Ethernet

Question

Consider two Station communicating via 1 Mbps Satellite link with propagation delay of 270 msec. The Satellite merely serves to retransmit data received from one station to another,with negligible switch delay. If Ethernet frame size is 1024 bits with 3 bit sequence number,then max. possible data throughput is.................

Comments

transmission time is 1 ms is right. why taking 1Mbps as 10^6 bps instead of 2^20 bps? 2^20 bps is more logical and intuitive, as we generally deal with bits and bytes.

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what will be efficiency in this case???

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correct me if i am wrong for efficiency (for GBN)

sender window size=7

maximum possible window size=1+2*$T_{P}/T_{t}$

now putting this value

efficiency=7/528.34

now for throughput=efficiency*bandwidth

Answer 1

In the Above question, the maximum windows Size is 1+2a

if we are using the max sender windows size(stop and wait ARQ, GBN ARQ, Selective Repeat ARG)

windows Size of GBN is (using 3 sequence bit ) 7

than efficiency = 7/(1+2a)        (a= tp/t_t )

tt =1024/1*10^6 = 1024 micro second 

a =270/1024=0.26

1+2a =0.26*2+1=1.52

efficiency =7/1.52  =4.6

 

 

 

Answer 2

Caption

Is the solution correct??

Comments

Hey what kind of answer is this ??