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ClaimsProblems (version 1.0.0)

MO: Minimal overlap rule

Description

This function returns the awards vector assigned by the minimal overlap rule rule (MO) to a claims problem.

Usage

MO(E, d, name = FALSE)

Value

The awards vector selected by the MO rule. If name = TRUE, the name of the function (MO) as a character string.

Arguments

E

The endowment.

d

The vector of claims.

name

A logical value.

Details

Let \(N=\{1,\ldots,n\}\) be the set of claimants, \(E\ge 0\) the endowment to be divided and \(d\in \mathbb{R}_+^N\) the vector of claims such that \(\sum_{i \in N} d_i\ge E\).

The truncated claim of a claimant \(i\in N\) is the minimum of the claim and the endowment, that is, \(t_i(E,d)=t_i=\min\{d_i,E\},\ i=1,\dots,n.\)

Suppose that each agent claims specific parts of \(E\) equal to her/his claim. After arranging which parts agents claim so as to “minimize conflict”, equal division prevails among all agents claiming a specific part and the minimal overlap rule (MO) assigns the sum of the compensations she/he gets from the various parts that he claimed.

Let \(d_0=0\). For each problem \((E,d)\) and each claimant \(i\in N\),

1) If \(E\le d_n\) then $$\text{MO}_i(E,d)=\frac{t_1}{n}+\frac{t_2-t_1}{n-1}+\dots+\frac{t_i-t_{i-1}}{n-i+1}.$$

2) If \(E>d_n\), let \(s'\in (d_{k'},d_{k'+1}]\), with \(k' \in \{0,1,\dots,n-2\}\), be the unique solution to the equation \(\underset{j \in N}{\sum} \max\{d_j-s,0\} =E-s\). Then, $$\text{MO}_i(E,d)= \begin{cases} \frac{d_1}{n}+\frac{d_2-d_1}{n-1}+\dots+\frac{d_i-d_{i-1}}{n-i+1}& \text{if } i\in \{1,\dots , k'\} \\[4pt] \text{MO}_i(s',d)+d_i-s' & \text{if } i\in \{k'+1,\dots, n\} \end{cases}.$$

References

Mirás Calvo, M.Á., Núñez Lugilde, I., Quinteiro Sandomingo, C., and Sánchez-Rodríguez, E. (2023). Refining the Lorenz‐ranking of rules for claims problems on restricted domains. International Journal of Economic Theory 19(3), 526-558.

O'Neill, B. (1982). A problem of rights arbitration from the Talmud. Mathemathical Social Sciences. 2, 345-371.

Thomson, W. (2019). How to divide when there isn't enough. From Aristotle, the Talmud, and Maimonides to the axiomatics of resource allocation. Cambridge University Press.

See Also

allrules, CD.

Examples

Run this code
E=10
d=c(2,4,7,8)
MO(E,d)

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