APLICACIÓN DE LA CADENA DE MARKOV PARA MAXIMIZAR LAS GANANCIAS Y MINIMIZAR LOS COSTOS EN EL TRASLADO DE MERCADERIA.
Enviado por tolero • 25 de Marzo de 2018 • 1.026 Palabras (5 Páginas) • 445 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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