# AIW Numerical || BIM STUDY NOTES

Numerical  Type 1 Examples

1. Change the multicast IP address 232.43.14.7 to an Ethernet multicast physical address.
• Covert 23 bits of the IP address in hexadecimal notation. This can be done by changing the rightmost 3 bytes (i.e. 43.14.7 in our example) to hexadecimal and then subtracting 8 from the leftmost digit if it is greater than or equal to 8. Therefore rightmost 3 bytes in hex notation are: 2B:0E:07. Since right most digit is 2 (<=8) so no need to subtract with 8.
• Add the result obtained in above steps to the starting multicast address (i.e. 01:00:5E:00:00:00). Hence the result is: 01:00:5E:2B:0E:07

2.Change the multicast IP address 238.212.24.9 to an Ethernet multicast address

• The rightmost 3 bytes in hex notation are: D4:18:09. We need to subtract 8 from the leftmost digit, resulting in 54:18:09 (Note: D-8=5)
• Add result 54:18:09 to the Ethernet multicast starting address. The result is: 01:00:5E:54:18:09

## Numerical Type 2

1. From the following information, map IP multicast address with Ethernet address. (2016 Model)

IP multicast Address: 229.122.54.99

• Convert physical address into binary notation: 01101000 10010100  00100011  01000000  11101001  10011100
• Convert IP multicast Address into binary notation: 11100101 01111010  00110110  01100011
• Insert rightmost 23 bit of IP multicast address to the right most 23 bit of physical address. Therefore 01101000  10010100  00100011  01111010  00110110  01100011 (68:94:23:7A:36:63)

## Numerical Type 3

1. Identify the first four /36 address blocks out of 2001:db8::/32

Solution:
Subnet 2001:db8::/32 into /36s
Number of Subnetting bits = 36 – 32       = 4
Therefore there are 2^4 = 16 subnets are possible
First four /36 address blocks are:
2001:DB8:0000 0000 0000 0000::/36      —–>         2001:DB8:0000::/36
2001:DB8:0001 0000 0000 0000::/36         —->       2001:DB8:1000::/36
2001:DB8:0010 0000 0000 0000::/36     —->          2001:DB8:2000::/36
2001:DB8:0011 0000 0000 0000::/36     —->          2001:DB8:3000::/36
2. Identify the first six /35 address blocks out of 2406:6400::/32
Solution:
Subnet 2406:6400::/32 into /35s
Number of Subnetting bits = 35 – 32       = 3
Therefore there are 2^3 = 8 subnets are possible
First six /35 address blocks are:
2406:6400:0000 0000 0000 0000::/35     ——–>      2406:6400:0000::/35
2406:6400:0010 0000 0000 0000::/35     ——–>    2406:6400:2000::/35
2406:6400:0100 0000 0000 0000::/35     ——–>      2406:6400:4000::/35
2406:6400:0110 0000 0000 0000::/35     ——–>      2406:6400:6000::/35
2406:6400:1000 0000 0000 0000::/35    ——–>       2406:6400:8000::/35
2406:6400:1010 0000 0000 0000::/35   ——–>        2406:6400:A000::/35

3.Identify the first six /42 address blocks out of ABCD:EFAB::/32
Subnet ABCD: EFAB::/32 into /42s
Number of Subnetting bits = 42 – 32       = 10
Therefore there are 2^10 subnets are possible
First six /42 address blocks are:
ABCD: EFAB: 0000  0000  0000  0000::/42 ——->    ABCD: EFAB:0000::/42
ABCD: EFAB: 0000  0000  0100  0000::/42 ——->   ABCD: EFAB:0004::/42
ABCD: EFAB: 0000  0000  1000  0000::/42 ——-> ABCD: EFAB:0008::/42
ABCD: EFAB: 0000  0000   1100  0000::/42 ——-> ABCD: EFAB:000C::/42
ABCD: EFAB: 0000  0001   0000  0000::/42   ——-> ABCD: EFAB:0010::/42
ABCD: EFAB: 0000  0001   0100  0000::/42 ——-> ABCD: EFAB:0014::/42

Numerical Type 4

#### IPv4 Subnetting (Variable length Subnetting)

1. An organization is granted a block of addresses with the beginning address 14.24.74.0/24. The organization needs to have 3 subblocks of addresses to use in its three subnets: one subblock of 10 addresses, one subblock of 60 addresses, and one subblock of 120 addresses. Design the subblocks

Solution:
There are 232-24  = 256 addresses in this block.
The first address is 14.24.74.0/24;
the last address is 14.24.74.255/24
a. The number of addresses in the largest subblock, which requires 120 addresses. To satisfy the requirement we have to allocate 128 address (since 120 is not a power of 2 so go to the upper nearest value that is power of two).
The subnet mask for this subnet (n1) = 32-logl2128     =25
Therefore the first address in this block is 14.24.74.0/25 and
the last address is 14.24.74.127/25
First host address: 14.24.74.1/25
Last host address: 14.24.74.126/25

b.The number of addresses in the second largest subblock, which requires 60 addresses. To satisfy the requirement we have to allocate 64 address (since 60 is not a power of 2 so go to the upper nearest value that is power of two).
The subnet mask for this subnet (n1) = 32-logl264     =26
Therefore the first address in this block is 14.24.74.128/26 and
the last address is 14.24.74.191/26
First host address: 14.24.74.129/26
Last host address: 14.24.74.190/26
c. The number of addresses in the last subblock, which requires 10 addresses. To satisfy the requirement we have to allocate 16 address (since 10 is not a power of 2 so go to the upper nearest value that is power of two).
The subnet mask for this subnet (n1) = 32-logl216     =28
Therefore the first address in this block is 14.24.74.192/28 and
the last address is 14.24.74.207/28
First host address: 14.24.74.193/28
Last host address: 14.24.74.206/28
2. If you are assigned an IP address 92.16.1.0/24 and plans to deploy CIDR.
Here are some requirement which you have to fulfill for Subnet A = 120 Host, Subnet B = 60 Host, Subnet C = 30 Host, Subnet D = 10 Host Subnet E=5. You are also required to calculate subnet-mask, range, net-id, broadcast id for each subnet.
There are 232-24 = 256 addresses in this block.
The first address is 92.16.1.0/24; the last address is 92.16.1.255/24
a. The number of addresses in the largest subblock, which requires 120 addresses. To satisfy the requirement we have to allocate 128 address
The subnet mask for this subnet (n1) = 32-logl2128     =25
Therefore the first address in this block is 92.16.1.0/25 and
the last address is 92.16.1.127/25
First host address: 92.16.1.1/25
Last host address: 92.16.1.126/25
b. The number of addresses in the second largest subblock, which requires 60 addresses. To satisfy the requirement we have to allocate 64 address
The subnet mask for this subnet (n1) = 32-logl264     =26
Therefore the first address in this block is 92.16.1.128/26 and
the last address is 92.16.1.191/26
First host address: 92.16.1.129/26
Last host address: 92.16.1.190/26

c. The number of addresses in the third largest subblock, which requires 30 addresses.
To satisfy the requirement we have to allocate 32 address
The subnet mask for this subnet (n1) = 32-logl232     =27
Therefore the first address in this block is 92.16.1.192/27 and
the last address is 92.16.1.223/27

First host address: 92.16.1.193/27
Last host address: 92.16.1.222/27
d. The number of addresses in the fourth largest subblock, which requires 10 addresses. To satisfy the requirement we have to allocate 16 address
The subnet mask for this subnet (n1) = 32-logl216     =28
Therefore the first address in this block is 92.16.1.224/28 and
the last address is 92.16.1.239/28