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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?

1. $(p+h)x/b \hspace{0.1cm} bits$
2. $(p+h)x/pb \hspace{0.1cm} bits$
3. $px/b \hspace{0.1cm} bits$
4. $hx/pb \hspace{0.1cm} bits$

### 1 comment

its ok and thanks
Prateek

Nice explnation
Thanks
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.