More on Ethernet, CSMA/CD

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##### Understanding CSMA CD in Ethernet
Content covered:

CSMA CD in Ethernet

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Illustration on finding frame length in Ethernet

##### Understanding Backoff Algorithm in CSMA/CD
Content covered:

Backoff Algorithm in CSMA/CD

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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

##### 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 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

• This quiz contains 5 questions on the topic Ethernet
• Lean well before you attempt the quiz
• You can attempt the quiz unlimited number of times.

Difficulty Level:  basic
No. of Questions:  5

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Utkarsh Kumar 9 Jan 2017 06:36 pm

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

Ashish Kumar Goyal 24 Jan 2017 12:50 pm

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?

Jatin Dhankhar 6 Feb 2017 03:17 pm