{"id":8062,"date":"2022-01-25T19:30:32","date_gmt":"2022-01-26T00:30:32","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=8062"},"modified":"2026-04-05T20:48:59","modified_gmt":"2026-04-06T01:48:59","slug":"s2eva2021paoii_t1-promedio-de-precipitacion-de-lluvia-en-area","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2021paoii_t1-promedio-de-precipitacion-de-lluvia-en-area\/","title":{"rendered":"s2Eva2021PAOII_T1 Promedio de precipitaci\u00f3n de lluvia en \u00e1rea"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2021PAOII_T1 Promedio de precipitaci\u00f3n de lluvia en \u00e1rea<\/a><\/p>\n\n\n\n

Los datos de la tabla corresponden a las medidas de precipitaci\u00f3n lluviosa en cada cuadr\u00edcula. El \u00e1rea observada es de 300x250, los tramos horizontales 6 y los verticales 4.<\/p>\n\n\n\n

Los tama\u00f1os de paso son:
h = 300\/6 = 50
k = 250\/4 = 62.5<\/p>\n\n\n\n

i<\/mark><\/strong> \\ j<\/mark><\/strong><\/th>1<\/mark><\/th>2<\/mark><\/th>3<\/mark><\/th>4<\/mark><\/th>5<\/mark><\/th>6<\/mark><\/th><\/tr><\/thead>
1<\/strong><\/mark><\/td>0.02<\/td>0.36<\/td>0.82<\/td>0.65<\/td>1.7<\/td>1.52<\/td><\/tr>
2<\/strong><\/mark><\/td>3.15<\/td>3.57<\/td>6.25<\/td>5<\/td>3.88<\/td>1.8<\/td><\/tr>
3<\/strong><\/mark><\/td>0.98<\/td>0.98<\/td>2.4<\/td>1.83<\/td>0.04<\/td>0.01<\/td><\/tr>
4<\/strong><\/mark><\/td>0.4<\/td>0.04<\/td>0.03<\/td>0.03<\/td>0.01<\/td>0.08<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Para obtener el integral doble, usando la forma compuesta de Simpson, primero desarrolla por filas.<\/p>\n\n\n\n

i=1<\/p>\n\n\n I_1 = \\frac{3}{8}(50) ( 0.02+3(0.36)+3(0.82)+0.65) <\/span>\n\n\n + \\frac{1}{3}(50) (0.65+4(1.7)+1.52) = 228.43 <\/span>\n\n\n\n

i=2<\/p>\n\n\n I_2 = \\frac{3}{8}(50) ( 3.15+3(3.57)+3(6.25)+5) <\/span>\n\n\n + \\frac{1}{3}(50) (5+4(3.88)+1.8) = 1077.18 <\/span>\n\n\n\n

i=3<\/p>\n\n\n I_3 = \\frac{3}{8}(50) ( 0.98+3(0.98)+3(2.4)+1.83) <\/span>\n\n\n + \\frac{1}{3}(50) (1.83+4(0.04)+0.01) = 276.14 <\/span>\n\n\n\n

i=4<\/p>\n\n\n I_4 = \\frac{3}{8}(50) ( 0.4+3(0.04)+3(0.03)+0.03) <\/span>\n\n\n + \\frac{1}{3}(50) (0.03+4(0.01)+0.08) = 14.5 <\/span>\n\n\n\n

con los resultados, luego se desarrolla por columnas:<\/p>\n\n\n\n

columnas = [ 228.43,  1077.18,  276.14,  14.5 ]<\/p>\n\n\n I_{total} = \\frac{3}{8}(62.5) ( 228.43+3(1077.18)+3(276.14)+14.5) <\/span>\n\n\n I_{total} = 100850.09<\/span>\n\n\n\n

para el promedio se divide el integral total para el \u00e1rea de rect\u00e1ngulo.<\/p>\n\n\n Precipitacion_{promedio} = \\frac{100850.09}{(300)(250)} = 1.34<\/span>\n\n\n\n

\n\n\n\n

Soluci\u00f3n con algoritmo<\/h2>\n\n\n\n

Resultados usando algoritmos con Python:<\/p>\n\n\n\n

integral por fila: [ 228.4375     1077.1875      276.14583333   14.5       ]\nintegral total: 100850.09765625\npromedio precipitaci\u00f3n:  1.34466796875\n>>>\n<\/code><\/pre>\n\n\n\n
Instrucciones con Python<\/h2>\n\n\n
\n# 2Eva_2021PAOII_T1 Promedio de precipitaci\u00f3n de \n#  lluvia en \u00e1rea\nimport numpy as np\n\ndef simpson_compuesto(vector,h):\n    m = len(vector)\n    suma = 0\n    j = 0\n    while (j+3)<m:  # Simpson 3\/8\n        f = vector[j:j+4]\n        s = (3*h\/8)*(f[0]+3*f[1]+3*f[2]+f[3])\n        suma = suma + s\n        j = j + 3\n    while (j+2)<m: # Simpson 1\/3\n        f = vector[j:j+3]\n        s = (h\/3)*(f[0]+4*f[1]+f[2])\n        suma = suma + s\n        j = j + 2\n    while (j+1)<m: # Trapecio\n        f = vector[j:j+2]\n        s = (h\/2)*(f[0]+f[1])\n        suma = suma + s\n        j = j + 1\n    return(suma)\n    \n\n# INGRESO\nA = [[0.02, 0.36, 0.82, 0.65, 1.7 , 1.52],\n     [3.15, 3.57, 6.25, 5.  , 3.88, 1.8 ],\n     [0.98, 0.98, 2.4 , 1.83, 0.04, 0.01],\n     [0.4 , 0.04, 0.03, 0.03, 0.01, 0.08]]\nbase = 300\naltura = 250\n\nh = base\/6\nk = altura\/4\n\n# PROCEDIMIENTO\nA = np.array(A)\n[n,m] = np.shape(A)\ncolumna = np.zeros(n,dtype=float)\nfor i in range(0,n,1):\n    vector = A[i]\n    integralfila = simpson_compuesto(vector,h)\n    columna[i] = integralfila\n\ntotal = simpson_compuesto(columna,k)\npromedio = total\/(base*altura)\n    \n# SALIDA\nprint('integral por fila:', columna)\nprint('integral total:', total)\nprint('promedio precipitaci\u00f3n: ', promedio)\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2021PAOII_T1 Promedio de precipitaci\u00f3n de lluvia en \u00e1rea Los datos de la tabla corresponden a las medidas de precipitaci\u00f3n lluviosa en cada cuadr\u00edcula. El \u00e1rea observada es de 300x250, los tramos horizontales 6 y los verticales 4. Los tama\u00f1os de paso son:h = 300\/6 = 50k = 250\/4 = 62.5 i \\ j 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":[49],"tags":[58,54],"class_list":["post-8062","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8062","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=8062"}],"version-history":[{"count":4,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8062\/revisions"}],"predecessor-version":[{"id":23903,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/8062\/revisions\/23903"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=8062"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=8062"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=8062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}