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Home > Archive > CCNA > December 2002 > Subnet Help
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| flah161 2002-11-24, 5:06 am |
| I'm a CCNA student who understands almost everything about basic internetworking except for subnet ranges. I am looking for help with the following subnetting question :
With an IP address of 172.16.0.0 how many bits should be borrowed to create 258 subnets and what would be the network / broadcast id and host range of the 1st, 10th, 57th & 253rd subnets.
I have borrowed 9 bits to create 510 subnets and a subnet mask of 255.255.255.128, with host ranges from 0-127 and 129-254. Can anyone help out with working out the full subnet ranges ?
Grateful for any feedback / patience you can provide
Cheers
John | |
| twister166 2002-11-24, 8:02 am |
| 172.16.0.0/16 is the default network and mask. To create 258 subnets, you will need to 9 bits to the right, so that will 172.16.0.0/25 which is 255.255.255.128. So, you got it pretty good.
1st: 172.16.0.0-172.16.0.127
2nd: 172.16.0.128-172.16.0.255
...
10th: 172.16.5.128-172.16.5.255
you basically get two network ranges per the last 8 bits.
57th, 57/2 = 28, remainer = 1, so the 3rd octect is 29. 172.16.29.0-172.16.29.127
253rd, 253/2 = 126, remainer = 1, so the 3rd octect is 127. 172.16.127.0-172.16.127.127.
I think that I am right, you can get an ip calculator to make sure... | |
| davidbeecken 2002-11-24, 1:43 pm |
| www.ciscotrack.com/subnetting.html
has helped some people before, its not a direct answer to your question, but mabey it will help you overall | |
| dbowen 2002-12-01, 1:07 am |
| I spelled out the solution, but the post-processor strained out all my spacing (formatting) so I pasted it into a text attachment.
Check it out.
Although my method looks like a real pain, it's actually very simple and very elegant. I've had many (terribly confused) Cisco Academy students pick this up faster than the (superficial and sometimes inadequate) method Cisco Academy teaches.
Anyway, this is how the Queen subnets. Say hi to Johnny Rotten and Elvis Costello for me.
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| nickperjak 2002-12-03, 5:47 pm |
| Your Question Is:
With an IP address of 172.16.0.0 how many bits should be borrowed to create 258 subnets and what would be the network / broadcast id and host range of the 1st, 10th, 57th & 253rd subnets.
My explanation is this:
In your example, you are using 2 byte subnetting. Since you know subnetting mostly, I'm not going to explain the basics.
Your FIRST subnet with that IP and a mask of 255.255.255.128 (which you had correct)could be 172.16.0.0 (using the zero subnet). The next subnet could be 172.16.0.128, the following subnet could be 172.16.1.0, followed by the subnet of 172.16.1.128, next could be 172.16.2.0, followed by the subnet number of 172.16.2.128 and so on and so on. Basically, as mentioned in a previous post, you get two subnets per each third octet subnet number (thus 510 subnets). The third octet can have subnet numbers from 0-255 and the fourth octet can have either 0 or 128 for each corresponding third octet number. Your third octet can be any number, but your fouth octet can only be a 0 or 128. That's where you get 510 subnets.
Then you may ask, "Well what about the hosts?" For each subnet you are allowed 126 hosts (2 to the power of 7, minus 2), therefore if using the first subnet of 172.16.0.0, your hosts would be 172.16.0.1-126 (can't use 0 because that's the subnet number, can't use 127 because that's the broadcast for that subnet). Another example, say using the subnet of 172.16.10.128, your usuable IPs for that subnet only would be 172.16.10.129 - 172.16.10.254 (thus 126 hosts). The usuable IP's for the last subnet of 172.16.255.128 would be 172.16.255.129-254
I hope I didn't make any typos.
Here's a good link for practicing your exact question, as mentioned above...www.learntosubnet.com go to the "view and print practice problems" for subnetting. Good luck. |
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