M/M/1 Queues
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M/M/1 Queues
• Customers arrive according to a Poisson process with rate .
• There is only one server.
• Service time is exponential with rate .
0j λ,λ j 1j μ,μ
0μ
j
0
0 1 2 j-1 j j+1
...
M/M/1 Queues
• We let = /, so • From Balance equations:
• As the stationary probabilities must sum to 1, therefore:
0
02
201 , , ,
j
j
) 1(1
12
0
210
j
j
j
j
jjc
21
110
M/M/1 Queues• But for <1,
• Therefore:
1/1 1 2 j
1 ),1(
)1(0
jjj
M/M/1 Queues• L is the expected number of entities in the system.
1)
1()1(
)1()1(
)1(
system)in entities j(
11
1
00
0
d
d
d
dj
jj
PjL
j
j
j
j
j
j
j j
j
M/M/1 Queues
• Lq is the expected number of entities in the queue.
LL
j
j
PjL
j jj j
j j
jq
)1(
)1(
system)in entities j()1(
0
11
1
1
)(1
22
qL
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)
qq WL
WL
1LW
M/M/1 queuing model Summary
2
1
1
q
WL
WW
WL
W
1
)(
1)0(0
n
n nNP
NP
M/M/s Queues
• There are s servers.• Customers arrive according to a Poisson process
with rate ,• Service time for each entity is exponential with
rate .
• Let = /s
0 , jj
sjs
sjj
j
j
,
,
00
M/M/s Queues
• Thus
1
0
0
)1(!)(
!)(
1s
j
sj
ss
js
0 1 2 s-1 s s+1
..
. j-1 j j+1...
2 s ss s
)1( s
M/M/s Queues
2,...s1,s s,j ,!
)(
sj ,!
)(
0
0
sj
j
j
j
j
ss
s
j
s
M/M/s Queues
• All servers are busy with probability
• This probability is used to find L,Lq, W, Wq
• The following table gives values of this probabilities for various values of and s
0)1(!
)()(
s
ssjP
s
M/M/s Queues
1)(1
)(
1
)(
s
sjPLLW
LL
s
sjPLW
sjPL
q
q
q
M/M/s queuing systemNeeded 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
cj
cj
j
j
,0
1 ,
1 ,
00
jj
cj ,0
cj ,
1
1
0
10
j
jj
c
M/M/1/c Queues
• L is the expected number of entities in the system.
)1)(1(
))1(1(1
1
c
cc ccL
11
1
0
1
)1(
)1)(1(
))1(1(
)1(
cc
cc
q
cc
LL
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
)1(
)1(
c
c
LW
LW