P/P/M/N

P/P/M/N
• Arribos Po
• Servicio Po
• Régimen permanente
• M canales
• 1 cola
• FIFO
• Capacidad finita (N clientes)
• Sin impaciencia
• Población infinita
Parámetros
• λ
1
Ta =
λ
• µ
1
Ts =
µ
• M
• N
Variables
• p(n)
• Lc
• L
• H
• Wc
• W
λ
µ
µ
_
λ
λ
_
µ
_
R
M
λ
λn = 
0
0

µ n = n ⋅ µ
M ⋅ µ

para n < N
para n = N
para n = 0
para 1 ≤ n < M
para M ≤ n ≤ N
λ n−1
p(n) =
⋅ p(n − 1)
µn
Para n = 1:
λ0
λ
p(1) =
⋅ p(0) = ⋅ p(0)
µ1
µ
Para n = 2:
λ1
λ
λ
p( 2) =
⋅ p(1) =
⋅ p(0) ⋅
µ2
2µ
µ
•
•
Para n = 3:
•
Para n = 4:
•
⇒
p(1) = ρ ⋅ p(0)
ρ2
p( 2) =
⋅ p(0)
2
λ2
λ
λ2
p(3) =
⋅ p( 2) =
⋅ p(0) ⋅ 2 ⇒
3µ
2µ
µ3
ρ3
p (3) =
⋅ p(0)
6
λ3
λ
λ3
p (4) =
⋅ p (3) =
⋅ p(0) ⋅ 3 ⇒
µ4
4µ
6µ
ρ4
p( 4) =
⋅ p(0)
24
………..
Para n < M
⇒
ρn
p (n ) =
⋅ p(0)
n!
Para n = M:
ρM
p ( M) =
⋅ p (0 )
M!
Para n = M+1:
ρ M+1
p( M + 1) =
⋅ p(0)
M!⋅ M
•
•
Para n = M+2:
ρ M +2
p(M + 2) =
⋅ p(0)
2
M!⋅ M
Para n = M+3:
ρ M+3
p (M + 3) =
⋅ p(0)
3
M!⋅ M
•
•
……….
Para M
n
N
ρn
p( n ) =
p(0)
( n −M ) ⋅
M!⋅ M
M
ρ M
p(n ) =   ⋅
⋅ p(0)
 M  M!
n
N
∑ p( n) = 1
0
M −1
N
0
M
∑ p (n ) + ∑ p ( n ) = 1
N
ρn
 ρ  MM
⋅ p(0) = 1
  ⋅
∑0 n ! ⋅ p(0) + ∑
M!
M M
n
M −1
p (0) =
1
ρn M M N  ρ 
 
∑0 n ! + M! ⋅ ∑
M M
M −1
n
p(0) =
1
  ρ  N−M+1 
ρn
ρM
1
+
⋅
1
−
⋅




∑0 n ! (M − 1)!  M 

 (M − ρ)
M −1
M −1
N
ρn
MM
L = ∑ n ⋅ p(n ) = ∑ n ⋅ ⋅ p( 0) + ∑ n ⋅
n!
M!
0
0
M
N
ρ M
⋅ ρ ⋅ p(0)
L= M
2
ρ

M ! ⋅ 1 − 
 M
n
ρ
⋅   ⋅ p (0 )
M
N −M
N
M
  ρ  N −M



ρ
ρ
ρ
 


  M
⋅ 1 −  
− (N − M ) ⋅  
⋅ 1 −  + ρ ⋅ 1 −   ⋅
⋅ p (0) 
M!
M
 M 
  M 
  M 

M −1
N
0
M
H = ∑ n ⋅ p ( n ) + ∑ M ⋅ p (n )
  ρ N MM

H = ρ ⋅ 1 −   ⋅
⋅ p( 0) 
M!
  M 

N
L C = ∑ (n − M ) ⋅ p (n )
M
ρ M
⋅ρ
M
LC =
2
ρ

M ! ⋅ 1 − 
 M
N
λ = ∑ λ n ⋅ p( n )
0
N −M
  ρ  N −M
ρ 
ρ

⋅ 1 −  
− (N − M ) ⋅  
⋅ 1 −  ⋅ p( 0)
M
 M 
  M 
λ = λ ⋅ [1 − p( N)]
N
µ = ∑ µ n ⋅ p(n )
0
λ=µ
µ = µ⋅H
λ ⋅ [1 − p( N)] = µ ⋅ H
PA =
H = ρ ⋅ [1 − p( N)]
H ρ ⋅ (1 − p (n ))
=
M
M
R = λ −λ
WC =
LC
λ
W = WC + TS
R = λ ⋅ p( N)
Ejemplos de funcionales
Z = ce ⋅ L + cC ⋅ M
→ Min
Z = c L ⋅ ( N − M) + u ⋅ R
→ Min
Z = u ⋅ µ − cL ⋅ (N − M) →
Z = cs ⋅ M ⋅ µ + u ⋅ R
Max
→ Min
A un sistema P/P/2/4, sin impaciencia, arriba un cliente cada
5 minutos, en promedio. La duración media del servicio es de
10 minutos. Determinar:
Probabilidad de que un cliente que arriba al sistema no tenga
que esperar para recibir el servicio.
Porcentaje de ocupación del canal.
Probabilidad de que un cliente que arriba al sistema no
pueda ingresar.
Tasa promedio de clientes que se retiran sin ser atendidos.
Número promedio de clientes esperando ser atendidos.
Número promedio de clientes en el sistema.
Ingreso de caja esperado, si cada servicio se cobra $50.
Tiempo promedio de permanencia de un cliente dentro del
sistema.
Lucro cesante esperado.
1 cl
min
cl
λ=
⋅ 60
= 12
5 min
h
h
µ=
1 cl
min
cl
⋅ 60
=6
10 min
h
h
λ 12
ρ= =
=2
µ 6
λn = λ
λn = 0
para n = 0, 1, 2 y 3
para n = 4
µn = 0
µn = µ
µn = 2 µ
para n = 0
para n = 1
para n = 2, 3 y 4
λ
p(1) = ⋅ p(0) = ρ ⋅ p(0)
µ
p(1) = 0,2222
λ
ρ
ρ2
p (2) =
⋅ p(1) = ⋅ ρ ⋅ p (0) = ⋅ p(0)
2 ⋅µ
2
2
p(2) = 0,2222
λ
ρ ρ2
ρ3
p(3) =
⋅ p(2) = ⋅ ⋅ p(0) = ⋅ p(0)
2⋅µ
2 2
4
p(3) = 0,2222
λ
ρ ρ3
ρ4
p (4) =
⋅ p (3) = ⋅ ⋅ p(0) = ⋅ p(0)
2 ⋅µ
2 4
8
p(4) = 0,2222
ρ2
ρ3
ρ4
p(0) + ρ ⋅ p(0) + ⋅ p( 0) + ⋅ p (0) + ⋅ p(0) = 1
2
4
8
1
1
1
p(0) =
= 0,1111
2
3
4 =
2
3
4 =
ρ ρ ρ
2
2
2
9
1+ ρ + + +
1+ 2 + + +
2
4
8
2
4
8