{"id":2906,"date":"2017-11-09T06:00:51","date_gmt":"2017-11-09T11:00:51","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=2906"},"modified":"2026-04-05T19:48:07","modified_gmt":"2026-04-06T00:48:07","slug":"s1eva2009ti_t1-demanda-de-un-producto-alcanza-la-produccion","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva10\/s1eva2009ti_t1-demanda-de-un-producto-alcanza-la-produccion\/","title":{"rendered":"s1Eva2009TI_T1 Demanda de un producto alcanza la producci\u00f3n"},"content":{"rendered":"\n

Ejercicio<\/strong><\/em>: 1Eva2009TI_T1 Demanda de un producto alcanza la producci\u00f3n<\/a><\/p>\n\n\n\n

Desarrollo anal\u00edtico<\/h2>\n\n\n\n

- igualar la ecuaci\u00f3n al valor buscado, 80<\/p>\n\n\n 200 t e^{-0.75t} = 80 <\/span>\n\n\n\n

- forma est\u00e1ndar de la ecuaci\u00f3n para el m\u00e9todo f(x) = 0:<\/p>\n\n\n f(t) = 200 t e^{-0.75t} - 80<\/span>\n\n\n\n

- derivada de la ecuaci\u00f3n<\/p>\n\n\n f'(t) = 200 e^{-0.75t} + 200 t (-0.75) e^{-0.75t} <\/span>\n\n\n f'(t) = 200 e^{-0.75t}(1-0.75t) <\/span>\n\n\n\n

- f\u00f3rmula del m\u00e9todo de Newton-Raphson<\/p>\n\n\n t_{i+1} = t_i - \\frac{f(t)}{f'(t)} <\/span>\n\n\n\n

- Punto inicial. Como la variable es t, tiempo, el rango de an\u00e1lisis es t>0<\/p>\n\n\n\n

El valor inicial de b\u00fasqueda se selecciona t0<\/sub> = 1<\/p>\n\n\n\n

iteraci\u00f3n 0:<\/p>\n\n\nf(1) = 200 (1) e^{-0.75(1)} - 80 =14.4733 <\/span>\n\n\n f'(1) = 200 e^{-0.75(1)}(1-0.75(1)) = 23.6183 <\/span>\n\n\n t_{1} = 1 - \\frac{14.4733}{23.6183} =0.3872 <\/span>\n\n\n error = |0.3872 - 1| = 0.6128 <\/span>\n\n\n\n

iteraci\u00f3n 1:<\/p>\n\n\n f(0.3872)= 200 (0.3872) e^{-0.75(0.3872)} - 80 = -22.0776<\/span>\n\n\n f'(0.3872 )= 200 e^{-0.75(0.3872)}(1-0.75(0.3872)) = 106.1511 <\/span>\n\n\n t_{2} = 0.3872 - \\frac{-22.0776}{106.1511} = 0.5952 <\/span>\n\n\n error = |0.5952- 0.3872| = 0.208 <\/span>\n\n\n\n

iteraci\u00f3n 2:<\/p>\n\n\n f(0.5952)=200 (0.5952) e^{-0.75(0.5952)} - 80 = -3.8242 <\/span>\n\n\n f'(0.5952) = 200 e^{-0.75(0.5952)}(1-0.75(0.5952)) = 70.855 <\/span>\n\n\n t_{3} = 0.5952 - \\frac{-3.8242}{70.855} = 0.64916 <\/span>\n\n\n\n

error = |0.64916-0.5952| = 0.053972 <\/span>
...
tabla de iteraciones<\/p>\n\n\n\n

i ['xi', 'fi', 'dfi', 'xnuevo', 'tramo']\n0 [ 1.     14.4733 23.6183  0.3872  0.6128]\n1 [  0.3872 -22.0776 106.1511   0.5952   0.208 ]\n2 [ 5.9518e-01 -3.8242e+00  7.0855e+01  6.4916e-01  5.3972e-02]\n3 [ 6.4916e-01 -2.1246e-01  6.3069e+01  6.5252e-01  3.3686e-03]\n4 [ 6.5252e-01 -7.9031e-04  6.2600e+01  6.5254e-01  1.2625e-05]\n5 [ 6.5254e-01 -1.1069e-08  6.2599e+01  6.5254e-01  1.7683e-10]\nra\u00edz en:  0.6525363029069534\n<\/code><\/pre>\n\n\n\n
Se obtiene el valor de la ra\u00edz con 5 iteraciones, y un error de 1.2625e-05<\/p>\n\n\n\n
\n\n\n\n
Desarrollo con Python<\/h2>\n\n\n\n
\"1E2009TI_T1_<\/figure>\n\n\n
\n# 1Eva_IT2009_T1 Demanda de un producto alcanza la producci\u00f3n\nimport numpy as np\n\ndef newton_raphson(fx,dfx,x0, tolera, iteramax=100,\n                   vertabla=False, precision=4):\n    '''fx y dfx en forma num\u00e9rica lambda\n    xi es el punto inicial de b\u00fasqueda\n    si no converge hasta iteramax iteraciones\n    la respuesta es NaN (Not a Number)\n    '''\n    itera=0\n    xi = x0\n    tramo = abs(2*tolera)\n    if vertabla==True:\n        print('m\u00e9todo de Newton-Raphson')\n        print('i', ['xi','fi','dfi', 'xnuevo', 'tramo'])\n        np.set_printoptions(precision)\n    while (tramo>=tolera):\n        fi = fx(xi)\n        dfi = dfx(xi)\n        xnuevo = xi - fi\/dfi\n        tramo = abs(xnuevo-xi)\n        if vertabla==True:\n            print(itera,np.array([xi,fi,dfi,xnuevo,tramo]))\n        xi = xnuevo\n        itera = itera + 1\n \n    if itera>=iteramax:\n        xi = np.nan\n        print('itera: ',itera,\n              'No converge,se alcanz\u00f3 el m\u00e1ximo de iteraciones')\n \n    return(xi)\n\n# INGRESO\nfx  = lambda t: 200*t*np.exp(-0.75*t) - 80\ndfx = lambda t: 200*np.exp(-0.75*t) + 200*t*(-0.75)*np.exp(-0.75*t)\n\n#fx  = lambda t: 200*np.exp(-0.75*t) + 200*t*(-0.75)*np.exp(-0.75*t)\n#dfx = lambda t: (112.5*t - 300.0)*np.exp(-0.75*t)\n\nx0 = 1\ntolera = 0.00001\n\n# PROCEDIMIENTO\nxi = newton_raphson(fx,dfx,x0, tolera, vertabla=True)\n# SALIDA\nprint('ra\u00edz en: ', xi)\n<\/pre><\/div>\n\n\n
Para la gr\u00e1fica se a\u00f1ade:<\/p>\n\n\n
\n# GRAFICA\nimport matplotlib.pyplot as plt\na = 0\nb = 2\nmuestras = 21\n \nxj = np.linspace(a,b,muestras)\nfj = fx(xj)\nplt.plot(xj,fj, label='f(x)')\nplt.plot(xi,0, 'o')\nplt.axhline(0)\nplt.xlabel('x')\nplt.ylabel('f(x)')\nplt.grid()\nplt.legend()\nplt.show()\n<\/pre><\/div>\n\n\n
\n\n\n\n
Literal b<\/h2>\n\n\n\n
Para el caso de encontrar el m\u00e1ximo se usar\u00eda la expresi\u00f3n fb<\/sub>(t) cuando f'(t)= 0.<\/p>\n\n\n\n
Con lo que para el algoritmo la expresi\u00f3n nueva f(t) es :<\/p>\n\n\n\n
y fb<\/sub>'(t) es f''(t)<\/code>:<\/p>\n\n\n f_b(t) = f'(t) = 200 e^{-0.75t} + 200 t (-0.75) e^{-0.75t} <\/span>\n\n\n f_b'(t) = f''(t) = (112.5t-300)e^{-0.75t} <\/span>\n\n\n\n
se desarrolla las expresiones con Sympy :<\/p>\n\n\n\n
>>> import sympy as sym\n>>> t = sym.Symbol('t')\n>>> f = 200*t*sym.exp(-0.75*t)-80\n>>> sym.diff(f,t,1)\n-150.0*t*exp(-0.75*t) + 200*exp(-0.75*t)\n>>> sym.diff(f,t,2)\n(112.5*t - 300.0)*exp(-0.75*t)\n>>> <\/code><\/pre>\n\n\n\n
Se actualiza el algoritmo con las funciones:<\/p>\n\n\n\n
fx  = lambda t: 200*np.exp(-0.75*t) + 200*t*(-0.75)*np.exp(-0.75*t)\ndfx = lambda t: (112.5*t - 300.0)*np.exp(-0.75*t)<\/code><\/pre>\n\n\n\n
y se obtiene como resultado:<\/p>\n\n\n\n
m\u00e9todo de Newton-Raphson\ni ['xi', 'fi', 'dfi', 'xnuevo', 'tramo']\n0 [  1.      23.6183 -88.5687   1.2667   0.2667]\n1 [  1.2667   3.8674 -60.9117   1.3302   0.0635]\n2 [ 1.3302e+00  1.7560e-01 -5.5445e+01  1.3333e+00  3.1671e-03]\n3 [ 1.3333e+00  4.1611e-04 -5.5183e+01  1.3333e+00  7.5406e-06]\nra\u00edz en:  1.3333333332906878<\/code><\/pre>\n\n\n\n
 <\/strong><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio: 1Eva2009TI_T1 Demanda de un producto alcanza la producci\u00f3n Desarrollo anal\u00edtico - igualar la ecuaci\u00f3n al valor buscado, 80 - forma est\u00e1ndar de la ecuaci\u00f3n para el m\u00e9todo f(x) = 0: - derivada de la ecuaci\u00f3n - f\u00f3rmula del m\u00e9todo de Newton-Raphson - Punto inicial. Como la variable es t, tiempo, el rango de an\u00e1lisis […]<\/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":[44],"tags":[58,54],"class_list":["post-2906","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva10","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2906","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=2906"}],"version-history":[{"count":5,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2906\/revisions"}],"predecessor-version":[{"id":23800,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/2906\/revisions\/23800"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=2906"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=2906"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=2906"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}