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GATE CSE 2005 | Question: 74

• edited by
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113 113 votes

Suppose the round trip propagation delay for a $10\text{ Mbps}$ Ethernet having $48\text{-bit}$ jamming signal is $46.4\ \mu s$. The minimum frame size is:

  1. $94$
  2. $416$
  3. $464$
  4. $512$

14 Answers

Best answer
158 158 votes

The sender must be able to detect a collision before completely sending a frame.
So, the minimum frame length must be such that, before the frame completely leaves the sender any collision must be detected.

Now, the worst case for collision detection is when the start of the frame is about to reach the receiver and the receiver starts sending. Collision happens and a jam signal is produced and this signal must travel to the sender.

The time for this will be the time for the start of the frame to reach near the receiver $+$ time for the jam signal to reach the sender $+$ transmission time for the jam signal.

(We do not need to include transmission time for the frame as as soon as the first bit of the frame arrives, the receiver will have detected it). Time for the start of the frame to reach near the receiver $+$ Time for the jam signal to reach the sender $=$ Round trip propagation delay $= 46.4 \mu s$. So, 

$46.4 +\dfrac{48}{10}\text{(48 bits at 10 Mbps takes 4.8 micro sec.) = 51.2 μs.}$

Now, the frame length must be such that its transmission time must be
more than $51.2 \mu s$.

So, minimum frame length$= 51.2\times 10^{-6}\times 10\times 10^6=512\text{ bits}$.

A reference question from Peterson Davie:

43.  Suppose the round-trip propagation delay for Ethernet is $46.4 \mu s$. This yields a minimum packet size of $512$ bits $(464$ bits corresponding to propagation delay $+ 48$ bits of jam signal$).$

  1. What happens to the minimum packet size if the delay time is held constant, and the signaling rate rises to $100$ Mbps?
  2. What are the drawbacks to so large a minimum packet size?
  3. If compatibility were not an issue, how might the specifications be written so as to permit a smaller minimum packet size?

Another reference for requiring jam signal bits to be included for minimum frame size.

Can collision be detected by the source without getting the full jam signal (by a change in current)?

Probably yes. But to be safe (from signal loss) the source waits for the entire jam signal. See below link

Correct Answer: D.

• edited by
26 26 votes

We know, 

Frame size>2*B*Propagation delay

Hence, 

Frame size >10*10^6*46.4*10^(-6)

Frame size min=464 bits

But in question its mentioned that 48 bits jamming signal is present.

Hence, 

Frame size min= 464+48=512 bits

Hence correct option is D

3 3 votes
Transmission delay>= 2*Prropagation delay

L/B.W=46.4 microsec

L=46.4 microsec*10Mbps

L=464b

note:-Use of Jamming signal:-If a collision is heard, both of the senders will send a jam signal over the Ethernet. This jam signal indicates to all other devices on the Ethernet segment that there has been a collision, and they should not send data onto the wire. After sending the jam signal, each of the senders will wait a random amount of time(decided by backoff algorithm) before beginning the entire process over. The random time helps to ensure that the two devices don't transmit simultaneously again.
2 2 votes
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L=464 Kbits
It has nothing to do with jamming signal.
2 2 votes
Tt >= 2*td + time for sending jamming signal

RTT = 2* Td =46.6 µs

time for sending jamming signal = 48 bits / 10Mbps =4.8 µs

Now,

Tt >= 2*td + time for sending jamming signal

Tt >= 46.6 µs + 4.8 µs

Tt >= 51.2 µs

L >= 51.2 µs * 10 Mbps

L >= 512
1 1 vote

To detect collision in worst case(collision occur at the end of link) transmission time must be greater than Tp(for data) + Tt(jamming) + Tp(jamming).

so 

Tt ≥ 2Tp + Tt(jamming)

hence

L ≥ (46.4+4.8)*10 bits

so L=512bits 

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