=
ac
c! p0
= ap0
a2
= 2·1
p0
3
= a3! p0
c
pc+1 = ρ · ac! p0
...
c
pn
= ρn−c · ac! p0
1=
k=0
∞
p k = p0
k=0
∞
k=c
ac k−c
ρ
+
k!
c!
c−1
ak
=:S
⎞
c−1 k
⎜
ac 1 ⎟
a
⎜
⎟
= p0 ⎜
+
⎟.
k!
c! 1 − ρ ⎠
⎝
k=0
⎛
2
In order to get an expression for p0, we use the condition, that the overall
sum of probabilities must be 1. This gives:
p1
p2
p3
...
pc
for n ≥ c.
Stat 330 (Spring 2015): slide set 23
The M/M/c queue: balance equations
Last update: March 22, 2015
Stat 330 (Spring 2015)
Slide set 23
Stat 330 (Spring 2015): slide set 23
an
n! p0
an
p
c!cn−c 0
for n ≥ c
for 0 ≤ n ≤ c − 1
Stat 330 (Spring 2015): slide set 23
1
3
The formula for this probability is known as Erlang’s C formula or Erlang’s
delay formula and written as C(c, a).
A key descriptor for the system is the probability that an entering customer
must queue for service - this is equal to the probability that all servers are
busy.
pn =
The other probabilities pn are given as:
p0 = S −1.
This system has a steady state, if ρ < 1; in that case,
The balance equations for steady state are:
Let a = λ/μ and ρ = a/c = λ/(cμ).
Clearly, the critical thing here in terms of whether or not a steady state
exists is whether or not λ/(cμ) < 1.
The transition state diagram for the X(t) is:
Again, X(t) the number of individuals in the queueing system can be
modeled as a birth & death process.
The M/M/c queue
Stat 330 (Spring 2015): slide set 23
= p0
= p0
1
−
p0
k=0
k!
=
1
ρ
1
·
C(c, a) =
C(c, a).
λ 1−ρ
cμ(1 − ρ)
ρ
C(c, a).
L=W ·λ=a+
1−ρ
6
and the expected overall number of individuals in the system is on average
1
W = Wq + W s = Wq + ,
μ
Thus the expected overall time in system is then
Wq = Lq /λ =
k=c
∞
(k − c)pk =
k=c
∞
(k − c)
ak
p0 =
c!ck−c
λ
μ
= 2, and ρ =
a
c
= 2/3
1
μ
Hence the expected number of people in the bank is L = W λ = 26/9
7
= 2 minutes using the service time
This gives the total waiting time as W = Ws + Wq = 26/9 min.
Expected service time is Ws =
distribution.
Calculate the expected waiting time in the queue: Wq = Lq /λ = 8/9 min.
ρ
Thus expected length of the queue is Lq = p0 · ac! · (1−ρ)
2 = 8/9
c
The probability that no customer is in the bank then is
−1 −1
3
c−1 ak
ac 1
1
p0 =
= 1 + 2 + 42 + 23! · 1−ρ
= 19 .
k=0 k! + c! 1−ρ
For this queue, λ = 1, μ = 0.5, c = 3, a =
Bank: A bank has three tellers. Customers arrive at a rate of 1 per minute
and stay in a single queue. Each teller needs on average 2 min to deal with
a customer. What are the specifications of this queue?
Example
∞
ac k
ac
ρ
kρ = p0
= p0
c!
c! (1 − ρ)2
k=1
k ρ( ∞
k=1 ρ )
ρ
=
C(c, a).
1−ρ
=
Stat 330 (Spring 2015): slide set 23
Lq
Stat 330 (Spring 2015): slide set 23
pk =
The steady state expected number of individuals in the queue Lq is
5
k=0
c−1
Stat 330 (Spring 2015): slide set 23
The M/M/c queue: properties
4
pk = 1 −
c−1 k
a
ac
.
c!(1 − ρ)
k=c
∞
By Little’s Law, the expected waiting time in the queue Wq is
t→∞
lim P (X(t) ≥ c) = C(c, a) =
Obviously, in a M/M/c queue, an entering individual must queue for service
exactly when c or more individuals are already in the system.
The M/M/c queue: Erlang’s C Formula