DISEÑO DE REDES TELEMÁTICAS
PROCESOS ESTOCASTICOS TALLER No. 3
CADENAS DE MARKOV
Presentado a:
Carlos Lesmes
Presentado por:
Jorge Andrés Chinome Reyes
Fredy Reinaldo Pineda Pedraza
library(expm)## Loading required package: Matrix##
## Attaching package: 'expm'## The following object is masked from 'package:Matrix':
##
## expm1. Dadas las siguientes matrices de transición de un paso de una cadena de Markov, determine las clases y clasifique los estados. Encuentre las matrices de transiciÓn de varios pasos para ver si converge o no. \[a.\begin{pmatrix} 0 & 0 & \frac{1}{3} & \frac{2}{3}\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\end{pmatrix} \; b.\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \frac{1}{2} & \frac{1}{2} & 0\\ 0 & \frac{1}{2} & \frac{1}{2} & 0\\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{pmatrix} \; c.\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\\ \frac{1}{2} & \frac{1}{2} & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}\] \[d.\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\\ \end{pmatrix} \; e.\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\ \frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3}\\ \frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0\\ \end{pmatrix} \; f.\begin{pmatrix} 0 & 0 & 1\\ \frac{1}{2} & \frac{1}{2} & 0\\ 0 & 1 & 0\\ \end{pmatrix}\] \[g.\begin{pmatrix} \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0\\ \frac{3}{4} & \frac{1}{4} & 0 & 0 & 0\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0\\ 0 & 0 & 0 & \frac{3}{4} & \frac{1}{4}\\ 0 & 0 & 0 & \frac{1}{4} & \frac{3}{4}\\ \end{pmatrix}\]
a)
a=matrix(c(0,1,0,0,0,0,1,1,1/3,0,0,0,2/3,0,0,0),nrow=4)
a## [,1] [,2] [,3] [,4]
## [1,] 0 0 0.3333333 0.6666667
## [2,] 1 0 0.0000000 0.0000000
## [3,] 0 1 0.0000000 0.0000000
## [4,] 0 1 0.0000000 0.0000000for (i in 1:5)
{
show(a%^%i)
}## [,1] [,2] [,3] [,4]
## [1,] 0 0 0.3333333 0.6666667
## [2,] 1 0 0.0000000 0.0000000
## [3,] 0 1 0.0000000 0.0000000
## [4,] 0 1 0.0000000 0.0000000
## [,1] [,2] [,3] [,4]
## [1,] 0 1 0.0000000 0.0000000
## [2,] 0 0 0.3333333 0.6666667
## [3,] 1 0 0.0000000 0.0000000
## [4,] 1 0 0.0000000 0.0000000
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.0000000 0.0000000
## [2,] 0 1 0.0000000 0.0000000
## [3,] 0 0 0.3333333 0.6666667
## [4,] 0 0 0.3333333 0.6666667
## [,1] [,2] [,3] [,4]
## [1,] 0 0 0.3333333 0.6666667
## [2,] 1 0 0.0000000 0.0000000
## [3,] 0 1 0.0000000 0.0000000
## [4,] 0 1 0.0000000 0.0000000
## [,1] [,2] [,3] [,4]
## [1,] 0 1 0.0000000 0.0000000
## [2,] 0 0 0.3333333 0.6666667
## [3,] 1 0 0.0000000 0.0000000
## [4,] 1 0 0.0000000 0.0000000a%^%10## [,1] [,2] [,3] [,4]
## [1,] 0 0 0.3333333 0.6666667
## [2,] 1 0 0.0000000 0.0000000
## [3,] 0 1 0.0000000 0.0000000
## [4,] 0 1 0.0000000 0.0000000a%^%15## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.0000000 0.0000000
## [2,] 0 1 0.0000000 0.0000000
## [3,] 0 0 0.3333333 0.6666667
## [4,] 0 0 0.3333333 0.6666667a%^%30## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.0000000 0.0000000
## [2,] 0 1 0.0000000 0.0000000
## [3,] 0 0 0.3333333 0.6666667
## [4,] 0 0 0.3333333 0.6666667a%^%60## [,1] [,2] [,3] [,4]
## [1,] 1 0 0.0000000 0.0000000
## [2,] 0 1 0.0000000 0.0000000
## [3,] 0 0 0.3333333 0.6666667
## [4,] 0 0 0.3333333 0.6666667a%^%100## [,1] [,2] [,3] [,4]
## [1,] 0 0 0.3333333 0.6666667
## [2,] 1 0 0.0000000 0.0000000
## [3,] 0 1 0.0000000 0.0000000
## [4,] 0 1 0.0000000 0.0000000En la matriz del punto a se puede observar que es periodo 3.
No es reflexiva, ya que ningún estado se relaciona con sigo mismo.
No es simétrica, ya que su matriz no corresponde a su transpuesta.
Es transitiva, ya que existe probabilidad positiva de que cada estado se comunique con otro que no se comunique de regreso.
No es irreducible, ya que no es posible acceder a cualquier estado desde cualquier punto.
No es transitorio, ya que no existen estados absorbentes donde se quede en este estado indefinidamente.
No es recurrente, ya que al ir de un estado i a j, este no regresa.
No es absorbente, ya que no existe un estado donde se quede ahí de forma infinita.
No es ergodico, ya que no es recurrente y aperiódico al mismo tiempo.
b)
b=matrix(c(1,0,0,1/2,0,1/2,1/2,0,0,1/2,1/2,0,0,0,0,1/2),nrow=4)
b## [,1] [,2] [,3] [,4]
## [1,] 1.0 0.0 0.0 0.0
## [2,] 0.0 0.5 0.5 0.0
## [3,] 0.0 0.5 0.5 0.0
## [4,] 0.5 0.0 0.0 0.5for (i in 1:5)
{
show(b%^%i)
}## [,1] [,2] [,3] [,4]
## [1,] 1.0 0.0 0.0 0.0
## [2,] 0.0 0.5 0.5 0.0
## [3,] 0.0 0.5 0.5 0.0
## [4,] 0.5 0.0 0.0 0.5
## [,1] [,2] [,3] [,4]
## [1,] 1.00 0.0 0.0 0.00
## [2,] 0.00 0.5 0.5 0.00
## [3,] 0.00 0.5 0.5 0.00
## [4,] 0.75 0.0 0.0 0.25
## [,1] [,2] [,3] [,4]
## [1,] 1.000 0.0 0.0 0.000
## [2,] 0.000 0.5 0.5 0.000
## [3,] 0.000 0.5 0.5 0.000
## [4,] 0.875 0.0 0.0 0.125
## [,1] [,2] [,3] [,4]
## [1,] 1.0000 0.0 0.0 0.0000
## [2,] 0.0000 0.5 0.5 0.0000
## [3,] 0.0000 0.5 0.5 0.0000
## [4,] 0.9375 0.0 0.0 0.0625
## [,1] [,2] [,3] [,4]
## [1,] 1.00000 0.0 0.0 0.00000
## [2,] 0.00000 0.5 0.5 0.00000
## [3,] 0.00000 0.5 0.5 0.00000
## [4,] 0.96875 0.0 0.0 0.03125b%^%10## [,1] [,2] [,3] [,4]
## [1,] 1.0000000 0.0 0.0 0.0000000000
## [2,] 0.0000000 0.5 0.5 0.0000000000
## [3,] 0.0000000 0.5 0.5 0.0000000000
## [4,] 0.9990234 0.0 0.0 0.0009765625b%^%15## [,1] [,2] [,3] [,4]
## [1,] 1.0000000 0.0 0.0 0.000000e+00
## [2,] 0.0000000 0.5 0.5 0.000000e+00
## [3,] 0.0000000 0.5 0.5 0.000000e+00
## [4,] 0.9999695 0.0 0.0 3.051758e-05b%^%30## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 0.0 0.000000e+00
## [2,] 0 0.5 0.5 0.000000e+00
## [3,] 0 0.5 0.5 0.000000e+00
## [4,] 1 0.0 0.0 9.313226e-10b%^%60## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 0.0 0.000000e+00
## [2,] 0 0.5 0.5 0.000000e+00
## [3,] 0 0.5 0.5 0.000000e+00
## [4,] 1 0.0 0.0 8.673617e-19b%^%100## [,1] [,2] [,3] [,4]
## [1,] 1 0.0 0.0 0.000000e+00
## [2,] 0 0.5 0.5 0.000000e+00
## [3,] 0 0.5 0.5 0.000000e+00
## [4,] 1 0.0 0.0 7.888609e-31Periodo=1 (Aperiódica).
Es reflexiva, en todos sus estados.
No es simétrica.
Es transitiva, en su estado [4,4].
No es irreducible.
Es transitorio, en su estado [1,1].
Es recurrente, en su estado [3,2] y [2,3].
Es absorbente, en su estado [1,1].
Es ergodico.
c)
c=matrix(c(0,0,1/2,0,0,0,1/2,0,0,0,0,1,1,1,0,0),nrow=4)
c## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 0 1
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 0.0 0.0 1 0for (i in 1:5)
{
show(c%^%i)
}## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 0 1
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 0.0 0.0 1 0
## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 1 0
## [2,] 0.0 0.0 1 0
## [3,] 0.0 0.0 0 1
## [4,] 0.5 0.5 0 0
## [,1] [,2] [,3] [,4]
## [1,] 0.5 0.5 0 0
## [2,] 0.5 0.5 0 0
## [3,] 0.0 0.0 1 0
## [4,] 0.0 0.0 0 1
## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 0 1
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 0.0 0.0 1 0
## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 1 0
## [2,] 0.0 0.0 1 0
## [3,] 0.0 0.0 0 1
## [4,] 0.5 0.5 0 0c%^%10## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 0 1
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 0.0 0.0 1 0c%^%15## [,1] [,2] [,3] [,4]
## [1,] 0.5 0.5 0 0
## [2,] 0.5 0.5 0 0
## [3,] 0.0 0.0 1 0
## [4,] 0.0 0.0 0 1c%^%30## [,1] [,2] [,3] [,4]
## [1,] 0.5 0.5 0 0
## [2,] 0.5 0.5 0 0
## [3,] 0.0 0.0 1 0
## [4,] 0.0 0.0 0 1c%^%60## [,1] [,2] [,3] [,4]
## [1,] 0.5 0.5 0 0
## [2,] 0.5 0.5 0 0
## [3,] 0.0 0.0 1 0
## [4,] 0.0 0.0 0 1c%^%100## [,1] [,2] [,3] [,4]
## [1,] 0.0 0.0 0 1
## [2,] 0.0 0.0 0 1
## [3,] 0.5 0.5 0 0
## [4,] 0.0 0.0 1 0Periodo=3.
No es reflexiva.
No es simétrica.
Es transitiva.
No es irreducible.
No es transitorio.
No es recurrente.
No es absorbente.
No es ergodico.
d)
d=matrix(c(0,0,1,1,0,0,0,1,0),nrow=3)
d## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0for (i in 1:5)
{
show(d%^%i)
}## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0
## [,1] [,2] [,3]
## [1,] 0 0 1
## [2,] 1 0 0
## [3,] 0 1 0
## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1
## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0
## [,1] [,2] [,3]
## [1,] 0 0 1
## [2,] 1 0 0
## [3,] 0 1 0d%^%10## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0d%^%15## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1d%^%30## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1d%^%60## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1d%^%100## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0Periodo=3.
No es reflexiva.
No es simétrica.
Es transitiva.
Es irreducible.
No es transitorio.
No es recurrente.
No es absorbente.
No es ergodico.
e)
e=matrix(c(0,1/3,1/3,1/3,1/3,0,1/3,1/3,1/3,1/3,0,1/3,1/3,1/3,1/3,0),nrow=4)
e## [,1] [,2] [,3] [,4]
## [1,] 0.0000000 0.3333333 0.3333333 0.3333333
## [2,] 0.3333333 0.0000000 0.3333333 0.3333333
## [3,] 0.3333333 0.3333333 0.0000000 0.3333333
## [4,] 0.3333333 0.3333333 0.3333333 0.0000000for (i in 1:5)
{
show(e%^%i)
}## [,1] [,2] [,3] [,4]
## [1,] 0.0000000 0.3333333 0.3333333 0.3333333
## [2,] 0.3333333 0.0000000 0.3333333 0.3333333
## [3,] 0.3333333 0.3333333 0.0000000 0.3333333
## [4,] 0.3333333 0.3333333 0.3333333 0.0000000
## [,1] [,2] [,3] [,4]
## [1,] 0.3333333 0.2222222 0.2222222 0.2222222
## [2,] 0.2222222 0.3333333 0.2222222 0.2222222
## [3,] 0.2222222 0.2222222 0.3333333 0.2222222
## [4,] 0.2222222 0.2222222 0.2222222 0.3333333
## [,1] [,2] [,3] [,4]
## [1,] 0.2222222 0.2592593 0.2592593 0.2592593
## [2,] 0.2592593 0.2222222 0.2592593 0.2592593
## [3,] 0.2592593 0.2592593 0.2222222 0.2592593
## [4,] 0.2592593 0.2592593 0.2592593 0.2222222
## [,1] [,2] [,3] [,4]
## [1,] 0.2592593 0.2469136 0.2469136 0.2469136
## [2,] 0.2469136 0.2592593 0.2469136 0.2469136
## [3,] 0.2469136 0.2469136 0.2592593 0.2469136
## [4,] 0.2469136 0.2469136 0.2469136 0.2592593
## [,1] [,2] [,3] [,4]
## [1,] 0.2469136 0.2510288 0.2510288 0.2510288
## [2,] 0.2510288 0.2469136 0.2510288 0.2510288
## [3,] 0.2510288 0.2510288 0.2469136 0.2510288
## [4,] 0.2510288 0.2510288 0.2510288 0.2469136e%^%10## [,1] [,2] [,3] [,4]
## [1,] 0.2500127 0.2499958 0.2499958 0.2499958
## [2,] 0.2499958 0.2500127 0.2499958 0.2499958
## [3,] 0.2499958 0.2499958 0.2500127 0.2499958
## [4,] 0.2499958 0.2499958 0.2499958 0.2500127e%^%15## [,1] [,2] [,3] [,4]
## [1,] 0.2499999 0.2500000 0.2500000 0.2500000
## [2,] 0.2500000 0.2499999 0.2500000 0.2500000
## [3,] 0.2500000 0.2500000 0.2499999 0.2500000
## [4,] 0.2500000 0.2500000 0.2500000 0.2499999e%^%30## [,1] [,2] [,3] [,4]
## [1,] 0.25 0.25 0.25 0.25
## [2,] 0.25 0.25 0.25 0.25
## [3,] 0.25 0.25 0.25 0.25
## [4,] 0.25 0.25 0.25 0.25e%^%60## [,1] [,2] [,3] [,4]
## [1,] 0.25 0.25 0.25 0.25
## [2,] 0.25 0.25 0.25 0.25
## [3,] 0.25 0.25 0.25 0.25
## [4,] 0.25 0.25 0.25 0.25e%^%100## [,1] [,2] [,3] [,4]
## [1,] 0.25 0.25 0.25 0.25
## [2,] 0.25 0.25 0.25 0.25
## [3,] 0.25 0.25 0.25 0.25
## [4,] 0.25 0.25 0.25 0.25Periodo=1 (Aperiódica).
No es reflexiva.
Es simétrica.
No es transitiva.
Es irreducible.
No es transitorio.
Es recurrente.
No es absorbente.
Es ergodico.
f)
f=matrix(c(0,1/2,0,0,1/2,1,1,0,0),nrow=3)
f## [,1] [,2] [,3]
## [1,] 0.0 0.0 1
## [2,] 0.5 0.5 0
## [3,] 0.0 1.0 0for (i in 1:5)
{
show(f%^%i)
}## [,1] [,2] [,3]
## [1,] 0.0 0.0 1
## [2,] 0.5 0.5 0
## [3,] 0.0 1.0 0
## [,1] [,2] [,3]
## [1,] 0.00 1.00 0.0
## [2,] 0.25 0.25 0.5
## [3,] 0.50 0.50 0.0
## [,1] [,2] [,3]
## [1,] 0.500 0.500 0.00
## [2,] 0.125 0.625 0.25
## [3,] 0.250 0.250 0.50
## [,1] [,2] [,3]
## [1,] 0.2500 0.2500 0.500
## [2,] 0.3125 0.5625 0.125
## [3,] 0.1250 0.6250 0.250
## [,1] [,2] [,3]
## [1,] 0.12500 0.62500 0.2500
## [2,] 0.28125 0.40625 0.3125
## [3,] 0.31250 0.56250 0.1250f%^%10## [,1] [,2] [,3]
## [1,] 0.2695312 0.4726562 0.2578125
## [2,] 0.2470703 0.5166016 0.2363281
## [3,] 0.2363281 0.4941406 0.2695312f%^%15## [,1] [,2] [,3]
## [1,] 0.2471924 0.5054932 0.2473145
## [2,] 0.2500305 0.4972229 0.2527466
## [3,] 0.2527466 0.5000610 0.2471924f%^%30## [,1] [,2] [,3]
## [1,] 0.2500007 0.4999692 0.2500302
## [2,] 0.2500074 0.5000080 0.2499846
## [3,] 0.2499846 0.5000147 0.2500007f%^%60## [,1] [,2] [,3]
## [1,] 0.25 0.5 0.25
## [2,] 0.25 0.5 0.25
## [3,] 0.25 0.5 0.25f%^%100## [,1] [,2] [,3]
## [1,] 0.25 0.5 0.25
## [2,] 0.25 0.5 0.25
## [3,] 0.25 0.5 0.25Periodo=1 (Aperiódica).
Es reflexiva, en su estado [2,2].
No es simétrica.
Es transitiva.
No es irreducible.
No es transitorio, solo si en su estado [2,2] fuera 1.
No es recurrente.
No es absorbente, solo si en su estado [2,2] fuera 1.
No es ergodico.
g)
g=matrix(c(1/4,3/4,1/3,0,0,3/4,1/4,1/3,0,0,0,0,1/3,0,0,0,0,0,3/4,1/4,0,0,0,1/4,3/4),nrow=5)
g## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2500000 0.7500000 0.0000000 0.00 0.00
## [2,] 0.7500000 0.2500000 0.0000000 0.00 0.00
## [3,] 0.3333333 0.3333333 0.3333333 0.00 0.00
## [4,] 0.0000000 0.0000000 0.0000000 0.75 0.25
## [5,] 0.0000000 0.0000000 0.0000000 0.25 0.75for (i in 1:5)
{
show(g%^%i)
}## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2500000 0.7500000 0.0000000 0.00 0.00
## [2,] 0.7500000 0.2500000 0.0000000 0.00 0.00
## [3,] 0.3333333 0.3333333 0.3333333 0.00 0.00
## [4,] 0.0000000 0.0000000 0.0000000 0.75 0.25
## [5,] 0.0000000 0.0000000 0.0000000 0.25 0.75
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.6250000 0.3750000 0.0000000 0.000 0.000
## [2,] 0.3750000 0.6250000 0.0000000 0.000 0.000
## [3,] 0.4444444 0.4444444 0.1111111 0.000 0.000
## [4,] 0.0000000 0.0000000 0.0000000 0.625 0.375
## [5,] 0.0000000 0.0000000 0.0000000 0.375 0.625
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.4375000 0.5625000 0.00000000 0.0000 0.0000
## [2,] 0.5625000 0.4375000 0.00000000 0.0000 0.0000
## [3,] 0.4814815 0.4814815 0.03703704 0.0000 0.0000
## [4,] 0.0000000 0.0000000 0.00000000 0.5625 0.4375
## [5,] 0.0000000 0.0000000 0.00000000 0.4375 0.5625
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.5312500 0.4687500 0.00000000 0.00000 0.00000
## [2,] 0.4687500 0.5312500 0.00000000 0.00000 0.00000
## [3,] 0.4938272 0.4938272 0.01234568 0.00000 0.00000
## [4,] 0.0000000 0.0000000 0.00000000 0.53125 0.46875
## [5,] 0.0000000 0.0000000 0.00000000 0.46875 0.53125
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.4843750 0.5156250 0.000000000 0.000000 0.000000
## [2,] 0.5156250 0.4843750 0.000000000 0.000000 0.000000
## [3,] 0.4979424 0.4979424 0.004115226 0.000000 0.000000
## [4,] 0.0000000 0.0000000 0.000000000 0.515625 0.484375
## [5,] 0.0000000 0.0000000 0.000000000 0.484375 0.515625g%^%10## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.5004883 0.4995117 0.000000e+00 0.0000000 0.0000000
## [2,] 0.4995117 0.5004883 0.000000e+00 0.0000000 0.0000000
## [3,] 0.4999915 0.4999915 1.693509e-05 0.0000000 0.0000000
## [4,] 0.0000000 0.0000000 0.000000e+00 0.5004883 0.4995117
## [5,] 0.0000000 0.0000000 0.000000e+00 0.4995117 0.5004883g%^%15## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.4999847 0.5000153 0.000000e+00 0.0000000 0.0000000
## [2,] 0.5000153 0.4999847 0.000000e+00 0.0000000 0.0000000
## [3,] 0.5000000 0.5000000 6.969172e-08 0.0000000 0.0000000
## [4,] 0.0000000 0.0000000 0.000000e+00 0.5000153 0.4999847
## [5,] 0.0000000 0.0000000 0.000000e+00 0.4999847 0.5000153g%^%30## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.5 0.5 0.000000e+00 0.0 0.0
## [2,] 0.5 0.5 0.000000e+00 0.0 0.0
## [3,] 0.5 0.5 4.856936e-15 0.0 0.0
## [4,] 0.0 0.0 0.000000e+00 0.5 0.5
## [5,] 0.0 0.0 0.000000e+00 0.5 0.5g%^%60## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.5 0.5 0.000000e+00 0.0 0.0
## [2,] 0.5 0.5 0.000000e+00 0.0 0.0
## [3,] 0.5 0.5 2.358982e-29 0.0 0.0
## [4,] 0.0 0.0 0.000000e+00 0.5 0.5
## [5,] 0.0 0.0 0.000000e+00 0.5 0.5g%^%100## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.5 0.5 0.000000e+00 0.0 0.0
## [2,] 0.5 0.5 0.000000e+00 0.0 0.0
## [3,] 0.5 0.5 1.940325e-48 0.0 0.0
## [4,] 0.0 0.0 0.000000e+00 0.5 0.5
## [5,] 0.0 0.0 0.000000e+00 0.5 0.5Periodo=1 (Aperiódica).
Es reflexiva, en su estado [1,1], [2,2], [3,3], [4,4] y [5,5].
No es simétrica.
Es transitiva, en su estado [3,1].
No es irreducible.
No es transitorio, solo si en sus estados [1,1], [2,2], [3,3], [4,4] y [5,5] fueran 1.
Es recurrente, en su estado [3,1] y [3,2].
No es absorbente, solo si en sus estados [1,1], [2,2], [3,3], [4,4] y [5,5] fueran 1.
Es ergodico.
2. Determine el periodo de los estados en la cadena de Markov que tiene las siguientes matrices de transición de un paso: \[a.\begin{matrix} Estado & | & 0 & 1 & 2 & 3 & 4 & 5\\ - & - & - & - & - & - & - & -\\ 0 & | & 0 &0 & 0 & \frac{2}{3} & 0 & \frac{1}{3}\\ 1 & | & 0 & 0 & 1 & 0 & 0 & 0\\ 2 & | & 1 & 0 & 0 & 0 & 0 & 0\\ 3 & | & 0 & \frac{1}{4} & 0 & 0 & \frac{3}{4} & 0\\ 4 & | & 0 & 0 & 1 & 0 & 0 & 0\\ 5 & | & 0 & \frac{1}{2} & 0 & 0 & \frac{1}{2} & 0 \end{matrix}\] \[b.\begin{matrix} Estado & | & 0 & 1 & 2 & 3 & 4\\ - & - & - & - & - & - & -\\ 0 & | & 0 & \frac{4}{5} & 0 & \frac{1}{5} & 0\\ 1 & | & \frac{1}{4} & 0 & \frac{1}{2} & \frac{1}{4} & 0\\ 2 & | & 0 & \frac{1}{2} & 0 & \frac{1}{10} & \frac{2}{5}\\ 3 & | & 0 & 0 & 0 & 1 & 0\\ 4 & | & \frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} & 0\\ \end{matrix}\]
a=matrix(c(0,0,1,0,0,0,0,0,0,1/4,0,1/2,0,1,0,0,1,0,2/3,0,0,0,0,0,0,0,0,3/4,0,1/2,1/3,0,0,0,0,0),nrow=6)
a## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.00 0 0.6666667 0.00 0.3333333
## [2,] 0 0.00 1 0.0000000 0.00 0.0000000
## [3,] 1 0.00 0 0.0000000 0.00 0.0000000
## [4,] 0 0.25 0 0.0000000 0.75 0.0000000
## [5,] 0 0.00 1 0.0000000 0.00 0.0000000
## [6,] 0 0.50 0 0.0000000 0.50 0.0000000for (i in 1:5)
{
show(a%^%i)
}## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.00 0 0.6666667 0.00 0.3333333
## [2,] 0 0.00 1 0.0000000 0.00 0.0000000
## [3,] 1 0.00 0 0.0000000 0.00 0.0000000
## [4,] 0 0.25 0 0.0000000 0.75 0.0000000
## [5,] 0 0.00 1 0.0000000 0.00 0.0000000
## [6,] 0 0.50 0 0.0000000 0.50 0.0000000
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [2,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [3,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [4,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [5,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [6,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [2,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [3,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [4,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [5,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [6,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [2,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [3,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [4,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [5,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [6,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [2,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [3,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [4,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [5,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [6,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000a%^%10## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [2,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [3,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [4,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [5,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [6,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000a%^%15## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [2,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [3,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [4,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [5,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [6,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000a%^%30## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [2,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [3,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [4,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [5,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [6,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000a%^%60## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [2,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [3,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [4,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [5,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [6,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333a%^%100## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 0.0000000 0 0.0000000 0.0000000 0.0000000
## [2,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [3,] 0 0.0000000 1 0.0000000 0.0000000 0.0000000
## [4,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333
## [5,] 0 0.3333333 0 0.0000000 0.6666667 0.0000000
## [6,] 0 0.0000000 0 0.6666667 0.0000000 0.3333333Periodo=4.
b=matrix(c(0,1/4,0,0,1/3,4/5,0,1/2,0,0,0,1/2,0,0,1/3,1/5,1/4,1/10,1,1/3,0,0,2/5,0,0),nrow=5)
b## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0000000 0.8 0.0000000 0.2000000 0.0
## [2,] 0.2500000 0.0 0.5000000 0.2500000 0.0
## [3,] 0.0000000 0.5 0.0000000 0.1000000 0.4
## [4,] 0.0000000 0.0 0.0000000 1.0000000 0.0
## [5,] 0.3333333 0.0 0.3333333 0.3333333 0.0for (i in 1:5)
{
show(b%^%i)
}## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0000000 0.8 0.0000000 0.2000000 0.0
## [2,] 0.2500000 0.0 0.5000000 0.2500000 0.0
## [3,] 0.0000000 0.5 0.0000000 0.1000000 0.4
## [4,] 0.0000000 0.0 0.0000000 1.0000000 0.0
## [5,] 0.3333333 0.0 0.3333333 0.3333333 0.0
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2000000 0.0000000 0.4000000 0.4000000 0.0000000
## [2,] 0.0000000 0.4500000 0.0000000 0.3500000 0.2000000
## [3,] 0.2583333 0.0000000 0.3833333 0.3583333 0.0000000
## [4,] 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000
## [5,] 0.0000000 0.4333333 0.0000000 0.4333333 0.1333333
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0000000 0.3600000 0.0000000 0.4800000 0.1600000
## [2,] 0.1791667 0.0000000 0.2916667 0.5291667 0.0000000
## [3,] 0.0000000 0.3983333 0.0000000 0.4483333 0.1533333
## [4,] 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000
## [5,] 0.1527778 0.0000000 0.2611111 0.5861111 0.0000000
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.1433333 0.0000000 0.2333333 0.6233333 0.0000000
## [2,] 0.0000000 0.2891667 0.0000000 0.5941667 0.1166667
## [3,] 0.1506944 0.0000000 0.2502778 0.5990278 0.0000000
## [4,] 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000
## [5,] 0.0000000 0.2527778 0.0000000 0.6427778 0.1044444
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.00000000 0.2313333 0.0000000 0.6753333 0.09333333
## [2,] 0.11118056 0.0000000 0.1834722 0.7053472 0.00000000
## [3,] 0.00000000 0.2456944 0.0000000 0.6541944 0.10011111
## [4,] 0.00000000 0.0000000 0.0000000 1.0000000 0.00000000
## [5,] 0.09800926 0.0000000 0.1612037 0.7407870 0.00000000b%^%10## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.03486730 0.00000000 0.05748892 0.9076438 0.00000000
## [2,] 0.00000000 0.07079787 0.00000000 0.9004577 0.02874446
## [3,] 0.03712826 0.00000000 0.06121639 0.9016554 0.00000000
## [4,] 0.00000000 0.00000000 0.00000000 1.0000000 0.00000000
## [5,] 0.00000000 0.06227966 0.00000000 0.9124345 0.02528581b%^%15## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.000000000 0.02219068 0.00000000 0.9687998 0.009009561
## [2,] 0.010688570 0.00000000 0.01762316 0.9716883 0.000000000
## [3,] 0.000000000 0.02362953 0.00000000 0.9667767 0.009593745
## [4,] 0.000000000 0.00000000 0.00000000 1.0000000 0.000000000
## [5,] 0.009402531 0.00000000 0.01550275 0.9750947 0.000000000b%^%30## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0003218993 0.0000000000 0.0005307428 0.9991474 0.0000000000
## [2,] 0.0000000000 0.0006536135 0.0000000000 0.9990810 0.0002653714
## [3,] 0.0003427714 0.0000000000 0.0005651564 0.9990921 0.0000000000
## [4,] 0.0000000000 0.0000000000 0.0000000000 1.0000000 0.0000000000
## [5,] 0.0000000000 0.0005749714 0.0000000000 0.9991916 0.0002334422b%^%60## [,1] [,2] [,3] [,4] [,5]
## [1,] 2.855426e-07 0.000000e+00 4.707984e-07 0.9999992 0.000000e+00
## [2,] 0.000000e+00 5.797916e-07 0.000000e+00 0.9999992 2.353992e-07
## [3,] 3.040573e-07 0.000000e+00 5.013252e-07 0.9999992 0.000000e+00
## [4,] 0.000000e+00 0.000000e+00 0.000000e+00 1.0000000 0.000000e+00
## [5,] 0.000000e+00 5.100316e-07 0.000000e+00 0.9999993 2.070762e-07b%^%100## [,1] [,2] [,3] [,4] [,5]
## [1,] 2.433728e-11 0.000000e+00 4.012695e-11 1 0.000000e+00
## [2,] 0.000000e+00 4.941663e-11 0.000000e+00 1 2.006348e-11
## [3,] 2.591532e-11 0.000000e+00 4.272880e-11 1 0.000000e+00
## [4,] 0.000000e+00 0.000000e+00 0.000000e+00 1 0.000000e+00
## [5,] 0.000000e+00 4.347086e-11 0.000000e+00 1 1.764946e-11Periodo=2.
3. Suponga que la probabilidad de que llueva mañana es 0.5 si llueve hoy y que la probabilidad de que no llueva mañana es 0.9 si hoy no llueve. Suponga, además, que estas probabilidades no cambian. Formule la evolución del clima como una cadena de Markov, de su matriz de transición de un paso y haga el diagrama de transición de estados. Obtenga las probabilidades de estado estable.
\[\begin{bmatrix}&LL&NLL\\LL&0,5&0,5\\NLL&0,1&0,9\end{bmatrix}\]
Probabilidad de estado estable: \[(1)P_{i 0}=0.5P_{i 0}+0.1P_{i 1}\] \[(2)P_{i 1}=0.5P_{i 0}+0.9P_{i 1}\] \[(3)P_{i 0}+P_{i 1}=1\] Se despeja la ecuación 3 y se reemplaza en la 1 \[1-P_{i 1}=P_{i 0}\] \[1-P_{i 1}=0.5(1-P_{i 1})+0.1P_{i 1}\] \[1=0.5-0.5P_{i 1}+0.1P_{i 1}+P_{i 1}\] \[1=0.5-0.5P_{i 1}+0.1P_{i 1}+P_{i 1}\] \[1-0.5=0.6P_{i 1}\] \[0.5=0.6P_{i 1}\] \[\frac{0.5}{0.6}=P_{i 1}\] \[\frac{\frac{1}{2}}{\frac{3}{5}}=P_{i 1}\] \[\frac{5}{6}=P_{i 1}=0.83\] se sustituye en la ecuación 2 \[0.83=0.5P_{i 0}+0.9(0.83)\] \[\frac{5}{6}-\frac{9}{10}\left(\frac{5}{6}\right)=\frac{1}{2}P_{i 0}\] \[\frac{1}{12}=\frac{1}{2}P_{i 0}\] \[\frac{1}{6}=P_{i 0}=0.17\]
Otro metodo
a=matrix(c(-0.5,1, 0.1,1),nrow=2)
a## [,1] [,2]
## [1,] -0.5 0.1
## [2,] 1.0 1.0b=matrix(c(0.5,1, -0.1,1),nrow=2)
b## [,1] [,2]
## [1,] 0.5 -0.1
## [2,] 1.0 1.0c=matrix(c(0, 1),nrow=2)
c## [,1]
## [1,] 0
## [2,] 1solve(a, c)## [,1]
## [1,] 0.1666667
## [2,] 0.8333333solve(b, c)## [,1]
## [1,] 0.1666667
## [2,] 0.83333334. En el problema anterior substituya 0.5 por alfa y 0.9 por beta y halle las probabilidades de estado estable en términos de alfa y beta.
\[\begin{bmatrix}&LL&NLL\\LL&\alpha&1-\alpha\\NLL&1-\beta&\beta\end{bmatrix}\]
Probabilidad de estado estable: \[(1)P_{i 0}=\alpha P_{i 0}+(1-\beta)P_{i 1}\] \[(2)P_{i 1}=(1-\alpha)P_{i 0}+\beta P_{i 1}\] \[(3)P_{i 0}+P_{i 1}=1\] Se despeja la ecuación 3 y se reemplaza en la 1 \[1-P_{i 1}=P_{i 0}\] \[1-P_{i 1}=\alpha(1-P_{i 1})+(1-\beta)P_{i 1}\] \[1=\alpha-\alpha P_{i 1}+P_{i 1}-\beta P_{i 1}+P_{i 1}\] \[1-\alpha=P_{i 1}(-\alpha-\beta+2)\] \[\frac{1-\alpha}{-\alpha-\beta+2}=P_{i 1}\] se sustituye en la ecuación 2 \[\frac{1-\alpha}{-\alpha-\beta+2}=(1-\alpha)P_{i 0}+\beta \left(\frac{1-\alpha}{-\alpha-\beta+2}\right)\] \[\frac{1-\alpha}{-\alpha-\beta+2}-\beta \left(\frac{1-\alpha}{-\alpha-\beta+2}\right)=(1-\alpha)P_{i 0}\] \[\frac{\frac{1-\alpha}{-\alpha-\beta+2}-\beta \left(\frac{1-\alpha}{-\alpha-\beta+2}\right)}{1-\alpha}=P_{i 0}\] \[\frac{1-\alpha-\beta(-\alpha+1)}{(2-\beta-\alpha)(-\alpha+1)}=P_{i 0}\]
5. La cervecería A está preocupada por su mayor competidor B. Suponga que el cambio de marca se puede modelar como una cadena de Markov incluyendo tres estados: A y B representan los clientes de las mencionadas cervezas y el estado C representa las demás marcas. Los datos se toman cada mes y se construye la siguiente matriz de transición de un paso con datos históricos. \[\begin{matrix} & A & B & C\\ A & 0.7 & 0.2 & 0.1\\ B & 0.2 & 0.75 & 0.05\\ C & 0.1 & 0.1 & 0.8\\ \end{matrix}\] Haga el diagrama de transición de estados. ¿Cuáles son los porcentajes de participación del mercado de cervezas?
porcentajes de participación del mercado de cervezas \[A=0,7A+0,2B+0,1C (1)\] \[B=0,2A+0,75B+0,05C (2)\] \[C=0,1A+0,1B+0,8C (3)\] \[1=A+B+C (4)\] Se realiza eliminacion con la ecuacion 1 y 2 \[0=-0,3A+0,2B+0,1C\] \[(-2)0=(-2)(0,2A-0,25B+0,05C)\] \[-----------------------\] \[0=-0,7A+0,7B+0C (5)\] se despeja la ecuacion 5 y se reemplaza en la ecuacion 3 \[0,7A=0,7B\] \[A=B\] \[C=0,1(A,B)+0,1(A,B)+0,8C\] \[0=0,1(A,B)+0,1(A,B)-0,2C\] \[0,2C=0,1(A,B)+0,1(A,B)\] \[C=\frac{0,2}{0,2}(A,B) (6)\] Entonces \[A=B=C=\frac{1}{3}=0.33\] Otro metodo
a=matrix(c(-0.3,0.2,1, 0.2,-0.25,1, 0.1,0.05,1),nrow=3)
a## [,1] [,2] [,3]
## [1,] -0.3 0.20 0.10
## [2,] 0.2 -0.25 0.05
## [3,] 1.0 1.00 1.00b=matrix(c(0.2,0.1,1, -0.25,0.1,1, 0.05,-0.2,1),nrow=3)
b## [,1] [,2] [,3]
## [1,] 0.2 -0.25 0.05
## [2,] 0.1 0.10 -0.20
## [3,] 1.0 1.00 1.00c=matrix(c(-0.3,0.1,1, 0.2,0.1,1, 0.1,-0.2,1),nrow=3)
c## [,1] [,2] [,3]
## [1,] -0.3 0.2 0.1
## [2,] 0.1 0.1 -0.2
## [3,] 1.0 1.0 1.0d=matrix(c(0, 0, 1),nrow=3)
d## [,1]
## [1,] 0
## [2,] 0
## [3,] 1solve(a, d)## [,1]
## [1,] 0.3333333
## [2,] 0.3333333
## [3,] 0.3333333solve(b, d)## [,1]
## [1,] 0.3333333
## [2,] 0.3333333
## [3,] 0.3333333solve(c, d)## [,1]
## [1,] 0.3333333
## [2,] 0.3333333
## [3,] 0.33333336. Suponga que una red de comunicaciones transmite dígitos binarios (0,1), en donde cada digito se transmite un número indefinido de veces. Durante cada transmisión, la probabilidad de que ese digito se transmita correctamente es de 0.99. En otras palabras, se tiene una probabilidad de 0.01 de que el digito transmitido se registre con el valor opuesto al final de la transmisión. Determine la matriz de un paso. Resuelva las ecuaciones de estado estable. ¿Qué sucede si se rediseña la red para mejorar la probabilidad de la exactitud de una transmisión de 0.99 a 0.999?
a=matrix(c(0.99,0.01,0.01,0.99),nrow=2)
a## [,1] [,2]
## [1,] 0.99 0.01
## [2,] 0.01 0.99\[P i_{0}=0.99 P i_{0}+0.01P i_{1}\] \[P i_{1}=0.01 P i_{0}+0.99P i_{1}\] \[P i_{0}+P i_{1}=1\] Se igualan a cero las ecuaciones \[0=-0.01 P i_{0}+0.01P i_{1}\] \[0=0.01 P i_{0}-0.01P i_{1}\] Se tiene que: \[P i_{0}=P i_{1}\] Entonces: \[P i_{0}=0.5 \; P i_{1}=0.5\] Se rediseña la red para mejorar la probabilidad
a1=matrix(c(0.999,0.001,0.001,0.999),nrow=2)
a1## [,1] [,2]
## [1,] 0.999 0.001
## [2,] 0.001 0.999\[P i_{0}=0.999 P i_{0}+0.001P i_{1}\] \[P i_{1}=0.001 P i_{0}+0.999P i_{1}\] \[P i_{0}+P i_{1}=1\] Se igualan a cero las ecuaciones \[0=-0.001 P i_{0}+0.001P i_{1}\] \[0=0.001 P i_{0}-0.001P i_{1}\] Se tiene que: \[P i_{0}=P i_{1}\] Entonces: \[P i_{0}=0.5 \; P i_{1}=0.5\] Otro metodo
Normal
a=matrix(c(-0.01,1, 0.01,1),nrow=2)
a## [,1] [,2]
## [1,] -0.01 0.01
## [2,] 1.00 1.00b=matrix(c(0.01,1, -0.01,1),nrow=2)
b## [,1] [,2]
## [1,] 0.01 -0.01
## [2,] 1.00 1.00c=matrix(c(0, 1),nrow=2)
c## [,1]
## [1,] 0
## [2,] 1solve(a, c)## [,1]
## [1,] 0.5
## [2,] 0.5solve(b, c)## [,1]
## [1,] 0.5
## [2,] 0.5Mejorado
d=matrix(c(-0.001,1, 0.001,1),nrow=2)
d## [,1] [,2]
## [1,] -0.001 0.001
## [2,] 1.000 1.000e=matrix(c(0.001,1, -0.001,1),nrow=2)
e## [,1] [,2]
## [1,] 0.001 -0.001
## [2,] 1.000 1.000f=matrix(c(0, 1),nrow=2)
f## [,1]
## [1,] 0
## [2,] 1solve(d, f)## [,1]
## [1,] 0.5
## [2,] 0.5solve(e, f)## [,1]
## [1,] 0.5
## [2,] 0.57. Un proceso de producción incluye una máquina que se deteriora con rapidez, tanto en cantidad como en calidad de producción con el trabajo pesado, por lo que se inspecciona al final de cada día. Después de la inspección se clasifica la condición de la maquina en uno de cuatro estados:
0: tan buena como nueva
1: operable con deterioro mínimo
2: operable con deterioro mayor
3: inoperable y reemplazada por nueva
El proceso se puede modelar como una cadena de Markov con matriz de transición de un paso dada por: \[\begin{matrix} Estado & | & 0 & 1 & 2 & 3\\ - & - & - & - & - & -\\ 0 & | & 0 & \frac{7}{8} & \frac{1}{16} & \frac{1}{16}\\ 1 & | & 0 & \frac{3}{4} & \frac{1}{8} & \frac{1}{8}\\ 2 & | & 0 & 0 & \frac{1}{2} & \frac{1}{2}\\ 3 & | & 1 & 0 & 0 & 0\\ \end{matrix} \]
(a) Encuentre las probabilidades de estado estable. \[(1)\Rightarrow (0)=0(0)+\frac{7}{8}(1)+\frac{1}{16}(2)+\frac{1}{16}(3)\] \[(2)\Rightarrow (1)=0(0)+\frac{3}{4}(1)+\frac{1}{8}(2)+\frac{1}{8}(3)\] \[(3)\Rightarrow (2)=0(0)+0(1)+\frac{1}{2}(2)+\frac{1}{2}(3)\] \[(4)\Rightarrow (3)=(0)+0(1)+0(2)+0(3)\] \[(5)\Rightarrow 1=(0)+(1)+(2)+(3)\] Se agrupan las ecuaciones \[0=-(0)+\frac{7}{8}(1)+\frac{1}{16}(2)+\frac{1}{16}(3)\] \[0=-\frac{1}{4}(1)+\frac{1}{8}(2)+\frac{1}{8}(3)\] \[0=-\frac{1}{2}(2)+\frac{1}{2}(3)\] \[0=(0)-(3)\] \[1=(0)+(1)+(2)+(3)\] Se despeja la ecuación 3 y 4: \[(3)=(0)\] \[(2)=(3)\] Reemplazar en la ecuación 2 \[0=-\frac{1}{4}(1)+\frac{1}{8}(2,3)+\frac{1}{8}(2,3)\] \[\frac{1}{4}(1)=\frac{1}{8}(2,3)+\frac{1}{8}(2,3)\] \[\frac{1}{4}(1)=\frac{1}{4}(2,3)\] \[\frac{1}{4}4(1)=(2,3)\] \[(1)=(2)=(3)\] Entonces: \[1=\frac{1}{4}(0)+\frac{1}{4}(1)+\frac{1}{4}(2)+\frac{1}{4}(3)\] \[\frac{1}{4}=(0)=(1)=(2)=(3)\] Otro metodo
a=matrix(c(-1,0,0,1, 7/8,(3/4)-1,0,1, 1/16,1/8,(1/2)-1,1, 1/16,1/8,1/2,1),nrow=4)
a## [,1] [,2] [,3] [,4]
## [1,] -1 0.875 0.0625 0.0625
## [2,] 0 -0.250 0.1250 0.1250
## [3,] 0 0.000 -0.5000 0.5000
## [4,] 1 1.000 1.0000 1.0000b=matrix(c(0, 0, 0, 1),nrow=4)
b## [,1]
## [1,] 0
## [2,] 0
## [3,] 0
## [4,] 1solve(a, b)## [,1]
## [1,] 0.25
## [2,] 0.25
## [3,] 0.25
## [4,] 0.25(b) Si los costos respectivos por estar en los estados 0, 1, 2 y 3 son $0, $1000, $3000 y $6000, ¿Cuál es el costo diario esperado a la larga?
\[($0)\Rightarrow (0)=0(0)+\frac{7}{8}(1)+\frac{1}{16}(2)+\frac{1}{16}(3)\] \[($1000)\Rightarrow (1)=0(0)+\frac{3}{4}(1)+\frac{1}{8}(2)+\frac{1}{8}(3)\] \[($3000)\Rightarrow (2)=0(0)+0(1)+\frac{1}{2}(2)+\frac{1}{2}(3)\] \[($6000)\Rightarrow (3)=(0)+0(1)+0(2)+0(3)\] Se reemplaza el valor para cada ecuación: \[(0)=\frac{7}{8}1000+\frac{1}{16}3000+\frac{1}{16}6000=$1437.5\] \[(1)=\frac{3}{4}1000+\frac{1}{8}3000+\frac{1}{8}6000=$1875\] \[(2)=\frac{1}{2}3000+\frac{1}{2}6000=$4500\] \[(3)=$0\]
8. Una partícula se mueve sobre un círculo por puntos marcados 0, 1, 2, 3 y 4 que están en el sentido de las agujas del reloj. En cada paso la partícula tiene una probabilidad de 0.5 de moverse en el sentido de las agujas del reloj y de 0.5 de moverse en sentido opuesto. Sea Xn la localización de la partícula después de n pasos.
(a) Encuentre la matriz de transición de un paso.
a=matrix(c(0,0.5,0,0,0.5,0.5,0,0.5,0,0,0,0.5,0,0.5,0,0,0,0.5,0,0.5,0.5,0,0,0.5,0),nrow=5)
a## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0 0.5 0.0 0.0 0.5
## [2,] 0.5 0.0 0.5 0.0 0.0
## [3,] 0.0 0.5 0.0 0.5 0.0
## [4,] 0.0 0.0 0.5 0.0 0.5
## [5,] 0.5 0.0 0.0 0.5 0.0(b) Discuta a cerca de la convergencia de la matriz.
for (i in 1:5)
{
show(a%^%i)
}## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.0 0.5 0.0 0.0 0.5
## [2,] 0.5 0.0 0.5 0.0 0.0
## [3,] 0.0 0.5 0.0 0.5 0.0
## [4,] 0.0 0.0 0.5 0.0 0.5
## [5,] 0.5 0.0 0.0 0.5 0.0
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.50 0.00 0.25 0.25 0.00
## [2,] 0.00 0.50 0.00 0.25 0.25
## [3,] 0.25 0.00 0.50 0.00 0.25
## [4,] 0.25 0.25 0.00 0.50 0.00
## [5,] 0.00 0.25 0.25 0.00 0.50
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.000 0.375 0.125 0.125 0.375
## [2,] 0.375 0.000 0.375 0.125 0.125
## [3,] 0.125 0.375 0.000 0.375 0.125
## [4,] 0.125 0.125 0.375 0.000 0.375
## [5,] 0.375 0.125 0.125 0.375 0.000
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.3750 0.0625 0.2500 0.2500 0.0625
## [2,] 0.0625 0.3750 0.0625 0.2500 0.2500
## [3,] 0.2500 0.0625 0.3750 0.0625 0.2500
## [4,] 0.2500 0.2500 0.0625 0.3750 0.0625
## [5,] 0.0625 0.2500 0.2500 0.0625 0.3750
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.06250 0.31250 0.15625 0.15625 0.31250
## [2,] 0.31250 0.06250 0.31250 0.15625 0.15625
## [3,] 0.15625 0.31250 0.06250 0.31250 0.15625
## [4,] 0.15625 0.15625 0.31250 0.06250 0.31250
## [5,] 0.31250 0.15625 0.15625 0.31250 0.06250a%^%10## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2480469 0.1611328 0.2148438 0.2148438 0.1611328
## [2,] 0.1611328 0.2480469 0.1611328 0.2148438 0.2148438
## [3,] 0.2148438 0.1611328 0.2480469 0.1611328 0.2148438
## [4,] 0.2148438 0.2148438 0.1611328 0.2480469 0.1611328
## [5,] 0.1611328 0.2148438 0.2148438 0.1611328 0.2480469a%^%15## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.1833496 0.2134705 0.1948547 0.1948547 0.2134705
## [2,] 0.2134705 0.1833496 0.2134705 0.1948547 0.1948547
## [3,] 0.1948547 0.2134705 0.1833496 0.2134705 0.1948547
## [4,] 0.1948547 0.1948547 0.2134705 0.1833496 0.2134705
## [5,] 0.2134705 0.1948547 0.1948547 0.2134705 0.1833496a%^%30## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2006931 0.1994393 0.2002142 0.2002142 0.1994393
## [2,] 0.1994393 0.2006931 0.1994393 0.2002142 0.2002142
## [3,] 0.2002142 0.1994393 0.2006931 0.1994393 0.2002142
## [4,] 0.2002142 0.2002142 0.1994393 0.2006931 0.1994393
## [5,] 0.1994393 0.2002142 0.2002142 0.1994393 0.2006931a%^%60## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2000012 0.1999990 0.2000004 0.2000004 0.1999990
## [2,] 0.1999990 0.2000012 0.1999990 0.2000004 0.2000004
## [3,] 0.2000004 0.1999990 0.2000012 0.1999990 0.2000004
## [4,] 0.2000004 0.2000004 0.1999990 0.2000012 0.1999990
## [5,] 0.1999990 0.2000004 0.2000004 0.1999990 0.2000012a%^%73## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.1999999 0.2000001 0.2000000 0.2000000 0.2000001
## [2,] 0.2000001 0.1999999 0.2000001 0.2000000 0.2000000
## [3,] 0.2000000 0.2000001 0.1999999 0.2000001 0.2000000
## [4,] 0.2000000 0.2000000 0.2000001 0.1999999 0.2000001
## [5,] 0.2000001 0.2000000 0.2000000 0.2000001 0.1999999a%^%74## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2000001 0.2000000 0.2000000 0.2000000 0.2000000
## [2,] 0.2000000 0.2000001 0.2000000 0.2000000 0.2000000
## [3,] 0.2000000 0.2000000 0.2000001 0.2000000 0.2000000
## [4,] 0.2000000 0.2000000 0.2000000 0.2000001 0.2000000
## [5,] 0.2000000 0.2000000 0.2000000 0.2000000 0.2000001a%^%75## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2 0.2 0.2 0.2 0.2
## [2,] 0.2 0.2 0.2 0.2 0.2
## [3,] 0.2 0.2 0.2 0.2 0.2
## [4,] 0.2 0.2 0.2 0.2 0.2
## [5,] 0.2 0.2 0.2 0.2 0.2a%^%100## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.2 0.2 0.2 0.2 0.2
## [2,] 0.2 0.2 0.2 0.2 0.2
## [3,] 0.2 0.2 0.2 0.2 0.2
## [4,] 0.2 0.2 0.2 0.2 0.2
## [5,] 0.2 0.2 0.2 0.2 0.2Luego de un tiempo la matriz tiende a estabilizarse en un punto, este punto de equilibrio da para cada estado un valor de 0.2 o se puede entender por el hecho de ser una matriz de 5X5, que el equilibrio es la quinta parte para cada uno y de esta forma el equilibrio se respeta dando 1.
(c) Halle las probabilidades de estado estable.
\[(1)\Rightarrow(0)=0(0)+\frac{1}{2}(1)+0(2)+0(3)+\frac{1}{2}(4)\] \[(2)\Rightarrow(1)=\frac{1}{2}(0)+0(1)+\frac{1}{2}(2)+0(3)+0(4)\] \[(3)\Rightarrow(2)=0(0)+\frac{1}{2}(1)+0(2)+\frac{1}{2}(3)+0(4)\] \[(4)\Rightarrow(3)=0(0)+0(1)+\frac{1}{2}(2)+0(3)+\frac{1}{2}(4)\] \[(5)\Rightarrow(4)=\frac{1}{2}(0)+0(1)+0(2)+\frac{1}{2}(3)+0(4)\] \[(6)\Rightarrow 1=(0)+(1)+(2)+(3)+(4)\]
b=matrix(c(-1,0.5,0,0,0.5, 0.5,-1,0.5,0,0, 0,0.5,-1,0.5,0, 0,0,0.5,-1,0.5, 0.5,0,0,0.5,-1),nrow=5)
b## [,1] [,2] [,3] [,4] [,5]
## [1,] -1.0 0.5 0.0 0.0 0.5
## [2,] 0.5 -1.0 0.5 0.0 0.0
## [3,] 0.0 0.5 -1.0 0.5 0.0
## [4,] 0.0 0.0 0.5 -1.0 0.5
## [5,] 0.5 0.0 0.0 0.5 -1.0Se despejan las ecuaciones \[0=-(0)+\frac{1}{2}(1)+\frac{1}{2}(4)\] \[0=\frac{1}{2}(0)-(1)+\frac{1}{2}(2)\] \[0=\frac{1}{2}(1)-(2)+\frac{1}{2}(3)\] \[0=\frac{1}{2}(2)-(3)+\frac{1}{2}(4)\] \[0=\frac{1}{2}(0)+\frac{1}{2}(3)-(4)\] \[1=(0)+(1)+(2)+(3)+(4)\] Se realizan las matrices
c=matrix(c(-1,0.5,0,0,1, 0.5,-1,0.5,0,1, 0,0.5,-1,0.5,1, 0,0,0.5,-1,1, 0.5,0,0,0.5,1),nrow=5)
c## [,1] [,2] [,3] [,4] [,5]
## [1,] -1.0 0.5 0.0 0.0 0.5
## [2,] 0.5 -1.0 0.5 0.0 0.0
## [3,] 0.0 0.5 -1.0 0.5 0.0
## [4,] 0.0 0.0 0.5 -1.0 0.5
## [5,] 1.0 1.0 1.0 1.0 1.0d=matrix(c(0, 0, 0, 0, 1),nrow=5)
d## [,1]
## [1,] 0
## [2,] 0
## [3,] 0
## [4,] 0
## [5,] 1solve(c, d)## [,1]
## [1,] 0.2
## [2,] 0.2
## [3,] 0.2
## [4,] 0.2
## [5,] 0.2