[EM] STV and having quotas to eliminate losers

[Craig Carey]

At 17:56 13.10.99 , Craig Carey wrote:
>
>At 11:53 13.10.99 , David Catchpole wrote:
>>On Wed, 13 Oct 1999, Craig Carey wrote:
....

The message I sent was flawed and had wrong numbers in it.

A rather major matter is that if STV is modified so that it satisfies
 the duality rule, its individual stages do not.
That is about where I went wrong; the values for Quotas to eliminate
 losers in a modified STV are of quite wrong. That can be seen below.
 Note that 1/2 isn't equal to zero. I attempted to fix the error while
 being sure that others were proving that an error existed. However
 the fix will take a uncertain amount of time before devised and sent
 to the mailing list.


Tab = (a+Zca<b+Zcb), Tba = (b+Zcb<a+Zca)

In IFPP 3 candidate 1 winner: c<b<a:
 (C in W) = False
 (B in W) = (1/3<b)Tab
 (A in W) = (b<1/3) or Tba
    Upper Quota: accept X if (x>z), (z>=1/2)
    Lower Quota: reject X if (x<1/3)

In the 2 winner 3 candidate problem: a<b<c:
 (C in W) = True
 (B in W) = (1/3<b) or Tab
 (A in W) = (b<1/3)Tba
    Upper Quota: accept X if (x>1/3)
    Lower Quota: reject X if (x<z), where z<=0


>Craig Carey

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Mr G. A. Craig Carey                   E-mail: research at ijs.co.nz
Auckland, Nth Island, New Zealand
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