1. No category

Solutions to COMP9334 Week 6 Sample Problems
Problem 1: For Week 3’s sample problem 1, find the response time in each option using
queueing analysis.
Solution:
(a) This option is an M/M/1 queue with λ = 0.02 customers per msec and µ = 0.05
customers per msec. Using the M/M/1 mean response time formula, we have
T =
1
1
=
= 33.3 msec
µ−λ
0.05 − 0.02
(b) In this option, the arrival rate at each queue is half of the system arrival rate. Effectively, we have two M/M/1 queues, each with λ = 0.01 customers per msec and
µ = 0.025 customers per msec. Using the M/M/1 mean response time formula, we
have
1
1
T =
=
= 66.7 msec
µ−λ
0.025 − 0.01
(c) This option is effectively an M/M/2 queue with λ = 0.02 customers per msec, µ =
λ
0.025 customers per msec, A = λ/µ = 0.8, and ρ = 2µ
= 0.4. Using the M/M/m
mean response time formula and letting m = 2, we have
C(A, 2) =
and therefore
T =
A2
2!
(1 − ρ)(1 + A) +
A2
2!
C(A, 2) 1
+ = 47.6 msec
mµ − λ µ
1
Problem 2: Show that the Erlang C formula and the Erlang B formula have the following
relationship:
E(A, m)
.
C(A, m) =
1 − ρ [1 − E(A, m)]
Solution:
(1 − ρ)
1
=
C(A, m)
Pm−1 Ak
k=0
Am
m!
k!
+
Am
m!
=
1−ρ
+ρ
E(A, m)
Problem 3: An Internet Service Provider has 4 dial-up ports. Connection requests follow
a Poisson arrival process with mean arrival rate of 3 requests per hour. The session duration
of each connection request is exponentially distributed with a mean of 1.5 hours. What is
the probability that a connection request will be rejected?
Solution: This is effectively an M/M/4/4 system with λ = 3 requests per hour, 1/µ = 1.5
hours, and A = λ/µ = 4.5. The probability that a connection request is blocked is given
by the Erlang B formula
E(A, m) =
Am
m!
Pm Ak
k=0 k!
=
4.54
4!
P4 4.5k
k=0 k!
= 0.3567
Alternatively, we have
1
=
E(A, m)
Ak
k=0 k!
Am
m!
Pm−1 Ak
Pm
=
k=0
k!
Am
m!
+
Am
m!
=
m
+1
AE(A, m − 1)
Therefore, we can compute E(A, m) using the following recursive formula
E(A, k) =
AE(A, k − 1)
, k = 1, 2, . . . , m
k + AE(A, k − 1)
(1)
with the initial condition E(A, 0) = 1. Note that the initial condition E(A, 0) = 1 holds
because if the number of servers in the system is 0, all customers will be blocked, so the
blocking probability is 1.
Using (1), for A = 4.5, m = 4, we have:
E(A, 1) =
AE(A, 0)
4.5
=
= 0.8182
1 + AE(A, 0)
1 + 4.5
AE(A, 1)
4.5 × 0.8182
=
= 0.6480
2 + AE(A, 1)
2 + 4.5 × 0.8182
AE(A, 2)
4.5 × 0.6480
E(A, 3) =
=
= 0.4929
3 + AE(A, 2)
3 + 4.5 × 0.6480
AE(A, 3)
4.5 × 0.4929
E(A, 4) =
=
= 0.3567
4 + AE(A, 3)
4 + 4.5 × 0.4929
E(A, 2) =
2