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APLICACIÓN DE LA CADENA DE MARKOV PARA MAXIMIZAR LAS GANANCIAS Y MINIMIZAR LOS COSTOS EN EL TRASLADO DE MERCADERIA.

Enviado por   •  25 de Marzo de 2018  •  1.026 Palabras (5 Páginas)  •  515 Visitas

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DESARROLLO DEL PROBLEMA

[pic 10]

[pic 11]

d_5

F_5 (S_5 )=R(S_5,d_5 )+ 〖F_6〗^∗ (S_6)

〖F_5〗^∗ (S_5)

〖d5〗^∗

S_5

d5=N

L

121.87 + 0

121.87

N

M

6.77 + 0

6.77

N

[pic 12]

d_4

F_4 (S_4 )=R(S_4,d_4 )+ 〖F_5〗^∗ (S_5)

〖F_4〗^∗ (S_4)

〖d_4〗^∗

S_4

d4=L

d4=M

I

24.37 + 121.87 = 146.24

97.50 + 6.77 = 104.27

104.27

M

J

151.66 + 121.87 = 273.53

54.17 + 6.77 = 60.94

60.94

M

K

78.54 + 121.87 = 200.41

24.37 + 6.77 = 31.14

31.14

M

[pic 13]

d_3

F_3 (S_3 )=R(S_3,d_3 )+ 〖F_4〗^∗ (S_4)

〖F_3〗^∗ (S_3)

〖d_3〗^∗

S_3

d_3=I

d_3=J

d_3=K

E

14.90 + 104.27 = 119.17

119.16 + 60.94 = 180.10

48.75 + 31.14 = 79.89

79.89

K

F

78.54 + 104.27 = 182.81

48.75 + 60.94 = 109.69

93.44 + 31.14 = 124.58

109.69

J

G

51.46+ 104.27= 155.73

51.46 + 60.94 = 112.40

29.79 + 31.14 = 60.93

60.93

K

H

59.58 + 104.27 = 163.85

65 + 60.94 = 125.94

36.56 + 31.14= 67.70

67.7

K

[pic 14]

d_2

F_2 (S_2 )=R(S_2,d_2 )+ 〖F_3〗^∗ (S_3)

〖F_2〗^∗ (S_2)

S_2

S_2

d_2=E

d_2=F

d_2=G

d_2=H

B

10.83 + 79.89 = 90.72

60.94 + 109.69 = 170.63

23.02 + 60.93 = 83.95

17.60 + 67.70 = 85.30

83.95

G

C

16.25 + 79.89 =96.14

16.25 + 109.69= 125.94

6.77 + 60.93 = 67.70

6.77 + 67.70 = 74.47

67.7

G

D

16.25 + 79.89 = 96.14

21.67 + 109.69 = 131.36

8.13 + 60.93 = 69.06

8.13 + 67.70 = 75.83

69.06

G

[pic 15]

d_1

F_1 (S_1 )=R(S_1,d_1 )+ 〖F_2〗^∗ (S_2)

〖F_1〗^∗ (S_1)

(S)1

S_1

d_1=B

d_1=C

d_1=D

A

10.83 + 83.95 =94.78

17.60 + 67.70 = 85.30

27.08 + 69.06 = 96.14

85.3

C

[pic 16]

Etapas

d1

R1

1

C

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