in Computer Networks edited by
5,807 views
28 votes
28 votes

The minimum frame size required for a CSMA/CD based computer network running at $1\text{Gbps}$ on a $200m$ cable with a link speed of $2 \times10^{8}\text{m/sec}$ is:

  1. $125 \text{bytes}$
  2. $250 \text{bytes}$
  3. $500 \text{bytes}$
  4. None of the above
in Computer Networks edited by
5.8k views

3 Answers

37 votes
37 votes
Best answer
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, $S  \geq 2\times \text{bandwidth} \times t_d$

$\geq 2 \times 10^9 \times \dfrac{ 200}{ 2 \times 10^8}$

$\geq 2000$ bits

$\geq 250$ bytes

Correct Answer: $B$
edited by
by
2 votes
2 votes
B) 250 bytes
0 votes
0 votes

Check the answer of the similar question below given by me there, one can easily find solution of this question.

https://gateoverflow.in/3834/gate2005-it-71

Answer:

Related questions