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Ejercicios de Teor&Atilde;&shy;a de colas

# Ejercicios de Teor&Atilde;&shy;a de colas -...

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que est´ a siendo reparada en ese momento). Para una cola M/M/1, como la distribuci´ on del n´umero de m´ aquinas en el sistema es geom´ etrica con par´ ametro ρ = λ/μ , tenemos que E [ N ] = k =0 kP ( N = k ) = k =0 k (1 - ρ ) ρ k = ρ 1 - ρ . Por tanto, el ahorro vendr´ a dado por A = 2500( E [ N 1 ] - E [ N 2 ]) = 2500 ρ 1 1 - ρ 1 - ρ 2 1 - ρ 2 , donde ρ 1 = 0 , 2 1 / 4 = 0 , 8 , ρ 2 = 0 , 2 1 / 3 = 0 , 6 , con lo que obtenemos A = 2500 0 , 8 0 , 2 - 0 , 6 0 , 4 = 6250 .
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