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TCOM501 – Networking:
Theory & Fundamentals
Final Examination
Professor Yannis A. Korilis
April 26, 2002
Problem 1 [30 points]:
Consider a ring network with nodes 1,2,…,
K
. In this network, a customer
that completes service at node
i
exits the network with probability
p
, or it is routed to node
i
+1
with probability 1
p
, for
i
= 1,2,…,
K
1. Customers that complete service at node
K
, either exit the
network, or are routed to node 1, with respective probabilities
p
and 1
p
.
At each node, external customers arrive according to a Poisson process with rate
γ
. The service
times at each node are exponentially distributed with rate μ. The arrival processes and the service
times at the various nodes are independent.
1.
[7 points] Find the aggregate arrival rates
,
1,2,.
..,
i
iK
λ
=
.
2.
[2 points] Under what conditions does the ring network have a stationary distribution?
3.
[7 points] Assuming that the conditions of question 2 are satisfied, find the stationary
distribution of the network.
4.
[7 points] Find the average time that a customer spends in the network.
5.
[7 points] Is this ring network reversible? Justify your answer.
Solution 1:
The network is shown in the following figure. Since there are external arrivals, it is
an open Jackson network.
γ
γ
γ
p
p
p
1
p
−
1
p
−
1
p
−
i
1
i
+
1.
The aggregate arrival rates satisfy:
1 of 15
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21
32
1
(1
)
)
)
)
K
KK
p
p
p
p
−
λ =γ+
−
λ
−
λ
−
λ
−
λ
#
Since all nodes are symmetric – same arrival rates, service rates, and routing probabilities –
the aggregate arrival rates must be equal:
12
K
λ =λ =
=λ =λ
…
. The above equations, then,
give:
)
p
p
γ
λ=γ+
−
λ⇒λ=
2.
The network has a stationary distribution if the aggregate arrival rate at each node is less than
the service rate at the node. Therefore, we must have:
p
γ
< µ
3.
From Jackson’s theorem for open networks, the stationary distribution is:
1
1
11
(, , )
()
(
1 )
(
,
i
K
n
nn
K
Ki
i
ii
pn
n
p n
p
++
==
γ
=
=
−ρ ρ
=
−ρ
ρ
ρ=
µ
∏∏
"
…
4.
The average number of customers at each queue is:
,1
,
,
1
i
N
iK
ρ
…
and the average number of customers in the network:
1
NK
ρ
=
Since the arrival rate at the network is
, Little’s theorem implies that the average time a
customer spends in the network is:
K
γ
1
N
T
Kp
ρ
=
γγ−
ρ µ
−
γ
5.
The network is not time reversible. To see this, note that a transition of a customer from node
i
to node
i
+1 in the original network corresponds to a transition from node
i
+1 to node
i
in the
reversed network. Considering state n with n
i
>0, in the original network, we have:
1
(,
)
(
1
) 0
qnn e
e
p
+
−
+=
µ
−
>
while in the reversed network:
*
1
qnne e
+
−
since customers cannot move from node
i
to node
i
+1 in the reversed network.
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Problem 2 [20 points]:
Consider a closed Jackson network with
K
nodes and
M
customers. The
normalization constant for the network is:
12
1
()
K
K
nn
n
K
M
GM
++ =
=
ρρ
ρ
∑
…
"
(1)
1.
[3 points] Show that the number of terms in the summation on the righthand side of eq. (1) is
1
MK
M
+
−
2.
[7 points] Show that the normalization constant can be calculated based on the following
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