INSTITUTO POLITECNICO NACIONAL

ESFM

OPTIMIZACION NO LINEAL JUAREZ VALENCIA JESUS GERARDO

Optimizacin no lineal Tarea 2

Prof. Adriana Lara Lpez

16/abril/2015 Fecha lmite para la entrega: mircoles 29 de abril, 21:00 hrs.

Instrucciones Deben desarrollarse los puntos de esta tarea y entregarse un reporte con lo que se pide. El reporte deber entregarse de manera electrnica, usando la plataforma Moodle, en un solo archivo con formato PDF (si est en cualquier otro formato se considerara como no entregada, con calificacin de cero).

1. Informacin de primer orden (20 puntos) Encontrar los puntos ptimos globales (dentro del intervalo correspondiente) para cada uno de los siguientes problemas. Aplicar las condiciones de optimalidad de primer y segundo orden (teoremas 1 y 2 vistos en clase), utilizando cada uno de los 3 mtodos: (a) Mtodo de Newton, (b) Mtodo de biseccin y (c) Mtodo de la secante, para encontrar las races de la funcin derivada. Explique sus razonamientos, justifique la pertinencia o no de aplicar los teoremas mencionados en cada caso, y los pasos que uso para obtener sus resultados. . () = + + + , [ ] SOLUCION Encontremos la primera derivada de la funcion: () = 283 + 62 + 3 () 283 + 62 + 3 . , .

+1 = ()

() 0 = 10

1 = 10 28(10)3+6(10)2+3

84(10)2+12(10)= 6.69118357

2 = 10 28(1)

3+6(1)2+3

84(1)2+12(1)= 4.48593309

3 = 10 28(2)

3+6(2)2+3

84(2)2+12(2)= -3.01704788

4 = 10 28(3)

3+6(3)2+3

84(3)2+12(3)= -2.04047674

5 = 10 28(4)

3+6(4)2+3

84(4)2+12(4)= -1.3951434

6 = 10 28(5)

3+6(5)

2+3

84(5)2

+12(5)= -0.97706306

7 = 10 28(6)

3+6(6)2+3

84(6)2+12(6)= -0.72307961

8 = 10 28(7)

3+6(7)2+3

84(7)2+12(7)=-0.59685058

9 = 10 28(8)

3+6(8)2+3

84(8)2+12(8)=-0.56100545

10 = 10 28(9)

3+6(9)2+3

84(9)2+12(9)=-0.55819303

11 = 10 28(10)

3+6(10)2+3

84(10)2+12(10)=-0.55817635

12 = 10 28(11)

3+6(11)2+3

84(11)2+12(11)=-0.55817635

= 1 [1 2

(1) (2)] (1) 0 = 10 1 = 10

2 = 10 [(10 + 10)(28(10)3 + 6(10)2 + 3)

((28(10)3 + 6(10)2 + 3) (28(10)3 + 6(10)2 + 3))] = .21535714

3 = 2 [(.21535714 10)(28(2)

3 + 6(2)2 + 3)

((28(2)3 + 6(2)2 + 3) (28(1)3 + 6(1)2 + 3))] = .21642819

4 = 3 [(.21642819 + .21535714)(28(3)

3 + 6(3)2 + 3)

((28(3)3 + 6(3)2 + 3) (28(2)3 + 6(2)2 + 3))] = 2.479302

5 = 4 [(2.479302 + .21642819)(28(4)

3 + 6(4)2 + 3)

((28(4)3 + 6(4)2 + 3) (28(3)3 + 6(3)2 + 3))] = .23382579

6 = 5 [(.23382579 + 2.479302)(28(5)

3 + 6(5)2 + 3)

((28(5)3 + 6(5)2 + 3) (28(4)3 + 6(4)2 + 3))] = .25093472

7 = 6 [(.25093472 + .23382579)(28(6)

3 + 6(6)2 + 3)

((28(6)3 + 6(6)2 + 3) (28(5)3 + 6(5)2 + 3))] = 1.69812445

8 = 7 [(1.69812445 + .25093472)(28(7)

3 + 6(7)2 + 3)

((28(7)3 + 6(7)2 + 3) (28(6)3 + 6(6)2 + 3))] = 1.69812445

9 = .28641123

10 = .31985498

11 = .9795784

12 = .40769816

13 = .46877386

14 = .600352

15 = .54975278

16 = .5574651

17 = .5581891

18 = .55817663

19 = .55817635

20 = .55817635

21 = .55817635

= +

2

() () < 0 [, ] () () > 0 [,] () () = 0 Intervalo [-10,10]

1=10 + 10

2= 0

(10) (0) = (27397)(3) < 0 [10,0]

2=10 + 0

2= 5

(10) (5) = (27397)(3347) > 0 [5,0]

3. 3=5 + 0

2= 2.5

(5) (2.5) = (3347)(397) > 0 [2.5,0]

4. 4=2.5 + 0

2= 1.25

(2.5) (1.25) = (397) (667

16) > 0 [1.25,0]

5. 5=1.25 + 0

2= .625

(1.25) (.625) = (667

16) (

191

128) > 0 [

5

8, 0]

6. 6=0.625 0

2= 0.3125

(5

8) (

5

16) = (

191

128) (

2797

1024) < 0 [

5

8,

5

16]

7. 7=

58

516

2= .

15

32

(5

8) (

15

32) = (

191

128) (1.434448242) < 0 [

5

8,

15

32]

8. 8=

58

1532

2= .546875

(5

8) (0.546875) = (

191

128) (0.2148895264) < 0 [

5

8, .546875]

9. 9=

58 , .546875

2= .5859375

(5

8) (0.5859375) > 0 [.5859375, .546875]

10. 10= .56640625 (.56640625) (0.5859375) < 0 [.56640625, .546875] 11. 11= .55664063

12. 12= .55152364 13. 13= .55908203 14. 14= .55786133 15. 15= .55847168 16. 16= .5581665 17. 17= .55831909 18. 18= .5582428 19. 19= .55820565 20. 20= .55818858 21. 21= .55817604 22. 22= .55818081 23. 23= .55817842 24. 24= .55817723 25. 25= .55817664 26. 26= .55817634 27. 27= .55817649

28. 28= .55817641 29. 29= .55817338 30. 30= .55817636 31. 31= .55817635 32. 32= .55817635 33. 33= .55817635 34. 34= .55817635 35. 35= .55817635

Encontramos los valores critico = .55817635 veamos si es un mximo o minimo con la segunda derivada

() = 842 + 12 (.55817635 ) = 84(.55817635 )2 + 12(.55817635 ) = 19.4732 > 0

= .55817635

. () = + + +

, [ ], .

SOLUCION Eliminemos el termino de abajo multiplicando por x2, entonces: 7x6 + 2x5 + 3x3 + 2 encontremos la primera derivada de la funcion: () = 425 + 104 + 92 () () = 2(423 + 102 + 9) 423 + 102 + 9 . , .

+1 = ()

() 0 = 10

1 = 10 42(10)3 + 10(10)2 + 9

126(10)2 + 20(10)= 6.69427

2 = 6.69427 42(6.69427)3 + 10(6.69427)2 + 9

126(6.69427)2 + 20(6.69427)= 4.49158

3 = 4.49158 42(4.49158)3 + 10(4.49158)2 + 9

126(4.49158)2 + 20(4.49158)= 3.02548

4 = 3.02548 42(3.02548)3 + 10(3.02548)2 + 9

126(3.02548)2 + 20(3.02548)= 2.05314

5 = 2.05314 42(2.05314)3 + 10(2.05314)2 + 9

126(2.05314)2 + 20(2.05314)= 1.41580

6 = 1.41580 42(1.41580)3 + 10(1.41580)2 + 9

126(1.41580)2 + 20(1.41580)= 1.01380

7 = 0.78963 8 = 0.70291 9 = 0.68952 10 = 0.68921 11 = 0.68921

= 1 [1 2

(1) (2)] (1) 0 = 10 1 = 10

2 = 10 [10 + 10

(42(10)3 + 10(10)2 + 9) (42(10)3 + 10(10)2 + 9)] 42(10)3 + 10(10)2 + 9

= .24024 3= .24024

[.24024 10

(42(.24024)3 + 10(.24024)2 + 9) (42(10)3 + 10(10)2 + 9)] 42(.24024)3

+ 10(.24024)2 + 9 = .24239 4= .24239

[.24239 + .24024

(42(.24239)3 + 10(.24239)2 + 9) (42(.24024)3 + 10(.24024)2 + 9)] 42(.24239)3

+ 10(.24239)2 + 9 = 3.82260 5= 3.82260

[3.82260 + .24239

(42(3.82260)3 + 10(3.82260)2 + 9) (42(.24239)3 + 10(.24239)2 + 9)] 42(3.82260)3

+ 10(3.82260)2 + 9 = .25701 6= .25701

[.25701 + 3.82260

(42(.25701)3 + 10(.25701)2 + 9) (42(3.82260)3 + 10(3.82260)2 + 9)] 42(3.82260)3

+ 10(.25701)2 + 9 = .27151 7= .27151

[.27151 + .25701

(42(.27151)3 + 10(.27151)2 + 9) (42(.25701)3 + 10(.25701)2 + 9)] 42(.27151)3

+ 10(.27151)2 + 9 = 2.80175

8= 2.80175

[2.80175 + .27151

(42(2.80175)3 + 10(2.80175)2 + 9) (42(.27151)3 + 10(.27151)2 + 9)] 42(2.80175)3

+ 10(2.80175)2 + 9 = .29815

9= .29815

[.29815 + 2.80175

(42(.29815)3 + 10(.29815)2 + 9) (42(2.80175)3 + 10(2.80175)2 + 9)] 42(.29815)3

+ 10(.29815)2 + 9 = .32415 10= .32415

[.32415 + .29815

(42(.32415)3 + 10(.32415)2 + 9) (42(.29815)3 + 10(.29815)2 + 9)] 42(.32415)3

+ 10(.32415)2 + 9 = 1.76502 11 = .38644 12 = .44238 13 = .98939 14 = .57948 15 = .64527 16 = .69889 17 = .68847 18 = .68919 19 = .689211 20 = .689211

= +

2

() () < 0 [, ] () () > 0 [,] () () = 0

1. 1=10 + 10

2= 0

(10) (0) = (40991)(9) = 368929 < 0 [10,0]

2. 2=10 + 0

2= 5

(10) (5) = (40991)(4991) > 0 [5,0]

3. 3=5 + 0

2= 2.5

(5) (2.5) = (4491)(584.75) > 0 [2.5,0]

4. 4=2.5 + 0

2= 1.25

(2.5) (1.25) = (584.75)(57.406) > 0 [1.25,0]

5. 5=1.25 + 0

2= .625

(1.25) (.625) = (57.406)(2.65) < 0 [1.25, .625]

6. 6=1.25 .625

2= .9375

(1.25) (.9375) = (57.406)(16.817) > 0 [.9375, .625]

7. 7=.9375 .625

2= .78125

(.9375) (.78125) = (16.817)(4.9236) > 0 [.78125, .625]

8. 8=.78125 .625

2= .703125

(.78125) (.703125) = (4.9236)(.6559) > 0 [.703125, .625]

9. 9=.703125 .625

2= .6640625

(.703125) (.6640625) = (.6559)(1.1106) < 0 [.703125, .66406]

10. 10=.703125 .66406

2= .683593

(.703125) (.683593) = (.6559)(. 2563) < 0 [.703125, .683593]

11. 11=.703125 .683593

2= .693359

(.703125) (.693359) = (.6559)(.1924) > 0 [.693359, .683593]

12. 12=.693359 .683593

2= .688476

(.693359) (.688476) = (.1924)(. 03383) < 0 [.693359, .688476]

13. 13=.693359 .688476

2= .689211

(.693359) (.689211) = (.1924)(. 00001) = 0 13= .689211 Encontramos los valores critico = .689211, = 0 veamos si es un mximo o minimo con la segunda derivada

() = 1262 + 20

(0) = 126(0)2 + 20(0) = 0 = 0 (.689211) = 126(.689211)2 + 20(.689211) = 46.06 > 0

= .689211

. () = + , [ ] SOLUCION Encontremos la primera derivada de la funcion: (x) = 6x2 () () = 0 , .

+1 = ()

() 0 = 5

1 = 5 6(5)2

12(5)= 2.5

2 = 2.5 6(2.5)2

12(2.5)= 1.25

3 = 1.25 6(1.25)2

12(1.25)= 0.625

4 = 0.625 6(0.625)2

12(0.625)= 0.3125

5 = 0.3125 6(0.3125)2

12(0.3125)= 0.15625

6 = 0.15625 6(0.15625)2

12(0.15625)= 0.078125

7 = 0.078125 6(0.078125)2

12(0.078125)= 0.0390625

8 = 0.0390625 6(0.0390625)2

12(0.0390625)= 0.01953125

9 = 0.00979563 10 = 0.00488281

11 = 0.0024141 12 = 0.0012207 13 = 0.00061035 14 = 0.00030518 15 = 0.00015259 16 = 0.000076294 17 = 0.000038147 18 = 0.000019073 19 = 0.0000095367 20 = 0.0000047684 21 = 0.0000023842 22 = 0.0000011921 23 = 0.00000059605 24 = 0.00000029802 25 = 0.00000014901 26 = 0.000000074506

= +

2

() () < 0 [, ] () () > 0 [,] () () = 0

1. 1=5 + 10

2= 2.5

(5) (2.5) = (150)(37.5) > 0 [2.5,10]

2. 2=2.5 + 10

2= 6.25

(2.5) (6.25) = (37.5)(234.375) > 0 [6.25,10]

3. 3=6.25 + 10

2= 8.125

(6.25) (8.125) = (396.09375)(234.375) > 0 [8.125,10]

4. 4=8.125 + 10

2= 9.0625

(9.0625) (8.125) = (396.09375)(492.7734374) > 0 [9.0625,10]

5. 5=9.0625 + 10

2= 9.53125

(9.0625) (9.53125) = (545.0683594)(492.7734374) > 0 [9.53125,10]

4. 4=9.765625 (9.0625) (8.125) = (396.09375)(492.7734374) > 0 [9.765625,10]

5. 5=9.8828125 (9.0625) (8.125) > 0 [9.882812510]

6. 6=9.94140625 (9.0625) (8.125) > 0 [9.94140625, 10]

7. 7=9.97070313 (9.0625) (8.125) > 0 [9.97070313, 10]

8. 8=9.98535156 (9.0625) (8.125) > 0 [9.98535156, 10]

9. 9=9.99267578 (9.0625) (8.125) > 0 [9.9926757, 10]

10. 10=9.99633789 (9.0625) (8.125) > 0 [9.99633789, 10]

11. 11=9.99816895 (9.0625) (8.125) > 0 [9.99816895, 10]

12. 12=9.99908447 (9.0625) (8.125) > 0 [9.99908447, 10]

13. 13=9.99954224 (9.0625) (8.125) > 0 [9.99954224, 10]

14. 14=9.99977112 (9.0625) (8.125) > 0 [9.99977112, 10]

15. 15=9.99988556E-3 (9.0625) (8.125) > 0 [9.99988556, 10]

16. 16=9.99994278E-4 (9.0625) (8.125) > 0 [9.99994278, 10]

17. 17=9.99997139E-4 (9.0625) (8.125) > 0 [9.99997139, 10]

18. 18=9.99998569E-6 (9.0625) (8.125) > 0 [9.99998569, 10]

19. 19=9.99999285E^-6 (9.0625) (8.125) > 0 [9.99999285, 10]

= 1 [1 2

(1) (2)] (1) 0 = 5 1 = 10

2 = 10 [5 600

600 150] = 10

3 = 10 [0 600

600 600] =

Tomemos un intervalo ms pequeos 0 = 2 1 = 5

2 = 5 [7 150

150 24] = 3.33333333

3 = 10

3

253

2003

2003 150

= 10

4 = 10 [

203 600

600 200

3

] = 2.5

5 = 2.5 [7.5 37.5

37.5 600] = 2

6 = 2 [0.5 24

24 37.5] = 1.11111111

7 =-0.71428571 8 =-0.43478261 9 =-0.27027027 10 =-0.16666667 11 =-0.10309278 12 =-0.06369427 13 =-0.03937008 14 =-0.0243309 15 =-0.01503759 16 =-0.00929368 17 =-0.00574383 18 =-0.00354988 19 =-0.00219394 20 =-0.00135593

21 =-0.00083801 22 =-0.00051792 23 =-0.00032009 24 =-0.00019783 25 =-0.00012226 26 =-7.5564E-05 Encontramos los valores critico = 0 veamos si es un mximo o minimo con la segunda derivada f(x) = 12x (0 ) = 12(0) = 0 = 0 = 0

. () = ( + ) , [ ] SOLUCION Encontremos los valores crticos de la funcin con la primera derivada (Teorema 1) () = 6 cos(2 + 1) Nos damos cuenta que para que la funcin sea cero el argumento tiene que valer

2,

3

2,

5

2,

Resolvamos las siguientes ecuaciones para encontrar las races de la funcin.

1. 2 + 1 =

2 =.28539816

2. 2 + 1 =

2 =-1.28539816

3. 2 + 1 =3

2 =1.85619449

4. 2 + 1 = 3

2 = 2.85619449

5. 2 + 1 =5

2 = 3.42699081

6. 2 + 1 = 5

2 = 4.42699081

7. 2 + 1 =7

2 = 4.99778714

8. 2 + 1 = 7

2 = 5.99778714

9. 2 + 1 =9

2 = 6.56858347

10 2 + 1 = 9

2 = 7.56858347

11. 2 + 1 =11

2 = 8.13937979

12. 2 + 1 = 11

2 = 9.13937979

13. 2 + 1 =13

2 = 9.71017612

Veamos si son mximos o mnimos con la segunda derivada (Teorema 2) () = 12sin (2 + 1)

(. 28539816) = 12 sin(2(. 28539816) + 1) = 12 < 0 = .28539816 (1.2853981) = 12 sin(2(1.2853981) + 1) = 12 > 0 = 1.28539816 (1.85619449) = 12 sin(2(1.85619449) + 1) = 12 > 0 = 1.85619449 (2.8561944) = 12 sin(2(2.8561944) + 1) = 12 < 0 = 2.8561944 (3.42699081) = 12 sin(2(3.42699081) + 1) = 12 < 0 = 3.42699081 (4.4269908) = 12 sin(2(4.4269908) + 1) = 12 > 0 = 4.42699081 (4.99778714) = 12 sin(2(4.99778714) + 1) = 12 > 0 = 4.99778714 (5.9977871) = 12 sin(2(5.9977871) + 1) = 12 < 0 = 5.99778714 (6.56858347) = 12 sin(2(6.56858347) + 1) = 12 < 0 = 6.56858347 (7.5685834) = 12 sin(2(7.5685834) + 1) = 12 > 0 = 7.56858347 (8.13937979) = 12 sin(2(8.13937979) + 1) = 12 > 0 = 8.13937979 (9.1393797) = 12 sin(2(9.1393797) + 1) = 12 < 0 = 9.13937979 (9.71017612) = 12 sin(2(9.71017612) + 1) = 12 < 0 = 9.71017612

. () = + , [ ] SOLUCION Encontremos la primera derivada de la funcion: () = 12x3 12x2 () = 12x2(x 1) () () = 0. , .

+1 = ()

() 0 = 2

1 = 2 (2) 1

1= 1

2 = 1 (1) 1

1= 1

3 = 1 (1) 1

1= 1

= +

2

() () < 0 [, ] () () > 0 [,] () () = 0

1. 1=2 + 2

2= 0

(2) (0) = (3)(1) = 3 > 0 [0,2]

2. 1=0 + 2

2= 1

(0) (1) = (1)(0) = 0 1

= 1 [1 2

(1) (2)] (1) 0 = 2 1 = 2

2 = 2 [(2 + 2)(2 1)

(2 1) (2 1)] = 1

3 = 1 [(1 2)(1 1)

(1 1) (2 1)] = 1

Encontramos los valores critico = 0 , x=1 veamos si es un mximo o minimo con la segunda derivada

f (x) = 36x2 24x (0 ) = 36(0)2 24(0) = 0 = 0 = 0 (1 ) = 36(1)2 24(1) = 12 > 0 = 1

2. Mtodos de eliminacin de regiones (30 puntos) Resolver los siguientes problemas, utilizando los mtodos a) bsqueda exhaustiva, b) divisin de intervalos por la mitad y c) Seccin dorada para aproximar el punto ptimo con una precisin de 0.01 en todos los casos. Mencione el nmero de evaluaciones de la funcin objetivo necesarias al terminar cada algoritmo. Indicar pas por paso los valores encontrados siguiendo el procedimiento a mano. . () = ( ) + [, ]. SOLUCION: DIVICION DE INTERVALOS POR LA MITAD

1. a=1, b=5, =.01, L=4, =3

1 = 2 2 = 4 (1) = 10 (2) = 12 () = 10

2. Entonces a=2, b=4, L=2, =3

1 = 2.5 2 = 3.5 (1) = 9.75 (2) = 10.75 () = 10

3. Entonces a=2, b=3, L=1, =2.5

1 = 2.25 2 = 2.75 (1) = 9.8125 (2) = 9.8125 () = 9.75

4. Entonces a=2.25, b=2.75, L=.5, =2.5

1 = 2.375 2 = 2.625 (1) = 9.765625 (2) = 9.765625

() = 9.75 5. Entonces a=2.375, b=2.625, L=.25, =2.5 1 = 2.4375 2 = 2.5625 (1) = 9.753906 (2) = 9.753906 () = 9.75

6. Entonces a=2.4375, b=2.5625, L=.125, =2.5 1 = 2.46875 2 = 2.53125 (1) = 9.750976 (2) = 9.750976 () = 9.75

7. Entonces a=2.46875, b=2.53125, L=.0625, =2.5 1 = 2.484375 2 = 2.515625 (1) = 9.750244 (2) = 9.750244 () = 9.75

8. Entonces a=2.484375, b=2.515625, L=.03125, =2.5 1 = 2.4921875 2 = 2.5078125 (1) = 9.750061 (2) = 9.750061 () = 9.75

9. Entonces a=2.484375, b=2.515625, L=.015625, =2.5 1 = 2.49609375 2 = 2.50390625 (1) = 9.750015 (2) = 9.750015 () = 9.75

Calculemos L=0.0078 (2) > (3) 2.

1 = 2.41 2 = 2.415 3 = 2.42 (1) = 9.7581 (2) = 9.757225 (3) = 9.7564 (1) > (2) > (3) 3. 1 = 2.415 2 = 2.42 3 = 2.425 (1) = 9.757225 (2) = 9.7564 (3) = 9.755625 (1) > (2) > (3) 4. 1 = 2.42 2 = 2.425 3 = 2.43 (1) = 9.7564 (2) = 9.755625 (3) = 9.7549 (1) > (2) > (3) 5. 1 = 2.425 2 = 2.43 3 =2.435 (1) = 9.755625

(2) = 9.7549 (3) = 9.754225 (1) > (2) > (3) 6. 1 = 2.43 2 = 2.435 3 = 2.44 (1) = 9.7549 (2) = 9.754225 (3) = 9.7536 (1) > (2) > (3) 7.

1 = 2.435 2 = 2.44 3 = 2.445 (1) = 9.754225 (2) = 9.7536 (3) = 9.753025 (1) > (2) > (3) 8.

1 = 2.44 2 = 2.445 3 = 2.45 (1) = 9.7536 (2) = 9.753025 (3) = 9.7525 (1) > (2) > (3) 9.

1 = 2.445 2 = 2.45 3 = 2.455 (1) = 9.753025 (2) = 9.7525 (3) = 9.752025 (1) > (2) > (3) 10.

1 = 2.45 2 = 2.455 3 = 2.46 (1) = 9.7525 (2) = 9.752025 (3) = 9.7516 (1) > (2) > (3)

11. 1 = 2.455 2 = 2.46 3 = 2.465 (1) = 9.752025 (2) = 9.7516 (3) = 9.751225 (1) > (2) > (3) 12. 1 = 2.46 2 = 2.465 3 = 2.47 (1) = 9.7516 (2) = 9.751225 (3) = 9.7509 (1) > (2) > (3) 13.

1 = 2.465 2 = 2.47 3 = 2.475 (1) = 9.751225 (2) = 9.7509 (3) = 9.750625 (1) > (2) > (3) 14.

1 = 2.47 2 = 2.475 3 = 2.48 (1) = 9.7509 (2) = 9.750625 (3) = 9.7504 (1) > (2) > (3) 15. 1 = 2.475 2 = 2.48 3 = 2.485 (1) = 9.750625 (2) = 9.7504 (3) = 9.750225 (1) > (2) > (3) 16. 1 = 2.48 2 = 2.485

3 = 2.49 (1) = 9.7504 (2) = 9.750225 (3) = 9.7501 (1) > (2) > (3) 17.

1 = 2.485 2 = 2.49 3 = 2.495 (1) = 9.750225 (2) = 9.7501 (3) = 9.750025 (1) > (2) > (3) 18. 1 = 2.49 2 = 2.495 3 = 2.5 (1) = 9.7501 (2) = 9.750025 (3) = 9.75 (1) > (2) > (3) 19. 1 = 2.495 2 = 2.5 3 = 2.55 (1) = 9.750025 (2) = 9.75 (3) = 9.7525 como (1) > (2) < (3) Entonces [2.49609375, 2.50390625] SECCION DORADA 1. a=1, b=5, = .610833, L=4 1 = 2.5566 2 = 3.4433 (1) = 9.7532 (2) = 10.6398 (1) < (2)

2. a=1, b=3.4433, = .610833, L=2.4433 1 = 1.9508 2 = 2.5566 (1) = 10.0515 (2) = 9.7532 (1) > (2)

3. a=1.9508, b=3.4433, = .610833, L=1.4924

1 = 2.5566 2 = 2.8625 (1) = 9.7532 (2) = 9.8814 (1) < (2) 4. a=1.9508, b=2.8625, = .610833, L=0.9116 1 = 2.3056 2 = 2.5566 (1) = 9.7877 (2) = 9.7532 (1) > (2) 5. a=2.3056, b=2.8625, = .610833, L=0.5568 1 = 2.5566 2 = 2.6457 (1) = 9.7532 (2) = 9.7712 (1) < (2)

6. a=2.3056, b=2.6457, = .610833, L=0.3407 1 = 2.4376 2 = 2.5566 (1) = 9.7538 (2) = 9.7532 (1) > (2)

7. a=2.4376, b=2.6457, = .610833, L=0.2081 1 = 2.5566 2 = 2.5647 (1) = 9.7532 (2) = 9.7541 (1) < (2) 8. a=2.4376, b=2.5647, = .610833, L=0.1274 1 = 2.4871 2 = 2.5566 (1) = 9.7501 (2) = 9.7532 (1) < (2) 9. a=2.4376, b=2.5566, = .610833, L=0.1190 1 = 2.4839 2 = 2.4871 (1) = 9.7502 (2) = 9.7501 (1) > (2) 10. a=2.4839, b=2.5566, = .610833, L=0.0726 1 = 2.4871

2 = 2.5283 (1) = 9.7501 (2) = 9.7508 (1) < (2)

a=2.4839, b=2.5283, = .610833, L=0.014> [2.4839, 2.5283] . () = ( + ) [. , . ]. SOLUCION DIVICION DE INTERVALOS POR LA MITAD

1. L= 2 1 = 4.4 (1) = 0.5106284975 2 = 5.4 (2) = 0.6134881555 = 4.9 ()= 0.5621439438 (1) < () 2. Entonces a=3.9, b=4.9, L = 1 1 = 4.15 (1) = 0.4848114633 2 = 4.65 (2) = 0.5364066453 = 4.4 () = 0.5106284975 (1)

(1) = 0.4654243511 2 = 4.11875 (2) = 0.4815816977 = 4.025 ()= 0.4718889781 (1)< () 5. Entonces a=3.9, b=4.025, L = 0.125 X1 = 3.93125 (1) = 0.4848114633 2= 3.99375 (2)= 0.4686569434 = 3.9625 ()= 0.4654243511 (1)< () 6. Entonces a=3.9, b=3.9625, L = 0.0625 1 = 3.915625 (1)= 0.4605744254 2 = 3.946875 (2) = 0.463807847 = 3.93125 ()= 0.462191205 (1)< () 7. Entonces a=3.9, b=3.93125, L = 0.03125 1= 3.9078125 (1) = 0.4597659842 2 = 3.9234375 (2) = 0.4613828323 = 3.915625 ()= 0.4605744254 (1)< () 8. Entonces a=3.9, b=3.9078125 L = 0.0078125 Entonces no se cumple la condicin de L > , podemos concluir que [3.9, 3.9078125] BUSQUEDA EXHAUSTIVA 1. 1= 3.9 (1) = 1.754751579 2 = 3.905

(2) = 1.73033146 3 = 3.91 (3) = 1.7057383 (1)> (2)> (3) 2

1 =3.905 (1)= 1.730331457 2 = 3.91 (2)= 1.7057383 3 = 3.915 (3)= 1.68097458 (1)> (2)> (3) 3

1 =3.91 (1)= 1.7057383 2 = 3.915 (2)= 1.68097458 3 = 3.92 (3) = 1.656042756 (1)> (2)> (3) 4 1 =3.915 (1)= 1.68097458 2 = 3.92 (2) = 1.656042756 3 = 3.925 (3)= 1.630945331 (1)> (2)> (3) 5

1 =3.92 (1)= 1.656042756 2 = 3.925 (2)= 1.630945331 3= 3.93 (3)= 1.605684813 (1)> (2)> (3)

6 1 =3.925 (1)= 1.630945331 2 = 3.93 (2) = 1.605684813 3= 3.935 (3)= 1.580263728 (1)> (2)> (3) 7 1 =3.93 (1) = 1.605684813 2= 3.935 (2) = 1.580263728 3 = 3.94 (3) = 1.554684618 (1)> (2)> (3) 8 1 =3.935 (1) = 1.580263728 2 = 3.94 (2) = 1.554684618 3 = 3.945 (3)= 1.52895004 (1)> (2)> (3) 9 1=3.94 (1) = 1.554684618 2 = 3.945 (2)= 1.52895004 3= 3.95 (3) = 1.503062569 (1)> (2)> (3) 10

1 =3.945 (1)= 1.52895004 2 = 3.95 (2)= 1.503062569 3 = 3.955 (3) = 1.477024793 (1)> (2)< (3) Entonces [3.945, 3.955] SECCION DORADA

1. L = 2

=1 + 5

2

1 = b L = 4.663932023 (1)= 0.5378420288 2 = a + L = 5.136067977 (2)= 0.5864076876 Como se cumple la condicin (1)< (2) b = 5.136067977

2 = 4.663932023 L = 1.236067977

1= 4.372135955 2 (1)= 0.507752919 (2)= 0.5378420288 b = 4.663932023 2 = 4.372135955 L = 0.763932023 1 = 4.191796068 3. (1)= 0.4891302889 (2)= 0.507752919 b = 4.372135955

2 = 4.191796068 L = 0.472135955

1= 4.080339887

4. (1) = 0.477611132 (2)= 0.4891302889 b = 4.191796068 2= 4.080339887 L = 0.291796068 1= 4.01145618 5. (1)= 0.4704882758 2= 0.477611132 b = 4.080339887

2= 4.01145618 L = 0.180339887

1 = 3.968883707 6. (1) = 0.4660847458 (2)= 0.4704882758 b = 4.01145618

2= 3.968883707 L = 0.11145618 1= 43.942572473 7. (1)= 0.4633626994 (2)= 0.4660847458 b = 3.968883707 2= 3.942572473 L = 0.068883707

1 = 3.926311235 8. (1)= 0.4616801866 (2)= 0.4633626994 b = 3.942572473 2 = 3.926311235 L = 0.042572473 1= 3.916261238 9. (1)= 0.4606402621 (2)= 0.4616801866 b = 3.926311235 2 = 3.916261238 L = 0.026311235

1 = 3.910049997 10. (1)= 0.459997525 (2)= 0.4606402621 b = 3.916261238

2= 3.910049997 L = 0.016261238

1= 3.90621124 11. (1) = 0.4596002809 (2)= 0.459997525 b = 3.910049997 2 = 3.90621124 L = 0.010049997 1= 3.903838757 12. (1)= 0.4593547663 (2)= 0.4596002809 b = 3.90621124

2= 3.903838757

L = 6.21124 x 103 1= 3.902372483 Como ya no se cumple L > decimos que el [3.9, 3.90621124]

. () = + [, ]. SOLUCION De clculo sabemos que para cambiar de maximizar a minimizar una funcin la multiplicbamos por -1 entonces

Minimizar f(x) = 3x4 7x3 en el intervalo [0, 3]. DIVICION DE INTERVALOS POR LA MITAD 1. a=0, b=3, =.01, L=3, =1.5 1 = .75 2 = 2.25 (1) = 2.003 (2) = 2.847 () = 8.4375

2. Entonces a=.75, b=2.25, L=1.5, =1.5 1 = 1.125 2 = 1.875 (1) = 5.1613 (2) = 9.0637 () = 8.4375 3. Entonces a=1.5, b=2.25, L=.75, =1.875 1 = 1.6875 2 = 2.0625 (1) = 9.3105 (2) = 7.1286 () = 9.0637 4. Entonces a=1.5, b=1.875, L=.375, =1.6875

1 = 1.59375 2 = 1.78125 (1) = 8.9819 (2) = 9.3605 () = 9.3105

5. Entonces a=1.6875, b=1.875, L=.1875, =1.78125 1 = 1.734375 2 = 1.828125 (1) = 9.3744 (2) = 9.2599 () = 9.3605 6. Entonces a=1.6875, b=1.78125, L=.09375, =1.734375 1 = 1.7109375 2 = 1.7578125 (1) = 9.3516 (2) = 9.3777 () = 9.3744 7. Entonces a=1.734375, b=1.78125, L=.046875, =1.7578125 1 = 1.74609375 2 = 1.76953125 (1) = 9.3786 (2) = 9.3717 () = 9.3777 8. Entonces a=1.734375, b=1.7578125, L=.0234375, =1.74609375 1 = 1.740234375 2 = 1.751953125 (1) = 9.3771 (2) = 9.3788 () = 9.3717 Entonces a=1.740234375, b=1.751953125, L=.011, por lo tanto el intervalo [1.740234375, 1.751953125] BUSQUEDA EXHAUSTIVA Encontremos un intervalo mas pequeo al establecido ya que n=600 entonces 1 = 1.7 , 2 = 1.8 a=1.7, b=1.8 y x=0.005 entonces 1.

1 = 1.705 2 = 1.71 3 = 1.715

(1) = 9.3429 (2) = 9.3503 (3) = 9.3569 (1) > (2) > (3) 2. 1 = 1.71 2 = 1.715 3 = 1.72 (1) = 9.3503 (2) = 9.3569 (3) = 9.3627 (1) > (2) > (3) 3. 1 = 1.715 2 = 1.72 3 = 1.725 (1) = 9.3569 (2) = 9.3627 (3) = 9.3676 (1) > (2) > (3) 4. 1 = 1.72 2 = 1.725 3 = 1.73 (1) = 9.3627 (2) = 9.3676 (3) = 9.3716 (1) > (2) > (3) 5. 1 = 1.725 2 = 1.73 3 = 1.735 (1) = 9.3676 (2) = 9.3716 (3) = 9.3748 (1) > (2) > (3) 6. 1 = 1.73 2 = 1.735 3 = 1.74 (1) = 9.3716 (2) = 9.3748 (3) = 9.3770 (1) > (2) > (3) 7.

1 = 1.735 2 = 1.74 3 = 1.745 (1) = 9.3748 (2) = 9.3770 (3) = 9.3784 (1) > (2) > (3) 8. 1 = 1.74 2 = 1.745 3 = 1.75 (1) = 9.3770 (2) = 9.3784 (3) = 9.3789 (1) > (2) > (3) 9.

1 = 1.745 2 = 1.75 3 = 1.755 (1) = 9.3784 (2) = 9.3789 (3) = 9.3784 Como (1) > (2) < (3)

[1.745, 1.755] SECCION DORADA 1. a=0, b=3, = .610833, L=3 1 = 1.1459 2 = 1.8540 (1) = 7.8125 (2) = 9.1636 (1) > (2) 2. a=1.1459, b=3, = .610833, L=1.854099 1 = 1.8540 2 = 2.2917 (1) = 9.1636 (2) = 1.5000 (1) < (2) 3. a=1.1459, b=2.2917, = .610833, L=1.1458 1 = 1.5835 2 = 1.8540 (1) = 8.9322 (2) = 9.1636 (1) > (2) 4. a=1.5835, b=2.2917, = .610833, L=0.7082

1 = 1.8540 2 = 2.0212 (1) = 9.1636 (2) = 7.7308 (1) < (2) 5. a=1.5835, b=2.0212, = .610833, L=0.4376 1 = 1.7507 2 = 1.8540 (1) = 9.3786 (2) = 9.1636 (1) < (2) 6. a=1.5835, b=1.8540, = .610833, L=0.2705 1 = 1.6869 2 = 1.7507 (1) = 9.3092 (2) = 9.3786 (1) > (2) 7. a=1.6869, b=1.8540, = .610833, L=0.1671 1 = 1.7507 2 = 1.7890 (1) = 9.3786 (2) = 9.3500 (1) < (2) 8. a=1.6869, b=1.7890, = .610833, L=0.1021 1 = 1.7266 2 = 1.7507 (1) = 9.3690 (2) = 9.3786 (1) > (2) 9. a=1.7266, b=1.7890, = .610833, L=0.0623 1 = 1.7507 2 = 1.7647 (1) = 9.3786 (2) = 9.3748 (1) < (2) 10. a=1.7266, b=1.7647, = .610833, L=0.0381 1 = 1.7414 2 = 1.7507 (1) = 9.3775 (2) = 9.3786 (1) > (2) 11. a=1.7414, b=1.7647, = .610833, L=0.0232 1 = 1.7507

2 = 1.7557 (1) = 9.3786 (2) = 9.3783 (1) < (2)

a=1.7414, b=1.7557, = .610833, L=0.014> [1.7414, 1.7557]

3. Investigacin (20 puntos) Investigue el mtodo de eliminacin de regiones basado en la sucesin de nmeros de Fibonacci, enumere y discuta las ideas principales, describa el algoritmo y d un ejemplo numrico. SOLUCION El principio de la bsqueda de Fibonacci es que de dos puntos requeridos para el uso de la propiedad de eliminacin de regiones, uno es siempre el punto previo y el otro punto es nuevo. Por ende, solo se requiere una evaluacin de la funcin a cada iteracin. En la iteracin k, se eligen dos puntos intermedios, cada uno de los cuales est a una distancia Lk* de cada extremo del espacio de bsqueda (L = b a). Cuando la propiedad de eliminacin de regiones elimina una porcin del espacio de bsqueda, el espacio de bsqueda restante es Lk. Definiendo Lk*= (Fn-k+1/Fn+1)/L y Lk = (Fn-k+2/ Fn+1)/L se puede mostrar que: Lk Lk*= Lk+1*lo que significa que uno de los dos puntos usados en la iteracin k se mantiene como uno de los puntos de referencia en la iteracin (k + 1). Puesto que los dos primeros nmeros de Fibonacci son los mismos, el algoritmo usualmente empieza con k = 2. ALGORITMO: 1: Elegir un lmite inferior a y un lmite superior b. L = b a Elegir un nmero deseado de iteraciones N k = 2

2: Lk*= (Fnk+1/Fn+1) L X1 = a + Lk*; x2 = b Lk* 3: Calcular f(x1) o f(x2) (el que no se haya evaluado antes) Usar la propiedad de eliminacin De Regiones. Establecer nuevos valores de a y b 4: Es k > N? i no, k = k + 1, GOTO Paso 2 ELSE TERMINAR

4. Programacin

(10 puntos) Escribir en forma de algoritmo (puede ser en pseudocdigo) el mtodo para la fase de acotamiento del intervalo (visto en clase), suponiendo como parmetros de entrada la funcin objetivo f(x), el incremento y el punto inicial x0, y como salida el intervalo [a, b] tal que a x b. SOLUCION. Entrada: f(x) funcin objetivo, x0 punto inicial y incremento. Salida: El intervalo [a, b] donde el ptimo est a x b. 1: Elegir un punto inicial x (0) y un incremento Hacer k = 0. 2: IF f(x (0) ||) f(x (0)) f(x (0) + ||), THEN es positivo. ELSE IF f(x (0) ||) f(x (0)) f(x (0) + ||), THEN es negativo. ELSE GOTO Paso 1. 3: x (k+1) = x (k) + 2k . 4: IF f(x (k+1)) < f(x (k) ) THEN k = k + 1 y GOTO Paso 3. ELSE a= x (k1) b= x(k+1) El mnimo se encuentra en el intervalo [a, b]. END.