ࡱ> VXU[ bjbj 2*jj %lDDD8|4I:"444+++3I5I5I5I5I5I5I$J L`YI++^+++YI?/44SnI?/?/?/+v443I?/+3I?/?/+0H03I4 $|C-b D[,H3II0IH`&M-&&M3I?/CIS 617 Chapter 6 Jun Young Park Instructor: Dr. Latecki April 21, 2004 14. Why does UDP exist? Would it not have been enough to just let user processes send raw IP packets? By using UDP, a segment will be delivered correctly to the specified application because UDP uses source and destination ports while raw IP packet does not include ports. That is, a segment cannot be delivered to a specified application as a raw IP packet. 18. Both UDP and TCP use port numbers to identify the destination entity when delivering a message. Give two reasons for why these protocols invented a new abstract ID (port numbers), instead of using process Ids, which already existed when these protocols were designed. Process ID is not static. Since process ID is assigned dynamically to the process when a process is created (forked), it cannot be used for well-known service such as HTTP. In some situations, one process manages multiple sockets on different TSAPs. If process ID is used, it cannot be implemented since one process can have one process ID. 39. What is the fastest line speed at which a host can blast out 1500 bytes TCP payloads with a 120 sec maximum packet lifetime without having the sequence numbers wrap around? Take TCP, IP, and Ethernet overhead into consideration. Assume that Ethernet frames may be sent continuously. There are two possibilities: one without segmentation and one with segmentation. Case 1: Assume that there is no segmentation needed. IP header = 20 bytes TCP header = 20 bytes TCP payload can be 1500 bytes. Ethernet header = 8 + 6 + 6 + 2 bytes Ethernet checksum = 4 bytes Therefore, the size of first frame is 1566 bytes. The time to live = 120 sec. The length of sequence number = 32 bits. Therefore, the maximum number of bytes that a host can send out without wrapping around the sequence number is 2 ^32/ 1500. Since Ethernet frames may be sent continuously, the maximum line speed, V can be: V <= 1566 * 8 * (2^32 / 1500) (bits)/120 (sec) = 298929723.8016 bits/sec = 298.9 Mbps The maximum line speed is 298.9 Mbps.  SHAPE \* MERGEFORMAT  Case 2: Since the maximum size of Ethernet payload is 1500 bytes, the TCP payload needs to be segmented as followings. Frame 1: IP header = 20 bytes TCP header = 20 bytes Since the maximum Ethernet Payload is 1500 bytes including IP header and TCP header, TCP payload can be 1460 bytes. Ethernet header = 8 + 6 + 6 + 2 bytes Ethernet checksum = 4 bytes Therefore, the size of first frame is 1526 bytes Frame 2: IP header = 20 bytes TCP header = 20 bytes TCP payload remained after segmentation = 40 bytes (= 1500 1460 bytes) Ethernet header = 8 + 6 + 6 + 2 bytes Ethernet checksum = 4 bytes Therefore, the size of second frame is 106 bytes The time to live = 120 sec. The length of sequence number = 32 bits. Therefore, the maximum number of bytes that a host can send out without wrapping around the sequence number is 2 ^32/ 1500. Since Ethernet frames may be sent continuously, the maximum line speed, V can be: V <= (1526 + 106) * 8 * (2^32 / 1500) (bits)/120 (sec) = 1632 * 8 * (2^32 /1500) (bits)/ 120 (sec) = 311528294.5365 bits/sec = 311.5 Mbps The maximum line speed is 311.5 Mbps. 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