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CSMA CD in Ethernet
Illustration on finding frame length in Ethernet
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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.
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
Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48bit 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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More CommentsBut according to iit2005 anssheet the ans of this que is 512 bits. So Can anyone will clarify that what should be the answer ?
should we consider the transmission time of jamming signal to be negligible?
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/RLet 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 = ⌈log_{2}((2LtR+k)/k) ⌉
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More CommentsHow we can say no. Of bit needed for n frame is logn .
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 nbit storage, we can store 2^{n }numbers... just reverse this...
n> 2^{n } , so let 2^{n }= k => n= log_{2 }k... 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)
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 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
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More CommentsBecause that's what <b>it</b> means!
What is probability of collisons? is it 4/64.
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.
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Please explain the solution for the 3rd question of the quiz.
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?
@dashish @limitless This might help http://gateoverflow.in/3278/gate2008it18
Responses
More CommentsThis 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.
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?
If adapter detects another transmission while trnsmitting. It aborts and send Jamming signal.(Jamming signal is nothing but have different frequency than data)
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.
yeah Thankyou for correcting me.