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1、Chapterfive第五章习题38.ConverttheIPaddresswhosehexadecimalrepresentationisC22F1582todotteddecimalnotation.(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:Theaddressis194.47.21.130.解答：先写成二进制：11000010,0010101111,0001010,10000010所以，它的点分十进制为：194.47.21.13039.AnetworkontheInternethasasub
2、netmaskof255.255.240.0.Whatisthemaximumnumberofhostsitcanhandle?(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:Themaskis20bitslong,sothenetworkpartis20bits.Theremaining12bitsareforthehost,so4096hostaddressesexist.Normally,thehostaddressis4096-2=4094.Becausethefirstad
3、dressbeusedfornetworkandthelastoneforbroadcast.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.AlargenumberofconsecutiveIPaddressareavailablestartingat198.16.0.0.Supposethatfourorganizations,A,B,C,andD,request4000,2000,4000,and8000addresses,re
4、spectively,andinthatorder.Foreachofthese,givethefirstIPaddressassigned,thelastIPaddressassigned,andthemaskinthew.x.y.z/snotation.(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:Tostartwith,allthereque
5、stsareroundeduptoapoweroftwo.Thestartingaddress,endingaddress,andmaskareasfollows:A:198.16.0.0–198.16.15.255writtenas198.16.0.0/20B:198.16.16.0–198.23.15.255writtenas198.16.16.0/21C:198.16.32.0–198.16.47.255writtenas198.16.32.0/20D:198.16.64.0–198.16.95.255writtenas198.16.64.0/19解答
6、：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最新范本,供参考！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A198.16.0.0198.16.15.255198.16.0.0/20B198.16.16.0198.16.23.255198.16.16.0/21C198.16.32.0198.16.47.255198.16.32.0/20D198.16.64.0198.16.95.255198.16.64.0/1941.Arouterha
7、sjustreceivedthefollowingnewIPaddresses:57.6.96.0/21,57.6.104.0/21,57.6.112.0/21,and57.6.120.0/21.Ifallofthemusethesameoutgoingline,cantheybeaggregated?Ifso,towhat?Ifnot,whynot?(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
8、聚集起来吗?如果可以的话,它们被聚集到那个地址上?如
