Exhibit 2

Proof That LSM Avoids Row Preference
Reversals (RPRs) for Strong Form Row Dominance

I. Definitions

(1)	LSM Model:	Minimize ESS	=	å i å j ( rij - wi / wj ) 2
		1	=	å i wi
		0	<	wi for i = 1, ...,n
(2)	Eij = ( rij - wi / wj ) 2 + ( rji - wj / wi ) 2 + å k ¹ i, j (( rik - wi / wk ) 2 + ( rjk - wj / wk ) 2 + ( rki - wk / wi ) 2 + ( rkj - wk / wj ) 2 )
(3)	ESS = Eij - Eij º (ESS components having a subscript i or j) + (All other ESS error components) for solution vector [w] = [w1,...,wn]
(4)	Ei/j = ( rij - wj / wi ) 2 + ( rji - wi / wj ) 2 + åk ¹ i, j (( rik - wj / wk ) 2 + ( rjk - wi / wk ) 2 + ( rki - wk / wj ) 2 + ( rkj - wk / wi ) 2 )
(5)	ESS/ = Ei/j + Eij º (ESS/ components having a subscript i or j) + (All other ESS/ error components) when wi and wj are transposed in [w]
(6)	DEij = ESS - ESS/ = Eij - Ei/j º Change in ESS resulting from transposing wi and wj in [w]
(7)	[w *] = [w1 *,..., wn *] º Optimal LSM solution vector
(8)	wij = wi / wj º LSM model surrogate for the rij response judgment

II. Theorem

If r_ik > r_jk for k = 1, ..., n

Then w_i * > w_j *

Interpretation: If Row i is dominant in the strong form over Row j, then the optimal LSM solution will not reverse the Row i dominance over Row j.

III. Proof

DEij	=	Eij - Ei/j
	=	- 2rij ( wij - wji ) - 2rji ( wji - wij ) + åk ¹ i, j ( -2rik ( wik - wjk ) - 2rjk ( wjk - wik ) - 2rki ( wki - wkj ) - 2rkj (wkj - wki ) )
	=	2 [ ( rij - rji ) (wji - wij ) + åk ¹ i, j ( ( rik - rjk ) (wjk - wik ) + ( rkj - rki ) ( wki - wkj ) ) ]
If wi < wj when rik > rjk for k = 1,...,n,
Then ( rij - rji ) ( wji - wij ) > 0, ( rik - rjk ) ( wjk - wik ) > 0, and (rkj - rki ) ( wki - wkj ) > 0
\ DEij = 0 if rik = rjk for all k = 1,...,n
\ DEij = 0 if rik > rjk for at least one k among k = 1,...,n
Conclusion: Conditions rik > rjk for all k, rik > rjk for at least one k, and wi < wj Þ wi ¹ wi * and/or wi * ¹ wj *. Q. E. D.

IV. Interpretation

In any response matrix [R] whose r_ik elements have a reciprocal property and r_ik > r_jk for all k = 1,...,n, then the optimal LSM solution vector will always have components in which w_i * > w_j *. In other words, if Row i strongly dominates Row j, then any solution in which w_i < w_j cannot be LSM optimal, because a reduction in ESS can be achieved by simply transposing w_i and w_j in the solution vector.

Whenever Row i and Row j have identical r_ik = r_jk values for all k = 1,...,n, then no error change can be achieved by transposing w_i and w_j since DE_ij = 0. However, in this case w_i * = w_j * is a LSM optimal solution.