1)The computer which is on the same subnet as IP address 207.21.102.105/27 is
a} 207.21.102.254
b} 207.22.102.106
c} 207.21.102.0
d} 207.21.102.99
2) Which of the following IP addresses are on the same network as 132.10.30.20/16
a} 132.0.0.21
b} 132.10.30.20
b} 132.10.31.21
b} 132.11.31.20
I would like to be shown how you work it out.
Thank you
2006-06-09 00:25:05 · 5 answers · asked by da a 1 in Computers & Internet ➔ Computer Networking
I think it is b} 132.10.30.20.
2006-06-09 00:30:17 · answer #1 · answered by Anonymous · 0⤊ 1⤋
The method that I think is easiest is the "block" method. Here's how it works. You subtract the last octet of the subnet mask from 256 and that tells you your block size. Knowing the block size you can easily calculate the network address, broadcast address and valid network addresses.
I'll use the first question as an example. The subnet mask is 255.255.255.224. Subtract the last octet from 256 and you get a block size of 32 (256 - 224 = 32). Now starting from 207.21.102.0 (zero is always the first network address) you simply count by 32's to get your network addresses:
207.21.102.0
207.21.102.32
207.21.102.64
207.21.102.96
207.21.102.128
207.21.102.160
207.21.102.192
207.21.102.224
The broadcast addresses are simply one less than the following network addresses. For example the network after .0 is .32, therefore .0's broadcast address is .31. Of course 255 requires a little thought. :)
207.21.102.31 = broadcast address for .0
207.21.102.63 = broadcast address for .32
207.21.102.95 = broadcast address for .64
207.21.102.127 = broadcast address for .96
207.21.102.159 = broadcast address for .128
207.21.102.191 = broadcast address for .160
207.21.102.223 = broadcast address for .192
207.21.102.255 = broadcast address for .224
Finally the valid host addresses are all the addresses in between the network address and broadcast address. For example if .0 is the network address and .31 is its boradcast address then everything between is a valid host address.
207.21.102.1 - 207.21.102.30
207.21.102.33 - 207.21.102.62
207.21.102.65 - 207.21.102.94
207.21.102.97 - 207.21.102.126
207.21.102.129 - 207.21.102.158
207.21.102.161 - 207.21.102.190
207.21.102.193 - 207.21.102.222
207.21.102.225 - 207.21.102.254
Using this chart you can easily see that the answer is b} 207.22.102.106.
2006-06-09 08:01:25 · answer #2 · answered by emmittnervend 4 · 0⤊ 0⤋
for first answer is d) 207.21.102.99
just subnet using 27bit so your host about 5 bits only means is about 2 rest to 5 means = 32
so in your network
total 8 bits
3 bits = subnet
5 bits = host
so 2rest to 3 is equal to 8 subnets
and 2 rest to 5 equal to 32 hosts
so first subnet is 0->31
second is 32->63
third is 64->95
fourth is 96->127
and yoiur ip is coming in fourth subnet and answer (d) is alsko in same subnet.....
so your answer is D.
fokr second same thing and your answers are 132.10.30.20 and
132.10.31.21.
2006-06-09 07:34:34 · answer #3 · answered by Atil 2 · 0⤊ 0⤋
your question is still not clear
islaminamdar@yahoo.com
inamdarinfotech.com
2006-06-09 07:30:28 · answer #4 · answered by islaminamdar@yahoo.com 3 · 0⤊ 0⤋
answer 1
207.21.102.99
answer 2
132.10.30.20
132.10.31.21
2006-06-09 07:45:26 · answer #5 · answered by Jitendar Sharma 3 · 0⤊ 0⤋