ACE DOUBT

Question

Comments

34.13 kbps ??

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maximum data that can senders send  = 2 x Tp x bandwidth

                                                                  = 2 x 256 x 10 ^-3 x 64 x 1000

                                                                  = 32768 bits

but senders sends only 4096 bits

 senders window size require  to send 4096 bits  =  ceil (32768 / 4096 ) = 8

but actually they used 15

therefore , they didn't utilize the throughput 100%

therefore , throughput = (8 /15) *64 Kbps= 34.13 Kbps

Answer 1

Transmission TIme = $\frac{Frame\,size}{Bandwidth} =\frac{4096}{64*10^{3}}=64\,ms$

Round Trip Time = $2*Propogation\,delay = 2*256 = 512\,ms$

For maximum throughput,

$Window\,size = 1+\frac{Round\,Trip\,Time}{Transmission\,time}=1+\frac{512}{64}=9$

As, window size is $15\geq9$, So the throughput will be maximum

Throughput = Bandwidth = 64 kbps

Correct me if I'm wrong

Comments

@Magma https://gateoverflow.in/59740/cn 

I think, this is a same type of question.

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yes 64kbps is the ans.

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Got it !

@Shobhit Joshi thanks