计算机网络课后习邈指导 陕西师范大学 p=(9)×p”×(1-pm)120- d.有21个或者更多用户在传输的概率为： P=1-∑（C20)p(1-p120- 我们用中心极限定理来求此概率，令X为独立随机变量，则P(X=1)=p。 P(21个或更多的用户")=1- 信-（品 =P≤3)=P≤270=097 当Z为标准正态随机变量时，所求概率P(21个或更多的用户")≈0.003。 P9.Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an isprovided witha1 Mbps link.Us swhen bus but are busy generating data only with probabilityp.1.Suppose that the 1 Mbps link is replaced by a I Gbps linl a.What is N.the maximum number of users that can be supported simultaneously under circuit switching? b Now onsider packet switchingand auser popuation formua (n terms ofp an Nu are sending dat 含：a链客布变为1Gps.用户传输速率为10OKPs,则可支持提大的用户数量 1Gps 10bps N=100Kbps=10X10bps=10000台： b.令传输用户数量为，则多于N用户发送数据的概率为： 或者 1-20ra-r P10.Consider a packet of length L which begins at end system A and travels over three links to a destination end system.These three links are acket switches.Let ds and dente the lnsth propon pd.and the rin1.2.3.The pcke switch delays each packet by dAssuming no queuing delays,in terms of d.s.R.(i=1.2.3).and L.what is the total end-to-end delay for the packet?Suppose now the packet is 1.500 bytes,the propagation 5 计算机网络课后习题指导 陕西师范大学 5 p = ￾ ￾￾￾ ￾ ￾ × ￾ ￾ × (1 − ￾) ￾￾￾￾￾ d. 有 21 个或者更多用户在传输的概率为： P=1 − ￾ ￾ ￾￾￾ ￾ ￾￾￾ (1 − ￾) ￾￾￾￾￾ ￾￾ ￾ 我们用中心极限定理来求此概率，令X￾为独立随机变量，则P￾￾￾ = 1￾ = ￾。 P￾"21 个或更多的用户"￾ = 1 − P￾￾ ￾￾ ￾￾￾ ￾￾￾ ≤ 21￾ P ￾￾ ￾￾ ￾￾￾ ￾￾￾ ≤ 21￾ = ￾ ￾ ∑ ￾￾ −12 ￾￾￾ ￾￾￾ √120 × 0.1 × 0.9 ≤ 9 √120 × 0.1 × 0.9￾ ≈ P ￾￾ ≤ 9 3.286￾ = ￾(￾ ≤ 2.74) = 0.997 当 Z 为标准正态随机变量时，所求概率P￾"21 个或更多的用户"￾ ≈ 0.003。 P9. Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability p = 0.1. Suppose that the 1 Mbps link is replaced by a 1 Gbps link. a. What is N, the maximum number of users that can be supported simultaneously under circuit switching? b. Now consider packet switching and a user population of M users. Give a formula (in terms of p, M, N) for the probability that more than N users are sending data. 答：a. 链路带宽为 1Gbps，用户传输速率为 100Kbps，则可支持最大的用户数量为： N = 1￾￾￾ 100￾￾￾￾ = 10￾￾￾￾ 100 × 10￾￾￾￾ = 10000 台； b. 令传输用户数量为 n，则多于 N 用户发送数据的概率为： ￾ ￾ ￾ ￾ ￾ ￾￾(1 − ￾) ￾￾￾ ￾ ￾￾￾￾￾ 或者 1− ￾ ￾ ￾ ￾ ￾ ￾￾ ￾ ￾￾￾ (1 − ￾) ￾￾￾ P10. Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2, 3. The packet switch delays each packet by dproc. Assuming no queuing delays, in terms of di, si, Ri,(i = 1,2,3), and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation