Hint: Suppose at time t = 0 bit times, A begins transmitting a frame. In the worst case, A transmits a minimum-sized frame of 512 + 64 bit times. So A would finish transmitting the frame at t = 512 + 64 bit times. Thus, the answer is no, if B’s signal reaches A before bit time t = 512 + 64 bits. In the worst case, when does B’s signal reach A?
因为A最快传输时间t = 512 + 64 bit time，看了答案，微积分神马的可以让我shi了……
Let Y be a random variable denoting the number of slots until a success: P(Y = m) = β (1 -β)^(m - 1)
Where β is the probability of a success.
This is a geometric distribution, which has mean 1 /β. The number of consecutive wasted slots is X = Y – 1 that x = E[ X ] = E[ Y ] -1 = (1 -β) /β β = Np(1 - p)^(N - 1)
x = {1 – Np * (1 - p)^(N - 1)} / {Np * (1 - p)^(N - 1)}
efficiency = k / (k + x) = k / {k + [1 – Np * (1 - p)^(N - 1)] / [Np(1 - p)^(N - 1)]}
19. Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable, and that they each have one frame of 1,000 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are four repeaters between A and B,
each inserting a 20-bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol. Ignore the jam signal and the 96-bit time delay.
a. What is the one-way propagation delay (including repeater delays) between A and B in seconds? Assume that the signal propagation speed is 2·10^8 m/sec.
传输时间t0 = 900 m / [2 * (10^8) m / sec] = 4.5 μsec Repeater 传输延迟 t1 = 4 * 20 bit / 10 Mbps = 8 μsec T = t0 + t1 = 12.5 μsec
b. At what time (in seconds) is A’s packet completely delivered at B? （为什么这里答案没有96个bit time的等待？？？） 1 bit time = 1 / 10 Mbps = 0.1 μsec
当t = 12.5 μsec 时，A、B检测到冲突，停止传输。
A的K值 = 0，于是A立即开始检测信道是否空闲，当B传输延迟结束后，信道空闲，此时t2 = 12.5 +12.5 = 25 μsec
A等待两个96 bit time，即 25 + 9.6 = 34.6 μsec时，A开始重传
B在A的传输延迟：34.6 + 12.5 μsec = 47.1 μsec之后，检测到A的第一个bit
B等待的K值 = 1，即B在512 * 0.1 μsec = 51.2 μsec之后重传，此时A已
经传输，信道忙，B侦听信道，等待信道空闲 A的传输时间tA = 1000 bit / 10 Mbps
A传输完毕的时间T = 1000 bit / 10 Mbps + 47.1 μsec= 147.1 μsec
c. Now suppose that only A has a packet to send and that the repeaters are replaced with switches. Suppose that each switch has a 20-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A’s packet delivered at B? （repeater和switch和router区别在哪里？？？）
t = 12.5 μsec + 5 * 100 μsec = 512.5μsec
21. Suppose now that the leftmost router in Figure 5.38 is replaced by a switch. Hosts A, B, C, and D and the right router are all star-connected into this switch. Give the source and destination MAC addresses in the frame encapsulating this IP datagram as the frame is transmitted (i) from A to the switch, (ii) from the switch to the right router, (iii) from the right router to F. Also give the source and destination IP addresses in the IP datagram encapsulated within the frame at each of these points in time.
i.
A → switch：
source IP:111.111.111.12 source MAC: aa-aa-aa-aa-aa-aa destination IP:111.111.111.10
destination MAC:gg-gg-gg-gg-gg-gg
ii. switch → router source IP:122.122.122.10
source MAC: hh-hh-hh-hh-hh-hh destination IP:122.122.122.20 destination MAC:ii-ii-ii-ii-ii-ii iii.
right router → F:
source IP:133.133.133.10 source MAC: jj-jj-jj-jj-jj-jj
destination IP:133.133.133.12 destination MAC:ff-ff-ff-ff-ff-ff
23. Consider Figure 5.26. Suppose that all links are 10 Mbps. What is the maximum total aggregate throughput that can be achieved among the 14 end systems in this network? Why?
因为所有节点的最大传输速度都是10Mbps，所以最大吞吐量为14 * 10Mbps = 140 Mbps
24. Suppose the three departmental switches in Figure 5.26 are replaced by hubs. All links are 10 Mbps. What is the maximum total aggregate throughput that can be achieved among the 14 end system in this network? Why?
All of the 14 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of 10Mbps is possible among the 14 end systems.