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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. $hx/pb \hspace{0.1cm} bits$
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first of all in the first step you have taken number of packets=x/p but packet size is p+h so you should take it as x/(p+h) and we know that in case of packet switching total time taken=k*tt +k*pt generally pt is neglisible if not given and hence total time =k*tt when k is number of hops and tt is transmission time.Please let me know if I am wrong
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@Anil ji First of all  At (k-1) hops data will be transmitted,at $k_{th}$ hop data will reach .header is overhead of hop ,so overhead will be considered in transmission time.packet switching works like pipelining in which first packet will be take all delays but remaining packet will be reach out in unit time means one tt time(propagation delay negligible here).
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Everything is fine but according to me 1'st packet will be transmitted in k hops
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Transmission time means time taken for putting data on link ,you have to put data only on (k-1) hops at kth hop data will be reach ,,I hope u understand what I mean
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suppose we have  2 intermediate nodes between sender and reciever and we have n packets to send then in that case (packet switching) first packet will take time=3*TT and other n-1 packets will take TT each means total time=3*TT+(n-1)*TT.So why are u considering k-1 hops ?Data will be transmitted in k hops .Please let me know if i am wrong
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I m sorry ,I was assuming hope as node ,hope means link between two router or node so now you are right for k hope time will be for first packet "k×Tt" in my solution only one thing will be change ,in place of k-1 that will be only k.

And final answer will be (k+x/p-1)×(p+h)/b.
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its ok and thanks
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Prateek

Nice explnation
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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.
answered ago by Junior (537 points)

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