Conference rusure::math

607.0. "Dubious Magazine Mathematics" by COMET::ROBERTS (Dwayne Roberts) Wed Nov 05 1986 11:33

    I just read a magazine article that contained some pretty dubious
    mathematics.  They attempted to gain an over-all ranking of six objects
    after ranking them in each of nine categories.  Their categorical
    rankings looked like this: 
    
    
    		A	B	C	D	E	F
    
    a		1	2	3	4	5	6
    b		5	6	2	1	4	3
    c		6	5	2	1	3	4
    d		1	4	2	3	5	6
    e		1	3	4	2	6	5
    f		6	5	4	3	1	2
    g		6	4	5	3	1	2
    h		2	4	1	3	5	6
    i		1	3	2	4	5	6
    
    where A-F are the objects being ranked and a-i are the categories. A
    ranking of "1" means that it was judged to be the best of the objects
    in that category. 
    
    They then computed the over-all ranking by assigning a value of 21 for
    a "1" rank, 15 for a "2", 10 for a "3", 6 for a "4", 3 for a "5", and 1
    for a "6"; did the arithmetic, and came up with the ordering: 
    
    
    RANK	OBJECT	POINTS
    ====	======	======
    1		C	110
    2		A,D	105	(tie)
    3		E	71
    4		B	60
    5		F	53
    
    My first comment is that because of the tie, the ranks should be
    1,2,4,5,6. 
    
    Second, they have not weighted the categories by importance.  The
    assumption is that category "a" is as important as category "b". 
    
    Third, they should check their arithmetic.  My calculator tells me that
    object C has a total of 106 points, and object D totals 109 points.
    This would make a dramatic difference in the ranking:  instead of
    C(AD)EBF, it would be DCAEBF. 
    
    Fourth, their choice of values for the rankings is suspicious.  These
    numbers could have been chosen to fix the over-all rank in a desired
    way. 

    I intend to write to the magazine about this disaster.  In my letter,
    I'll suggest a substitute valuing of ranks such that val("1") >
    val("2") > val("3") ...  Can you suggest some values that would make
    the biggest difference between DCAEBF and yours?  Is FBEACD possible? 
    

Let val("1") = u, val("2") = v, ..., val("6") = z, with u > v > ... > z.

Then we have:

    X       points(X)
    =   =================
    A   4u+ v      + y+3z
    B       v+2w+3x+2y+ z
    C    u+4v+ w+2x+ y
    D   2u+ v+4w+2x
    E   2u   + w+ x+4y+ z
    F      2v+ w+ x+ y+4z

We see that points(C) > points(B), since

	(u+4v+w+2x+y) - (v+2w+3x+2y+z) = u+3v-w-x-y-z
		= (u-w) + (v-x) + (v-y) + (v-z)
		> 0

Similarly, we find that (using a briefer notation) D>E, E>F, C>B, A>F,
D>B, and C>F.  Thus, we have the following partial ordering:

	A-----\
	       \
	D--E----F
	 \     /
	  >B  /
	 /   /
	C---'

Thus, there are 37 (out of 6!=720) orderings that *might* be possible.