For detailed information about the preferential voting system
used in General Faculty elections, see the
General Faculty Election Rules. This preferential system
is a version of the Hare-Cla rk system, which is a refinement
of the Hare system used at UT Austin for many years. The Hare
system has the inherent disadvantage that a random proce ss
is involved in transferring any excess votes from candidates
who have been elected; different choices can affect the results
of the election. The Hare-Clark system removes this defect
by using fractional weighting for transferred excess votes.
In addition, fractional weighting takes into account more
of the information provided by the voters' indicated preferences.
Counting votes electronically makes fractional weighting easy.
In the Hare-Clark system, you need not rank all the candidates
on the ballot, but skipping a number for those you do rank
is not allowed. As a matter of strategy, there is one basic
rule: if you prefer one candidate to another, then you should
rank the one above the other, or at least rank the one who
is perferred.
Hare-Clark Example
This example illustrates the preferential voting system
used in General Faculty elections. The process, a variation
of the Hare-Clark system, is described in Paragraph 3 of
Policy Memorandum 1.301 (By-Laws of the Faculty Council),
revised. The example uses the following assumptions:
That 100 voters cast final ballots to fill 5 positions
from 10 nominees.
That those casting ballots are divided clearly into two
groups, 60 conservatives and 40 liberals, whose members
vote along strict party lines. The final ballot has 6 conservatives
(A, B, C, D, E, F) and 4 liberals (W, X, Y, Z).
To make the details simple, assume:
- 30 vote ACDEFB (That is, A first, C second, and so
on.)
- 30 vote BCDEFA
- 10 vote WXYZ
- 10 vote XYZW
- 10 vote YZWX
- 10 vote ZWXY
(The conservatives are evenly split with strong differing
opinions on A and B, and are in agreement on the others.
The liberals vote in such a way that the final result is
bound to depend on random selection.)
Because of the 60-40 split, we should expect that 3 conservatives
and 2 liberals will be elected.
Quota
If n denotes the number of ballots cast and p
denotes the number of positions to be filled, then the electoral
quota is
q = (n/(p + 1)) + 1 = (100/(5 + 1)) + 1 = 16 2/3 + 1
rounded down, which gives q = 17.
|
After the first count
|
Candidate
|
Vote
|
|
A |
30 (Candidate A elected) |
|
B |
30 (Candidate B elected) |
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
|
C |
0 |
|
D |
0 |
|
E |
0 |
|
F |
0 |
|
After first transfer (down from A and B)
|
Candidate
|
Vote
|
|
C |
26 [(13/30)30 = 13 from each of A and B.]
(Candidate C elected) |
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
|
D |
0 |
|
E |
0 |
|
F |
0 |
|
After second transfer (down from C)
|
Candidate
|
Vote
|
|
D |
9 [(9/26)(13/30)60 = 9] |
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
|
E |
0 |
|
F |
0 |
Choose one of E or F randomly to eliminate.
Assume it is E. (Candidate E eliminated.)
|
After third transfer (up from E)
|
Candidate
|
Vote
|
|
D |
9 |
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
|
F |
0 |
(Candidate F eliminated.)
|
After fourth transfer (up from F)
|
Candidate
|
Vote
|
|
D |
9 |
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
(Candidate D eliminated.)
|
After fifth transfer (up from D)
|
Candidate
|
Vote
|
|
W |
10 |
|
X |
10 |
|
Y |
10 |
|
Z |
10 |
Choose one randomly to eliminate.
Assume it is Z.
(Candidate Z eliminated.)
|
After sixth transfer (up from Z)
|
Candidate
|
Vote
|
|
W |
20 [10 + 10 (from Z)]
(Candidate W elected) |
|
X |
10 |
|
Y |
10 |
|
After seventh transfer (down from W)
|
Candidate
|
Vote
|
|
X |
13 [10 + (3/20)20 (from W)] |
|
Y |
10 |
(Candidate Y eliminated.)
|
Final result: Candidates A, B, C, W,
and X elected.
|
|
