02-03-2009 09:04 AM - edited 03-06-2019 03:50 AM
I am working toward my CCNA and studying on Cisco website for NAT translation. According to the scenario presented, is this correct?
S0 - 172.16.10.64/24
The available valid addressed which we can use (for NAT) are 172.16.10.1 - 172.16.10.62.
I would think the correct valid addresses for this network would be:
172.16.10.65 - 172.16.10.126
Why is this incorrect?
Thanks for any help.
02-03-2009 12:18 PM
Something looks wierd. Does this mean that the IP address of the serial (S0) interface is 172.16.10.64? If this is correct, with a 24-bit subnet mask, then .1-.254 are available (except for .64 since it is already in use)for NATing purposes. The range that they have specified would be valid for the network 172.16.10.0 /26, though then the S0 interface could not have the .64 address.
The range that you have specified (.65-.126) is 172.16.10.64 /26
They are actually both wrong. Can you post the link where you read this?
HTH,
Paul
02-03-2009 01:34 PM
Paul,
Here is the link:
http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080094837.shtml
Scroll down to the section entitled "Example: Allowing Internal Users to Access the Internet" and see the diagram below it. Perhaps I am just misreading what I am seeing?
Thanks,
Robb
02-03-2009 01:34 PM
Let me break this down ...
S0 - 172.16.10.64 /24
Mask: 255.255.255.0 (/24)
Subnet: 172.16.10.0
Broadcast: 172.16.10.255
Ergo: Useable valid addresses are from
172.16.10.1 to 172.16.10.254 where 172.16.10.64 is a member thereof.
Does this help?
02-03-2009 01:42 PM
If you go the link I posted above and view the example, now that I reconsider it, the following statement threw me off:
"The available valid addresses which we can use are in the range of 172.16.10.1 through 172.16.10.63"
With a /24 network, I don't think 172.16.10.64 can be a subnet ID --- as Paul said above, a .64 ID would require a /26 mask.
02-03-2009 01:58 PM
I looked at the link - they are just defining a pool of available addresses for NAT (not overloading) to be .1 - .63. So, in effect, this example is correct. They could, however use the whole /24 network for NAT if they wanted to, minus the .64 (S0) address.
By the way, just defining a network of 172.16.10.64 does not necessarily mean that it has to be a /26 - this could be variably subnetted to whatever you need. /26 just means that you are using .65-.126 as available addresses, with .127 being your broadcast address. You could alternatively, for example, use 172.16.10.64 /27 or /28, /29, /30
for networks of 172.16.10.65 - 94, 172.16.10.65 - .78, etc.
HTH,
Paul
02-03-2009 02:01 PM
That is correct...thanks for your help.
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