Introduction to flow control, Stop and Wait, GBN, Selective repeat, Error Control (CRC), Ethernet Frame Format: Header.
Example on Example on Sliding Window Protocol
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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.
13Comments
Flow Control and Ethernet Frame Format (IEEE 802.3)
\(ɑ = { Propagation delay (Tp)\over Transmission delay (Tt)} \)

Stop & Wait 
GBN 
SR 
Efficiency 
\( {1 \over1+ 2a}\) 
\( {N \over 1+2a}\) 
\( {N \over 1+2a}\) 
Sequence Numbers 
(1+1) 
(N+1) 
(N+N) 
Retransmissions (for 1 packet lost ) 
1 
N 
1 
Buffers 
(1+1) 
(N+1) 
(N+N) 
Bandwidth 
Low 
High 
Medium 
CPU 
Low 
Medium 
High 
Implementation 
Low 
Medium 
High 
In CSMA/CD, minimum length of packet to detect collision, L >= 2 * Tp * Bandwidth
Ethernet Frame Format (IEEE 802.3):
(Img.Source cse.iitk.ac.in/users/dheeraj/cs425/lec06.html/)
0Comment
Backoff Algorithm
After some transmission on Ethernet, current frame of A's is collided 2 times and B's is collided 4 times. The probability that A wins the present back off race is____.
Number of times A has been involved in collision = 2
Number of times B has been involved in collision = 4
A can choose any number between (0 to 2^{2}1 ) , i.e, (0,1,2,3)
B can choose any number between (0 to 2^{4}1 ) , i.e, (0,1,2,3,......,15)
It is given that A wins the race, so choices for (A,B) will be:
(0,1) , (0,2) ,(0,3)................ , (0,15)
(1,2), (1,3),...........................,(1,15)
(2,3),(2,4),............................,(2,15)
(3,4),(3,5),.............................,(3,15)
Total choices = 15 + 14 + 13 + 12 = 54 out of 64 choices.
So probability = 54 / 64 = 0.84
7Comments
SWP
Consider SWP with error probability 0.2. Suppose if sender wants to send 800 packets then how many packets have to be transmitted in total ?
Assume error probability is 'p' and number of packets that we want to send are 'n'.
So total transmission needed will be :
n+n*p + n*p*p + n*p*p*p.......
= n(1+p + p^{2} + p^{3} +..... )
= n(1/1p)
Here n=800 and p =0.2
So no. of transmissions = 800* (1/(10.2)
=800 * (1/0.8)
=1000
Hence total 1000 tranmission will take place.
4Comments
GBN
In GB4, if every 6th packet being transmitted is lost and if we have to send 10 packets, then how many transmissions are required?
So total 17 transmissions are required.
8Comments
Ethernet
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.
T_{t} >=2*T_{p}
L(length of the cable)<=(Data size*Signal speed)/(2*Bandwidth)
L<=(10,000 bits * 2 *10^{8}m/s)/(2*500*10^{6}b/s)
L<=2000m
Hence L<=2Km.
4Comments
 This quiz contains 5 questions on the topic Data Link Layer
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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.
Now you can attempt the quiz.
I can't see Quiz icon,why?
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(1p)^(n1)?? 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 =(1p)^{(n1)}.
for n stations required probability =n*( p * (1p)^{(n1)}).
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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