Solutions to Chapter 6 - Communication Networks

Communication Networks (2 nd Edition)
Chapter 6 Solutions
c. Non-persistent CSMA-CD.
collision
A
0 1τ 2τ 3τ 4τ 5τ 6τ 7τ 8τ
A detects collision, aborts
collision
B
0 1τ 2τ 3τ 4τ 5τ 6τ 7τ 8τ
B detects collision, aborts
C
0 1τ 2τ 3τ 4τ 5τ 6τ 7τ 8τ
Station C senses channel busy and backs off
11. Estimate the maximum throughput of the CDPD system assuming packets are 1096 bytes in length.
Hint: What is a for this system
Solution:
During the contention interval, it takes two microblock delays (2t µblock ) for a station to know if it has
captured the reverse channel. Divide the time into minislots of 2t µblock .
As in the CSMA-CD system, the maximum probability of successful transmission is
the average number of minislots in a contention period is
E J ]
1
P
[ .
= max
success
= e
max 1
P success
= , and
e
The maximum throughput occurs when the channel time is spent in packet transmission followed by a
contention period. After the packet transmission E[X], each station spends one t µblock to determine if
the channel is idle or busy.
The maximum throughput is given by
max
E[
X ]
1
=
=
E[
X ] + tublock + e(2tublock
) 1+
(2e
+ 1)
a
ρ where
t ublock
a = .
E[X ]
Calculation:
60bits
t ublock
= = 3. 1ms
19,200bps
Assume the packet is already in HDLC format.
1096 bytes = 8768 bits
Packet length after segmentation and header insertion:
Leon-Garcia/Widjaja 5