Help with understanding ip address problem

[dan_track]

Hi

Can someone please help me with this. I enetered the following in my pix 515e ver (6.2):

ip address dmz xxx.xxx.210.72 255.255.255.248

But I got the following reply:

IP and subnetmask may be an invalid pair

I don't see why this came up. Can someone please explain it to me. I can't see anything wrong with the ip and subnet.

Thanks

Dan

[rais]

72=0100 1000

With 248 or /29 as a subnet mask, the host bits are all zeros which is not valid. Make it 73.

Thanks.

[sundar.palaniappan]

Dan,

IP/72 - 0100 1000

SM/248 - 1111 1000

In your case all but the last 3/32 bits have to match. Last 3 bits can be either 0 or 1.

As such your subnet address is 72 and the broadcast address is 79. You could use an address between 73-78 for the interface.

Pls. rate the post if it helped.

HTH,

Sundar

[dan_track]

Hi

Thanks for the replies.

I need a subnet with 8 ip's available. Could you please suggest one for me.

I have the following range:

xxx.xxx.210.64/26

Thanks

Dan

[jaoua · Accepted solution]

Amigo, To solve that, it's very simple: (Recall CCNA subnetting course)

your subnet mask is 255.255.255.192 (26 in decimal)

So --> 192 is the interesting octet in mask

and --> 64 is interesting octet of your Address

1).Find the Magic Number, By Substracting:

256 - Interesting Mask octet -> 256 - 192 = 64

2) Find a Multiple of Magic Number (Here :64) which is less or equal than interestzing octet of Address (also 64).

Then First Multiple of 64 = 64 ,

Then your working subnets start from 64

1st subnet 255.255.255.64,

-> Add 64 to find the following subnets :Always < 256

2nd subnet 255.255.255.128

3rd subnet 255.255.255.192

SO YOUR AVAILABLE IP ADDRESSES in 1st subnet are From : xxx.xxx.210.65 TO : xxx.xxx.210.126

Remember xxx.xxx.210.64 and xxx.xxx.210.127 are not allowed as IP address (Subnet Number and Broadcast)

Now, if you want subnets with 8 hosts each , then you have to solve (2^y - 2 = 8) or (2 ^y = 10)

you will have y = 4

(because 2^4 = 16 and 2^3 =8 which is unsuffisant)

So you have to reserve 4 bits for subnetting.

So your Netmask is 32-4 = 28 decimal

Meaning it is 255.255.255.240

240 = 128+64+32+16 (2^7 + 2^6 + 2^5 + 2^4)

To find the available Ip addresses for /28, same reasoning than above (magic number is 16 here)

So They are FROM xxx.xxx.210.65 TO xxx.xxx.210.78

xxx.xxx.210.79 is not allowed (broadcast) and in

this case you have 14 available addresses

[dan_track]

Hi

Thanks for the reply.

Could you please tell me the following:

1) Why are there 4 bits reserved for subnetting?

2) Would my new command be:

"ip address dmz xxx.xxx.210.73 255.255.248.0" ?

2) How to remove the ip address command from teh pix. When I enter "no" as a prefix to the sttement teh pix says "no" not supported for "ip address".

Thanks in advance

Dan

[Richard Burts]

To answer your question: the number of bits reserved for subnetting is variable and the number of bits used depends on how many IP addresses you need to have in the subnet. When you established a requirement for 8 addresses, then the smallest subnet that has that many addresses uses 4 bits for subnetting. There is a solution that would work using 5 bits - but that would not be as efficient.

Your new command would be

ip address dmz xxx.xxx.210.73 255.255.255.240

when you enter the new command it should over-write (or erase) the old command.

HTH

Rick

[dan_track]

Hi

I appreciate this may have not fit into the question, but just as a sanity check, could someone please verify my calculations.

If I want a subnet with a total of 8 ip's (so 6 usable ip's, one network, one broadcast)

Would the following sets of subnets be correct?

xxx.xxx.210.64-xxx.xxx.210.72

xxx.xxx.210.73-xxx.xxx.210.81

xxx.xxx.210.82-xxx.xxx.210.90

Thanks for your help

Dan

[grant.maynard]

.72 is the network number for that subnet.