Jefferson’s Method
This is a Parliamentary Allocation method.
The task is:
- given
- $p_i$ - the number of voters in favor of party $i$
- $N$ - total number of parties, $i \in { 1, 2, …, N} \equiv P$
- $n$ - total number of voters
- the quota of $i$ is $q_i = S \cdot \cfrac{p_i}{n}$. Note that $q_i$ is a read number, not integer
- allocate $S$ seats in parliament
- $(s_1, …, s_N)$ s.t. $\sum s_i = S$
- $s_i$ must be an integer
The main idea of this method is to satisfy the following constraint:
- $s_i \ne 0, \cfrac{p_i}{s_i} \geqslant \cfrac{p_j}{s_j + 1}$
If this constraint is not respected, we have:
- $\cfrac{p_i}{s_i} < \cfrac{p_j}{s_j + 1}$
- but party $P_j$ won’t be happy about it: they may need more people to allocate one seat than $P_i$
So we give a place to a party $i$ with maximal $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ score
Meaning:
- suppose $S=10, p_1 = 6373, q_1 = 6.4$
- allocate $s_1 = 6$ seats to $P_1$
- $\cfrac{p_1}{s_1 + 1}$ means “to get one additional seat they need 910 people in the worst case”
Example 1
$S = 10$
| | $p_i$ | $q_i$ | $\lfloor q_i \rfloor$ | $s_i$ | $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ | $P_1$ | 6373 | 6.373 | 6 | 6 | 910.42 || $P_2$ | 2505 | 2.505 | 2 | 2 | 835 || $P_3$ | 602 | 0.602 | 0 | 0 | 602 || $P_4$ | 520 | 0.520 | 0 | 0 | 502 || | | | 8 | 8 | | $P_1$ has the highest $\cfrac{p_1}{\lfloor s_1 \rfloor + 1}$ score
- so allocating additional seat to them
- note that we need to re-calculate the value $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ for $P_1$ after we allocate the seat to them
| | $p_i$ | $q_i$ | $\lfloor q_i \rfloor$ | $s_i $ | $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ | $P_1$ | 6373 | 6.373 | 6 | 7 | 796 || $P_2$ | 2505 | 2.505 | 2 | 2 | 835 || $P_3$ | 602 | 0.602 | 0 | 0 | 602 || $P_4$ | 520 | 0.520 | 0 | 0 | 502 || | | | 8 | 9 | | $P_2$ has the highest score now
- so allocating the 10th seat to them
Example 2
$S = 10$
| | $p_i$ | $q_i$ | $\lfloor q_i \rfloor$ | $s_i$ | $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ | $P_1$ | 6373 | 6.4 | 6 | 6 | 910.42 || $P_2$ | 2505 | 2.505 | 2 | 2 | 768.33 || $P_3$ | 702 | 0.602 | 0 | 0 | 702 || $P_4$ | 620 | 0.520 | 0 | 0 | 620 || | | | 8 | 8 | | Give the seat to $P_1$, recalculate the score:
| | $p_i$ | $q_i$ | $\lfloor q_i \rfloor$ | $s_i$ | $\cfrac{p_i}{\lfloor s_i \rfloor + 1}$ | $P_1$ | 6373 | 6.4 | 6 | 7 | 796.6 || $P_2$ | 2505 | 2.505 | 2 | 2 | 768.33 || $P_3$ | 702 | 0.602 | 0 | 0 | 702 || $P_4$ | 620 | 0.520 | 0 | 0 | 620 || | | | 8 | 9 | | Again allocate the seat to $P_1$
| | $p_i$ | $q_i$ | $\lfloor q_i \rfloor$ | $s_i$ | $P_1$ | 6373 | 6.4 | 6 | 7 || $P_2$ | 2505 | 2.505 | 2 | 2 || $P_3$ | 702 | 0.602 | 0 | 0 || $P_4$ | 620 | 0.520 | 0 | 0 || | | | 8 | 10 | This shows that the method is still not perfect.
Properties
Consistency
Show that
- $\forall i, j \in P: p_i < p_j \Rightarrow s_i \leqslant s_j$
Solution
- Jefferson Rule is: $\cfrac{p_i}{s_i} \geqslant \cfrac{p_j}{s_j + 1}$
- or $\cfrac{p_i}{p_j} \geqslant \cfrac{s_i}{s_j + 1}$
- $\cfrac{p_i}{p_j} < 1$ always (by the hypothesis $p_i < p_j$)
- thus, $\cfrac{s_i}{s_j + 1} < 1$ or $s_i < s_j + 1$
Monotonicity
Also respected
Links
- http://www.math.colostate.edu/~spriggs/m130/apportionment2.pdf