{"id":10647,"date":"2025-01-29T06:10:13","date_gmt":"2025-01-29T11:10:13","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=10647"},"modified":"2026-04-05T20:40:25","modified_gmt":"2026-04-06T01:40:25","slug":"s2eva2024paoii_t2-mayoria-entre-grupos-azules-y-rojos","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2024paoii_t2-mayoria-entre-grupos-azules-y-rojos\/","title":{"rendered":"s2Eva2024PAOII_T2 EDO Mayor\u00eda entre grupos Azules y Rojos"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2024PAOII_T2 EDO Mayor\u00eda entre grupos Azules y Rojos<\/a><\/p>\n\n\n\n

literal a<\/h2>\n\n\n\n

La poblaci\u00f3n anual del pa\u00eds se describe con x(t), con tasas de natalidad a = 0.018 y mortalidad b = 0.012,<\/p>\n\n\n f(t,x,,y) = \\frac{\\delta x}{\\delta t} = 0.018 x - 0.012 x^2 <\/span>\n\n\n\n

La poblaci\u00f3n de Rojos es minor\u00eda y se describe con y(t). Los valore iniciales son: x(0)=2 , y(0)=0.5 y tama\u00f1o de paso h=0.5<\/p>\n\n\n g(t,x,y) = \\frac{\\delta y}{\\delta t} = 0.026x - 0.017 y^2 +0.19 (0.012) (x-y) <\/span>\n\n\n\n

el algoritmo de Runge-Kutta para sistemas de ecuaciones aplicado al ejercicio:<\/p>\n\n\n K1x = h f(t,x,y) = h (0.018 x - 0.012 x^2) <\/span>\n\n\n K1y = h g(t,x,y) = h \\Big(0.026x - 0.017 y^2 +0.19 (0.012) (x-y)\\Big) <\/span>\n\n\n K2x = h f(t+h,x+K1x,y+K1y) <\/span>\n\n\n = h (0.018 (x+K1x) - 0.012 (x+K1x)^2) <\/span>\n\n\n K2y = h g(t+h,x+K1x,y+K1y) <\/span>\n\n\n = h \\Big(0.026(x+K1x) - 0.017 (y + K1y)^2<\/span>\n\n\n+0.19 (0.012) ((x+K1x)-(y + K1y))\\Big) <\/span>\n\n\n x[i+1] = x[i] + \\frac{K1x+K2x}{2} <\/span>\n\n\n y[i+1] = y[i] + \\frac{K1y+K2y}{2} <\/span>\n\n\n t[i+1] = t[i] + h <\/span>\n\n\n\n

literal b<\/h2>\n\n\n\n

Desarrolle tres iteraciones con expresiones completas para x(t), y(t) con tama\u00f1o de paso h=0.5.<\/p>\n\n\n\n

itera = 0<\/p>\n\n\n K1x = 0.5 (0.018 (2) - 0.012 (2)^2) = -0.006 <\/span>\n\n\n K1y = 0.5 \\Big(0.026(2) - 0.017 (0.5)^2 +0.19 (0.012) ((2)-(0.5))\\Big) <\/span>\n\n\n= 0.006085<\/span>\n\n\n K2x = 0.5 (0.018 (2-0.006) - 0.012 (2-0.006)^2) = -0.00591 <\/span>\n\n\n K2y = 0.5 \\Big(0.026(2-0.006) - 0.017 (0.5 + 0.006085)^2 <\/span>\n\n\n+0.19 (0.012) ((2-0.006)-(0.5 + 0.006085))\\Big) = 0.006098 <\/span>\n\n\n x[1] = 2 + \\frac{-0.006+-0.00591}{2} = 1.9940 <\/span>\n\n\n y[1] = 0.5 + \\frac{0.006085+0.006098}{2} = 0.5060<\/span>\n\n\n t[1] = 0 + 0.5 =0.5<\/span>\n\n\n\n

itera = 1<\/p>\n\n\n K1x = 0.5 (0.018 (1.9940) - 0.012 (1.9940)^2) <\/span>\n\n\n=-0.005911 <\/span>\n\n\n K1y = 0.5 \\Big(0.0261.9940 - 0.017 (0.5060)^2 +0.19 (0.012) (1.9940-y)\\Big)<\/span>\n\n\n= 0.006098 <\/span>\n\n\n K2x = 0.5 (0.018 (1.9940-0.005911) - 0.012 (1.9940-0.005911)^2<\/span>\n\n\n= -0.005823 <\/span>\n\n\n K2y = 0.5 \\Big(0.026(1.9940-0.005911) - 0.017 (0.5060 + 0.006098)^2 <\/span>\n\n\n +0.19 (0.012) ((1.9940-0.005911)-(0.5060 + 0.006098))\\Big)<\/span>\n\n\n = 0.006111 <\/span>\n\n\n x[2] = 1.9940 + \\frac{-0.005911-0.005823}{2} = 1.988178 <\/span>\n\n\n y[2] = 0.5060 + \\frac{0.006098+0.006111}{2} = 0.512196 <\/span>\n\n\n t[2] = 0.5 + 0.5 = 1 <\/span>\n\n\n\n

itera = 2<\/p>\n\n\n K1x = 0.5 (0.018 (1.988178) - 0.012 (1.988178)^2) =-0.005824 <\/span>\n\n\n K1y = 0.5 \\Big(0.026(0.512196) - 0.017 (0.512196)^2<\/span>\n\n\n+0.19 (0.012) (1.988178-0.512196)\\Big) = 0.006111 <\/span>\n\n\n K2x = 0.5 (0.018 (1.988178-0.005824) - 0.012 (1.988178-0.005824)^2)<\/span>\n\n\n=-0.005737 <\/span>\n\n\n K2y = 0.5 \\Big(0.026(0.512196+0.006111) - 0.017 (0.512196 + 0.006111 )^2 <\/span>\n\n\n +0.19 (0.012) ((1.988178-0.005911)-(0.512196 + 0.006111)\\Big)<\/span>\n\n\n=0.006124<\/span>\n\n\n x[3] = (1.988178) + \\frac{-0.005824-0.005737}{2} = 1.982398 <\/span>\n\n\n y[3] = 0.512196 + \\frac{0.006111+0.006124}{2} = 0.518314 <\/span>\n\n\n t[3] = 1 + 0.5 = 1.5<\/span>\n\n\n\n

literal c<\/h2>\n\n\n\n

Los resultados para x(t) de las primeras iteraciones indican que la poblaci\u00f3n total del pa\u00eds disminuye en el intervalo observado. Se puede comprobar al usar el algoritmo para mas iteraciones como se muestra en la gr\u00e1fica.<\/p>\n\n\n\n

\"Mayor\u00eda<\/figure>\n\n\n\n

literal d<\/h2>\n\n\n\n

Los resultados para y(t) muestran que la poblaci\u00f3n clasificada como Rojo aumenta en el intervalo observado. Sin embargo la mayor\u00eda sigue siendo Azul para las tres iteraciones realizadas.<\/p>\n\n\n\n

literal e<\/h2>\n\n\n\n

La poblaci\u00f3n y(t) alcanza la mitad de la poblaci\u00f3n cuanto t cambia de 30.5 a 31. Tiempo en el que de realizar elecciones, ganar\u00edan los Rojos.<\/p>\n\n\n\n

Usando el algoritmo, se a\u00f1ade una columna con la condici\u00f3n que MayoriaY = yi>xi\/2, que se convierte a 1 o verdadero cuando se cumple la condici\u00f3n. Con lo que se encuentra el cambio de mayor\u00eda a Rojo.<\/p>\n\n\n\n

Runge-Kutta Segundo Orden\ni  [ ti,  xi,  yi ]\n   [ K1x,  K1y,  K2x,  K2y , MayoriaY]\n0 [0.  2.  0.5]\n   [0. 0. 0. 0. 0.]\n1 [0.5      1.994045 0.506092]\n  [-0.006     0.006085 -0.00591   0.006098  0.      ]\n2 [1.       1.988178 0.512196]\n  [-0.005911  0.006098 -0.005823  0.006111  0.      ]\n3 [1.5      1.982398 0.518314]\n  [-0.005824  0.006111 -0.005737  0.006124  0.      ]\n4 [2.       1.976702 0.524443]\n\u2026\n   [-0.002761  0.005927 -0.002727  0.005907  0.      ]\n60 [30.        1.755801  0.869617]\n   [-0.002728  0.005907 -0.002695  0.005887  0.      ]\n61 [30.5       1.753123  0.875494]\n   [-0.002695  0.005887 -0.002662  0.005867  0.      ]\n62 [31.        1.750476  0.88135 ]\n   [-0.002663  0.005867 -0.002631  0.005846  1.      ]<\/strong>\n63 [31.5       1.747861  0.887185]\n   [-0.002631  0.005846 -0.002599  0.005824  1.      ]\n64 [32.        1.745277  0.892998]\n   [-0.002599  0.005824 -0.002568  0.005802  1.      ]\n65 [32.5       1.742724  0.898789]\n   [-0.002568  0.005802 -0.002538  0.00578   1.      ]\n66 [33.        1.740201  0.904558]\n   [-0.002538  0.00578  -0.002508  0.005757  1.      ]\n67 [33.5       1.737708  0.910303]\n   [-0.002508  0.005757 -0.002478  0.005734  1.      ]<\/code><\/pre>\n\n\n\n
 <\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2024PAOII_T2 EDO Mayor\u00eda entre grupos Azules y Rojos literal a La poblaci\u00f3n anual del pa\u00eds se describe con x(t), con tasas de natalidad a = 0.018 y mortalidad b = 0.012, La poblaci\u00f3n de Rojos es minor\u00eda y se describe con y(t). Los valore iniciales son: x(0)=2 , y(0)=0.5 y tama\u00f1o de paso h=0.5 […]<\/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-10647","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\/10647","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=10647"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/10647\/revisions"}],"predecessor-version":[{"id":23884,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/10647\/revisions\/23884"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=10647"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=10647"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=10647"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}