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Ejercicios de optimización
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Problema 1. Encuentre el mínimo la función
y=2x2+16x
por cada uno de los siguientes métodos:
a) Método analítico
b) Método de la sección dorada
c) Método de Newton
d) Método de Quasi-Newton
e) Método de la secante
f) Método de interpolación cuadrática
Método Analítico
y=2x2+16x
dydx
=4 x−16x2
=0
4 x=16x2
x3=4
k x1 f1 s x3 f3 t s' x2 f2 x4 f4 error
1 1.4 -15.3485714 0.1 1.5 -15.1666667 0.618034 0.1618034 1.6618034 -15.1512755
2 1.5 -15.1666667 1.6618034 -15.1512755 0.26180339 1.92360679 -15.7182344 1.7618034 -15.289506
3 1.7618034 -15.289506 1.6618034 -15.1512755 1.5 -15.1666667 1.6 -15.12 1.12107127
4 1.6618034 -15.1512755 1.6 -15.12 1.5 -15.1666667 1.5618034 -15.123027 0.02001581
5 1.5618034 -15.123027 1.6 -15.12 1.6618034 -15.1512755 1.6236068 -15.1268008 0.02494785
6 1.6236068 -15.1268008 1.6 -15.12 1.5618034 -15.123027 1.58541019 -15.1190764 0.05109048
7 1.6 -15.12 1.58541019 -15.1190764 1.5618034 -15.123027 1.5763932 -15.119783 0.00467348
8 1.5763932 -15.119783 1.58541019 -15.1190764 1.6 -15.12 1.590983 -15.1191295 0.0043227
9 1.590983 -15.1191295 1.58541019 -15.1190764 1.5763932 -15.119783 1.58196601 -15.1192302 0.00066655
x=3√4x=1.587401
Método directo Sección DoradaEncuentre el mínimo de la función f(x)=2x^2+16/x
Métodos indirectos: Newton
Encuentre el mínimo de la función f(x)=2x^2+16/x
f(x)=2x^2+16/x f'(x)=4x-16/x^2 f''(x)=4+32/x^3
k xk f'xk f''xk error
1 1.4 -2.5632653115.661807
6
2 1.56366344 -0.2891974912.369897
426.612267
9
3 1.58704257 -0.0043027112.005422
33.0359203
7
4 1.58740097 -9.7159E-0712.000001
2 0.0451756
5 1.58740105 -4.885E-14 12 1.0201E-05
Métodos indirectos: Quasi Newtonh= 0.001k xk f(x) x+h x-h f(x-h) f(x+h) error
1 1.4 15.3485714 1.401 1.399 15.3511425 15.346016 2 1.56316364 15.1226137 1.56416364 1.56216364 15.1229153 15.1223245 1.479213333 1.58652745 15.1190572 1.58752745 1.58552745 15.1190737 15.1190527 0.021640614 1.58690078 15.1190541 1.58790078 1.58590078 15.1190661 15.1190541 9.2744E-065 1.5869011 15.1190541 1.5879011 1.5859011 15.1190661 15.1190541 1.2821E-08
Métodos indirectos: La secante
f(x)=2x^2+16/x f'(x)=4x-16/x^2
k xp f'xp xq f'xq error1 1 -12 1.4 -2.56326531
2 1.5086505
2 -0.9951934 7.20183486
3 1.5546491
1 -0.40136126 2.95877662
4 1.5738422
9 -0.16411094 1.21951051
5 1.5817989
2 -0.067464 0.5030115
6 1.5850882
8 -0.02779375 0.20751944
7 1.5864465
8 -0.01146058 0.085618758 1.5870072 -0.00472742 0.03532562
9 1.5872385
4 -0.00195032 0.0145752210 1.587334 -0.00080467 0.0060137
11 1.5873733
9 -0.000332 0.00248124
12 1.5873896
4 -0.00013698 0.00102376
13 1.5873963
4 -5.6518E-05 0.0004224
Métodos directosInterpolación Cuadráticaf(x)=2x^2+16/x
k x1 x2 x3 f1 f2 f3 x* f* error1 1.4 1.5 1.6 15.3485714 15.1666667 15.12 1.58450704 15.1191029 2 1.4 1.5 1.58450704 15.3485714 15.1666667 15.1191029 1.58358719 15.11914 0.058086433 1.4 1.5 1.58358719 15.3485714 15.1666667 15.11914 1.58353241 15.1191425 0.003459334 1.4 1.5 1.58353241 15.3485714 15.1666667 15.1191425 1.58352915 15.1191427 0.000206055 1.4 1.5 1.58352915 15.3485714 15.1666667 15.1191427 1.58352896 15.1191427 1.2273E-05
Problema 2. Determine los puntos críticos de la función
f (x1 , x2 )=27 x1−19x13−2x2
2+x24
y clasifíquelos para determinar si son máximos o mínimos relativos.
∇ f=⌊
dfd x1dfd x2
⌋= 0
dfd x1
=27−13x12
dfd x2
=−4 x2+4 x23
x1=[9 ,−9 ]
x2=[1 ,−1,0 ]
Di=⌊
∂2 f∂ x1
2=−23x1
∂2 f∂ x2∂ x1
=0
∂2 f∂ x1∂ x2
=0 ∂2 f∂ x2
2=−4+12 x22
⌋
D1=⌊−23x1 ⌋ D2=⌊
−23x1 0
0 −4+12 x22
⌋
Para x1=9 y x2=1
D1=−6D 2=−48
Para x1=9 y x2=−1
D1=−6D 2=−48
Para x1=9 y x2=0
D1=−6D 2=24 Máximo
Para x1=−9 y x2=1
D1=6D2=48 Mínimo relativo
Para x1=−9 y x2=−1
D1=6D2=48 Mínimo relativo
Para x1=−9 y x2=0
D1=6D2=−24 Máximo
Problema 3 Minimizar la función
f (x1 x2 )=x13+x23−3x1 x2
por cada uno de los siguientes métodos:
a) Método de Newton
b) Método de Gradiente.
Método del gradientef(x1,x2)=x1^3+x2^3-3x1x2 alpha= 0.05
df/dx1= 3x1^2-3x2 df/dx2= 3x2^2-3x1
k x1 x2 df/dx1 df/dx2 x1(i+1) x2(i+1) ||dF||1 1.4 1.8 0.48 5.52 1.376 1.524 5.540830262 1.376 1.524 1.108128 2.839728 1.3205936 1.3820136 3.048278663 1.3205936 1.3820136 1.08586157 1.76810397 1.26630052 1.2936084 2.074918554 1.26630052 1.2936084 0.92972583 1.22136652 1.21981423 1.23254008 1.53496785
5 1.21981423 1.23254008 0.76622004 0.89802242 1.18150323 1.18763895 1.180481866 1.18150323 1.18763895 0.62493277 0.68694917 1.15025659 1.1532915 0.928676557 1.15025659 1.1532915 0.50939618 0.53947405 1.12478678 1.12631779 0.741968148 1.12478678 1.12631779 0.41648253 0.43141497 1.10396265 1.10474704 0.599647049 1.10396265 1.10474704 0.34195949 0.34951013 1.08686468 1.08727154 0.48897201
10 1.08686468 1.08727154 0.28200988 0.28588415 1.07276419 1.07297733 0.4015710711 1.07276419 1.07297733 0.233537 0.2355485 1.06108734 1.06119991 0.3316965912 1.06108734 1.06119991 0.19411928 0.19517371 1.05138137 1.05144122 0.2752727313 1.05138137 1.05144122 0.1618847 0.1624418 1.04328714 1.04331913 0.2293338114 1.04328714 1.04331913 0.13538676 0.13568301 1.0365178 1.03653498 0.1916753815 1.0365178 1.03653498 0.1135025 0.11366089 1.03084267 1.03085193 0.1606288216 1.03084267 1.03085193 0.09535405 0.09543911 1.02607497 1.02607998 0.1349111517 1.02607497 1.02607998 0.0802496 0.08029546 1.02206249 1.02206521 0.113522518 1.02206249 1.02206521 0.06763959 0.06766438 1.01868051 1.01868199 0.0956743619 1.01868051 1.01868199 0.05708399 0.05709744 1.01582631 1.01582711 0.0807384620 1.01582631 1.01582711 0.04822794 0.04823525 1.01341491 1.01341535 0.0682097821 1.01341491 1.01341535 0.04078331 0.04078729 1.01137575 1.01137599 0.0576791222 1.01137575 1.01137599 0.03451475 0.03451692 1.00965001 1.00965014 0.0488127723 1.00965001 1.00965014 0.02922901 0.02923019 1.00818856 1.00818863 0.041336924 1.00818856 1.00818863 0.02476663 0.02476727 1.00695023 1.00695027 0.0350257625 1.00695023 1.00695027 0.02099549 0.02099584 1.00590046 1.00590048 0.0296923626 1.00590046 1.00590048 0.01780575 0.01780594 1.00501017 1.00501018 0.0251812727 1.00501017 1.00501018 0.01510577 0.01510588 1.00425488 1.00425489 0.0213628728 1.00425488 1.00425489 0.01281893 0.01281899 1.00361393 1.00361394 0.0181287529 1.00361393 1.00361394 0.01088097 0.010881 1.00306988 1.00306989 0.0153880430 1.00306988 1.00306989 0.00923792 0.00923794 1.00260799 1.00260799 0.013064431 1.00260799 1.00260799 0.00784437 0.00784438 1.00221577 1.00221577 0.0110936232 1.00221577 1.00221577 0.00666204 0.00666204 1.00188267 1.00188267 0.0094215533 1.00188267 1.00188267 0.00565864 0.00565864 1.00159974 1.00159974 0.0080025234 1.00159974 1.00159974 0.00480689 0.00480689 1.00135939 1.00135939 0.0067979635 1.00135939 1.00135939 0.00408372 0.00408372 1.00115521 1.00115521 0.0057752536 1.00115521 1.00115521 0.00346962 0.00346962 1.00098172 1.00098172 0.0049067937 1.00098172 1.00098172 0.00294807 0.00294807 1.00083432 1.00083432 0.004169238 1.00083432 1.00083432 0.00250505 0.00250505 1.00070907 1.00070907 0.0035426839 1.00070907 1.00070907 0.00212872 0.00212872 1.00060263 1.00060263 0.0030104640 1.00060263 1.00060263 0.00180899 0.00180899 1.00051218 1.00051218 0.002558341 1.00051218 1.00051218 0.00153734 0.00153734 1.00043532 1.00043532 0.0021741242 1.00043532 1.00043532 0.00130652 0.00130652 1.00036999 1.00036999 0.001847743 1.00036999 1.00036999 0.00111038 0.00111038 1.00031447 1.00031447 0.0015703244 1.00031447 1.00031447 0.00094371 0.00094371 1.00026729 1.00026729 0.0013346145 1.00026729 1.00026729 0.00080207 0.00080207 1.00022718 1.00022718 0.001134346 1.00022718 1.00022718 0.0006817 0.0006817 1.0001931 1.0001931 0.0009640747 1.0001931 1.0001931 0.0005794 0.0005794 1.00016413 1.00016413 0.000819448 1.00016413 1.00016413 0.00049246 0.00049246 1.0001395 1.0001395 0.0006964549 1.0001395 1.0001395 0.00041857 0.00041857 1.00011858 1.00011858 0.0005919550 1.00011858 1.00011858 0.00035577 0.00035577 1.00010079 1.00010079 0.0005031351 1.00010079 1.00010079 0.00030239 0.00030239 1.00008567 1.00008567 0.00042765
52 1.00008567 1.00008567 0.00025702 0.00025702 1.00007282 1.00007282 0.0003634953 1.00007282 1.00007282 0.00021846 0.00021846 1.00006189 1.00006189 0.0003089654 1.00006189 1.00006189 0.00018569 0.00018569 1.00005261 1.00005261 0.0002626155 1.00005261 1.00005261 0.00015783 0.00015783 1.00004472 1.00004472 0.0002232156 1.00004472 1.00004472 0.00013416 0.00013416 1.00003801 1.00003801 0.0001897357 1.00003801 1.00003801 0.00011403 0.00011403 1.00003231 1.00003231 0.0001612658 1.00003231 1.00003231 9.6925E-05 9.6925E-05 1.00002746 1.00002746 0.0001370759 1.00002746 1.00002746 8.2386E-05 8.2386E-05 1.00002334 1.00002334 0.0001165160 1.00002334 1.00002334 7.0027E-05 7.0027E-05 1.00001984 1.00001984 9.9033E-05
Método de Newton
f(x1,x2)=x1^3+x2^3-3x1x2
df/dx1= 3x1^2-3x2 df/dx2= 3x2^2-3x1i x df/dx1 df/dx2 H inv H-1 ||df||
1 1.4 1.8 0.48 5.52 8.4 -30.1321585
90.0367107
25.5408302
6
-3 10.80.0367107
20.1027900
1
21.133920
7 1.2149780.2123945
71.0267523
16.8035242
3 -30.1795675
80.0738974
61.0484902
3
-37.2898678
40.0738974
60.1675877
3
31.019907
11.027211
50.0389972
80.1057687
8 6.1194428 -3 0.21463010.1044722
10.1127289
8
-36.1632687
90.1044722
1 0.2131039
41.000487
21.000597
6 0.00113140.0021248
96.0029234
6 -30.2220337
70.1109505
80.0024073
3
-36.0035855
50.1109505
80.2220092
9
51.000000
31.000000
3 7.1141E-07 1.0702E-066.0000016
6 -30.2222221
20.1111110
2 1.2851E-06
-3 6.00000190.1111110
20.2222221
1
Problema 4 Obtenga el máximo valor de la función por el método simplex
Z=f (x1 , x2)=6 x1+5 x2
s . a :2x1+5 x2≤20
−5 x1−x2≤−5
−3 x1−11 x2≤−33
Convertir las desigualdades en igualdades
2 x1+5 x2+s1=20
−5 x1−x2+s2=−5
−3 x1−11 x2+s3=−33
Igualar la función objetivo a cero y después agregar las variables de holgura del sistema anterior:
Z−6 x1−5 x2=0
Tablero Inicial
Base
Variable de decisión
Variable de holgura Solución
X1 X2 S1 S2 S3
S1 2 5 1 0 0 20
S2 -5 -1 0 1 0 -5
S3 -3 -11 0 0 1 -33
Z -6 -5 0 0 0 0
Tablero Inicial
Base
Variable de decisión
Variable de holgura Solución
X1 X2 S1 S2 S3
S1 2 5 1 0 0 20
S2 -5 -1 0 1 0 -5
S3 1 -11/3 0 0 1/6 -11
Z -1 -5 0 0 0 0