M/M/1 Queues
• Customers arrive according to a Poisson
process with rate .
λ j  λ,
μ0  0
j 0
μ j  μ,
j 1
• There is only one server.
• Service time is exponential with rate .


0
1

2



...
j-1
j

j+1

M/M/1 Queues
0 1   j 1
j
c



• We let  = /, so
j
1 2   j
• From Balance equations:
 1   0 ,  2    0 ,  ,
2
 j   0
j
• As the stationary probabilities must sum to 1,
therefore:
1   0  1   2     j  
1   0 (1     2     j  )
M/M/1 Queues
• But for  <1,
1           1/1  
2
j
• Therefore:
 0  (1   )
 j   (1   ),
j
j 1
M/M/1 Queues
• L is the expected number of entities in the system.
L   j 0 j  P( j entities in system)

  j 0 j j   j 0 j (1   )


 (1   )   j 1 j

j
j 1
d
 (1   ) 
d

d



 (1   ) 
(
)

d 1  
1    


j 1
j
M/M/1 Queues
• Lq is the expected number of entities in the
queue.
Lq   j 1 ( j  1)  P ( j entities in system)

  j 1 ( j  1) j

  j 1 j j   j 1  j


 L  (1   0 )  L  


Lq 

1    (   )
2
2
Little’s Formula
• W is the expected waiting time in the system
this includes the time in the line and the time in
service.
• Wq is the expected waiting time in the queue.
• W and Wq can be found using the Little’s
formula. (explain for the deterministic case)
L  W
Lq  Wq
L
1
W 
  
M/M/1 queuing model Summary



 0  P( N  0)  1   
 n  P ( N  n)
  1   
n
1
W
 
L  W 

 

Wq  W  
     
1
2
Lq  Wq 
    
M/M/s Queues
• There are s servers.
• Customers arrive according to a Poisson
process with rate ,  j   , j  0
• Service time for each entity is exponential
0  0
with rate .
 j  j , j  s
• Let  = /s
  s , j  s
j
M/M/s Queues

0


1

2
2

..
..
s
s-1
s+1
.
.
( s  1) 
s
s
j
j-1

s
• Thus
0 
1
( s )
( s )
 j 0 j!  s! (1   )
s 1
j


s
j+1
s
M/M/s Queues
( s )
j 
0, j  s
j!
j
( s )
 j  j  s  0 , j  s, s  1, s  2,...
s! s
j
M/M/s Queues
• All servers are busy with probability
(s )
P( j  s ) 
0
s!(1   )
s
• This probability is used to find L,Lq, W, Wq
• The following table gives values of this
probabilities for various values of  and s
M/M/s Queues
P ( j  s )
Lq 
1 
P( j  s)
Wq 


s  
Lq

L  Lq 

P( j  s)
1
W 







s  

L
Lq
1
M/M/s queuing system
Needed for steady state
• Steady state occurs only if the arrival rate
is less than the maximum service rate of
the system
– Equivalent to traffic intensity  = /s < 1
• Maximum service rate of the system is
number of servers times service rate per
server
M/M/1/c Queues
• Customers arrive according to a Poisson
process with rate .
• The system has a finite capacity of c
customers including the one in service.
• There is only one server.
• Service times are exponential with rate .
M/M/1/c Queues
• The arrival rate is
 j  ,
j  c 1
 j  0,
jc
0


0  0
 j  , j  1
1 

1   c 1
j
  j 0 ,
j c
j
 0,
j c
M/M/1/c Queues
• L is the expected number of entities in the
system.
 (1  ( c  1)  c  c c 1 )
L
(1   c 1 )(1   )
Lq  L  (1   0 )
 (1  (c  1)  c  c c 1 )  (1   )


c 1
(1  
)(1   )
1   c 1
M/M/1/c Queues
• We shall use Little’s formula to find W
and Wq. Note that:
– Recall that  was the arrival rate.
– But if there are c entities in the system,
any arrivals find the system full, cannot
“arrive”.
– So of the  arrivals per time unit, some
proportion are turned away.
– c is the probability of the system being
full.
– So (1- c) is the actual rate of arrivals.
M/M/1/c Queues
L
W
 (1   c )
Wq 
Lq
 (1   c )