Tanenbaum -A 1-km-long, 10-Mbps CSMA/CD LAN

Question

A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of
200 m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long,
including 32 bits of header, checksum, and other overhead. The first bit slot after a
successful transmission is reserved for the receiver to capture the channel in order to
send a 32-bit acknowledgement frame. What is the effective data rate, excluding
overhead, assuming that there are no collisions?

ans=3.8 mbps     please explain??

Answer 1

Here we have to consider the keyword "effective data rate" which means that data size that needs to be considered is excluding headers and ACKs..But in total time calculation , in transmission time calculation we need to take the frame size as a whole as it is transmitted in reality..So the question is solved as :

Comments

You can also proceed as follows:

1. Calculate the efficiency using Habibkhan's diagram. (0.435)

2. Calculate throughput = efficiency * Bandwidth = 0.435 * 10 = 4.35 Mbps

3. Effective throughput (Excluding headers) = 4.35 * (256-32)/256 = 3.8 Mbps.

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So there is no contention here?

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please explain how to calculate efficiency using Habibkhan's diagram. i tried but couldn’t understand

Answer 2

qno. 8  https://lms.ksu.edu.sa/bbcswebdav/users/mdahshan/Courses/CEN444/Tutorials/Tutorial-04.pdf

Comments

Thanku  Aashish S

very well

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The above link isn't working - here's the new link :-

http://webhome.csc.uvic.ca/~wkui/Courses/networks/midterm-F05.pdf