a n ( ε ) = 0 where a k ( ε )

det (A + B - I) = n + (

is a polynomial in ( ε

of degree (n-k) , such that a k ε ) → a k as ε → 0.

Let λ 1 be the maximum simple eigenvalue corresponding to the characteristic equation of A . It is known in matrix theory that for small ε , there exists an eigenvalue of A + ε B which can be expressed as

the sum of a convergent power series, i.e.,

Let w 1 denote the eigenvector of A corresponding to λ 1 and let w 1 ( ε )

be the eigenvector of

A + ε B corresponding to λ 1 ( ε ).

The elements of w 1 ( ε ) are polynomials in λ ( ε ) and ε , and, since the power series for λ 1 ( ε ) is convergent for small ε , each element of w 1 ( ε ) can be represented as a

convergent power series in ε . We may write

If the matrix A has linear elementary divisors, then there exist complete sets of right and left

eigenvectors w 1 ,w 2 , .., w n and v 1 , v 2 , ..., v n , respectively, such that

Note that w j and v j are the j th eigenvectors (right and left), and not the j th components of the vectors. The vectors z i can be expressed in terms of the w j as

z i = ∑ s ij w j

j=1

w i ( ε which, when substituted in the formula for ) , gives

1 ( ε ) = w 1 + ∑ ∑ t ij ε w i

i = 2 j=1 where the t ij are obtained by dividing the s ij by the coefficient of w 1 .

k 1 λ 1 ( The first order perturbations of the eigenvalues are given by the coefficient ). of ε We now derive the expression for the first order perturbations of the corresponding eigenvectors. Normalizing the vectors w j and v j by using the euclidean metric we have

We know that

we obtained above and use Aw 1 = λ 1 w 1 , we have

If we substitute the expressions for

and

Multiplying across by

and simplifying, we obtain

and

where, as noted above, k sub 1 is the first order perturbation of λ 1 and where, as noted above, k sub 1 is the first order perturbation of λ 1 and

Thus for sufficiently small ε the sensitivity of λ 1 depends primarily on v 1 w 1 . v 1 w 1 might be arbitrarily small.

The first order perturbation of w 1 is given by

j= 2

ε ∑ ( v j B w 1 /( λ 1 - λ j ) v j w j ) w j

j= 2

v j ( ∆ A) w 1 /( λ 1 - λ j ) v j w j ) w j where ∆ A ≡ ε B

j= 2 The eigenvector w 1 will be very sensitive to perturbations in A if λ 1 is close to any of the other

eigenvalues. When λ T

1 is well separated from the other eigenvalues and none of the w v i i is small, the eigenvector w i corresponding to the eigenvalue λ 1 will be comparatively insensitive to perturbations in A .

This is the case, for example, with skew-symmetric matrices ( a ji = -a ij ).

are interdependent in a way which precludes the possibility that just one 1/ T v i w i i = 1,2,..., n is large. Thus if one of them is arbitrarily large, they are all arbitrarily large.

The w T v

ii

However, we want them to be small, i.e., near unity. To see this let

w i = ∑ c ij v j and v j = ∑ d ij w j

| w i = | | b i = | 1, i = 1, 2, ..., n. where It is easy to verify by substitution that

i w i = ∑ d ij w j ∑ c ij v j

i w i = ( v i w i ) + ∑ ( w j w i )( v j v i )( v j w j )

Since

we have

which must be true for all i = 1, 2, .., n . This proves that all the must be of the same order. We now show that for consistent matrices

cannot be arbitrarily large. We have in the case of consistency

Therefore

1 w 1 ) = [(1/ w 11 ,...,1/ w 1n )( w 11 ,..., 1 1n ) / ∑ 1/ w 1i ]

i =1

- = 1 [n/ ∑ 1/

w 1i ] > n

i =1

n/ ∑ 1/ w 1i < ∑ w 1i /n.

since

i =1 i =1

∑ w 1i = 1.

T -1

Now ( v 1 w 1 ) is minimized when all w 1i are equal since i =1

In practice, to keep v 1 w 1 ) near its minimum we must deal with relatively comparable activities so that no single w 1i is too small. To improve consistency the number n must not be too large. On the other hand, if we are to make full use of the available information and produce results which are valid in practice, n should also not be too small.

( -1 T

If, for example, we reject the values v 1 w 1 ≤ 0.1, then we must have n 9.

Under the assumption that the number of activities being compared is small and that they are relatively comparable, i.e., their weights differ by a multiple of their number, we can show that none of the components of w 1 is arbitrarily small and none of those of v 1 is arbitrarily small, and hence the scalar product of the two normalized vectors cannot be arbitrarily small. With large inconsistency one cannot guarantee that none of the w 1i is arbitrarily small. Thus, near-consistency is a sufficient condition for stability. Note also that we need to keep the number of elements relatively small, so that the values of all the w 1i are of the same order.

Illustrative Matrices

Here are examples of inconsistent matrices with their inconsistency indices:

12 = 2, a 1) a 13 = 9, a 23 should be 9/2 but is 9 (twice what it should be)  1 2 9   .582 

λ max = 3.054, C.I. = .027, C.R. = .046 (acceptable).

2) If we change just a 12 from 2 to 3, we have λ max = 3.136, C.I. = .068, C.R. = 0.117.

3) Interchange an element (a 12 ) with its reciprocal in

max λ = 4 C.R. = 0 λ

max 4.250 = C.R. = .092