Sunday, December 30, 2007
CCIE BGP ORF
Wednesday, December 26, 2007
Mary Christmas and Happy New Year
Wednesday, December 05, 2007
CCIE EIGRP METRIC
so I would like to start by giving you a little review and may be more indepth look over EIGRP Metric, as we all should know EIGRP is a Hybrid Routing Protocol that use a Composite Metric in his DUAL Diffusing Update Algorithm. the Composite metric is composed (good word selection :-)!?) from K Values.
K1 K2 K3 K4 K5
K1 = Bandwidth = 1 meaning it is used in the calculation by default.
K2 = load = 0 you gust it, meaning that is not participating in the calculation by default.
K3 = Delay = 1 you can take it from here
K4 = Reliability = 0
K5 = MTU = 0
the K values tells us if the value is participating in the calculation or not and what is the weight of the value, meaning that if you set K1 to 3 then it means that the weight of K1 will be 3 times more "important" or in other words it will take the BW * 3.
So now we know what are the "K" Values we need to know how to calculate:
well I will not endolge you with long version of the formula as it is hardly ever use (by hardly I mean I never seen it fully used).
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
in the attached picture I draw a small network to demonstrate what I mean to get from R1 to R3 netwrok 3.3.3.0/24
you have the following options:
R1 -> R2 -> R3
The lowes bandwith in that path is 64Kbps
The Delay is 20000 for the 64Kbps link 100 for the 100 Mbps linke and another 100 for the 3.3.3.0/24 network link also 100Mbps = 20200 microseconds delay.
so now lets put all into the formula:
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/64) + 256 * (20000 + 100 + 100)]
Metric = [256 * (156250) + 256 * (20200)]
Metric = [40000000 + 5171200]
Metric = 45171200 = FD Feasable Distance (I will Talk about this in a bit)
R1 -> R4 -> R3
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/64) + 256 * (20000 + 1000 + 100)]
! note that now you have 1000 in the path as it is a regular 10 Mbps Ethernet
Metric = [256 * (156250) + 256 * (21100)]
Metric = [40000000 + 5401600]
Metric = 45401600 = FD
R1 -> R2 -> R4 -> R3
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/64) + 256 * (20000 + 20000 + 20000 + 100)]
Metric = [256 * (156250) + 256 * (60100)]
Metric = [40000000 + 15385600]
Metric = 55385600 = FD
R1 -> R4 -> R2 -> R3 I think you should do it now, you should get the point of the calculation
Now that have an idea how to calculate we need to see what route is preferd, in that senarion it is easy but in complex networks this basics can help us understand the proccess (leave the calc to the router).
FD as I was mentioning this Feasable Distance is the Metric that our router see from his outgoing interface to the point he wish to go, also caled end-to-end metric.
AD Advertised Distance that is the Distance we recive from our directly connected neighbor for the path we wish to go.
so for FD of R1 ->R2 ->R3 we get an AD of R2->R3.
FS Feasible Successor is the Router that is chosen by us to be the next hop to the Destination Prefix.
FC Feasible Condition is a rule that tells us in a very logical way to select a FS he must send to us an AD that is lower then the FD.
if we received from R4 for example a AD higher then the FD that we had already in our topology table to get to R3 then we would not consider him to be a Feasible Successor.
so final step lets take the Paths and FD and Calculate the AD and see what path sould we go as I am already exited where to go from now:
Now our AD for the Paths sent to R1:
R2 -> R3
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/100000) + 256 * (100 + 100)]
Metric = [256 * (100) + 256 * (200)]
Metric = [25600 + 51200]
Metric = 76800 = AD
Metric = 45171200 =FD
R4 -> R3
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/64) + 256 * (20000 + 100)]
Metric = [256 * (156250) + 256 * (20100)]
Metric = [40000000 + 5145600]
Metric = 45145600 = AD
Metric = 45401600 = FD
R1 -> R2 -> R4 -> R3
Metric = [256 * (10^7/BW) + 256 * (Accumulated Delay)]
Metric = [256 * (10^7/64) + 256 * (20000 + 20000 + 20000 + 100)]
Metric = [256 * (156250) + 256 * (60100)]
Metric = [40000000 + 15385600]
Metric = 55385600 Here we can clearly see that there is not way that this route can meet the FC.
So we have a winner:
Metric = 76800 = AD
Metric = 45171200 =FD
to get to R3 3.3.3.0/24 R1 will take the path R1 ->R2 -> R3.
if it will fail then the path R1 ->R4->R3 will be taken as he still qualified to meet then FC for FS
FS Feasible Successor the second path in the topology that is set only of FC are met.
I hope this was an intersting article :-)
Tuesday, November 13, 2007
IPV6 Tricks
Saturday, October 13, 2007
CCIE IP Precedence Vs DSCP Values
IPP |
IPP |
IPP | DropP | DropP |
0 |
ECN |
ECN |
Class1 | AF11 001010 | AF12 001100 | AF13 001110 |
Class2 | AF21 010010 | AF22 010100 | AF23 010110 |
Class3 | AF31 011010 | AF32 011100 | AF33 011110 |
Class4 | AF41 100010 | AF42 100100 | AF43 100110 |
Thursday, October 11, 2007
CCIE INTERNETWROKEXPERT
Monday, October 08, 2007
CCIE SecureCRT Tip
Sunday, October 07, 2007
CCIE VERIFY RPF FAIL
Friday, October 05, 2007
CCIE PIM AUTO RP
When using PIM Sparse in NBMA environment like Frame Relay you need to make sure you enter the ip pim nbma command to disable the split horizon rule that traffic coming into the interface is not going out that same interface. The IP pim nbma is working only for Sparse Mode Group Candidate RP need to be able to communicate only with the mapping agent and the Routers in the Domain need to be able to communicate with the Mapping agent so from that we can understand that we need to watch our Route to the Mapping Agent. Ok I think I will end it here for now, I hope this was good info.
Sunday, September 23, 2007
Web Design
Monday, August 27, 2007
CCIE Regular Expression in BGP
CCIE QinQ
why like Houdini because basically the technology make your middle switch go away like it never was there and sowing to the person you wanted to show like he is directly connected to a switch when he is not.
now that I made every one a little confused, i will start to explain, QinQ allow you to take a middle switch and turn it to transparent while connecting 2 other components and maid them fill like they are directly connected.
why do you need that, well the CCIE lab love it as it is confusing setup but there are more reason then that, with QinQ you can connect topology's while you migrating, you can also connect between clients that have there own topology that you do not want to encounter in your topology by transferring all of them with a single tag between there point of presents you do not need to create a trunk between the 2 branches and pass all the vlan id inside while limiting your self to 1 or 2 clients.
you can see a much more elaborate details in cisco site
here i will give you few steps to create QinQ simple scenario
you can see that R1 is connected to SW1 and SW1 connected to SW2, if you will do a show cdp nei on R1 you will see that you are connected on F0/0 to SW1 F0/1, I want you to make you see like you are connected to SW2 port F0/10, how to do that magic?!
Go into SW1
global config
! this is to allow another tag to pass the metro tag as it also
! called an additional of 4 byte
!
switch(config)#system mtu 1504
! i will create a vlan for access control between the ports
switch(config)#vlan 200
!
!now I will go under int F0/1 and F0/10 and type the same commands
switch(config-if)#sw mode dot1q-tunnel
switch(config-if)#sw acc vlan 200
switch(config-if)#l2protocol-tunnel cdp
switch(config-if)#no shut
now when you will go to R1 and do show cdp nei you will see like magic that you are directly connected to SW2 F0/10 instead of SW1 and I didn't switched the cable!!!
Read more on cisco site you will love it.
Tuesday, August 21, 2007
CCIE LAB with Dynamips
I am working today with IEWBv4 to get my number. during my study I have said to my self why not create also your own LAB topology with your own scenarios, and here we are.
you can download my .net and initial config
As for the Lab Scenarios rules:
1) Do NOT use Frame-Relay Inverse Arp
2) Do NOT use sub interface with int Frame-Relay
3) There should be connectivity trough out the network include routes advertised from BB1
4) Y represent the Router number
General
1) Set loopback interfaces on R1/R2/R3/R4 with 150.1.y.y
WAN
1) Set Connectivity on the Frame Relay cloud between R1 R2 and R3
2) set Connectivity on the Frame Relay cloud between R4 to BB1
3) Set Serial using ppp between R3 and R4
4) I am a little afraid from unwanted users so please make sure you add on MD5 authentication between R3 to R4 and to make it more efficient as most of my traffic is txt files please use compression.
IGP
OSPF
1) Router 1 , 2 , 3 should all be on Backbone area
2) Do not use DR/BDR
3) Set Timers as on Non-Brodcast Topology
4) Set Authentication MD5
Eigrp
1) Set Router 3 and 4 on AS 34
2) The Administrator have connectivity between R3 and 4 trogh the Serial and FastEthernet make sure that both path will be used.
3) Advertise loopback interfaces with out using the network command
4) please make R3 and R4 authenticate each other with "CCIEP3" password and make sure that on December 28 2008 12 AM (when I turn 30) the password will be changed to "CCIEP?" also i wont you to allow a R4 and R3 to be able to authenticate CCIEP3 until 3AM Dec 28 2008.
Redistribute
1) Simply redistribute between Eigrp and OSPF on R3
I will Continue this lab Scenario later, please review and advice if you have any suggestions or feedback I will be happy to get some.
Thank you
Sunday, July 29, 2007
CCIE PPP over Frame-Relay
ppp (point to point protocol) another layer 2 protocol design to give a better solution to hdlc (high level data link control) used to control and transport ip traffic in a reliable and secure way.
now people taught how do we give the old frame-relay more features without developing a new frame-relay, they said why not take the good old ppp and join them in a "holy matrimony" just kidding, but really why not take them both and enjoy both benefits one with sharing bandwidth and traffic shaping options and one with link integrity and security features.
so now I come to our scenario where you can see I have setup 2 routers, one with 1 serial dividing it to 2 sub interfaces and the other one with 2 serial joining them together to one multilink interface making them load balance packets and also secure using chap authentication.
the multilink is needed here only to enable load balancing and interleaving, you can do without but it is less efficient and way less cooler :-)
This is the show run on router 2:
!
username Rack1R1 password 0 cisco
!
!
interface Multilink1
ip address 192.168.1.2 255.255.255.0
ppp multilink
ppp multilink group 1
!
interface Serial1/0
no ip address
encapsulation frame-relay
serial restart-delay 0
no frame-relay inverse-arp
!
interface Serial1/0.203 point-to-point
frame-relay interface-dlci 203 ppp Virtual-Template1
!
interface Serial1/0.213 point-to-point
frame-relay interface-dlci 213 ppp Virtual-Template1
!
!
interface Virtual-Template1
no ip address
ppp authentication chap
ppp multilink group 1
!
that is the show run on router 1:
!
username Rack1R2 password 0 cisco
!
!
interface Multilink1
ip address 192.168.1.1 255.255.255.0
ppp multilink
ppp multilink group 1
!
interface Serial1/0
no ip address
encapsulation frame-relay
serial restart-delay 0
no frame-relay inverse-arp
!
interface Serial1/0
no ip address
encapsulation frame-relay
serial restart-delay 0
frame-relay interface-dlci 302 ppp Virtual-Template1
no frame-relay inverse-arp
!
interface Serial1/1
no ip address
encapsulation frame-relay
serial restart-delay 0
frame-relay interface-dlci 312 ppp Virtual-Template1
no frame-relay inverse-arp
!
!interface Virtual-Template1
no ip address
ppp authentication chap
ppp multilink group 1
!
Verify:
sh ppp multilink
Multilink1, bundle name is Rack1R1
.
.
.
Member links: 2 active,....
Vi1, since 07:53:31
Vi2, since 07:53:31
and good old ping:
ping 192.168.1.2
Type escape sequence to abort.
Sending 5, 100-byte ICMP Echos to 174.1.23.2, timeout is 2 seconds:
!!!!!
Success rate is 100 percent (5/5), round-trip min/avg/max = 24/82/128 ms