Suppose there are three parties A, B and C and three regions I, II and III and that there are 20 seats are to be distributed and that the Sainte-Laguë method is used. The votes for the regional party lists are as follows:
Upper apportionment
For the upper apportionment, the overall seat number for the parties and the regions are determined.
Since there are 3805 voters and 20 seats, there are 190 (rounded) voters per seat. Thus the results for the distribution of the party seats is:
More information Party, A ...
Party | A | B | C |
#votes | 983 | 2040 | 782 |
#votes/divisor | 5.2 | 10.7 | 4.1 |
#seats | 5 | 11 | 4 |
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Using the divisor 190, the results for the distribution of the region seats is:
More information Region, I ...
Region | I | II | III |
#votes | 1347 | 1014 | 1444 |
#votes/divisor | 7.1 | 5.3 | 7.6 |
#seats | 7 | 5 | 8 |
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Lower apportionment
Initially, regional divisors have to be found to distribute the seats of each region to the regional party lists. In the tables, for each regional party list, there are two cells, the first shows the number of votes and the second the number of seats allocated.
More information Party, region ...
Party | region |
I | II | III |
A | 123 | 1 | 45 | 0 | 815 | 5 |
B | 912 | 4 | 714 | 4 | 414 | 2 |
C | 312 | 2 | 255 | 1 | 215 | 1 |
total | 1347 | 7 | 1014 | 5 | 1444 | 8 |
regional divisor | 205 | 200 | 180 |
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Now, the party divisors are initialized with ones and the number of seats within each party is checked (that is, compared to the number computed in the upper apportionment):
More information Party, region ...
Party | region |
total |
party
divisor |
I | II | III |
A | 123 | 1 | 45 | 0 | 815 | 5 | 983 | 6 | 1 |
B | 912 | 4 | 714 | 4 | 414 | 2 | 2040 | 10 | 1 |
C | 312 | 2 | 255 | 1 | 215 | 1 | 782 | 4 | 1 |
total | 1347 | 7 | 1014 | 5 | 1444 | 8 | 3805 | 20 |
regional divisor | 205 | 200 | 180 |
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Since not all parties have the correct number of seats, a correction step has to be executed: For parties A and B, the divisors are to be adjusted. The divisor for A has to be raised and the divisor for B has to be lowered:
More information Party, region ...
Party | region |
total |
party
divisor |
I | II | III |
A | 123 | 1 | 45 | 0 | 815 | 4 | 983 | 5 | 1.1 |
B | 912 | 5 | 714 | 4 | 414 | 2 | 2040 | 11 | 0.95 |
C | 312 | 2 | 255 | 1 | 215 | 1 | 782 | 4 | 1 |
total | 1347 | 8 | 1014 | 5 | 1444 | 7 | 3805 | 20 |
regional divisor | 205 | 200 | 180 |
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Now, the divisors for regions I and III have to be modified. Since region I has one seat too much (8 instead of the 7 seats computed in the upper apportionment), its divisor has to be raised; in opposite, the divisor for region III has to be lowered.
More information Party, region ...
Party | region |
total |
party
divisor |
I | II | III |
A | 123 | 1 | 45 | 0 | 815 | 4 | 983 | 5 | 1.1 |
B | 912 | 5 | 714 | 4 | 414 | 3 | 2040 | 12 | 0.95 |
C | 312 | 1 | 255 | 1 | 215 | 1 | 782 | 3 | 1 |
total | 1347 | 7 | 1014 | 5 | 1444 | 8 | 3805 | 20 |
regional divisor | 210 | 200 | 170 |
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Again, the divisors for the parties have to be adjusted:
More information Party, region ...
Party | region |
total |
party
divisor |
I | II | III |
A | 123 | 1 | 45 | 0 | 815 | 4 | 983 | 5 | 1.1 |
B | 912 | 4 | 714 | 4 | 414 | 3 | 2040 | 11 | 0.97 |
C | 312 | 2 | 255 | 1 | 215 | 1 | 782 | 4 | 0.98 |
total | 1347 | 7 | 1014 | 5 | 1444 | 8 | 3805 | 20 |
regional divisor | 210 | 200 | 170 |
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Now, the numbers of seats for the three parties and the three regions match the numbers computed in the upper apportionment. Thus, the iterative process is completed.
The final seat numbers are:
More information #seats, region ...
#seats | region |
total |
Party | I | II | III |
A | 1 | 0 | 4 |
5 |
B | 4 | 4 | 3 |
11 |
C | 2 | 1 | 1 |
4 |
total | 7 | 5 | 8 | 20 |
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