Sunday, September 12, 2010

BSCI - Creating/Converting addresses

One easy way to pick up marks in the BSCI exam is to practise creating or converting addresses in one format to another, quickly.


So far I've spotted 4 situations where you would be asked to identify suitable converted addresses for a given address.


These are:
1) Identify the correct Multicast MAC address for a given Multicast IP addresses
- Multicast MAC addresses always start 01-00-5e. You need to find the last 23 bits to add to these first 25 bits to create a 48bit MAC.
- Take you Multicast IP and convert to binary
- section off the last 23 bits starting from the RIGHT in to 4 bit sections. the last section will contain only 3 bits so to get a Hex figure for this just tack a 0 on to the start of it. 
- Next convert your binary to Hex
- Finally add these to the 01-00-5e to get your Multicast MAC
e.g - 224.90.17.43
Binary = 11100000.0 101 1010. 0001 0001 0010 1011
Hex -                      | 5 |  A |  1   |  1  |   2   |   B
MAC = 01-00-5e-5a-11-2b


2) When regarding IPv6 6to4 tunnels, identify a suitable IPv6 address for a given IPv4 address that is assigned to a physical interface.
 Using a Global address you know 2 things A) the high order bits always start 001 = 2000::/3 and generally end with 0001 in the first 16bits and B) a Global prefix is 48 bits long ( or /48). With this in mind  you know your first 16 bits, 0010 0000 0000 0001: X:X:X:X:X:X:X, so you are looking for the remaining 32 bits to form the address
Copy out into binary the IPv4 address of the physical interface the tunnel will be associated with then convert it to hex, e.g.) 192.168.99.1
    1100 0000. 1010 1000. 0110 0011. 0000 0001
      c     0        a      8       6     3      0     1
- Combine this with your first 16 bits and you have your Global IPv6 to apply to your tunnel interface
     2001:c0a8:6301::/48

3) Given the ASN 5662 what would a suitable GLOP address be.
- For this type of question you know that the 1st octet is always 233 and you can choose what the last octet can be (1-255) so you need to calculate the 2nd and 3rd octet values.
- Take the ASN 5662
-Convert it to binary and pad the left of the binary figure with zero's until you have 16 bits (octets 2 and 3 combined) -0001011000011110
- Divide the 16 bits in 2 and you are left with 2 octets (8bits each)
- convert these to decimal - 00010110 = 22, 00011110 = 30 and add to you GLOP address starting 233.X.X.X
- GLOP addresses always have /24 subnet mask because the implementer can select 255 addresses to be assigned locally. As a result in this example the GLOP address will be 233.22.30.XXX/24

4) Finally there is the question that asked you about subnetting. Be it a host is not communicating (it has the wrong subnet mask), Which path would the router select (you need to work out the correct Network in a routing table for a given IP), will a packet be permitted or denied in the ACL (again you need to work out if your IP is with the range permitted or denied with in the subnet mask stated) 
- For all these types of questions you are looking at basic subnetting. Aim to get these calculations down to less the 20 SECONDS. 
- Imagine a slide ruler in your head. Slide it to the where the subnet mask stops and bang, you have your subnet increments. From here you know the Network address, 1st host in the subnet, last host in the subnet, broadcast address for the subnet, and the next network address.
- Practise this over and over until you can see in your head the octet with the bit values and where you stop for each subnet.

If you can get each of the situations above nailed, and nailed quickly you can buy time on the harder questions. Practice, practice, and practice again.

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