161 views

Assume all frames have size $\text{F bits}$, and the propagation delay on the link is $\text{P sec}$, and the bandwidth is $\text{B bps}$. Assume also that the receiver piggybacks acknowledgements on frames going to the sender.
What should be the minimum window size $n$ (in frames) at the sender in order to effectively use the bandwidth?

1. $2+\dfrac{2PB}{F}$
2. $1+\dfrac{2PB}{F}$
3. $\dfrac{2PB}{F}$
4. $1+\dfrac{PB}{F}$
edited | 161 views
+3

@Shaik Masthan

For effective use of bandwidth that means 100% efficiency

In RTT time RTT*BW data should be sent i.e

= [Tt + Tp + Tack + Tp] * BW

= [2F/B + 2P] * B

in frames ==> [2F + 2PB] / F

wht is wrong here ?

+2
Yes, thats correct. You'll get option A - I have fixed the typo
+1
Thanks sir

I left it in thought that it will be marks to all :)
0
1= w * Tx/ Tx + 2Tp

w>= 1 + 2PB/F

Both A and B are correct.
0

@Utkarsh Joshi Again formula :( See the one given by @jatin khachane 1 - he also used formula but it is clear how each term comes in each step and hence that is correct. In this question ACKs are piggybacked as in GATE2009 question. So, transmission time for packets no longer becomes negligible.

+4
Sir, I did not use any formula. forgot to take Tx for ack.

efficiency =  w * Tx /  ( Tx + Tp + Tx + Tp)

1 = w * Tx / ( 2 Tx + 2 Tp)

w = 2(Tx + Tp) / Tx

= 2( 1 + PB/F)

Now after considering Tx for ack, it is correct. :)
+2
Yes. Just read question and you'll derive the formula and less chance of mistakes too.
0
okay sir :)
0
ack is piggybacked why you are adding tp 2 times....i think total time should tx+tp+tx.....

correct me if am wrong

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