{"id":1372,"date":"2016-10-30T09:50:49","date_gmt":"2016-10-30T14:50:49","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=1372"},"modified":"2026-04-05T16:39:25","modified_gmt":"2026-04-05T21:39:25","slug":"s1eva2017tii_t3-multiplexor-colas","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-ejemplos\/s1eva2017tii_t3-multiplexor-colas\/","title":{"rendered":"s1Eva2017TII_T3 Multiplexor-Colas"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2017TII_T3 Multiplexor-Colas<\/a><\/p>\n\n\n\n

a) Determine el espacio de estados del sistema<\/p>\n\n\n\n

S={0,1,2,3,4,5,6}<\/p>\n\n\n\n

b) Dibuje y etiquete el diagrama de estados del sistema<\/p>\n\n\n\n

\"multiplexor<\/figure>\n\n\n\n
SALE = ENTRA\n(1) \u03bbp0<\/sub> = \u00b5p1<\/sub>\n(2) (\u03bb + \u00b5)p1<\/sub> = \u03bbp0<\/sub> + 2\u00b5p2<\/sub>\n(3) (\u03bb + 2\u00b5)p2<\/sub> = \u03bbp1<\/sub> + 3\u00b5p3<\/sub>\n(4) (\u03bb + 3\u00b5)p3<\/sub> = \u03bbp2<\/sub> + 4\u00b5p4<\/sub>\n(5) (\u03bb + 4\u00b5)p4<\/sub> = \u03bbp3<\/sub> + 5\u00b5p5<\/sub>\n(6) 5\u00b5p5<\/sub> = \u03bbp4<\/sub>\n(7) p0<\/sub> + p1<\/sub> + p2<\/sub> + p3<\/sub> + p4<\/sub> + p5<\/sub> = 1<\/code><\/pre>\n\n\n\n
(1) p1<\/sub> = (\u03bb\/\u00b5) p0<\/sub> \n\n(2) (\u03bb + \u00b5)(\u03bb\/\u00b5) p0<\/sub>  = \u03bbp0<\/sub> + 2\u00b5p2<\/sub> \n    p2<\/sub> = [1\/2\u00b5] [(\u03bb + \u00b5)(\u03bb\/\u00b5) - \u03bb] p0<\/sub> \n    p2<\/sub> = [\u03bb\/2\u00b52<\/sup>] [(\u03bb + \u00b5) - \u00b5] p0<\/sub> \n    p2<\/sub> = (1\/2)(\u03bb\/\u00b5)2<\/sup> p0<\/sub> \n\n(3) (\u03bb + 2\u00b5)p2<\/sub> = \u03bbp1<\/sub> + 3\u00b5p3<\/sub> \n    (\u03bb + 2\u00b5)(1\/2)(\u03bb\/\u00b5)2<\/sup> p0<\/sub>  = \u03bb (\u03bb\/\u00b5) p0<\/sub>  + 3\u00b5p3<\/sub> \n    [(\u03bb + 2\u00b5)(1\/2)(\u03bb\/\u00b5)2<\/sup> - (\u03bb2<\/sup>\/\u00b5)] p0<\/sub>  = + 3\u00b5p3<\/sub> \n    3\u00b5p3<\/sub> = (\u03bb\/\u00b5)2<\/sup> [(1\/2)(\u03bb + 2\u00b5) - \u00b5 ] p0<\/sub> \n    p3<\/sub> = (1\/3\u00b5)(\u03bb\/\u00b5)2<\/sup> (\u03bb + 2\u00b5 - 2\u00b5)\/2 p0<\/sub> \n    p3<\/sub> = [1\/(2*3)](\u03bb\/\u00b5)2<\/sup> (\u03bb\/\u00b5) p0<\/sub> \n    p3<\/sub> = (1\/3!)(\u03bb\/\u00b5)3<\/sup> p0<\/sub>\n\n(4) (\u03bb + 3\u00b5)(1\/(2*3))(\u03bb\/\u00b5)3<\/sup> p0<\/sub> = \u03bb(1\/2)(\u03bb\/\u00b5)2<\/sup> p0<\/sub>  + 4\u00b5p4<\/sub>\n    4\u00b5p4 <\/sub> = [(\u03bb + 3\u00b5)(1\/(2*3))(\u03bb\/\u00b5)3<\/sup> - \u03bb(1\/2)(\u03bb\/\u00b5)2<\/sup>] p0<\/sub> \n    4\u00b5p4 <\/sub> = (1\/2)(\u03bb\/\u00b5)3<\/sup>[(1\/3)(\u03bb + 3\u00b5) - \u00b5] p0<\/sub> \n    p4 <\/sub> = [1\/2*4\u00b5](\u03bb\/\u00b5)3<\/sup>((\u03bb + 3\u00b5) - 3\u00b5)\/3 p0<\/sub> \n    p4 <\/sub> = [1\/2*3*4](\u03bb\/\u00b5)3<\/sup>(\u03bb\/\u00b5) p0<\/sub> \n    p4 <\/sub> = (1\/4!)(\u03bb\/\u00b5)4<\/sup> p0<\/sub> \n\n(5) (\u03bb + 4\u00b5)(1\/4!)(\u03bb\/\u00b5)4<\/sup> p0<\/sub>  = \u03bb(1\/3!)(\u03bb\/\u00b5)3<\/sup> p0<\/sub> + 5\u00b5p5<\/sub>\n     5\u00b5p5<\/sub> = [(\u03bb + 4\u00b5)(1\/4!)(\u03bb\/\u00b5)4<\/sup> - \u03bb(1\/3!)(\u03bb\/\u00b5)3<\/sup>] p0<\/sub> \n     5\u00b5p5<\/sub> = [(1\/4!)(\u03bb\/\u00b5)4<\/sup>][(\u03bb + 4\u00b5-  4\u00b5] p0<\/sub>\n     p5<\/sub> = (1\/5!)(\u03bb\/\u00b5)5<\/sup> p0<\/sub> \n\n(7) p0<\/sub> + (\u03bb\/\u00b5) p0<\/sub> + (1\/2)(\u03bb\/\u00b5)2<\/sup> p0<\/sub> + [1\/3!](\u03bb\/\u00b5)3<\/sup> p0<\/sub> + (1\/4!)(\u03bb\/\u00b5)4<\/sup> p0<\/sub>  + (1\/5!)(\u03bb\/\u00b5)5<\/sup> p0<\/sub> = 1\n    p0<\/sub> [1 + (\u03bb\/\u00b5) + (1\/2)(\u03bb\/\u00b5)2<\/sup> + (1\/3!)(\u03bb\/\u00b5)3<\/sup> + (1\/4!)(\u03bb\/\u00b5)4<\/sup> + (1\/5!)(\u03bb\/\u00b5)5<\/sup> ] = 1\n    p0<\/sub>  = 1\/ [1 + (\u03bb\/\u00b5) + (1\/2)(\u03bb\/\u00b5)2<\/sup> + (1\/3!)(\u03bb\/\u00b5)3<\/sup> + (1\/4!)(\u03bb\/\u00b5)4<\/sup> + (1\/5!)(\u03bb\/\u00b5)5<\/sup> ]<\/code><\/pre>\n\n\n\n
d) Encuentre la probabilidad de p\u00e9rdidas de conexiones.<\/p>\n\n\n\n
p5<\/sub> =  (1\/5!)(\u03bb\/\u00b5)5<\/sup> p0<\/sub>  \n   = (1\/5!)(\u03bb\/\u00b5)5<\/sup> \/  [1 + (\u03bb\/\u00b5) + (1\/2)(\u03bb\/\u00b5)2<\/sup> + (1\/3!)(\u03bb\/\u00b5)3<\/sup> + (1\/4!)(\u03bb\/\u00b5)4<\/sup> + (1\/5!)(\u03bb\/\u00b5)5<\/sup> ]<\/code><\/pre>\n\n\n\n
e) \u00bfCu\u00e1l es el factor de utilizaci\u00f3n del enlace?<\/p>\n\n\n\n
Es el valor esperado de uso de capacidad relativo a la capacidad m\u00e1xima del multiplexor.<\/p>\n\n\n \\rho = \\frac{\\sum_{i=0}^{5}(i)(p_i )\\text{1MB}}{\\text{5MB}} <\/span>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2017TII_T3 Multiplexor-Colas a) Determine el espacio de estados del sistema S={0,1,2,3,4,5,6} b) Dibuje y etiquete el diagrama de estados del sistema d) Encuentre la probabilidad de p\u00e9rdidas de conexiones. e) \u00bfCu\u00e1l es el factor de utilizaci\u00f3n del enlace? Es el valor esperado de uso de capacidad relativo a la capacidad m\u00e1xima del multiplexor.<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-stp-ejercicios","format":"standard","meta":{"footnotes":""},"categories":[203],"tags":[58,237],"class_list":["post-1372","post","type-post","status-publish","format-standard","hentry","category-stp-ejemplos","tag-ejemplos-python","tag-pestocasticos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1372","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=1372"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1372\/revisions"}],"predecessor-version":[{"id":23538,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1372\/revisions\/23538"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1372"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1372"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}