# [EM] General PR question (from Andy Jennings in 2011)

Toby Pereira tdp201b at yahoo.co.uk
Sat Oct 4 09:41:05 PDT 2014

```Kathy
I'd already sent that last e-mail before seeing this, but it seems your method is essentially the Largest Remainder method http://en.wikipedia.org/wiki/Largest_remainder_method It comes with its own problems such as the Alabama paradox http://en.wikipedia.org/wiki/Apportionment_paradox#Alabama_paradox where increasing the seats available can actually cause a party to lose seats.
Toby

From: Kathy Dopp <kathy.dopp at gmail.com>
To: Toby Pereira <tdp201b at yahoo.co.uk>
Cc: Kristofer Munsterhjelm <km_elmet at t-online.de>; EM <election-methods at lists.electorama.com>
Sent: Saturday, 4 October 2014, 17:07
Subject: Re: [EM] General PR question (from Andy Jennings in 2011)

Toby,

After looking at how the Sainte-Laguë and D'Hondt methods work, a
similar algorithmic approach to implementing my method can be easily
shown to always provide proportionate results in terms of seat
allocation.

So instead of calculating the somewhat complex (for the average voter)
Sum over i of (v_i/v*Absolute(v_i/v - s_i/s))

The algorithm for my method, as you noticed, would be simply to:
(1)  multiple the overall ratio of the (total # seats)/(total #
voters) times the number of voters in each voting group
(2) the integer portion of each result is the number of seats assigned

to each group
(3) order the remainder decimal portion of each voting group's result
from greatest to least and beginning at the top (group with the
largest remainder) assign one more seat to each  group until the total
number of seats to be elected is achieved.

(4)  Although unlikely in most elections, some tie-breaking procedure
could be needed: E.g. If there are ties towards the end of the
allocation procedure, some random selection or asking tied groups at
the end of the allocation process to co-select a winner, or possibly,
the number of seats could be increased by the number of groups - 1 who
tie for the last seat allocation.

I am unclear why, exactly, either the Sainte-Laguë and D'Hondt methods
would always give exactly proportionate results in all cases, but it
is easy to understand, simply by the cancellation of the units of
analysis seats/voters * voters = seats (as physicists always do) that
the above algorithm always would given the most proportionate outcomes
in seats (disregarding exactly tie votes).

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