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A 1 KM long CSMA/CD with a propagation delay 5 microsec has 10 mbps bandwidth.Repeaters are not allowed in the system.Data bits of 224 ,header bits of 32 bits are in packet.One slot time is reserved for the sender to start its transmission and another slot time is reserved for receiver to capture the channel to send 32 bits acknowledgements frame.What is the effective data rate?

my approach-

total transmission time Tt- (224+32+32)/10x10^6

slot time-2*propagation delay= 2*5 microsec

now for data rate = total data/total time

total data=224

total time =2*slottime +2*Tp+Tt

is this approach correct ? in the solution they are only considering 1 slot time

1 Answer

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In this question , following courses of action will occur :

A) First prior to transmit the data , the sender needs to acquire the channel and we know for that ,

Time required  =  2 * Propogation delay  =  2 * 5  =   10 μs

B) Then data is placed into channel for that we need time = (Data + Header) / (Bandwidth)

                                                                                    = 256 / 10 *106

                                                                                    = 25.6 μs

C) Then the data goes through channel for that propogation delay  =  5  μs

D) Then another 2 P.T. time is required by receiver to acquire the channel..That is what they are referring to as 'slots' ..So time taken for that                                                                = 10 μs

E) Now for placing acknowledgement , we need time        =  32 / 10 * 106

                                                                                    =  3.2 μs

F) Now again propogation delay of ack is involved            =  5 μs

Hence effective data rate           =     Data size ( excluding header and acknowledgement) / Total time involved as found above

                                                =    224 / ( 10 + 25.6 + 5 + 10 + 3.2 + 5 )

                                                =    3.8 Mbps  

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