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+2 votes


asked in Computer Networks by Boss (15.4k points) | 648 views

2 Answers

+4 votes

125000 Bytes (or 125 KB) should be the right answer.

Jamming signal is like a negative acknowledgement used to inform all the nodes connected to a link that collision occurred.

Suppose a node P is transmitting frames to other node Q which is 200 km away from P.

Now suppose the frame sent by P to Q collides with another frame, at a point very near to Q after travelling 199.99 km, then to inform node P about collision, the jamming signal has to travel almost 200 km.

So in the worst case, P will get informed about collision only after two way propagation delay.

& if P receives jamming signal after it has completed its frame transmission, then will never know that which frame caused the collision & that the frame that it sent to Q has been collided & lost.

So the frame that P is going to transmit should be large enough to be transmitted until a longest two way propagation time.

That is frame transmission time must be at least as much as 2*Propagation time.

Here 2 * Propagation time = 2 * 10-3 seconds,

& Transmission rate = 500 Mega bits per second.

Let frame size be L bits.

Then (L / 500 * 106) >= 2 * 10-3

Hence L >= 10bits or L >= 1,25,000 Bytes or L >= 125 KB.

answered by Boss (14.1k points)

In the question you mentioned above, order of magnitude of transmission time for jamming signal & two way propagation time is almost same.

Here I did not considered the  transmission time of jamming signal , as it is almost 0.1 microseconds & around 20000 times smaller than the two way propagation time.

But to be very precise I should have considered it,

& then it will be like

 (L / 500 * 106) >= 2 * 10-3 + (48/(500*106)).

Here, if we would have considered jamming signal , answer would be 125006 Bytes.

So, 6 Bytes negligible? As the question asking in terms of Bytes..
Yes, it will be 125007 Bytes as we have to give answer in bytes, we must consider jamming signal transmission time.I am sorry, My bad.

But answer in test series given as 125000 , so, here my question is in Gate what should we consider - 125000 or 125006 ?

If the size of jamming signal is given then, then we must consider its transmission time & 125006 Bytes will be right answer to this question.
0 votes
TT ≦ 2PT  
   x/500*10^6 = 2*(200*10^3 / 2*10^8)

x= 2000 bits
x= 250 bytes
answered by Boss (16.2k points)
edited by

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