CCNA - Subnetting Problem

Author

Subnetting Problem

vschristopher

2002-09-23, 12:32 pm

Given the network 197.141.16.0 with subnet mask of 255.255.255.240 find the valid host address from the following (choose 3)

a.197.141.16.33
b.197.141.16.112
c.197.141.16.119
d.197.141.16.126
e.197.141.16.175

chris

edmonds_robert

2002-09-23, 1:13 pm

If you typed the question correctly, none of the listed answers are correct. As you have typed it, the valid host addresses on the network 197.141.16.0 255.255.255.240 are 197.141.16.1 - 197.141.16.14 (0 is the network address and 15 is the broadcast).
If, however, the question should have read 197.141.16.0 255.255.240.0, then ALL of the answers are correct, as the network would range from 197.141.16.1 - 197.141.31.254 (again, first and last for network and broadcast, respectively).

ANDRONDA

2002-09-23, 2:13 pm

The last guy is ringht about one thing- and that is the fact that the question appears to be written incorrectly.

Assuming that thre answer should all be in network 197.141.16.0 here is the correct solution.

Given the subnet mask of 255.255.255.240 you know that there are four bits reserved for subnets and four for hosts (in the last octet). In binary the 240 would be 11110000. So you do the power of two thing for the ones and for the zeros which is 16 for both.

Now you list out the different network and broadcast addresses:

197.141.16. 0 to 15
197.141.16. 16 to 31
197.141.16. 32 to 47
197.141.16. 48 to 63
197.141.16. 64 to 79
197.141.16. 80 to 95
197.141.16. 96 to 111
197.141.16. 112 to 127
197.141.16. 128 to 143
197.141.16. 144 to 159
197.141.16. 160 to 175
197.141.16. 176 to 191
197.141.16. 192 to 207
197.141.16. 208 to 223
197.141.16. 224 to 239
197.141.16. 240 to 255

Now compare

a.197.141.16.33
b.197.141.16.112
c.197.141.16.119
d.197.141.16.126
e.197.141.16.175

Answer a is OK
Answer b is invalid (it is a network address)
Answer c is OK
Answer d is OK
Answer e is invalid (it is a broadcast address)
Got it?

edmonds_robert

2002-09-23, 2:34 pm

Brainfart. I thought the question was asking only for the valid addresses on the 197.141.16.0 subnet, not the entire class C network. Sorry.

vschristopher

2002-09-23, 2:34 pm

yeah fellas the question was wrong, i confirmed it toooo, damn made me confused myself , lol good u guided me well,i think i shud work hard on my subnetting.

anyways i corrected the main problem now so other can see and understand if they have a problem with subnetting tooo

edmonds_robert

2002-09-23, 6:48 pm

In case the materials you were studying didn't provide this tip (none I read did, I had to learn it in a class), you can figure out how many numbers a subnet mask will give you by subtracting it from 265. For example, if you have the IP address 172.16.12.14 and the subnet mask 255.255.255.248, subtract the 248 from 256 and you get 8. Each network will have 8 IP addresses. So, the first is 0-7, the second is 8-15, and so on.
It works the same (only different) when doing it beyond the first byte. Instead of telling you how many hosts, it tells you how many numbers in that byte are in that network.
For example, if you have the same IP address of 172.16.12.14 and a subnet mask of 255.255.248.0, again subtract 248 from 256, and again you get 8. So your first network is 172.16.0.0 - 172.16.7.255, the second is 172.16.8.0 - 172.16.15.255, and so on.
I hope I explained it in a way that makes sense. If not, let me know, and maybe someone else can explain it better. But it does work, and it makes subnetting a lot easier.

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