![]() (b) In this case, after sending each data packet we must wait one RTT before sending the next packet. ![]() This will be added to the above transfer time equation to get the total transfer time. The initial handshaking time before data is sent is 2*50 ms = 100ms. Solution: Transfer time = RTT + ((1/Bandwidth) * Transfer size) (a) Bandwidth = 1.5 Mbps, RTT = 50 ms The RTT in the transfer time calculation will be the propagation delay, which in this case is 50ms /2 = 25ms. (d) The bandwidth is infinite, and during the first RTT we can send one packet (21−1), during the second RTT we can send two packets (22−1), during the third we can send four (23−1), and so on. (c) The bandwidth is “infinite,” meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. (b) The bandwidth is 1.5 Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. (a) The bandwidth is 1.5 Mbps, and data packets can be sent continuously. Q2] Calculate the total time required to transfer a 1,000-KB file in the following cases, assuming an RTT of 50 ms, a packet size of 1-KB data, and an initial 2×RTT of “handshaking” before data is sent. As an alternative, explore the whois interface at. Try whois and whois princeton, for starters. Read the man page documentation for whois and experiment with it. HOMEWORK 1 F14 Q1] The Unix utility whois can be used to find the domain name corresponding to an organization, or vice versa.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |