The allotment of seats to parties.
The number of representatives awarded to each party is determined by the method formulated by M. Victor d'Hondt, a professor of the University of Ghent. Its working may best be shown by an illustration. Let it be assumed that three lists have been presented; that they have obtained 8000, 7500, and 4500 votes respectively, and that there are five vacancies to be filled. The total number of votes for each list is divided successively by the numbers 1, 2, 3, and so on, and the resulting numbers are arranged thus:—
List No. 1. List No. 2. List No. 3. 8,000 7,500 4,500 4,000 3,750 2,250 2,666 2,500 1,500
The five highest numbers (five being the number of vacancies to be filled) are then arranged in order of magnitude as follows:—
8,000 7,500 4,500 4,000 3,750
The lowest of these numbers, 3750, is called the "common divisor"[1] or the "electoral quotient," and forms the basis for the allotment of seats. The number of votes obtained by each of the lists is divided by the "common divisor" thus:—
8,000 divided by 3,750 = 2 with a remainder of 500. 7,500 " 3,750 = 2 4,500 " 3,750 = 1 with a remainder of 750.
The first list contains the "electoral quotient" twice, the second twice, and the third once, and the five seats are allotted accordingly. Each party obtains one representative for every quota of voters which it can rally to its support, all fractions of "quotas" being disregarded.
The method of determining the electoral quotient may appear at first sight rather empirical, but the rule is merely the arithmetical expression, in a form convenient for returning officers, of the following train of reasoning. The three lists with 8000, 7500, and 4500 supporters are competing for seats. The first seat has to be allotted; to which list is it to go? Plainly to the list with 8000 supporters. Then the second seat has to be disposed of; to which list is it to go? If it is given to the first list, then the supporters of the first list will have two members in all, or one member for each 4000 votes. This would be unfair while 7500 supporters of the second list are unrepresented, therefore the second seat is allotted to the list with 7500 supporters. Similar reasoning will give the third seat to the list with 4500 supporters, the fourth to the list with 8000 supporters, which now will rightly have one representative for each 4000, and the fifth to the list with 7500. The question in each case is to what list must the seat be allotted in such a way that no one group of unrepresented electors is larger than a represented group. The separate allotment of seats one by one in accordance with the foregoing reasoning may be shown thus:—
8,000 (List No. 1) 7,500 ( " No. 2) 4,500 ( " No. 3) 4,000 ( " No. 1) 3,750 ( " No. 2)