Tighter price of anarchy for selfish task allocation on selfish machines

Given a set $$L = \{J_1,J_2,\ldots ,J_n\}$$ L = { J 1 , J 2 , … , J n } of n tasks and a set $$M = \{M_1,M_2, \ldots ,M_m\}$$ M = { M 1 , M 2 , … , M m } of m identical machines, in which tasks and machines are possessed by different selfish clients. Each selfish client of machine $$M_i \in M$$ M i ∈ M gets a profit equal to its load and each selfish client of task allocated to $$M_i$$ M i suffers from a cost equal to the load of $$M_i$$ M i . Our aim is to allocate the tasks on the m machines so as to minimize the maximum completion times of the tasks on each machine. A stable allocation is referred to as a dual equilibrium (DE). We firstly show that 4/3 is tight upper bound of the Price of Anarchy(PoA) with respect to dual equilibrium for $$m\in \{3,\ldots ,9\}$$ m ∈ { 3 , … , 9 } . And secondly $$(7m-6)/(5m-3)$$ ( 7 m - 6 ) / ( 5 m - 3 ) is an upper bound for $$m\ge 10$$ m ≥ 10 . The result is better than the existing bound of 7/5.