计算机网络课后习题习题四、五

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 194.47.21.130.

解答：

先写成二进制：

11000010,0010101111,0001010,10000010

所以，它的点分十进制为：

194.47.21.130

39. A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts it can handle?

(39.Interent上一个网络的子网掩码为255.255.240.0.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 address 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255.255.240.0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 198.16.0.0. Suppose that four organizations, 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198.16.0.0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4000,2000,4000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: 198.16.0.0 –

198.16.15.255 written as 198.16.0.0/20

B: 198.16.16.0 – 198.23.15.255 written as 198.16.16.0/21

C: 198.16.32.0 – 198.16. 47.255 written as 198.16.32.0/20

D: 198.16.64.0 – 198.16.95.255 written as 198.16.64.0/19

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：

1111 1111 1111 1111 1111 0000 0000 0000

128+64+32+16=240

198.16.0.0 198.16.16.0 198.16.32.0 198.16.48.0 198.16.64.0 1111 1000 0000 0000

248

1110 0000 0000 0000

224

198.16.0.0 198.16.32.0 198.16.64.0 198.16.96.0 ——————————————————————————————————————41. A router has just received the following new IP addresses: 57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21, and 57.6.120.0/21. If all of them use the same outgoing line, can they be aggregated? If 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:57.6.96.0/27,57.6.104.0/21,57.6.112.0/21和57.6.120.0/21.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)2

104=(0110 0100)2

112=(0110 1000)2

120=(0110 1110)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 57.6.96.0/19 上。

42. The set of IP addresses from 29.18.0.0 to 19.18.128.255 has been aggregated to 29.18.0.0/17. However, there is a gap of 1024 unassigned addresses from 29.18.60.0 to 29.18.63.255 that are now suddenly assigned to 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29.18.0.0到29.18.128.255之间的IP地址集合已经被聚集到29.18.0.0/17.然而,这

里有一段空隙地址,即从29.18.60.0到29.18.63.255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