计算机网络课后习题习题五

1、.chapter five第五章习题38. convert the ip address whose hexadecimal representation is c22f1582 to dotted decimal notation.(38. 如果一个 ip 地址的十六进制表示c22f1582, 请将它转换成点分十进制标记. )solution:the address is 30.解答：先写成二进制：11000010,0010101111,0001010,10000010所以，它的点分十进制为：3039. a network on the inter

2、net has a subnet mask of . what is the maximum numberof hosts it can handle?(39.interent上一个网络的子网掩码为.请问它最多能够处理多少台主机?)solution:the mask is 20 bits long, so the network part is 20 bits. the remaining 12 bits are for the host, so 4096 host addresses exist.normally, the host add

3、ress is 4096-2=4094. because the first address be used for network and the last one for broadcast.解答：从子网掩码 可知，它还有 12位用于作主机号。故它的容量有 2的 12次方，也即有4096地址。除去全 0 和全 1 地址，它最多能够处理4094 台主机40. a large number of consecutive ip address are available starting at . suppose that four organiza

4、tions, a, b, c, and d, request 4000, 2000, 4000, and 8000 addresses, respectively, and in that order. for each of these, give the first ip address assigned, the last ip address assigned, and the mask in the w.x.y.z/s notation.(40. 假定从 开始有大量连续的 ip 地址可以使用 . 现在 4个组织 a,b,c 和 d按照顺序依次申请 4000,20

5、00,4000 和 8000个地址 . 对于每一个申请 , 请利用 w.x.y.z/s 的形式写出所分配的第一个 ip 地址 , 以及掩码 . )solution:to start with, all the requests are rounded up to a power of two. the starting address, ending address, and mask are as follows:a: 55 written as /20b: 55 written a

6、s /21c: 198.16. 47.255 written as /20d: 55 written as /19解答：因为只能是 2的整数次方的, 故应分别借 4096,2048,4096,8192个 ip 地址。它们分别为2的 12;.次方 ,2 的 11次方 ,2 的11次方 ,2 的13次方 . 故可有如下分配方案：组织首地址末地址w.x.y.z/s 的形式a55/20b198.16.16.

7、055/21c55/20d55/1941. a router has just received the following new ip addresses: /21, /21, /21, and /21. if all of them use the same outgoing line, can they beaggregated? if

8、 so, to what? if not, why not?(41. 一台路由器刚刚接收到一下新的ip 地址 :/27,/21,/21和/21. 如果所有这些地址都使用同一条输出线路 , 那么 , 它们可以被聚集起来吗 ? 如果可以的话 , 它们被聚集到那个地址上 ?如果不可以的话 , 请问为什么 ?)solution:they can be aggregated to 57.6.96/19.解答：96=(0110 0000)2104=(0110 0100)2112=(0110 1000)2120=(0110 111

9、0)2可以看出，四个 ip 地址前 19位都是相同的（前面 57的 8位以及 6的 8位和后面 011这 3位，共 19 位）故得聚合到地址/19上。42. the set of ip addresses from to 55 has been aggregated to /17. however, there is a gap of 1024 unassigned addresses from to 55 that are now suddenly assigned to

10、a host using a different outgoing line. is it now necessary to split up the aggregate address into its constituent blocks, add the new block to the table, and then see if any reaggregation is possible? if not, what can be done instead?(42. 从 到 55之间的 ip 地址集合已经被聚集到 /17. 然而

11、 , 这里有一段空隙地址, 即从 到55 之间的 1024个地址还没有被分配 . 现在这段空隙地址突然要被分配给一台使用不同输出线路的主机. 请问是否有必要将聚集地址分割成几块 , 然后把新的地址块加入到路由表中, 再看一看是否可以重新聚集?如果没有必要的话, 那该怎么办 ?)solution:it is sufficient to add one new table entry: /22 for the new block. if an incoming packet matches both /17 and 29.

12、18.0.0./22, the longest one wins. this rule makes it possible to assign a large block to one outgoing line but make an exception for one or more small blocks within its range.解答：;.没有必要。只要在路由表中添加一项：/22就可以了。当有一个分组到来时，如果它既匹配 /17，又匹配 /22，那么它将被发送到掩码位数较大的目标地址，即 /22。这样做的

13、好处是使得一个大段的地址能够被指定到一个目标，但又允许其中少量的地址出现例外的情况。43. a router has the following (cidr) entries in its routing table:address/masknext hop/22interface 0/22interface 1/23router 1defaultrouter 2for each of the following ip addresses, what does the router do if a packet with tha

14、t address arrives?a. (a) 0b. (b) 4c. (c) d. (d) e. (e) 43.一台路由器的路由表中有以下的（cidr）表项：地址 / 掩码下一跳/22接口 0/22接口 1/23路由器 1默认路由器 2对于下面的每一个 ip 地址，请问，如果一个到达分组的目标地址为该 ip 地址， 那么路由器该怎么办？(a)135 46 63 10(b)135 46 57 14(c)135 46 5

15、2 2(d)192 53 40 7;.(e)192 53 56 7solution:the packets are routed as follows:(a) interface 1(b) interface 0(c) router 2(d) router 1解答：(a)0和 做与运算得到 ，故发送给接口1；(b)4和 做与运算得到 ，故发送给接口0；(c)和 做与运算得到 ，故发送给

16、路由器2；(d)和 做与运算得到 ，故发送给路由器1；(e)和 做与运算得到 ，故发送给路由器2。第四章习题24.some books quote the maximum size of an ethernet frame as 1518 bytes instead of 1500 bytes. are they wrong? explain your answer.（ 24有些书将以太网帧的最大长度说成是 1518 字节，而不是 1500 字节，这些书错

17、了吗？请说明你的理由。 ）solution:the payload is 1500 bytes, but when the destination address, source address,type/length, and checksum fields are counted too, the total is indeed 1518.解答：没错，以太网帧的最大净荷为 1500 字节，算上目标地址 6 字节、源地址 6 字节、类型 2 字节、校验和 4 字节，则为 1518字节。不同的书常可能出现题目中的两种不同的表述，但它们的实质是一样的。42 briefly describe th

18、e difference between store-and-forward and cut-through switches.（简略地描述一下存储-转发型交换机和直通型交换机之间的区别。）solution:a store-and-forward switch stores each incoming frame in its entirety, then examines it and forwards it. a cut-through switch starts to forward incoming frames before they have arrived completely.

19、 as soon as the destination address is in, the forwarding can begin.解答：;.存储转发：整个帧完整接收并存储到缓冲区，对整个帧进行差错检验，然后再查表找出目的端口并转发。优点是进行差错校验，错误不会扩散到目的网段，缺点是交换延迟比较大。直通式：因为转发仅依赖于目的地址目标地址，所以只要收到帧的前 6 个字节 (目标地址字段 )，就可查表找出目的端口并转发。 优点是交换延迟小， 缺点是无法进行差错校验， 帧错误会扩散到目的网段。43 store-and-forward switches have an advantage over cut-thro