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Question from a networking class:
"In a csma/cd lan of 2 km running at 100 megabits per second, what would be the minimum frame size to hear all collisions?"
Looked all over and can't find info anywhere on how to do this. Is there a formula for this problem? Thanks for any help.
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Since data travels at a finite speed a minimum packet size was required to ensure that if a collision happens it happened everywhere. The larger you make the minimum packet size the larger and/or faster you can make the network before CSMA/CD breaks down.
In the standard IEEE 802.3 Ethernet specification, the minimum frame size was 64 bytes and the maximum was 1518 bytes (since expanded to 1522 bytes). The preamble consists of 7 bytes followed by a single byte as a start frame delineator.
The minimum frame size on Ethernet is so that by the time the beginning of a frame gets all the way across a maximum-width network, if a collision is detected there on the far side of the network, there's still enough time for the jam signal (collision detection notification) to make it all the way back across the ...
Q #5) How does CSMA/CD detect collision? Answer: CSMA/CD detects collisions by sensing transmissions from other stations first and starts transmitting when the carrier is idle.
bandwidth delay product is the amount of data on transit. propagation delay is the amount of time it takes for the signal to propagate over network.
propagation delay=(lenght of wire/speed of signal). assuming copper wire i.e speed =2/3* speed of light
propagation delay =(2000/(2*3*10^8/3)) =10us
round trip time is the time taken for message to travel from sender to receiver and back from receiver to sender.
round trip time =2*propagation delay =20us
minimum frame size =bandwidth *delay (rtt)
frame size = bandwidth *rtt =100Mbps*20us=2000bits
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