{"id":429,"date":"2016-10-28T15:07:52","date_gmt":"2016-10-28T20:07:52","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=429"},"modified":"2026-04-05T16:40:57","modified_gmt":"2026-04-05T21:40:57","slug":"s1eva2017ti_t3-call-center-operadora-y-dos-tecnicos","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-ejemplos\/s1eva2017ti_t3-call-center-operadora-y-dos-tecnicos\/","title":{"rendered":"s1Eva2017TI_T3 Call Center Operadora y Dos T\u00e9cnicos"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2017TI_T3 Call Center Operadora y Dos T\u00e9cnicos<\/a><\/p>\n\n\n\n

Tema 3 <\/strong>
Usamos dos d\u00edgitos para representar (operadora,t\u00e9cnicos) en cantidades de ocupado=0,1,2<\/p>\n\n\n\n

Los estados de operadora libre ser\u00e1n: (00),(01),(02)
Los estados de operadora ocupada ser\u00e1n: (10),(11),(12)<\/p>\n\n\n\n

Diagrama de Transici\u00f3n
<\/p>\n\n\n\n

Ecuaciones de Balanceo<\/p>\n\n\n\n

\u03bb P00<\/sub> = \u03bcT<\/sub> P01<\/sub>\n(\u03bcT<\/sub> + \u03bb) P01<\/sub> = 2 \u03bcT<\/sub> P02<\/sub> + \u03bcR<\/sub> P10<\/sub>\n(2 \u03bcT<\/sub> + \u03bb) P02<\/sub> = \u03bcR<\/sub> P11<\/sub> + \u03bcR<\/sub> P12<\/sub>\n\u03bcR<\/sub> P10<\/sub> = \u03bb P00<\/sub> + \u03bcT<\/sub> P11<\/sub>\n(\u03bcR<\/sub> + \u03bcT<\/sub>) P11<\/sub> = \u03bb P01<\/sub> +  2 \u03bcT<\/sub> P12<\/sub>\n(\u03bcR<\/sub> + 2 \u03bcT<\/sub>) P12<\/sub> = \u03bb P02<\/sub>\n\nP00<\/sub> + P01<\/sub> + P02<\/sub> + P10<\/sub> + P11<\/sub>+  P12<\/sub> = 1\n\n\u03bb = 1\/10, \u03bcR<\/sub> = 1\/3, \u03bcT<\/sub> =1\/15<\/code><\/pre>\n\n\n\n
reemplazando<\/p>\n\n\n\n
1\/10 P00<\/sub> = 1\/15 P01<\/sub>\n(1\/15 + 1\/10) P01<\/sub> = 2 (1\/15) P02<\/sub> + 1\/3 P10<\/sub>\n(2 (1\/15) + 1\/10) P02<\/sub> = 1\/3 P11<\/sub> + 1\/3 P12<\/sub>\n1\/3 P10<\/sub> = 1\/10 P00<\/sub> + 1\/15 P11<\/sub>\n(1\/3 + 1\/15)P11<\/sub> = 1\/10 P01<\/sub> +  2 (1\/15) P12<\/sub>\n(1\/3 + 2 (1\/15)) P12<\/sub> = 1\/10 P02<\/sub>\n\nP00<\/sub> + P01<\/sub> + P02<\/sub> + P10<\/sub> + P11<\/sub> + P12<\/sub> = 1<\/code><\/pre>\n\n\n\n
reordenando para hacer la matriz que resuelve el sistema de ecuaciones:<\/p>\n\n\n\n
1\/10 P00<\/sub> - 1\/15 P01<\/sub> = 0\n    (1\/15 + 1\/10) P01<\/sub> - 2 (1\/15) P02<\/sub> - 1\/3 P10<\/sub> = 0\n(2 (1\/15) + 1\/10) P02<\/sub> - 1\/3 P11<\/sub> - 1\/3 P12<\/sub> = 0\n1\/10 P00<\/sub>  - 1\/3 P10<\/sub> + 1\/15 P11<\/sub>= 0\n             1\/10 P01<\/sub> -  (1\/3+1\/15) P11<\/sub> +  2 (1\/15) P12<\/sub> = 0\n             1\/10 P02<\/sub> - (1\/3 + 2 (1\/15)) P12<\/sub> = 0\n\nP00<\/sub> + P01<\/sub> + P02<\/sub> + P10<\/sub> + P11<\/sub> + P12<\/sub> = 1<\/code><\/pre>\n\n\n\n
\n\n\n\n
Resolver usando matrices, en python:<\/p>\n\n\n
\nimport numpy as np\nA=np.array([\n    [1\/10,        -1\/15,        0,   0,   0,   0],\n    [   0,(1\/15 + 1\/10),    -2\/15,-1\/3,   0,   0],\n    [   0,            0,2\/15+1\/10,   0,-1\/3,-1\/3],\n    [1\/10,            0,        0,-1\/3,1\/15,   0],\n    [   0,         1\/10,        0,   0,-(1\/3+1\/15),2\/15],\n    [   1,            1,        1,   1,   1,   1]\n    ])\n\n# Sustituyendo la ultima ecuaci\u00f3n por la de suma de probabilidades,\n# resolviendo por matrices y usando el vector de ceros excepto el \u00faltimo\nk=len(A)\nB=np.zeros(k,dtype=int)\nB[-1]=1\nPncalc=np.linalg.solve(A,B)\nprint('A= ')\nprint(A)\nprint('B=',B)\nprint('Solucion P:')\nprint(Pncalc)\n<\/pre><\/div>\n\n\n
A= \n[[ 0.1     -0.0666667  0.        0.         0.         0.       ]\n [ 0.       0.1666667 -0.133333 -0.3333333  0.         0.       ]\n [ 0.       0.         0.233333  0.        -0.3333333 -0.3333333]\n [ 0.1      0.         0.       -0.3333333  0.0666667  0.       ]\n [ 0.       0.1        0.        0.        -0.4        0.1333333]\n [ 1.       1.         1.        1.         1.         1.       ]]\nB= [0 0 0 0 0 1]\nSoluci\u00f3n P:\n[ 0.2259007 0.3388510 0.2044791 0.0876339 0.099318  0.0438169]<\/code><\/pre>\n\n\n\n
e) Encuentre la probabilidad que los t\u00e9cnicos est\u00e9n ocupados.<\/p>\n\n\n\n
Ocupados al menos uno e independiente si esta o no ocupada la recepcionista<\/p>\n\n\n\n
P01<\/sub>+P02<\/sub>+P11<\/sub>+P12<\/sub> =<\/p>\n\n\n\n
 0.33885102 + 0.20447907 + 0.0993184 + 0.04381694 = \n 0.68646543<\/code><\/pre>\n\n\n\n
f) \u00bfCu\u00e1l es la probabilidad que una llamada se pierda en la recepci\u00f3n?<\/p>\n\n\n\n
para ello la operadora tiene que estar ocupada<\/p>\n\n\n\n
P10<\/sub> + P11<\/sub> + P12<\/sub> =<\/p>\n\n\n\n
0.08763389 + 0.0993184  + 0.04381694 =\n0.23076923000000002<\/code><\/pre>\n\n\n\n
g) \u00bfCu\u00e1l es la tasa de clientes satisfechos? (salida del sistema, throughput)<\/p>\n\n\n\n
es la tasa de clientes que la operadora logra transferir a los t\u00e9cnicos:<\/p>\n\n\n\n
P10<\/sub> \u03bcR<\/sub> + P11<\/sub> \u03bcR =<\/sub><\/p>\n\n\n\n
0.08763389 * (1\/3) + 0.0993184 *(1\/3) =\n0.06231742999999999 <\/code><\/pre>\n\n\n\n
o tambi\u00e9n la tasa ponderada de partida de los atendidos por los t\u00e9cnicos:<\/p>\n\n\n\n
P01<\/sub> \u03bcT<\/sub> + P02<\/sub> 2 \u03bcT<\/sub> + P11<\/sub> \u03bcT<\/sub> + P12<\/sub> 2\u03bcT<\/sub> =<\/p>\n\n\n\n
0.33885102*(1\/15)+0.20447907*(2\/15)+0.0993184*(1\/15)+0.04381694 *(2\/15) =\n0.06231742933333334<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2017TI_T3 Call Center Operadora y Dos T\u00e9cnicos Tema 3 Usamos dos d\u00edgitos para representar (operadora,t\u00e9cnicos) en cantidades de ocupado=0,1,2 Los estados de operadora libre ser\u00e1n: (00),(01),(02)Los estados de operadora ocupada ser\u00e1n: (10),(11),(12) Diagrama de Transici\u00f3n Ecuaciones de Balanceo reemplazando reordenando para hacer la matriz que resuelve el sistema de ecuaciones: Resolver usando matrices, en […]<\/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-429","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\/429","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=429"}],"version-history":[{"count":5,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/429\/revisions"}],"predecessor-version":[{"id":23542,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/429\/revisions\/23542"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=429"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=429"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=429"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}