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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

See also

• Hamilton's Method

Sources

• Decision Engineering (ULB)