{"id":8586,"date":"2022-11-25T18:00:03","date_gmt":"2022-11-25T23:00:03","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8586"},"modified":"2026-04-05T20:00:50","modified_gmt":"2026-04-06T01:00:50","slug":"s1eva2022paoii_t2-admision-universitaria-cupos-por-recursos","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva30\/s1eva2022paoii_t2-admision-universitaria-cupos-por-recursos\/","title":{"rendered":"s1Eva2022PAOII_T2 Admisi\u00f3n universitaria - cupos por recursos"},"content":{"rendered":"\n

Ejercicios<\/strong>: 1Eva2022PAOII_T2 Admisi\u00f3n universitaria - cupos por recursos<\/a><\/p>\n\n\n\n

Se requiere determinar la distribuci\u00f3n de cupos en base a los costos relativos al promedio por estudiante para docencia, infraestructura y servicios mostrados en la tabla.<\/p>\n\n\n\n

Costo referencial \/carrera<\/th>Mecatr\u00f3nica<\/strong><\/th>Computaci\u00f3n<\/strong><\/th>Civil<\/strong><\/th>Matem\u00e1ticas<\/strong><\/th><\/tr><\/thead>
Docencia<\/em><\/strong><\/td>1.5<\/td>0.9<\/td>0.6<\/td>0.7<\/td><\/tr>
Infraestructura<\/em><\/strong><\/td>0.8<\/td>1.4<\/td>0.4<\/td>0.5<\/td><\/tr>
Servicios<\/em><\/strong><\/td>0.45<\/td>0.55<\/td>1.1<\/td>0.5<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Con los datos del total de recursos relativos al promedio por estudiante disponibles son docencia 271, infraestructura 250 y servicios 230.<\/p>\n\n\n 1.5 a + 0.9 b + 0.6 c + 0.7 d = 271 <\/span>\n\n\n 0.8 a + 1.4 b + 0.4 c + 0.5 d = 250 <\/span>\n\n\n 0.45 a + 0.55 b + 1.1 c + 0.5 d = 230 <\/span>\n\n\n\n

se indica que en carreras como matem\u00e1ticas de baja demanda, se establece el cupo de 10,<\/p>\n\n\n 1.5 a + 0.9 b + 0.6 c + 0.7 (10) = 271 <\/span>\n\n\n 0.8 a + 1.4 b + 0.4 c + 0.5(10) = 250 <\/span>\n\n\n 0.45 a + 0.55 b + 1.1 c + 0.5(10) = 230 <\/span>\n\n\n\n

el sistema se convierte en:<\/p>\n\n\n 1.5 a + 0.9 b + 0.6 c = 271 - 0.7 (10) <\/span>\n\n\n 0.8 a + 1.4 b + 0.4 c = 250 - 0.5(10) <\/span>\n\n\n 0.45 a + 0.55 b + 1.1 c = 230 - 0.5(10) <\/span>\n\n\n\n

Para usar un m\u00e9todo iterativo se convierte a matriz aumentada:<\/p>\n\n\n \\begin{pmatrix} 1.5 & 0.9 & 0.6 & \\Big| & 264 \\\\ 0.8 & 1.4 & 0.4 & \\Big| & 245 \\\\ 0.45 & 0.55 & 1.1 &\\Big| & 225 \\end{pmatrix} <\/span>\n\n\n\n

con pivoteo parcial por filas, la matriz aumentada se mantiene igual, pues los valores de la diagonal ya son los mayores posibles seg\u00fan el algoritmo.<\/p>\n\n\n\n

Para un m\u00e9todo iterativo se despeja una ecuaci\u00f3n por cada inc\u00f3gnita.<\/p>\n\n\n a = \\frac{1}{1.5}(264 - 0.9 b - 0.6 c)<\/span>\n\n\n b = \\frac{1}{1.4}(245 - 0.8 a - 0.4 c) <\/span>\n\n\n c = \\frac{}{1.1}(225 -0.45 a - 0.55 b) <\/span>\n\n\n\n

Para los valores iniciales se consideran n\u00fameros mayores que cero, pues existen recursos para los cupos. No se admiten cupos negativos.<\/p>\n\n\n X_0 = [50,50,50] <\/span>\n\n\n\n

Las iteraciones para el m\u00e9todo iterativo de Gauss-Seidel<\/p>\n\n\n\n

itera = 0<\/strong><\/p>\n\n\n a = \\frac{1}{1.5}(264 - 0.9 (50) - 0.6 (50)) = 126 <\/span>\n\n\n b = \\frac{1}{1.4}(245 - 0.8 (126) - 0.4 (50)) = 88.714 <\/span>\n\n\n c = \\frac{}{1.1}(225 -0.45 (126) - 0.55 (88.714)) =108.642 <\/span>\n\n\n diferencia = [126-50, 88.714-50, 108.42-50] <\/span>\n\n\n diferencia = [76, 38.714, 58.642] <\/span>\n\n\n errado = max|[76, 38.714, 58.642]| =76 <\/span>\n\n\n X = [126, 88.714, 108.42] <\/span>\n\n\n\n

itera = 1<\/strong><\/p>\n\n\n a = \\frac{1}{1.5}(264 - 0.9 (88.714) - 0.6 (108.42)) = 79.314 <\/span>\n\n\n b = \\frac{1}{1.4}(245 - 0.8 (79.314) - 0.4 (108.42)) = 98.637 <\/span>\n\n\n c = \\frac{}{1.1}(225 -0.45 (79.314) - 0.55 (98.637)) =122.780 <\/span>\n\n\n diferencia = [79.314-126, 88, 98.637-88.714, 122.780-108.42] <\/span>\n\n\n diferencia = [46.685,9.922, 14.137] <\/span>\n\n\n errado = max| [46.685,9.922, 14.137] | = 46.685 <\/span>\n\n\n\n

el error disminuye en la iteraci\u00f3n<\/p>\n\n\n X = [79.314 , 98.637, 122.780] <\/span>\n\n\n\n

itera = 2<\/strong><\/p>\n\n\n a = \\frac{1}{1.5}(264 - 0.9 (79.314) - 0.6 (122.780)) = 67.705 <\/span>\n\n\n b = \\frac{1}{1.4}(245 - 0.8 (67.705) - 0.4 (122.780)) = 101.230 <\/span>\n\n\n c = \\frac{}{1.1}(225 -0.45 (67.705) - 0.55 (101.230)) =126.232 <\/span>\n\n\n diferencia = [67.705-79.314, 101.230-98.637, 126.232-122.780] <\/span>\n\n\n diferencia = [-11.608, 2.594, 3.451] <\/span>\n\n\n errado = max| [-11.608, 2.594, 3.451] | = 11.608 <\/span>\n\n\n\n

el error disminuye en la iteraci\u00f3n, se considera que el m\u00e9todo converge<\/strong><\/p>\n\n\n X = [67.705 , 101.230, 126.232] <\/span>\n\n\n\n

con el algoritmo se tiene como resultado:<\/p>\n\n\n\n

[126.          88.71428571 108.64285714]\n[76.         38.71428571 58.64285714]\n\n[ 79.31428571  98.63673469 122.78033395]\n[46.68571429  9.92244898 14.13747681]\n\n[ 67.7058256  101.23086138 126.23218611]\n[11.60846011  2.59412669  3.45185216]\n\n[ 64.76860873 101.92302755 127.08769174]\n[2.93721688 0.69216617 0.85550564]\n\n[ 64.01110677 102.11145563 127.30336487]\n[0.75750196 0.18842808 0.21567312]\n\n[ 63.81178067 102.16373537 127.3587675 ]\n[0.1993261  0.05227973 0.05540263]\n\n[ 63.75825178 102.17849398 127.37328637]\n[0.05352889 0.01475862 0.01451887]\n\n[ 63.74358906 102.18272443 127.37716953]\n[0.01466272 0.00423045 0.00388316]\n\n[ 63.73949753 102.18395297 127.37822907]\n[0.00409153 0.00122854 0.00105954]\n\n[ 63.73833659 102.18431364 127.37852366]\n[0.00116094 0.00036067 0.0002946 ]\n\n[ 63.73800235 102.18442047 127.37860699]\n[3.34240232e-04 1.06824300e-04 8.33224905e-05]\n\nrespuesta X: \n[[ 63.73800235]\n [102.18442047]\n [127.37860699]]\nverificar A.X=B: \n[[264.00014614]\n [245.00003333]\n [225.        ]]\n>>><\/code><\/pre>\n\n\n\n
se interpreta la respuesta como la parte entera de la soluci\u00f3n:<\/p>\n\n\n\n
cupos = [ 63, 102 , 127]<\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicios: 1Eva2022PAOII_T2 Admisi\u00f3n universitaria - cupos por recursos Se requiere determinar la distribuci\u00f3n de cupos en base a los costos relativos al promedio por estudiante para docencia, infraestructura y servicios mostrados en la tabla. Costo referencial \/carrera Mecatr\u00f3nica Computaci\u00f3n Civil Matem\u00e1ticas Docencia 1.5 0.9 0.6 0.7 Infraestructura 0.8 1.4 0.4 0.5 Servicios 0.45 0.55 1.1 […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[58,54],"class_list":["post-8586","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8586","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=8586"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8586\/revisions"}],"predecessor-version":[{"id":23833,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8586\/revisions\/23833"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8586"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8586"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8586"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}