Made Easy Test Series

Question

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

Comments

ans should be 480  here given RTT so never add one for frame transsmission time

Answer 1

@LRU

 

Optimal window size =(1+2a) where a =Tp/Tt …..

Full duplex link,

so window size = 2* Tp * Band Width

= 60 * 10-3 * 512*103

=30720 bytes…

 

packet size = 64 bytes...

 

window size in terms of packets = 30720 / 64 = 480….

 

optimal window size = 480+1= 481 ...

 

1. https://gateoverflow.in/144462/Cn-optimal-windows-size

 

 

Comments

@aaa 1 Why you added 1 to 480 ?

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We have to consider the receiver's end as well or not? Like we are fine with only sender's window size. Please elaborate this.

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@Isha_99 @Vishal_kumar98

 

optimal window size is when link utilization is 100% …

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....

In case optimal window size, we can indirectly say that maximum throughput should be achieved. It means link utilization is 100% in case of optimal window...

And as you said , window size is defined as number of frames sent in 1 RTT--> This is window size in terms of packets ....

we can say , optimal window size is 1 greater than number of frames send in 1 RTT .

See this thread also https://gateoverflow.in/107261/optimal-window-size  …