More on Ethernet, CSMA/CD


Content covered: 

CSMA CD in Ethernet


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pritam's picture

This was just an example. The jamming signal can also be different on the basis of frequency. 

However, +15 to -15 is not a range. It is +ve and -ve voltage of the signal.

suman08513's picture

A doubt regd Jamming Signal:


If there is a collision in the middle of the channel (say exactly at the midpoint) who will create the Jamming signal. I mean at the mid point there is no node to generate the jamming signal.


Jamming signal is usually generated by a node to notify all the other nodes in the LAN that there a collision. If there is a collision takes place just before the receiver then the receiver generates the jamming signal to notify the sender that there is a collision.


In your video illustration you told that the collision happens at the middle then who will generate the jamming signal at the middle. There some node to generate the Jamming signal . Isn't it?

shraddhagami's picture

If adapter detects another transmission while trnsmitting. It aborts and send Jamming signal.(Jamming signal is nothing but have different frequency than data)

suman08513's picture

I don't think your assumption is correct Shraddhagami. Jamming Signal is to be generated by some node. If not how can it automatically be created.


But your assumption of Jamming Signal has different frequency is correct . It is to avoid the collision with data signal.

shraddhagami's picture

yeah Thankyou for correcting me.

Illustration on finding frame length in Ethernet


shiveshroy's picture

Sir Minimum Length of the packet is 277.7bits  i.e packet size should be greater than or at least equal to 277.7 bits , So I think after rounding off the packet size should be 278bits (Even if it comes to be 277.2 bits, After rounding off the packet size should be 278 technically). Please clear this confusion because for numerical type questions 277 is different from 278.

pritam's picture

Sivesh, You are absolutely correct. It will be 278 as

L/B >= RTT

I will add a comment in the video too. Thanks for pointing out.

hradeshpatel's picture

i think here TT >= 2*PT so here L>= 277.78 so here sir 277.78 is minmum so here 278 bit is right?? plz check sir??

In Practice. packet length or slot time is always a power of 2.

eg: in 10Mbps ethernet it will be 512 bit times

Content covered: 

Backoff Algorithm in CSMA/CD


suman08513's picture

Sir, here K is choosen randomly from 0 to 2^n-1 not 2^(n-1). where n=collision number. 

pritam's picture

Yes Suman, That way a writing mistake. I corrected it in the later part of the video.

Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is: 

  • Explanation

    Transmission Speed = 10Mbps.
    Round trip propagation delay = 46.4 ms
    The minimum frame size = (Round Trip Propagation Delay) * (Transmission Speed) = 10*(10^6)*46.4*(10^-3) = 464 * 10^3 = 464 Kbit


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But according to iit-2005 ans-sheet the ans of this que is 512 bits. So Can anyone will clarify that what should be the answer ?

arjunsinghra's picture

means u didnt try yourself first see here in given question RTT is 46.4ms

if RTT is 46.4mircro second then it will be 512bits

nishantgupta's picture

Sir what is the significance of jamming signal here???

should we consider the transmission time of jamming signal to be negligible?

amit0368's picture

jamming signal has no significance here because we only need enough transmitting time so that we can detect the collision. so TT>=2TP.

The distance between two stations M and N is L kilometers. All frames are k bits long. The propagation delay per
kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that the processing delay is negligible, what are the
number of bits for the sequence number field in a frame for maximum utilization, when the is used ?

  • Explanation

    Distance between stations = L KM
    Propogation delay per KM = t seconds
    Total propagation delay = Lt seconds
     Frame size = k bits
    Channel capacity = R bits/second
    Transmission Time = k/R

    Let n be the window size.
    UtiliZation = n/(1+2a) where a = Propagation time / transmission time
                = n/[1 + 2LtR/k]
                = nk/(2LtR+k) 
    For maximum utilization: nk = 2LtR + k
    Therefore, n = (2LtR+k)/k
    Number of bits needed for n frames is Logn.
    So Number of bits for sequence numbers =  ⌈log2((2LtR+k)/k) ⌉


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How we can say no. Of bit needed for n frame is logn . 

dashish's picture


for storing 4 no.'s (0, 1, 2, 3) we need 2 bits to store them in binary..(00,01,10,11)..

similar as in an n-bit storage, we can store 2numbers... just reverse this...

n-> 2n  , so let 2= k => n= logk... take ceil if k is not the power of 2

therefore storing 

if we are considering here go back n, then for sequence number , why we are not adding 1 (sender window)

dashish's picture


we consider here simply sliding window protocol as here nothing  as such GBN or SR is mentioned.

Sliding Window protocol is a theoritical protocol of which the practical applications are GBN and SR.

So, for max. utilization, we send as much of the packets as we can in RTT time of first packet, viz., 1+2a.

Thank u

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____.

  • Explanation

    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 22-1 ) , i.e, (0,1,2,3)
    B can choose any number between (0 to 24-1 ) , 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)
    Total choices = 15 + 14 + 13 + 12 = 54 out of 64 choices.

    So probability = 54 / 64 = 0.84


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Because that's what <b>it</b> means!

What is probability of collisons? is it 4/64.

jagmt's picture

1 Collision: (0,0)(1,1)(2,2)(3,3) {4/64}.
2 B wins the race: (1,0)(2,0)(3,0)(2,1)(3,1)(3,2) {6/64}

avdominic's picture

I think the ans given in this question is of B's winning the race which is 54/64

For A's winning the race it is 6/64......

and for collision it is 4/64....

Correct me If I'm wrong......:-)

For collision both A and B's waiting time must be equal  (0,0) (1,1) (2,2) (3,3)      - only 4 possibilities out of 64              for collision

For  B to win it has to transmit before A (1,0) (2,0) (2,1) (3,0) (3,1) (3,2)   -    only 6 possibilities out of 64                           for B to win the race. 

In remaining all cases, A will have lesser waiting time than B  so A will win the race  in  -  54 possibilities out of 64 .       for A to win the race.

  • This quiz contains 5 questions on the topic Ethernet
  • Lean well before you attempt the quiz
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Difficulty Level:  basic
No. of Questions:  5

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Please explain the solution for the 3rd question of the quiz.

dashish's picture

@techtud @pritam @sumitverma

pls explain how to solve this-

How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame?

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