Queueing Model based
on the birth-and-death
process
M/M/1
•an infinite number of
customers
•Infinite queue
•Inputs and outputs are at a
constant rate
•There is only one server
M/M/1 system
• For M/M/1 the probability of
exactly n customers arriving during
an interval of length t is given by
the Poisson law.
Poisson’s Law
( l t) - l t
Pn (t ) =
e
n!
n
(1)
l- Rate parameter
• =event per unit interval (time
l
distance volume...)
The M/M/1 queue in equilibrium
queue
server
State of the system:
• There are 4 people in the system.
• 3 in the queue.
• 1 in the server.
Memory of M/M/1:
• The amount of time the person in the
server has already spent being served is
independent of the probability of the
remaining service time.
Memoryless
• M/M/1 queues are memoryless (a
popular item with queueing
theorists, and a feature unique to
exponential pdfs).
.
P k = equilibrium prob
that there are k in system
Birth-death system
• In a birth-death system once
serviced a customer moves to the
next state.
• This is like a nondeterminis-tic
finite-state machine.
State-transition Diagram (Rate
diagram)
• The following state-transition diagram
is called a Markov chain model.
• Directed branches represent transitions
between the states.
• Exponential pdf parameters appear on
the branch label.
Single-server queueing system
lPo
2
1
0
 P1
l P k -1
l P1
P 2
...
lP k
k
k-1
P k
k+1
P k +1
State diagram: single-server model
λ
# in system
0
λ
●
2
1
μ
λ
μ
3
μ
# in system also called state.
●To get from one state to another, an arrival (λ) must occur or a service
completion (μ) must occur.
●
In long-run, for each state: Rate in = Rate out
Mean # λ = Mean # μ
Waiting Lines
13
Symbles:
l = mean arrival rate (cust. /sec)
l P0= mean number of transitions/ sec
from state 0 to 1
 = mean service rate (cust./sec)
 P1 = mean number of transitions/ sec
from state 1 to 0
States
• State 0 = system empty
• State 1 = cust. in server
• State 2 = cust in server, 1 cust in
queue etc...
Probalility of Given State
• Prob. of a given state is invariant if
system is in equilibrium.
• The prob. of k cust’s in system is
constant.
Similar to AC
• This is like AC current entering a node
• is called detailed balancing
• the number leaving a node must equal
the number entering
Balance equations for each state
State
0
Rate in
μP1
=
Rate out
λP0
Probability in
state 1
The only way
into state 0
is service
completion from 1
Probability in
state 0
The only way
out of state 0
is to have
an arrival
Waiting Lines
18
Balance equations for each state
(cont.)
State
1
Rate in
λP0 + μP2
=
=
Can arrive
state 1 by
arrival from 0
or service
completion from 2
2
3
λP1 + μP3
λP2 + μP4
Rate out
λP1 + μP1
Two ways
out of state 1,
arrival or
service completion
=
λP2 + μP2
=
λP3 + μP3
etc.
19
Derivation
State 0:
l P 0 = P 1
lP0
P1 =

Derivation
lP0
l

P2 =

State 1
2
l P0
P2 =
2

since
l P k = P k+1
then:
k
6
l P0
k
P k = k =  P0

l
where  =
= traffic intensity < 1

since all prob. sum to one


k=0
k
P0 = 1 = P0


k
=1
k =0
Note: the sum of a geometric series is

1
 = 1- 
k=0
k



k
k=0
1
=
1- 
• Suppose that it is right, cross multiply and
simplify:


k=0
k=0
k
k




 =1

So
Q.E.D.

k=0
k

-  =  = 1
k =1
k
0
subst 7 into 6a
6a

P0

k
=1
k =0
7a
P0
=1
1- 
7b
P0 = 1- 
and
=prob server is empty
subst into
6
k
l P0
k
P k = k =  P0

yields:
8
P k = (1 -  )
k
Thanking
You…!!