GATE CSE 2024 | Set 2 | Question: 45

Question

Consider an Ethernet segment with a transmission speed of $10^{8} \mathrm{bits} / \mathrm{sec}$ and a maximum segment length of $500$ meters. If the speed of propagation of the signal in the medium is $2 \times 10^{8} \mathrm{~meters} / \mathrm{sec}$, then the minimum frame size (in bits) required for collision detection is ___________.

Comments

is maximum segment length the same as in maximum segment size in TCP ? Why is MSL, the distance between two stations ?

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Actually, both are different things. 

Maximum Segment Size in TCP: the maximum payload we can have in an PDU at tcp. a good read on this . 

Maximum Segment Lifetime: the maximum time a TCP segment can exist in the network before being discarded. 

Maximum Segment Length in Ethernet: the maximum distance to which the host can detect collision.

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Similar Question : https://gateoverflow.in/1547/gate-cse-2013-question-36

Answer 1

Comments

Why you have taken the maximum segment length as a distance ?? please explain...

 

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Because that was the maximum distance in which collision can be possible in between !

 

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Maximum segment length is the longest distance in Ethernet where the sender can detect collision.

Answer 2

Given,

$Bw = 10^{8}\ \text{bits/sec}$

$d = 500\ m$

$v = 2\times10^{8}\ \text{m/sec}$

For ethernet,

$T_t\  \geq 2\ . \ T_p$

where $T_t$ is transmission time and $T_p$ is propagation time

$\implies\ \frac{f_s}{Bw}\ \geq\ 2\ \times \frac{d}{v}$

$\implies\ f_s\ \geq\ 2\ \times \frac{d}{v}\ \times Bw$

$\implies\ f_s\ \geq\ 2\ \times \frac{500\ m}{2\times10^{8}\ \text{m/sec}}\ \times 10^{8}\ \text{bits/sec}$

$\implies\ f_s \geq 500 \text{ bits}$

$\therefore\ f_{s_{min}} = 500\ \text{bit}$

Comments

ethernet use CSMA/CD as access control method to deal with collisions.

that's why to detect collision we use 

Transmision delay >= 2* propagation delay

Answer 3

Ans (C) : 500 Bits 
CSMA/CD is a network access method used in ethernet networks to handle collisions.

For CSMA/CD to work below condition need to be satisfied
T_x >= 2*T_p (T_x is Transmission Time while T_p is Propagation Delay)

$Frame Size / Transmission Speed >= 2* Distance / Signal Speed$

Put the values Ans is 500 Bits

Answer 4

Intuitive answer

The question is relatively easy, but we need to make sure just one important note. Here we can see. It is written. Consider an ethernet segment with a transmission speed of 10 to the power eight bits per second.

 

We can see here The value of throughput is given But if we Calculate By Assuming it to be throughput then we will never reach any ans so in these type of questions by Transmission speed, we mean to say bandwidth.

 

Now rest is easy just apply the condition to find minimum frame size, which is transmission time must be greater than equal to 2 into time for propagation

Answer 5

COMMENT BELOW IF YOU HAVE ANY DOUBT

Answer 6