On the grey Baker-Thompson rule

Olgun Mehmet Onur; Palanci Osman; Gök Sirma Zeynep Alparslan

doi:10.3934/jdg.2020024

Article Contents

2020, Volume 7, Issue 4: 303-315. Doi: 10.3934/jdg.2020024

This issue Previous Article On class of non-transferable utility cooperative differential games with continuous updating Next Article Real option value and poverty trap

On the grey Baker-Thompson rule

1.
Süleyman Demirel University, Faculty of Engineering, Department of Industrial Engineering, Isparta, 32260, Turkey
2.
Süleyman Demirel University, Faculty of Economics and Administrative Sciences, Department of Business Administration, Isparta, 32260, Turkey
3.
Süleyman Demirel University, Faculty of Arts and Sciences, Department of Mathematics, Isparta, 32260, Turkey

^* Corresponding author: zeynepalparslan@yahoo.com
^* Corresponding author: zeynepalparslan@yahoo.com

Received: December 2018

Revised: April 2020

Early access: August 2020

Published: October 2020

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Abstract

Cost sharing problems can arise from situations in which some service is provided to a variety of different customers who differ in the amount or type of service they need. One can think of and airports computers, telephones. This paper studies an airport problem which is concerned with the cost sharing of an airstrip between airplanes assuming that one airstrip is sufficient to serve all airplanes. Each airplane needs an airstrip whose length can be different across airplanes. Also, it is important how should the cost of each airstrip be shared among airplanes. The purpose of the present paper is to give an axiomatic characterization of the Baker-Thompson rule by using grey calculus. Further, it is shown that each of our main axioms (population fairness, smallest-cost consistency and balanced population impact) together with various combina tions of our minor axioms characterizes the best-known rule for the problem, namely the Baker-Thompson rule. Finally, it is demonstrated that the grey Shapley value of airport game and the grey Baker-Thompson rule coincides.

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Mathematics Subject Classification: Primary: 91A12.

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Table 1. Grey marginal vectors

$\sigma $	$m_{1}^{\sigma }\left(c^{\prime }\right) $	$m_{2}^{\sigma }\left(c^{\prime }\right) $	$m_{3}^{\sigma }\left(c^{\prime }\right) $
$\sigma _{1} = \left(1, 2, 3\right) $	$m_{1}^{\sigma _{1}}\left(c^{\prime }\right) \in \left[30, 36\right] $	$m_{2}^{\sigma _{1}}\left(c^{\prime }\right) \in \left[14, 18\right] $	$m_{3}^{\sigma _{1}}\left(c^{\prime }\right) \in \left[16, 26\right] $
$\sigma _{2} = \left(1, 3, 2\right) $	$m_{1}^{\sigma _{2}}\left(c^{\prime }\right) \in \left[30, 36\right] $	$m_{2}^{\sigma _{2}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{3}^{\sigma _{2}}\left(c^{\prime }\right) \in \left[30, 44\right] $
$\sigma _{3} = \left(2, 1, 3\right) $	$m_{1}^{\sigma _{3}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{2}^{\sigma _{3}}\left(c^{\prime }\right) \in \left[44, 54\right] $	$m_{3}^{\sigma _{3}}\left(c^{\prime }\right) \in \left[16, 26\right] $
$\sigma _{4} = \left(2, 3, 1\right) $	$m_{1}^{\sigma _{4}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{2}^{\sigma _{4}}\left(c^{\prime }\right) \in \left[44, 54\right] $	$m_{3}^{\sigma _{4}}\left(c^{\prime }\right) \in \left[16, 26\right] $
$\sigma _{5} = \left(3, 1, 2\right) $	$m_{1}^{\sigma _{5}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{2}^{\sigma _{5}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{3}^{\sigma _{5}}\left(c^{\prime }\right) \in \left[60, 80\right] $
$\sigma _{6} = \left(3, 2, 1\right) $	$m_{1}^{\sigma _{6}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{2}^{\sigma _{6}}\left(c^{\prime }\right) \in \left[0, 0\right] $	$m_{3}^{\sigma _{6}}\left(c^{\prime }\right) \in \left[60, 80\right] $

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