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3 Answers

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Packet=M

BW= R bsp

Header= 4h bits

Data= M*L/M => L

Packet Size = data+header =>L+4h

Transmission time(t) of one packet= Packet size/Bandwidth =>L+4h/R

In Packet Switching using the concept of pipelining,So

Time taken by first packet to reach destinantion = (No. of hops * transmission time of one packet)

                                                                      = Q*(L+4h)/R

Time taken by remaining packet = no. of remaining packet * transmission time of one packet

                                                = (M-1)*(L+4h)/R

Total time = Q*(L+4h)/R + (M-1)*(L+4h)/R

              =(Q+M-1)* (L+4h)/R
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There are M number of packets.

Since each packet contains L bits and 4h bits are for header, so,

Total number of bits in a packet are (L+4h).

But total number of packets are M.

Total number of bits to be transmitted becomes M*(L+4h).

When we need to transmit it only one link having bandwidth R bps,

We get, M*(L+4h)/R.

But since we have Q such links,

Time taken in second = Q*M*(L+4h)/R.

Correct me if I am wrong.

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