计算机网络课后习遐指导 陕西师范大学 speed on all three links is 2.510'8 m/s,the transmission rates of all three links are 2 Mbps.the end-to-end delay 答：由题意，设两个端系统为A和B,分组交换器为S,和S2,链路传输速率为R,由题意总 时延由A,S,和S2上的传输时延，在交换器S,和S的处理时延以及在三条链路上的传播时延 构成，则传输时延为： +L1500×8×3b =18ms 传搭时延为： 4ra=4+d+4_500+40+100)×10n =40ms 2.5×108m/s 则总时延为 dcotal drram +dprop +dproc=18+0+3+3=64ms P11.In the above problem,suppose R=Ra=Rs=R and dpoe=0.Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire nacket to a ive.Wha is the nd-ten delay? 答：由题意可知 分组链路 速率相 7点处理 延为0，路由 不再存储转发而是在 接收到后立即发送每一个比特b,所以在两个路由器上就不存在传输时延，类似的，传输 时延为： 1500×8bi drm=片-2x10ps6ms 传播时延不变 4rp-4+4+色-50o0+40ot10o0x102m=40nmg 2.5×108m/s 端到端的总时延为：6+40=46ms P1.Apacket switch receivesa packet and determnes the oubound link to which the packet should be forwarded.When the packet arrives,one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted.Packets are transmitted in bps.What is the queuing delay for the packet?More generally,what is the queuing delay when all packets have length L,the transmission rate isR.x bits of the currently-being-transmitted packet have been transmitted,and n packets are already in the queue? 答：由题意，新到达的分组必须等待它之前到达的分组传输完成才能传输，之前分组包括 等待传输的分组和正在传输的分组两 部分，则： a.新的分组需要等待1500×4+1500÷2=6750 bytesf的传输，即其排队时延为 1500×4+1500÷2)×86i=27ms 2×106bps b.更一般的情况，当前分组已传输bits,队列有n个分组等待传输，则新到达的分组排队计算机网络课后习题指导 陕西师范大学 6 speed on all three links is 2.5 ·10*8 m/s, the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay? 答：由题意，设两个端系统为 A 和 B，分组交换器为￾￾和￾￾，链路传输速率为 R，由题意总 时延由 A，￾￾和￾￾上的传输时延，在交换器￾￾和￾￾的处理时延以及在三条链路上的传播时延 构成，则传输时延为： ￾￾￾￾￾￾ = ￾￾ + ￾￾￾+￾￾￾ ￾ = 1500 × 8 × 3￾￾￾ 2 × 10￾￾￾￾ = 18ms 传播时延为： ￾￾￾￾￾ = ￾￾ + ￾￾ + ￾￾ ￾ = (5000 +4000 +1000) × 10￾￾ 2.5 × 10￾￾/￾ = 40￾￾ 则总时延为： d￾￾￾￾￾ = ￾￾￾￾￾ +￾￾￾￾￾ + d￾￾￾￾ = 18 + 40 +3 + 3 = 64￾￾ P11. In the above problem, suppose R1 = R2 = R3 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire packet to arrive. What is the end-to-end delay? 答：由题意可知，分组链路传输速率相同，节点处理时延为 0，路由器不再存储转发而是在 接收到后立即发送每一个比特 bit，所以在两个路由器上就不存在传输时延，类似的，传输 时延为： ￾￾￾￾￾￾ = ￾￾ ￾ = 1500 × 8￾￾￾ 2 × 10￾￾￾￾ = 6ms 传播时延不变： ￾￾￾￾￾ = ￾￾ + ￾￾ + ￾￾ ￾ = (5000 +4000 +1000) × 10￾￾ 2.5 × 10￾￾/￾ = 40￾￾ 端到端的总时延为：6 +40 = 46ms P12. A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue? 答：由题意，新到达的分组必须等待它之前到达的分组传输完成才能传输，之前分组包括 等待传输的分组和正在传输的分组两部分，则： a. 新的分组需要等待1500 × 4 + 1500 ÷ 2 = 6750bytes的传输，即其排队时延为： (1500 × 4 +1500 ÷2) × 8￾￾￾ 2 × 10￾￾￾￾ = 27￾￾ b. 更一般的情况，当前分组已传输 x bits，队列有 n 个分组等待传输，则新到达的分组排队