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

 

 

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3 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  …

 

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