1 votes 1 votes 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? $(p+h)x/b \hspace{0.1cm} bits$ $(p+h)x/pb \hspace{0.1cm} bits$ $px/b \hspace{0.1cm} bits$ $hx/pb \hspace{0.1cm} bits$ Computer Networks packet-switching + – Anil Ji asked May 23, 2018 • edited May 29, 2018 by srestha Anil Ji 3.8k views answer comment Share Follow See 1 comment See all 1 1 comment reply roh commented Jul 15, 2020 reply Follow Share Here is asked about the time taken by source to upload the total packets. So, B should be the answer. 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes Answer is wrong because asking about total time ,,answer is given in bits,,although acc. to me solution will be- Prateek Raghuvanshi answered May 24, 2018 • selected May 31, 2018 by Anil Ji Prateek Raghuvanshi comment Share Follow See all 9 Comments See all 9 9 Comments reply Anil Ji commented May 24, 2018 reply Follow Share 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 1 votes 1 votes Prateek Raghuvanshi commented May 29, 2018 reply Follow Share @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). 0 votes 0 votes Anil Ji commented May 30, 2018 reply Follow Share Everything is fine but according to me 1'st packet will be transmitted in k hops 0 votes 0 votes Prateek Raghuvanshi commented May 30, 2018 reply Follow Share 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 0 votes 0 votes Anil Ji commented May 30, 2018 reply Follow Share 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 0 votes 0 votes Prateek Raghuvanshi commented May 30, 2018 reply Follow Share 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. 0 votes 0 votes Anil Ji commented May 31, 2018 reply Follow Share its ok and thanks 1 votes 1 votes abhishekmehta4u commented May 31, 2018 reply Follow Share Prateek Nice explnation 1 votes 1 votes Prateek Raghuvanshi commented May 31, 2018 reply Follow Share Thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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. KULDEEP SINGH 2 answered Jan 21, 2019 KULDEEP SINGH 2 comment Share Follow See all 0 reply Please log in or register to add a comment.