| Author |
1st and last Subnets
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| justindu 2003-09-18, 11:24 am |
| Looks here:
Scenario 1
Required Number of Physical Segments:
5
Maximum Number of Hosts/Physical Segment:
5000
Network Address:
152.77.0.0
Proposed Custom Subnet Mask:
255.255.224.0
Number of Subnets Supported:
6
Maximum Number of Host ID per Subnet:
8190
Subnet IDs:
152.77.0.0 (Invalid)
152.77.32.0
152.77.64.0
152.77.96.0
152.77.128.0
152.77.160.0
152.77.192.0
152.77.224.0 (Invalid)
1st Host – Last Host on Subnet
Host ID Ranges per Subnet
152.77.32.0
152.77.64.0
152.77.96.0
152.77.128.0
152.77.160.0
152.77.192.0
152.77.32.1 – 152.77.63.254
152.77.64.1 – 152.77.95.254
152.77.96.1 – 152.77.127.254
152.77.128.1 – 152.77.159.254
152.77.160.1 – 152.77.191.254
152.77.192.1 – 152.77.223.254
Scenario 2
Required Number of Physical Segments:
100
Maximum Number of Hosts/Physical Segment:
88,000
Network Address:
39.0.0.0
Proposed Custom Subnet Mask:
255.254.0.0
Number of Subnets Supported:
126
Maximum Number of Host ID per Subnet:
131,070
Subnet IDs:
39.2.0.0
39.4.0.0
39.6.0.0
39.8.0.0
……
39.254.0.0
Why is the first and last subnet invalid in the scenario one but not two? | |
| martek 2003-09-18, 11:59 am |
| In theory both the first and last subnet are invalid, whatever the network class. Though both subnets are widely used anyway. So in case of both scenarios the first and last subnets are invalid, unless the two scenarios are based on different assumptions, such as the 'ip subnet-zero' command being used in one and not the other.  | |
| justindu 2003-09-18, 12:54 pm |
| I see...
But I was thinking maybe it is becuase the first scenario only supports 6 subnets.. so maybe thats why the first and second are not used?
On the exam should i assume the first and last subnets are not used? | |
| martek 2003-09-18, 5:49 pm |
| For the exam you should assume that both subnets are invalid, unless the 'ip subnet-zero' command was used on the router, which would make the first subnet valid. | |
| justindu 2003-09-18, 8:46 pm |
| Cool.. I think have this subnetting down for the most part... 
Only thing I am still struggling with is broadcat addresses.
When figuring out valid host ranges it seems sometimes the valid host is two before the next subnet.
For example if you had network 162.62.0.0 with a 255.255.224.0 mask the first netwrk would be 162.62.32.0 with host addresses 162.62.32.1 - 162.62.32.62 right? Because 63 would be the broadcast right? Yet sometimes I see the last host only one before the next subnet.
Does this make sense? How do you know whether or not to subract one or two before the next subnet?? | |
| joel316 2003-09-18, 10:15 pm |
| actually if you look at your Scenerio 1 you will be able to answer that questions
IP address: 162.62.0.0
Subnet mask: 255.255.224.0
So your subnets will be 2^3=8-2=6 Subnets
And since this is a Class B Ip, you still have 13 bits for host therefore 2^13=8192-2=8190.
Alright so your subnets will be
162.62.0.0 (Not valid)
162.62.32.0
162.62.64.0
162.62.96.0
162.62.128.0
162.62.160.0
162.62.192.0
162.62.224.0 (Not valid)
Host ranges:
162.62.32.1 - 162.62.63.254
162.62.64.1 - 162.62.95.254
162.62.96.1 - 162.62.127.254
162.62.128.1 - 162.62.159.254
162.62.160.1 - 162.62.191.254
162.62.192.1 - 162.62.223.254
So you better review your question since the range you gave 162.62.32.1 - 162.62.32.62 at all host addresses. I'm pretty sure that I'm right on this unless someone has a different opinion | |
| martek 2003-09-18, 10:20 pm |
| No, you got it wrong again. 162.62.0.0 with subnet mask 255.255.224.0 (which is 162.62.0.0/19) gives you subnets 162.62.32.0, 162.62.64.0, 162.62.96.0 and so on.
So for subnet 162.62.32.0 the last address before the next subnet is 162.62.63.255, which is the broadcast address. So the last valid IP is 162.62.63.254.
The trick is this:
1. To get the broadcast address for a subnet, substract 1 from the next subnet number. So in case of 162.62.32.0 that's 162.62.63.0 (the next subnet is 162.62.64.0).
2. Now replace all the zeros with 255 and you get the broadcast address. So 162.62.63.0 becomes 162.62.63.255. That's it!
3. Now substract 1 from the last 255 and you get the last valid IP: in this case that's 162.62.63.254.
This trick works for all subnets where the subnet mask ends on the third octet or before.
If the subnet mask ends on the fourth octet, like IP 192.168.1.33 with subnet mask 255.255.255.224, the trick is:
1. To get the broadcast address, substract 1 from the next subnet number: in this case that gives you 192.168.1.63 (the next subnet is 192.168.1.64 - subnets go up by 32 in this example).
2. Substract 1 from the broadcast address and you get the last valid IP: in this case that's 192.168.1.62.
There you go! Two easy tricks to always get the broadcast address and last valid IP right. | |
| justindu 2003-09-18, 10:35 pm |
| got it! man what would i do without you guys! | |
| martek 2003-09-18, 10:51 pm |
| Since you got it, tell me what the broadcast address and valid IP range are for IP 10.5.254.254 with subnet mask 255.252.0.0.
You better get it right or else. | |
| justindu 2003-09-18, 11:09 pm |
| Ok maybe I dont got it... But let me tkae a shot:
Network: 10.4.0.0
10.4.0.1 - 10.7.254.254
broadcast
10.7.255.255 | |
| martek 2003-09-18, 11:19 pm |
| You got the broadcast address right, but the last valid IP is 10.7.255.254 not 10.7.254.254. You substract 1 from the last 255 only. | |
| justindu 2003-09-18, 11:34 pm |
| Ok.. where can i get more practice questions? |
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