0110 0101 0110 0111 0110 1001 1111 1001
0110 0110 0110 1000 0110 1010 1111 1101
→→→→ 取反可得 0000 0110
6. Consider the previous problem, but suppose that D has the value a. 1001 0001. b. 1010 0011. c. 0101 0101.
前一题题目：Consider the 7-bit generator, G=10011, and suppose that D has the value 1010101010. What is the value of R？求余而已，记住不要做减法而是做与运算就好。 a. R = 001 b. R = 101 c. R = 101
12. Consider three LANs interconnected by two routers, as shown in Figure 5.38.
0000 0010
a. Redraw the diagram to include adapters.重新画图
b. Assign IP addresses to all of the interfaces. For Subnet 1 use addresses of the form 111.111.111.xxx; for Subnet 2 uses addresses of the form 122.122.122.xxx; and for Subnet 3 use addresses of the form 133.133.133.xxx.所有的接口分配IP地址。 c. Assign MAC addresses to all of the adapters. a.b.c如图
d. Consider sending an IP datagram from Host A to Host F. Suppose all of
the ARP tables are up to date. Enumerate all the steps, as done for the single-router example in Section 5.4.2.
1. host A发送一个数据报，通过转发表查询F的IP，向路由器1发送，其中destination IP为133.133.133.12，MAC未知，source IP为111.111.111.12，source MAC为aa-aa-aa-aa-aa-aa。
2. 适配器更改destination的IP为111.111.111.12，MAC地址变为gg-gg-gg-gg-gg-gg
3. 路由器1发现目标IP和MAC不属于子网1中任何host，属于子网3（图中忘了画了，意会一下）。于是根据转发表向路由器2进行转发。Destination的IP为122.122.122.20，MAC为ii-ii-ii-ii-ii-ii。
4. 路由器2收到了数据报，发现host F在自己的子网内，于是修改destination的IP为133.133.133.12，MAC地址为ff-ff-ff-ff-ff-ff。修改source IP为133.133.133.20，MAC为jj-jj-jj-jj-jj-jj，然后向F发送数据报。 5. F收到来自A的数据报。
e. Repeat (d), now assuming that the ARP table in the sending host is empty (and the other tables are up to date). 发送方的ARP表为空，首先需要建立ARP表
1. host A发送一个广播，destination IP是255.255.255.255，MAC为空。Source IP为111.111.111.12，MAC为aa-aa-aa-aa-aa-aa
2. 适配器收到了来自host A的数据报，更新自己的ARP表，同时发送一个ACK给host A，告诉host A自己的IP、MAC。
3. host A建立ARP表
4. 如d小问所答，开始进行数据发送。
14. Recall that with the CSMA/CD protocol, the adapter waits K·512 bit times after a collision, where K is drawn randomly. For K = 100, how long does the adapter wait until returning to Step 2 for a 10 Mbps Ethernet? For a 100 Mbps Ethernet?
当网速 = 10 Mbps时，bit time = 1 bit / 10 Mbps t = 100 * 512 * 1 / (10^6) = 5.12 msec
当网速 = 100 Mbps时，bit time = 1 bit / 100 Mbps t = 100 * 512 * 1 / (10^7) = 0.512 msec
15. Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t = 0 bit times. They both detect collisions at t = 225 bit times. They finish transmitting a jam signal at t = 225 + 48 = 273 bit times. Suppose KA = 0 and KB = 1. At what time does B schedule its retransmission? At what time does A begin transmission? (Note: The
nodes must wait for an idle channel after returning to Step 2—see protocol.) At what time does A’s signal reach B? Does B refrain from transmitting at its scheduled time?
1. 因为A的K值 = 0，所以A从273 bit time开始检测是否冲突
2. 由于传输延迟的问题，B的最后一个bit要等到273 + 225 = 498 bit time才能传到A，也就是说此时A检测到没有冲突。
3. A传输前先等待96个bit time，即t = 498 + 96 = 594 bit time的时候A开始传输数据
4. 因为propagation delay = 225 bit time，所以t = 594 +225 = 819 bit time的时候A传输完毕
5. 因为KB = 1，当B等待1 * 512 * 1 = 512 bit time的时候，B可以重传，此时B开始检测信道是否空闲，此时t = 273 + 512 = 785 bit time 6. 此时B检测到信道忙，因此不能重传，等待信道空闲
7. 当t = 819 bit time的时候信道空闲，此时B等待96 bit time，即t = 96+819= 915 bit time的时候B可以开始重传
16. Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes is 225 bit times. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision.
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