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+6 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 * 10^{6}) >= 2 * 10^{-3}

Hence L >= 10^{6 }bits or L >= 1,25,000 Bytes or L >= 125 KB.

+1

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*10^{6})).

0

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..

So, 6 Bytes negligible? As the question asking in terms of Bytes..

0

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.

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