计算机网络自顶向下方法第四版答案整理.doc

Problem 4a) bytes are sent every 20 msec. Thus the transmission rate is (160+h)*8/20 Kbps = (64+0.4h)Kbpsb)IP header:20bytesUDP header: 8bytesRTP header: 12bytesh=40 bytes (a 25% increase in the transmission rate!)Problem 10a) The delay of packet 2 is 7 slots. The delay of packet 3 is 9 slots. The delay of packet 4 is 8 slots. The delay of packet 5 is 7 slots. The delay of packet 6 is 9 slots. The delay of packet 7 is 8 slots. The delay of packet 8 is 8 slots.b) Packets 3, 4, 6, 7, and 8 will not be received in time for their playout if playout begins at t-8.c) Packets 3 and 6 will not be received in time for their playout if playout begins at t=9.d) No packets will arrive after their playout time if playout time begins at t=10.Problem 12a) Both schemes require 25% more bandwidth. The first scheme has a playback delay of 5 packets. The second scheme has a delay of 2 packets.b) The first scheme will be able to reconstruct the original high-quality audio encoding. The second scheme will use the low quality audio encoding for the lost packets and will therefore have lower overall quality.c) For the first scheme, many of the original packets will be lost and audio quality will be very poor. For the second scheme, every audio chunk will be available at the receiver, although only the low quality version will be available for every other chunk. Audio quality will be acceptable.Problem 20a) One possible sequence is 1 2 1 3 1 2 1 3 1 2 1 3 Another possible sequence is 1 1 2 1 1 3 1 1 2 1 1 3 1 1 2 1 1 3 b) 1 1 3 1 1 3 1 1 3 1 1 3 Problem 21a)PacketTime leaving the queueDelay10021132143255264276387298310921110212113Average Delay1.91b) PacketTime leaving the queueDelay10022231046554167573089495010103118012113Average Delay1.91c) PacketTime leaving the queueDelay10022234341053066475287299410114118012102Average Delay1.91d) PacketTime leaving the queueDelayNote100WFQ222WFQ310WFQ454WFQ530WFQ675Idealized WFQ scheduling741WFQ894WFQ961Idealized WFQ scheduling10103WFQ1180WFQ12113WFQAverage Delay1.91e) It can be noticed that the average delay for all four cases is the same (1.91 seconds).Problem 22a) PacketTime leaving the queueDelay10024435441053062076389497210103118012113Average Delay1.91b) PacketTime leaving the queueDelay10021133242156364275289497210103118012113Average Delay1.91c) PacketTime leaving the queueDelay10021132143259666474187295010811111312102Average Delay1.91Problem 23Time SlotPackets in the queueNumber of tokens in bucket01, 2, 3213, 4124,5135,614615-167, 8279, 1018101Time SlotPackets in output buffer01, 2132435465-67, 879810Problem 24Time SlotPackets in the queueNumber of tokens in bucket01, 2, 3213, 422523624-25-267, 8