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The round trip delay between x and y is given as 60 ms and  bandwidth of link between X and Y is 512 KBps. What is the optimal window size (in packets) if the packet size is 64 bytes and channel is full duplex

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Best answer
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8 votes

To find optimal window size for sender , 

 we consider N=1+2a  

=>( N/1+2a) = 1 .....  where N = sender window size and a = Tp / Tt

Given, RTT  between x and y  as 60 ms =  (60 * 10-3 ) seconds and  bandwidth of link between X and Y is 512 KBps = (512 * 103 ) Bytes/seconds.

It is Full duplex  channel,  so total data send from x to y is = (2* Tp) * Band Width

= RTT * band width

=( 60 * 10-3)  * ( 512 * 103)

= 60 * 512 

= 30720 Bytes

packet size = 64 bytes.

Number of window require  = 30720 / 64 = 480 

so optimal window size is N = 1+2a = 1+480 = 481

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3 votes
3 votes
As it is full duplex communication link, to achieve optimal window size , sender need to send data during RTT i.e during 60ms.

Now, Data bytes send during 60ms is = 512KB*60ms = 512*60 byte.

so, Window size = no of packet send = 512*60/64 = 480

 

Now if the channel is given as half duplex in that case:

For half duplex communication link,  to achieve optimal window size , sender need to send data during one way progagation time i.e during 60/2 =30ms .

Now, Data bytes send during 30ms is = 512KB*30ms = 512*30 bytes.

so, Window size = no of packet send = 512*30/64 = 240

 

 

@Bikram or Arjun Sir Can you please verify If the above procedure is correct or not?
edited by
1 votes
1 votes

window size (optimal) considering 100% efficiency

= 1 + 2 * (propogation time / transmission time)

= 1 + (60 * 10-3 * 512 * 10-3 ) / 64

= 1 + 60 * 8

= 1 + 480

= 481

 

Answer:

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