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A network with CSMA/CD protocol in the MAC layer is running at $1 \text{Gbps}$ over a $1 \text{km}$ cable with no repeaters. The signal speed in the cable is $2 \times  10^{8}\text{m/sec}$. The minimum frame size for this network should be:

  1. $10000 \text{bits}$
  2. $10000 \text{bytes}$
  3. $5000\text{ bits}$
  4. $5000 \text{bytes}$
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2 Answers

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Best answer

Answer is(A)

Minimum frame size is needed to ensure that collisions are detected properly. The minimum frame size ensures that before a frame is completely send, it would be notified of any possible collision and hence collision detection works perfectly.

In CSMA/CD a sender won't send a packet if it senses that another sender is using it. So, assume a sender A and a receiver B. When sender sends a packet, receiver might use the cable until it is notified that a packet is being send to it. The receiver will be notified as soon as the first bit arrives that a packet is coming and it won't send any packet after this until that packet is finished. So, in the worst case for collision, receiver will transmit a packet back to the sender just before the first bit of the packet reaches it. (If $t_d$ is the propagation delay of the channel, this time would be just $t_d$). In this case, surely there will be collision. But for the sender to detect it, it should be notified of B's packet before the sending of the first packet finishes. i.e., when B's packet arrives at A (takes another $t_d$ time), A shouldn't have finished transmission of the first packet for it to detect a collision. i.e., A should be still continuing the sending of the packet in this time interval of $2 \times t_d$. Thus,

The amount of bits that can be transmitted by A in $2\times t_d$ time should be less than the frame size (S) (sending of the frame shouldn't finish in this time)

Amount of bits transmitted in time $t$ is $\text{bandwidth} \times t$ and propagation delay- $t_d$ is $\dfrac{\text{distance}}{\text{link speed}}$

So, $S  \geq 2\times \text{bandwidth} \times t_d$

$\geq 2 \times 10^9 \times \dfrac{ 1000}{ 2 \times 10^8}$

$\geq 10000$ bits

edited by
how 10^9 ? I mean the Bandwidth
Oh..gbps it is, got it, sorry for the noob question


$1 Gbps = 1$ Gigabits per second $= 10^9$ bits per second

$1 GBps = 1$ GigaByte per Second $ = 10^9 $ Byte per second

But In some cases

$1Kilobits = 1000 bits$


$1 Kilobits = 1024 bits$

please explain me sir

in case of data its 1024 and for frequence its 1000 bits

@Arjun Veetran sir 1gb=10^9 byte so why you take here bits ??


@Shubham Aggarwal I think you might have understood the distinction between GB and Gb by now.

GB stands for "Gigabyte", and Gb stands for "Gigabit"


@Lakshman Patel RJIT

in case of bandwidth u will take  1Kilobits=1000bits

In case of data

1Kb = 1024 bits

In case of bandwidth

1Kbps = 1000 bits/second

Guys who are wondering why $A$ needs to keep transmitting(i.e. why it's $2 \times T_p$ and not just $T_p$) in order to detect collision is -

1. This protocol works at physical layer. 

2. It's not possible to identify whether a particular packet belongs to a host uniquely at physical layer.

3. There are no acknowledgements here, so once a package is lost, no way to know for sender if it's actually lost or sent successfully. So, if host $A$ sends a packet completely, and then there's a collision, there's no way this collision will be identified by $A$.

4. Carrier Sensing means, the hosts can only sense if the line is empty or not. So, even if the collision signal is sent from a point, $A$ will not be able to identify that if the collision was due to $A$'s packet or anyone else's. 

+1 vote

CSMA/CD can detect collisions always if below condition is satisfied:

$T.T \geq 2 * P.T$ where T.T is Transmission time and P.T is Propagation time.

If above condition is satisfied CSMA/CD can work always and can not stuck.

So, We know that $T.T = \frac{Data size}{Band width}$ and $P.T = \frac{Distance}{Velocity}$

Therefore,  $\frac{Data size}{Band width} \geqslant 2 * \frac{Distance}{Velocity}$

                    $\frac{Data size}{2\ast 10^{9}} \geqslant 2 * \frac{1000 }{2\ast 10^{8}}$

by solving,  Data size $\geqslant$ 10000 bits.

Hence, Option (A) is correct.


You can apply same procedure for below questions also by keeping $T.T \geq 2 * P.T$ condition satisfied. Same question as above, asked for 'Frame size' all other things are given. This question is also easily solvable by the same method, here we have to find 'velocity' all other things are given in question. This question also solvable by above formula, here we have to find 'Frame size' all other things are given in question.

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