May 2024
Intermediate to advanced
309 pages
13h 33m
English
Content preview from Data Communications and Computer Networks by Pearson
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Media Access Control 159
Thus, total channel throughput (S) can be determined as
S = S
1
+ S
2
+ S
3
+ S
4
+ S
5
= 0.0905 + 0.1291 + 0.1637 + 0.1947 + 0.2222
= 0.8002
21. Consider a CSMA/CD LAN running at 1 Gbps over a 1 km long cable with no repeaters.
The signal propagation speed is 200 m/µs. What is the minimum frame size?
Ans: Frame transmission time = 2 × propagation time
Here, propagation time = distance/propagation speed
That is, propagation time = 1000/200 = 1000/2 = 5 µs
So, frame transmission time = 2 × 5 = 10 µs
The minimum size of the frame is given as
data rate × frame transmission time
= 1 Gbps × 10 µs
= 1 × 10
9
× 10 × 10
–6
= 10,000 bits ...
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