0 votes 0 votes Computer Networks computer-networks + – Akshay Koli 4 asked Jan 17, 2018 Akshay Koli 4 568 views answer comment Share Follow See all 11 Comments See all 11 11 Comments reply joshi_nitish commented Jan 17, 2018 reply Follow Share it will take $7.05s$ 0 votes 0 votes Anu007 commented Jan 17, 2018 reply Follow Share Yes it will take 7050 msec = 7.05 sec 0 votes 0 votes Akshay Koli 4 commented Jan 17, 2018 reply Follow Share if propagation delay is 20msec per hop, packet size is same, data rate is 50kbps and all packets follow the same 4-hop path then what is delay in sending and 100000 bits in terms of packet?what will be the answer of this question? 0 votes 0 votes Anu007 commented Jan 17, 2018 reply Follow Share 2140 msec = 2.14 sec 0 votes 0 votes joshi_nitish commented Jan 17, 2018 reply Follow Share then it will be, $2140ms=2.14s$ 0 votes 0 votes joshi_nitish commented Jan 17, 2018 reply Follow Share i have assumed packet size=1000 B for your later qsn 0 votes 0 votes Anu007 commented Jan 17, 2018 reply Follow Share Nitish 1000 bits na not Byte. 0 votes 0 votes joshi_nitish commented Jan 17, 2018 reply Follow Share yes, typo! 1 votes 1 votes Anjan commented Jan 17, 2018 reply Follow Share Anu007 please give hint to solve this type of problems... 0 votes 0 votes Akshay Koli 4 commented Jan 17, 2018 reply Follow Share in second question, to send 1 packet delay is 160msec and to send remaining 99 packet delay is 1980. in first question, to send 1 packet delay is 2550 and to send remaining 9 packet what was the delay?isn't the delay is 90msec only(which means after 1 packet, all next packet will require only propagation delay)? 0 votes 0 votes vishal chugh commented Jan 17, 2018 reply Follow Share I solved it by calculating time for each packet. But is this method correct? Total time taken = (Propagation Time + Transmission Time)*Number of Hops + (Transmission Time)*Remaining Number of Packets 0 votes 0 votes Please log in or register to add a comment.