Transfer Transfer
1st of R's of T's
Candidates. Count. Surplus. Result. Votes. Result.
P 9,050 9,050 9,050 (Elected).
Party A. Q 8,950 8,950 8,950 (Elected).
R 10,000 -749 9,251 9,251 (Elected).
Party B. S 6,000 +500 6,500 +2,400 8,900
T 3,000 +249 3,249 -3,249
Exhausted +849 849
——— ——— ———
37,000 37,000 37,000
The members of the two parties recorded their votes as follows:—
Party A. Party B.
P. 9,050 R. 10,000
Q. 8,950 S. 6,000
T. 3,000
The total number of votes polled is 37,000, and the quota, therefore, is 9251. Candidate R, having received more than a quota would be declared elected, and his surplus of 749 votes carried forward. It may be assumed that candidates S and T, who are of the same party, received 500 and 249 as their shares of this surplus. The result of this transfer is shown in the table. T, the lowest candidate on the poll, would then be eliminated. Now, if the contingent of voters Supporting T are not fully loyal to their party, and as many as 849 have recorded no preference save for T, then 2400 would be available for transfer to S, whose total would be only 8900. S would be eliminated, and the three candidates elected would be P and Q of party A, and R of party B, although R and S between them represented 18,151 voters. This case can be met by providing that whenever votes are exhausted the quota should be counted afresh. The votes in play, ignoring those exhausted, would be in all 36,151, the new quota would be 9038, while an additional number of votes, viz. 213, would be available for transfer from R to S, with the result that the position of these candidates would be as follows:—
R 9,038 S 9,113 P 9,050 Q 8,950
Party B would obtain two seats, the party A only one seat.]
[Footnote 20: Address delivered on 6 September 1909.]