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Ethernet, CSMA / CD

Understanding CSMA CD in Ethernet
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CSMA CD in Ethernet

In CSMA/CD, minimum length of packet to detect collision, L >= 2 * Tp * Bandwidth

example: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.

Tt >=2*Tp
L(length of the cable)<=(Data size*Signal speed)/(2*Bandwidth)
L<=(10,000 bits * 2 *108m/s)/(2*500*106b/s)
L<=2000m
Hence L<=2Km.

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Understanding Backoff Algorithm in CSMA/CD
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Backoff Algorithm in CSMA/CD

Example: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 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)
(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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Minimum Frame Size

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: 

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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Consider the distance between two stations M and N is L kilometers

Consider 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 ?

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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Illustration on finding frame length in Ethernet

  • This quiz contains 5 questions on the topic Network Layer.
  • You can attempt the quiz unlimited number of times.

Difficulty Level:  basic