求解释这题
本帖最后由 心魅d 于 2018-6-3 05:26 编辑The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of 255.255.255.224. The routingprotocol in use is RIP version 1. Which address could be assigned to the FastEthernet interface on RouterA?
A. 192.168.1.31
B. 192.168.1.64
C. 192.168.1.127
D. 192.168.1.190
E. 192.168.1.192
答案为什么是D??? 求各位帮助解答
31, 64, 127, 190, 192中只有190不為{2^5*n}(network address)或{2^5*n-1}(broadcast address)。 byronyj 发表于 2018-6-3 13:54
31, 64, 127, 190, 192中只有190不為{2^5*n}(network address)或{2^5*n-1}(broadcast address)。
还是有点不懂呢 byronyj 发表于 2018-6-3 13:54
31, 64, 127, 190, 192中只有190不為{2^5*n}(network address)或{2^5*n-1}(broadcast address)。
192.168.1.0 with a subnet mask of 255.255.255.224 <---This is key
192.168.1.0 /27 = 32 (30 host address per subnet (1 Network / 1 Broadcast)
So..
192.168.1.0
192.168.1.32 -
192.168.1.64 - 95
192.168.1.96 - 127
192.168.1.128 - 159
192.168.1.160 - 191 (160 network, 191 Broadcast) last usable is 190.
192.168.1.192
找到比较容易理解的答案了谢谢anyway byronyj 发表于 2018-6-3 13:54
31, 64, 127, 190, 192中只有190不為{2^5*n}(network address)或{2^5*n-1}(broadcast address)。
192.168.1.0 with a subnet mask of 255.255.255.224 <---This is key
192.168.1.0 /27 = 32 (30 host address per subnet (1 Network / 1 Broadcast)
So..
192.168.1.0
192.168.1.32 -
192.168.1.64 - 95
192.168.1.96 - 127
192.168.1.128 - 159
192.168.1.160 - 191 (160 network, 191 Broadcast) last usable is 190.
192.168.1.192
找到比较容易理解的答案了谢谢anyway {:6_268:}{:6_268:}{:6_268:} Router A下30个主机192.168.1.32/27
Router B下40个主机192.168.1.33/26
Router C下50个主机192.168.1.96/26
以上得出选项ABC都是错的,而E是网络地址,故只能选D。如有不懂请提。
3Q! lanfengjing 发表于 2018-6-5 09:46
Router A下30个主机192.168.1.32/27
Router B下40个主机192.168.1.33/26
Router C下50个主机192.168.1.96 ...
Router A下30个主机192.168.1.0/27上面写错了。3Q! 子网划分 子网划分后,合理的可分配的主机ip只有 1-31 33-6297-126 129-158161-190193-222
这几段ip地址,所有上面所有的选项只有D选项符合条件 打错一个,是1-30 打错一个是1-30 打错一个是1-30 {:6_267:} {:6_267:}
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