##### Host A is sending data to host B over a f : GATE 2003

Please first try yourself this question and then look at the answer and explanation.

First , please write down the data and convert it into simplest unit possible. it eliminates calculation mistakes and give always uniform answer ,which you can convert later in required quantity.

My personal approach in to convert

1) bandwidth from mbps , gbps etc to bps,

2) Data size ( ack or frame) from bytes to bits ,

3) time ( PT,Transmission time, RTT etc ) from Microsec , millisec to sec.

Sliding window protocol, Window size W = 5, frame size F = 1000 bits , F/C that is transmission time of frame over link of capacity/bandwidth C = 50 * \(10^{-6}\)sec , ack A = 0 bits ( given negligible) , PT = 200*\(10^{-6}\)sec.

we dont have bandwidth C, but we can find out it by using Transmission time TT= 50 *\(10^{-6}\)sec

F/C= 50* \(10^{-6}\) sec=> 1000 bits/C= 50* \(10^{-6}\)sec => C = 1000 / 50 * \(10^{-6}\) bps = 20 * \(10^6\) bps => C = 20mbps

Efficiency of anything is defined as useful time out of total time.

Useful time will be transmission time TT of the frame which is 50 microsec * Window size = 50 * 5 * \(10^{-6}\)sec

total time will be transmission time + RTT + Ack transmission time,* not W *TT + RTT+ A/C. A/C is 0 as A = 0.*

**In sliding window protocol, the transmission of 2nd and subsequent frames can be overlapped with propagation time of 1st frame and so on. so total time does not include Window size * transmission time. Rather it is just transmission of 1st frame and thats it**.

So efficiency of the protocol will be => TT/ ( TT+RTT+A/C) => TT / TT + RTT => 5*50 *\(10^{-6}\)sec /( 50*\(10^{-6}\)+ 2*200*\(10^{-6}\)) sec => 250/(50+400) => 250 /450 => 0.5555 => 55.55 %

Throughput = Efficiency of link * bandwidth = 0.5555* 20 * \(10^6\) bps => 11.11.*\(10^6\) bps.

answer : 11.11 * \(10^6\)bps.

where is the quiz sir?

Quiz is live now.

pls paste the link of quiz. i am not able to see the quiz link on this topic.

Now you can see.

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Now you can attempt the quiz.

Where we can find quiz solutions?

Can you please share the solutions for the quiz

nyc quiz ... good work

Can someone clear doubt regarding 2D parity as written in NPTEL slides that 2D parity can detect a burst error of n bits and >n+1 bits is detected but probabilty is less compared to n bits?

According to me, 1bit errors can be corrected and detected

2 bit errors can't be corrected and detected.

@amit0368 @pritam

There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?

How is this np(1-p)^(n-1)?? pls explain...

@dashish

probability that only one station transmits in a given slot = probability that one station transmit * probability that others don't transmit.

probability that only one station transmits in a given slot = p.

probability that others don't transmit =(1-p)

^{(n-1)}.for n stations required probability =n*( p * (1-p)

^{(n-1)}).n is multiplied because we can choose the transmitting station in^{n}C_{1}ways.hope it helps.

@amit0368

Thanks! I missed the point u made bold......

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