Here is asked about the time taken by source to upload the total packets. So, **B **should be the answer.

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Suppose that $x$ bits of user data are to be transmitted over $k-$hop path in a packet-switched network as a series of packets each containing $p$ data bits and $h$ header bits with $x>>(p+h)$. The bit rate of lines is $b$ bps and propagation delay is negligible.What is the time taken by the source to transmit total bits?

- $(p+h)x/b \hspace{0.1cm} bits$
- $(p+h)x/pb \hspace{0.1cm} bits$
- $px/b \hspace{0.1cm} bits$
- $hx/pb \hspace{0.1cm} bits$

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The total number of packets needed is x/p. The total of data+header works out to (p+h)x/p bits. The source takes (p+h)x/(pb) sec to transmit this. The forwarding of the last packet by intermediate routers on the way takes up (k-1)(p+h)/b sec. Adding up the two gives the time to clear the full pipe (that is, from start at source of the first bit of first packet to move out till the receipt of the last bit of the last packet at the destination). Total time = (p+h)x/(pb) + (p+h)(k-1)/b.