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 解答：

先写成二进制：

,00,0001010,

所以，它的点分十进制为：

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

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 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 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 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4000,2000,4000和8000个地址.对于每一个申请,请利用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: –

written as – written as – . written as – written as 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：

128+64+32+16=240

1111 1000 0000 0000

248

1110 0000 0000 0000

224

——————————————————————————————————————41. A router has just received the following new IP addresses: and 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:和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 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）

故得聚合到地址上。

42. The set of IP addresses from to has been aggregated to However, there is

a gap of 1024 unassigned addresses from to 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.从到之间的IP地址集合已经被聚集到然而,这里有一段空隙地址,即从到之间的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: for the new block. If an incoming packet matches both and 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：就可以了。当有一个分组到来时，如果它既匹配，又匹配，那么它将被发送到掩码位数较大的目标地址，即。这样做的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。