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Assume a CSMA/CD network that transmit data at a rate of 20 Mbps over a 20 km cable with no repeater. If minimum frame size require for network is 5,000 KB then speed of the signal inside cable is __________ (in Km/sec).

What should be correct ans : 19.53 or 20

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Following the standard equation for CSMA/CD :

           T.T. = 2 * P.T.

==>   (5000 * 8 * 210) / (20 * 106)  =  (2 * 20 * 103) / x

==>   x   =   (20 * 2 * 20 * 109) / (5000 * 8 * 210)  m/s

==>   x   =   (8 * 1011) / (4 * 104 * 210)

==>   x   =   1011 / (104 * 29

==>   x   =    10000 / 512

==>   x   =    19.53 km /s

Hence the required speed is 19.53 km/s..The main issue is whenever we consider something related to system like packet size etc. so we take powers of 2 in such cases..And for something related to channel and not system like bandwidth etc are taken in powers of 10..

Reference of Unit System Consideration : Petersen Davie on CN

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