{"id":1352,"date":"2016-10-30T07:43:02","date_gmt":"2016-10-30T12:43:02","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/estg1003\/?p=1352"},"modified":"2026-04-05T16:40:00","modified_gmt":"2026-04-05T21:40:00","slug":"s1eva2017tii_t1-codigo-morse-con-markov-2","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/stp-ejemplos\/s1eva2017tii_t1-codigo-morse-con-markov-2\/","title":{"rendered":"s1Eva2017TII_T1 C\u00f3digo Morse con Markov"},"content":{"rendered":"\n

Ejercicio<\/strong>: 1Eva2017TII_T1 C\u00f3digo Morse con Markov<\/a><\/p>\n\n\n\n

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

s={0,1,2} = {'.- '}<\/code><\/p>\n\n\n\n

b) Realice el diagrama de estados<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

c) Escriba la matriz de transici\u00f3n y ubique los valores encontrados en el diagrama.<\/p>\n\n\n\n

\"\"<\/a><\/figure>\n\n\n\n
\"\"<\/a><\/figure>\n\n\n\n

d) Calcule la probabilidad de estado estable o largo plazo.<\/p>\n\n\n\n

Ecuaciones:<\/p>\n\n\n\n

0,2961 \u03c00<\/sub> + 0,4032 \u03c01<\/sub> + 0,4348 \u03c02<\/sub> = \u03c00<\/sub>\n0,2856 \u03c00<\/sub> + 0,2609 \u03c01<\/sub> + 0,2842 \u03c02<\/sub> = \u03c01<\/sub>\n0,4181 \u03c00<\/sub> + 0,3358 \u03c01<\/sub> + 0,2809 \u03c02<\/sub> = \u03c02<\/sub>\n\u03c00<\/sub> + \u03c01<\/sub> + \u03c02<\/sub> = 1<\/code><\/pre>\n\n\n\n
Tomando 3 ecuaciones, siempre usando que la suma de probabilidades es 1<\/p>\n\n\n\n
(0,2961-1) \u03c00<\/sub> + 0,4032 \u03c01<\/sub> + 0,4348 \u03c02<\/sub> = 0\n0,2856 \u03c00<\/sub> + (0,2609-1) \u03c01<\/sub> + 0,2842 \u03c02<\/sub> = 0\n0,4181 \u03c00<\/sub> + 0,3358 \u03c01<\/sub> + (0,2809-1) \u03c02<\/sub> = 0\n\u03c00<\/sub> + \u03c01<\/sub> + \u03c02<\/sub> = 1<\/code><\/pre>\n\n\n\n
-0,703817809\t0,403205408\t0,434822831\t\t0\n0,285650584\t-0,73908635\t0,284203177\t\t0\n0,418167225\t0,335880942\t-0,719026008\t\t0\n       1\t       1\t       1\t\t1<\/code><\/pre>\n\n\n\n
       1\t       1\t       1\t\t1\n0,285650584\t-0,73908635\t0,284203177\t\t0\n0,418167225\t0,335880942\t-0,719026008\t\t0<\/code><\/pre>\n\n\n\n
 1\t       1\t       1\t\t1\n0\t3,587379095\t0,005067055\t\t1\n0\t0,196778413\t2,719470024\t\t1<\/code><\/pre>\n\n\n\n
1\t1\t1\t\t   1\n0\t1\t0,001412467\t0,278755039\n0\t0\t1\t\t0,347583767<\/code><\/pre>\n\n\n\n
1\t0\t0\t\t0,3742\n0\t1\t0\t\t0,2783\n0\t0\t1\t\t0,3476<\/code><\/pre>\n\n\n\n
\u03c0 = [0,3742  0,2783  0,3476]<\/code><\/pre>\n\n\n\n
\u03c00<\/sub> = 0,3742
\u03c01<\/sub> = 0,2783
\u03c02<\/sub> = 0,3476<\/p>\n\n\n\n
\n\n\n\n
Usando Python:<\/p>\n\n\n\n
la suma de filas es: \n[123343  91720 114580]\nla matriz de transici\u00f3n es: \n[[ 0.29618219  0.28565058  0.41816722]\n [ 0.40320541  0.26091365  0.33588094]\n [ 0.43482283  0.28420318  0.28097399]]\nen estado estable o largo plazo: \n[[ 0.37415214  0.27826409  0.34758377]\n [ 0.37415214  0.27826409  0.34758377]\n [ 0.37415214  0.27826409  0.34758377]]\n>>> <\/code><\/pre>\n\n\n
\n# 1ra Evaluaci\u00f3n II T\u00e9rmino 2017\n# Tema 1. C\u00f3digo Morse-Cadena Markov\nimport numpy as np\n# Ingreso\nconteo = np.array([[36532, 35233, 51578],\n                   [36982, 23931, 30807],\n                   [49822, 32564, 32194]])\n# Procedimiento\nn = len(conteo)\np = np.zeros(shape=(n,n),dtype=float)\n\nsumafila = np.sum(conteo, axis=1)\nfor f in range(0,n,1):\n    p[f] = conteo[f,:]\/sumafila[f]\n\nk=50\npn = np.linalg.matrix_power(p,k)\n\n# Salida\nprint('la suma de filas es: ')\nprint(sumafila)\nprint('la matriz de transici\u00f3n es: ')\nprint(p)\nprint('en estado estable o largo plazo: ')\nprint(pn)\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2017TII_T1 C\u00f3digo Morse con Markov a) Determine el espacio de estados s={0,1,2} = {'.- '} b) Realice el diagrama de estados c) Escriba la matriz de transici\u00f3n y ubique los valores encontrados en el diagrama. d) Calcule la probabilidad de estado estable o largo plazo. Ecuaciones: Tomando 3 ecuaciones, siempre usando que la suma […]<\/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-1352","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\/1352","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=1352"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1352\/revisions"}],"predecessor-version":[{"id":23540,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/1352\/revisions\/23540"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=1352"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=1352"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=1352"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}