Question

I could use some help with this problem:


An ISP is granted the block 80.70.56.0/21. The ISP needs to allocate addresses for two organizations each with 500 addresses, two organizations each with 250addresses, and 3 organizations each with 50 addresses. Give a table that shows the range of addresses allocated to each organization. For each of the blocks (the 7organizations and the unused block(s)), indicate the size of the block, the first address in the block, and the last address in the block using CIDR notation.

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Answer 1

a) Given ISP as an address 80.70.56.0/21, therefore the prefix length is 21.

• The number of addresses in the given ISP are N= 2^32-21= 2¹¹= 2048 addresses are present in the given block

• The first address can be calculated by converting the given address into binary format and performing an AND operation with its subnet mask that is

01010000 01001000 00111000 00000000- given address

AND 11111111 11111111 11111000 00000000 subnet mask

Result 01010000 01001000 00111000 00000000 first Address

• Therefore the first address in decimal format is 80.70.56.0/21, by adding 2047 addresses to the first address we get the last address of the block that is 80.70.63.255/21.

• The range of the address is 80.70.56.0/21 to 80.70.63.255/21

b) Given that 2 organizations are to be assigned 500 addresses each, 2 organizations have to be assigned 250 addresses each and 3 organizations must be assigned 50 addresses each.

• According to CIDR rules each organization or in a block the number of addresses must be a power of 2, as 500, 250, 50 are not power of 2.

• To satisfy the CIDR restriction 1, assign 512, 256, 64 block of addresses for the organization instead of 500, 250, 64.

• After assigning 2 blocks of 512 addresses each, 2 blocks of 256 addresses each, 3 blocks of 64 addresses each from 2048 addresses, 320 addresses are left out unassigned.

• Block1: let the first address be 80.70.56.0, as there are 512 addresses, the prefix length is 23, then the last address is the first address + 511.

• For the unassigned block as there are 320 addresses, according CIDR restriction the number of addresses in a block should be a power of 2, divide the unassigned block into two parts, unassigned block1 contains 256 addresses and Block2 contains 64 addresses.

• Therefore the ranges for each organization are assuming that all the blocks are continuous

Block name	Block size	Prefix	Range
Block1	512	32-log2512= 23	80.70.56.0/23 to 80.70.57.255/23
Block2	512	32-log2512=23	80.70.58.0/23 to 80.70.59.255/23
Block3	256	32-log2256=24	80.70.60.0/24 to 80.70.60.255/24
Block4	256	32-log2256=24	80.70.61.0/24 to 80.70.61.255/24
Block5	64	32-log264= 26	80.70.62.0/26 to 80.70.62.63/26
Block6	64	32-log264= 26	80.70.62.64/26 to 80.70.62.127/26
Block7	64	32-log264= 26	80.70.62.128/26 to 80.70.62.191/26
Un B1	256	32-log2256=24	80.70.62.192/24 to 80.70.63.191/24
Un B2	64	32-log264= 26	80.70.63.192/26 to 80.70.63.255/26

c) The forwarding table of a router can be given as:

Subnet mask	Interface
01010000 01000110 0011100	m0
01010000 01000110 0011101	m1
01010000 01000110 00111100	m2
01010000 01000110 00111101	m3
01010000 01000110 00111110 00	m4
01010000 01000110 00111110 01	m5
01010000 01000110 00111110 10	m6
01010000 01000110 00111110 11	un b1
01010000 01000110 00111111	un b2

d) The diagrammatic representation of the router and the interfaces is given below: