Just a quick update to say that I have finally passed the CCIE!
Not the wireless lab but I wanted to get a good grounding in Routing and Switching first, and decided to dedicate the last 12 months purely to getting my number.
Its been hard, but now I can focus on wireless again. I am going to rebuild my lab and get back on the CCIEW case, but first a couple of months off!
Let me know if you wish to join in a study group!
Monday, 5 August 2013
Thursday, 10 January 2013
IP Prefix Lists - Explained
A normal access-list CANNOT check the subnet mask of a network. It can
only check bits to make sure they match, nothing more. A prefix-list
has an advantage over an access-list in that it CAN check BOTH bits and
subnet mask - both would have to match for the network to be either
permitted or denied.
For checking bits a prefix list ALWAYS goes from left to right and
CANNOT skip any bits. A basic example would be this:
172.16.8.0/24
If there is only a / after the network (no le or ge) then the number
after the / is BOTH bits checked and subnet mask. So in this case it
will check the 24 bits from left to right (won't care about the last 8
bits) AND it will make sure that it has a 24 bit mask. BOTH the 24 bits
checked and the 24 bit subnet mask must match for the network to be
permitted or denied.
Now we can do a range of subnet masks also that could be permitted or denied:
172.16.8.0/24 ge 25
If we use either the le or ge (or both le and ge) after the /, then the
number directly after the / becomes ONLY bits checked and the number
after the ge or le (or both) is the subnet mask. So in this case we are
still going to check the first 24 bits of the network from left to
right. If those match we are then going to check the subnet mask, which
in this case can be GREATER THAN OR EQUAL TO 25 bits - meaning that as
long as the first 24 bits of the network match the subnet mask could be
25,26,27,28,29,30,31,or 32 bits. They would all match.
We can also do:
172.16.8.0/24 le 28
Again this will check the first 24 bits of the network to make sure that
they match. Then it will check to make sure that the subnet mask is
LESS THAN OR EQUAL TO 28 bits. Now this isn't going to be 28 bits down
to 0 bits, the subnet mask can't be any lower than the bits we are
checking. So the valid range of subnet masks for this one would be 28
bits down to 24 bits (24,25,26,27,and 28). All of those would match.
We can also do both ge and le:
172.16.8.0/24 ge 25 le 27
Here again we are checking the first 24 bits to make sure they match.
Then our subnet mask must be GREATER THAN OR EQUAL TO 25 bits LESS THAN
OR EQUAL TO 27 bits. Meaning that 25,26,and 27 bit subnet masks would
match.
Now for a couple of examples:
If we have the following networks:
172.16.8.0/28
172.16.8.16/28
172.16.8.32/28
172.16.8.48/28
172.16.8.64/28
We could permit all of these networks with one prefix-list statement:
172.16.8.0/24 ge 28 le 28
This will check the first 24 bits to make sure they match. All of these
networks have 172.16.8 as the first 24 bits, and it won't care what is
in the last 8 bits. Then it will check to make sure that the subnet
mask is GREATER THAN OR EQUAL TO 28 bits LESS THAN OR EQUAL TO 28 bits -
the only number that works for this is 28 bits. So the first 24 bits in
the network must match and it has to have a 28 bit subnet mask. All 5
of our networks would match for this.
We could be even more precise with this and use:
172.16.8.0/25 ge 28 le 28
If we take a look at our 4th octects we will see that for all of them
the 128 bit is off so we can check that bit also (25 bits total we are
checking).
0 -- 0 0 0 0 0 0 0 0
16 - 0 0 0 1 0 0 0 0
32 - 0 0 1 0 0 0 0 0
48 - 0 0 1 1 0 0 0 0
64 - 0 1 0 0 0 0 0 0
This would be closer to permitting the 5 networks that we have.
We could also permit only the classful networks. The first thing that
we need to do is figure out exactly what a classful network is.
For a class A network we know that it has to have an 8 bit mask and must
be between 0 and 127 in the first octect. If we break down 0 and 127 we
get:
0 --- 0 0 0 0 0 0 0 0
127 - 0 1 1 1 1 1 1 1
For the first octect of a class A network the first bit has to be a 0,
it must be off. So we can do a prefix-list like this:
0.0.0.0/1 ge 8 le 8
In our first octet the first bit is a 0 (which is what it would need to
be to be class A), with the /1 we have we are ONLY checking the first
bit to make sure it's a 0 (meaning it would be a class A network 0 -
127). We are then making sure that this class A network actually has a
class A subnet mask of 8 bits, and only 8 bits would match.
For the class B's we need to make sure that they have a 16 bit subnet
mask and that they are in the range of 128 - 191 in the first octet. If
we break down 128 and 191 we get:
128 - 1 0 0 0 0 0 0 0
191 - 1 0 1 1 1 1 1 1
The first two bits are what we are going to care about. We need to make
sure that the first two bits in the first octet are 1 0 . The first
number that we can use as our standard we are checking against is 128 -
128 has a 1 0 as the first two bits in its first octet.
128.0.0.0/2 ge 16 le 16
So we are checking the first two bits to make sure the network has a 1
0, meaning that it must be in the range of 128 - 191. We are then going
to check to make sure that it has the classful 16 bit mask, and ONLY a
16 bit mask.
Finally we have the class C networks. Class C networks are in the range
of 192 - 223 and they must have a 24 bit mask. If we break down 192 and
223 we get:
192 - 1 1 0 0 0 0 0 0
223 - 1 1 0 1 1 1 1 1
The first 3 bits in the first octet are what we care about. 192 would
be the first number we can put in that first octect that will have 1 1 0
as its first 3 bits.
192.0.0.0/3 ge 24 le 24
We are going to check the first 3 bits of the octet and make sure that
its 1 1 0 meaning that it has to be in the range of 192 - 223 being
class C, then we are going to check to make sure it has a class C
classful subnet of 24 bits.
Finally how to permit or deny any could be very helpful since a
Prefix-list just like an Access-list has an implicit deny at the end:
0.0.0.0/0 le 32
This is 'any' for a prefix-list. It says check 0 bits; I don't care
what any of the bits are. It also says that the subnet mask can be 32
bits or less (down to the number of bits we are checking) down to 0. So
we aren't going to check any bits and the network can have a subnet mask
of anything between 0 and 32 bits. This would be 'any'.
Now for an example.
In the 3rd Octet we have 1, 4, and 5. We'll break these down to binary
to see if we can summarize these into one line:
1 - 0 0 0 0 0 0 0 1
4 - 0 0 0 0 0 1 0 0
5 - 0 0 0 0 0 1 0 1
For a Prefix-list we need to go from the left to the right and we can't
skip bits. So for these three networks we would need to stop at the 8
bit since it is the last bit from left to right that is the same. This
would give us 3 bits that are different, or 8 possible networks. We
only have 3 of the 8 possible networks and we should not permit or deny
more than we actually have. We should be as specific as possible.
If we leave the 91.86.1.0/24 alone by itself it will give us a
Prefix-list of:
91.86.1.0/24
This will check the first 24 bits from left to right to make sure that
they match, and it will also check to make sure that it has a 24-bit
subnet mask.
For the 4 and 5 networks we can permit or deny both of those with one
line. If we take a look at 4 and 5 again we can see that all of the
bit's match down to the 2 bit. This would leave 1 bit that doesn't
match, which would give us 2 possible networks, both of which we have.
The Prefix-list to permit or deny both 4 and 5 would be:
91.86.4.0/23 ge 24 le 24
This will check the first 23 bits from left to right. The 24th bit
could either be off, which would give us 4, or it could be on which
would give us 5. Since we have the ge and le involved the /23 is only
bits checked. The ge and le specify that our subnet mask must be
greater than or equal to 24-bits and less than or equal to 24-bits which
means that the subnet mask must be 24-bits for both possible networks.
only check bits to make sure they match, nothing more. A prefix-list
has an advantage over an access-list in that it CAN check BOTH bits and
subnet mask - both would have to match for the network to be either
permitted or denied.
For checking bits a prefix list ALWAYS goes from left to right and
CANNOT skip any bits. A basic example would be this:
172.16.8.0/24
If there is only a / after the network (no le or ge) then the number
after the / is BOTH bits checked and subnet mask. So in this case it
will check the 24 bits from left to right (won't care about the last 8
bits) AND it will make sure that it has a 24 bit mask. BOTH the 24 bits
checked and the 24 bit subnet mask must match for the network to be
permitted or denied.
Now we can do a range of subnet masks also that could be permitted or denied:
172.16.8.0/24 ge 25
If we use either the le or ge (or both le and ge) after the /, then the
number directly after the / becomes ONLY bits checked and the number
after the ge or le (or both) is the subnet mask. So in this case we are
still going to check the first 24 bits of the network from left to
right. If those match we are then going to check the subnet mask, which
in this case can be GREATER THAN OR EQUAL TO 25 bits - meaning that as
long as the first 24 bits of the network match the subnet mask could be
25,26,27,28,29,30,31,or 32 bits. They would all match.
We can also do:
172.16.8.0/24 le 28
Again this will check the first 24 bits of the network to make sure that
they match. Then it will check to make sure that the subnet mask is
LESS THAN OR EQUAL TO 28 bits. Now this isn't going to be 28 bits down
to 0 bits, the subnet mask can't be any lower than the bits we are
checking. So the valid range of subnet masks for this one would be 28
bits down to 24 bits (24,25,26,27,and 28). All of those would match.
We can also do both ge and le:
172.16.8.0/24 ge 25 le 27
Here again we are checking the first 24 bits to make sure they match.
Then our subnet mask must be GREATER THAN OR EQUAL TO 25 bits LESS THAN
OR EQUAL TO 27 bits. Meaning that 25,26,and 27 bit subnet masks would
match.
Now for a couple of examples:
If we have the following networks:
172.16.8.0/28
172.16.8.16/28
172.16.8.32/28
172.16.8.48/28
172.16.8.64/28
We could permit all of these networks with one prefix-list statement:
172.16.8.0/24 ge 28 le 28
This will check the first 24 bits to make sure they match. All of these
networks have 172.16.8 as the first 24 bits, and it won't care what is
in the last 8 bits. Then it will check to make sure that the subnet
mask is GREATER THAN OR EQUAL TO 28 bits LESS THAN OR EQUAL TO 28 bits -
the only number that works for this is 28 bits. So the first 24 bits in
the network must match and it has to have a 28 bit subnet mask. All 5
of our networks would match for this.
We could be even more precise with this and use:
172.16.8.0/25 ge 28 le 28
If we take a look at our 4th octects we will see that for all of them
the 128 bit is off so we can check that bit also (25 bits total we are
checking).
0 -- 0 0 0 0 0 0 0 0
16 - 0 0 0 1 0 0 0 0
32 - 0 0 1 0 0 0 0 0
48 - 0 0 1 1 0 0 0 0
64 - 0 1 0 0 0 0 0 0
This would be closer to permitting the 5 networks that we have.
We could also permit only the classful networks. The first thing that
we need to do is figure out exactly what a classful network is.
For a class A network we know that it has to have an 8 bit mask and must
be between 0 and 127 in the first octect. If we break down 0 and 127 we
get:
0 --- 0 0 0 0 0 0 0 0
127 - 0 1 1 1 1 1 1 1
For the first octect of a class A network the first bit has to be a 0,
it must be off. So we can do a prefix-list like this:
0.0.0.0/1 ge 8 le 8
In our first octet the first bit is a 0 (which is what it would need to
be to be class A), with the /1 we have we are ONLY checking the first
bit to make sure it's a 0 (meaning it would be a class A network 0 -
127). We are then making sure that this class A network actually has a
class A subnet mask of 8 bits, and only 8 bits would match.
For the class B's we need to make sure that they have a 16 bit subnet
mask and that they are in the range of 128 - 191 in the first octet. If
we break down 128 and 191 we get:
128 - 1 0 0 0 0 0 0 0
191 - 1 0 1 1 1 1 1 1
The first two bits are what we are going to care about. We need to make
sure that the first two bits in the first octet are 1 0 . The first
number that we can use as our standard we are checking against is 128 -
128 has a 1 0 as the first two bits in its first octet.
128.0.0.0/2 ge 16 le 16
So we are checking the first two bits to make sure the network has a 1
0, meaning that it must be in the range of 128 - 191. We are then going
to check to make sure that it has the classful 16 bit mask, and ONLY a
16 bit mask.
Finally we have the class C networks. Class C networks are in the range
of 192 - 223 and they must have a 24 bit mask. If we break down 192 and
223 we get:
192 - 1 1 0 0 0 0 0 0
223 - 1 1 0 1 1 1 1 1
The first 3 bits in the first octet are what we care about. 192 would
be the first number we can put in that first octect that will have 1 1 0
as its first 3 bits.
192.0.0.0/3 ge 24 le 24
We are going to check the first 3 bits of the octet and make sure that
its 1 1 0 meaning that it has to be in the range of 192 - 223 being
class C, then we are going to check to make sure it has a class C
classful subnet of 24 bits.
Finally how to permit or deny any could be very helpful since a
Prefix-list just like an Access-list has an implicit deny at the end:
0.0.0.0/0 le 32
This is 'any' for a prefix-list. It says check 0 bits; I don't care
what any of the bits are. It also says that the subnet mask can be 32
bits or less (down to the number of bits we are checking) down to 0. So
we aren't going to check any bits and the network can have a subnet mask
of anything between 0 and 32 bits. This would be 'any'.
Now for an example.
In the 3rd Octet we have 1, 4, and 5. We'll break these down to binary
to see if we can summarize these into one line:
1 - 0 0 0 0 0 0 0 1
4 - 0 0 0 0 0 1 0 0
5 - 0 0 0 0 0 1 0 1
For a Prefix-list we need to go from the left to the right and we can't
skip bits. So for these three networks we would need to stop at the 8
bit since it is the last bit from left to right that is the same. This
would give us 3 bits that are different, or 8 possible networks. We
only have 3 of the 8 possible networks and we should not permit or deny
more than we actually have. We should be as specific as possible.
If we leave the 91.86.1.0/24 alone by itself it will give us a
Prefix-list of:
91.86.1.0/24
This will check the first 24 bits from left to right to make sure that
they match, and it will also check to make sure that it has a 24-bit
subnet mask.
For the 4 and 5 networks we can permit or deny both of those with one
line. If we take a look at 4 and 5 again we can see that all of the
bit's match down to the 2 bit. This would leave 1 bit that doesn't
match, which would give us 2 possible networks, both of which we have.
The Prefix-list to permit or deny both 4 and 5 would be:
91.86.4.0/23 ge 24 le 24
This will check the first 23 bits from left to right. The 24th bit
could either be off, which would give us 4, or it could be on which
would give us 5. Since we have the ge and le involved the /23 is only
bits checked. The ge and le specify that our subnet mask must be
greater than or equal to 24-bits and less than or equal to 24-bits which
means that the subnet mask must be 24-bits for both possible networks.
Tuesday, 30 October 2012
Cisco WLC keeping web auth persistent
How would you keep client’s web authentication persistent even after client gets disconnected or de authenticated?
Device: Cisco WLC 5508
Code: 7.0.116.0
Recently after setting up the Wireless Network and Web Authentication Redirect option on a Cisco Wireless LANcontroller – 5508 I had an issue where after approximately an hour mobile clients specially mobile phones would disconnect and they would have to go through the Web Authentication Redirect page again and again. This was very annoying. Basically on Cisco WLC 5508 webauth devices timeout and they would have to re authenticate.
After doing lots of research and trying to change the time out settings under User Idle Timeout, ARP timeout, Session timeout nothing worked. Finally after working with Cisco TAC and doing a debug on the client “debugclient mac-id. I noticed that after an hour WLC sends the new EAP key to the client.
Updated broadcast key sent to mobile 00:23:76:D5:68:61
Cisco WLC 5508 tries this 3 times and after the 3rd time it gives up and considers the client not active any more and sends a de authentication packet, next Cisco WLC 5508 removes the client completely. Hence why when the clientcomes back they have to go through the Web Authentication Redirect Page again because key they have is old and is not valid any more.
Retransmit failure for EAPOL-Key M5 to mobile mac-id, retransmit count 3, mscb deauth count 0
Sent Deauthenticate to mobile on BSSID ap-mac-id slot 0(caller 1x_ptsm.c:534)
*apfReceiveTask: Jun 16 10:47:30.960: client-mac client-ip RUN (20) Deleted mobile LWAPP rule on AP [ap-mac]
Solution
Solution is to increase the broadcast key time interval. I used the following command to accomplish this. PS: This option was not available in the GUI with the code I am using so the only way for me to do it was via the Cisco WLC 5508 Command Line Interface, this applies globally to all the WLAN’s as of this code:
config advanced eap bcast-key-interval seconds (120 to 86400)
config advanced eap bcast-key-interval seconds (120 to 86400)
Wednesday, 19 September 2012
Installing a Web Authentication Certificate
This post details how to request and install a Web Authentication Certificate for the WLC. I am going to use Go Daddy to purchase the SSL certificate, this is most likely a chained cert.
This guide shows how to do this with Windows 7. Most documents on the web are only for Windows XP/2000.
You need to download and install openssl 0.9.8.
http://www.ie7pro.com/openssl/openssl-dll-0.9.8g_x64.zip
First you need to generate the Cert Request.
by default, openssl.exe is located at C:\ > openssl > bin.
After you submit the CSR to the third-party CA, the third-party CA digitally signs the certificate and sends back the signed certificate chain through e-mail. In case of chained certificates, you receive the entire chain of certificates from the CA. If you only have one intermediate certificate in our example, you receive these three certificates from the CA:
Once you have all the three certificates, copy and paste into another file the contents of each .pem file in this order:
Combine the All-certs.pem certificate with the private key that you generated along with the CSR (the private key of the device certificate, which is mykey.pem in this example), and save the file as final.pem.
Issue these commands in the OpenSSL application in order to create the All-certs.pem and final.pem files:
Note: In this command, you must enter a password for the parameters -passin and -passout . The password that is configured for the -passout parameter must match the certpassword parameter that is configured on the WLC. In this example, the password that is configured for both the -passin and -passout parameters is check123.
final.pem is the file that we need to download to the Wireless LAN Controller. The next step is to download this file to the WLC.
This guide shows how to do this with Windows 7. Most documents on the web are only for Windows XP/2000.
You need to download and install openssl 0.9.8.
http://www.ie7pro.com/openssl/openssl-dll-0.9.8g_x64.zip
First you need to generate the Cert Request.
by default, openssl.exe is located at C:\ > openssl > bin.
Issue this command: OpenSSL>req -config "C:\OpenSSL\bin\openssl.cfg" -new -newkey rsa:2048 -nodes -keyout mykey.pem -out myreq.pem
Provide all information. The important fiels is the "Common Name". This will need to match up with the DNS name that resolves to the virtail IP address of the WLC. Make sure to note the password that you set.
After you provide all the required details, two files are generated:
- a new private key that includes the name mykey.pem
- a CSR that includes the name myreq.pem
After you submit the CSR to the third-party CA, the third-party CA digitally signs the certificate and sends back the signed certificate chain through e-mail. In case of chained certificates, you receive the entire chain of certificates from the CA. If you only have one intermediate certificate in our example, you receive these three certificates from the CA:
- Root certificate.pem
- Intermediate certificate.pem
- Device certificate.pem
Once you have all the three certificates, copy and paste into another file the contents of each .pem file in this order:
------BEGIN CERTIFICATE------
*Device cert*
------END CERTIFICATE------
------BEGIN CERTIFICATE------
*Intermediate CA cert *
------END CERTIFICATE--------
------BEGIN CERTIFICATE------
*Root CA cert *
------END CERTIFICATE------
Save the file as All-certs.pem.Combine the All-certs.pem certificate with the private key that you generated along with the CSR (the private key of the device certificate, which is mykey.pem in this example), and save the file as final.pem.
Issue these commands in the OpenSSL application in order to create the All-certs.pem and final.pem files:
openssl>pkcs12 -export -in All-certs.pem -inkey mykey.pem -out All-certs.p12 -clcerts -passin pass:check123 -passout pass:check123 openssl>pkcs12 -in All-certs.p12 -out final-cert.pem -passin pass:check123 -passout pass:check123
Note: In this command, you must enter a password for the parameters -passin and -passout . The password that is configured for the -passout parameter must match the certpassword parameter that is configured on the WLC. In this example, the password that is configured for both the -passin and -passout parameters is check123.
final.pem is the file that we need to download to the Wireless LAN Controller. The next step is to download this file to the WLC.
Download the Third-Party Certificate to the WLC using the CLI
Complete these steps in order to download the chained certificate to the WLC using the CLI:
- Move the final.pem file to the default directory on your TFTP server.
- In the CLI, issue these commands in order to change the download settings:
>transfer download mode tftp >transfer download datatype webauthcert >transfer download serverip <TFTP server IP address> >transfer download path <absolute TFTP server path to the update file> >transfer download filename final.pem - Enter the password for the .pem file so that the operating system can decrypt the SSL key and certificate.
Note: Be sure that the value for certpassword is the same as the -passout parameter password that was set in step 4 of the Generate a CSR section. In this example, the certpassword must be check123.>transfer download certpassword password - Issue the transfer download start command in order to view the updated settings. Then enter y at the prompt in order to confirm the current download settings and start the certificate and key download. Here is an example:
(Cisco Controller) >transfer download start Mode............................................. TFTP Data Type........................................ Site Cert TFTP Server IP................................... 10.77.244.196 TFTP Packet Timeout.............................. 6 TFTP Max Retries................................. 10 TFTP Path........................................./ TFTP Filename.................................... final.pem This may take some time. Are you sure you want to start? (y/N) y TFTP EAP Dev cert transfer starting. Certificate installed. Reboot the switch to use new certificate. - Reboot the WLC in order for the changes to take effect.
Monday, 24 October 2011
I am still here!
OK, so I took the lab in April...I was nowhere near and due to the fact that the next available date was in September, I decided to wait for the V2.0 track. So I am still waiting for the material to be released and am currently working on becoming an CCSI (to keep me busy!!!). Looking forward to get back on track!
Monday, 18 April 2011
Restart the Discovery Process - Lightweight AP
You are sometimes in the position where you have just configured DHCP or DNS for AP discovery and you need the AP to get an IP address and start the discovery process.
You can reboot the AP, or you can release and renew to start the process without the need for a reboot...
release dhcp f0
renew dhcp f0
You can reboot the AP, or you can release and renew to start the process without the need for a reboot...
release dhcp f0
renew dhcp f0
Friday, 8 April 2011
Reboot your Autonomous AP's !
I spent the best part of an hour today trying to work out why a bridge link would not pass traffic. The config was right. I did a reboot, and it all started working.
So, if you think its set up and its not working...REBOOT!
So, if you think its set up and its not working...REBOOT!
Subscribe to:
Posts (Atom)