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Home > Archive > CCNA > June 2003 > Subnetting 101 - LAB 06/04/2003
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Subnetting 101 - LAB 06/04/2003
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| anchor40 2003-06-04, 8:49 am |
| Good Morning!
Here's today's practice question:
Given the following IP address and mask, provide the class,
network ID, # subnets, and # of hosts per net:
code:
Address/mask Class Net ID #subs # hosts
================== ===== ================== ======= ========
___.___.___.___/__ _ ___.___.___.___/__ _______ ________
64.133.97.2/24
193.246.6.253/30
10.100.100.100/28
192.168.252.253/26
Here's the 6/3 question and answers:
Given the following network address and subnet mask, how many
point-to-point subnets can be created? 6-host subnets?
code:
Address Mask P2P 6-host
------- ---- --- ------
4.136.61.224 /27 8 4
P2P = .224 (.225-.226)
.228 (.229-.230)
.232 (.233-.234)
.236 (.237-.238)
.240 (.241-.242)
.244 (.245-.246)
.248 (.250-.251)
.252 (.253-.254)
6host = .224 (.225-.230)
.232 (.233-.238)
.240 (.241-.246)
.248 (.249-.254)
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| jason892 2003-06-04, 9:26 am |
| hows this?
64.133.97.2/24 a 64.133.97.0 65,534 subs
254 hosts
193.246.6.253/30 c 193.246.6.252 62 subs
2 hosts
10.100.100.100/28 a 10.100.100.96 1,048,574 subs 14 hosts
192.168.252.253/26 c 192.168.252.192 2 subs
64 hosts
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| B4yaman3 2003-06-04, 1:23 pm |
| jason you on the ball..
I said I had to ask anchor40 this: I remember you posting a file for we to download on subnetting for the Class C subnet mask..For the subnets you had it like 2^n-1. And I can also remember you saying if not specified by cisco you only do -1. Can you fully explain this over again.. | |
| anchor40 2003-06-04, 3:11 pm |
| Initially, the RFCs restricted the use of all 0 and all 1 networks, so a /25 was not a valid network, since the network ID is either 0 (.0) or 1 (.128).
The latest RFCs allow you to use the all 1's networks, but not the all 0 network. But Cisco added the IP subnet-zero command to allow those addresses to be used.
HTH... | |
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| How did you come up with the answer? is there a formula ? or manually computing it starting from the Network address till end of the subnet block ?
because I got the answer manually by counting how many networks will I get.... a /27 subnet block will give me 32 IPs per block to work on. because I have 32 IPs and I know that the start of this block is .224, a P2P network will consume 4 IPs so I did my math adding 4 for each P2P network and get this
NA (IP's to use) BC
1> .224 .225 - .226 .227
+4
2> .228 .229 - .230 .231
+4
3> .232 .233 - .234 .235
+4
4> .236
+4
5> .240
+4
6> .244
+4
7> .248
+4
8> .252 .253 - .254 .255
=========================
computing the 6-host, to get a 6 host subnet i need 3 bits to work on so 2^3 = 8 IPs each to work on so i come up with this,
NA (IP's to use) BC
1> .224 .225 - .226 .231
+8
2> .232 .229 - .230 .239
+8
3> .240 .233 - .234 .247
+8
4> .248 .249 - .253 .254
==============================
========
Is there a formula or steps to take for this scenarios, like how to compute a P2P network given only a /27 subnet mask ?
thnks | |
| anchor40 2003-06-04, 4:44 pm |
| Good news and bad - you got it right, but there's not a formula.
This type of question requires a little brute-force computation, as it would if your boss asks you to take a /24 network and slice it into un-even portions for five P2Ps, five 10-host segments, and one 32-host segment for future use.
For any of these, get back to the bit-boundary basics. Find the network ID of the subnet, and work it just like you did in your post.
Ok, so I wasn't exactly "honest" in my boss example above - in the real world you'd use a cheat-sheet, but for an exam, you can't.
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