Host A is sending data to host B over a f : GATE 2003
GATE 2003 
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5   packets each. Data packets (sent only from A to B) are all 1000 bits long and the transmission time for such a packet is 50 microsec Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 microsec. What is the maximum achievable throughput in this communication?
(A) 7.69 × 106 bps (B) 11.11 × 106 bps
(C) 12.33 × 106 bps (D) 15.00 × 106 bp

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.

13Comments
Ashish Kumar Goyal dashish 21 Jan 2017 02:11 pm

@chetnawadhwa

but full duplex means that u can send and recieve at same time. the bandwidth of channel is shared

here, the packet first need to be received by the receiver then only it can send Ack. Both actions can't happen simultaneously.

so it is always 2*PT

Vivek Vikram Singh vivek14 12 Jan 2017 01:55 pm

RTT is always 2*PT.

Harshita Sharma harshitasharma 7 Feb 2017 08:51 am

Sliding Window uses full duplex only.