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An ISP granted the block of addresses starting with 10.80.0.0/16 . ISP want to  distribute it four groups of  customers .

A)first group has 64 customers,  each need 256 addresses

B) second gp has 128 customers,  each need 128 addresses

C) third gp has 64 customers,  each need 32 addresses

D) and last gp also  has  64 customers, each need 32 addresses.

Design the subblock and how many addresses still available after allocation ?
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1 Answer

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Given 10.80.0.0/16 is the starting address. So, we have 2^16 = 65,536 addresses with the ISP initially.

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For the 1st group, each customer needs 256 addresses. So, 8 bits are needed to define each host. The prefix length is 32 – 8 = 24.

Therefore the addresses of 1st group are:

1st customer =>  10.80.0.0/24 to 10.80 .0.255/24

2nd customer => 10.80.1.0/24 to 10.80.1.255/24

3rd customer => 10.80.2.0/24 to 10.80 .1.255/24

…………….

64th customer => 10.80.63.0/24 to 10.80.63.255/24

So, for 1st group the ISP has allocated 64*256 = 16384 addresses.

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For the 2nd group, each customer needs 128 addresses. So, 7 bits are needed to define each host. The prefix length is 32 – 7 = 25.

Therefore the addresses of 2nd group are:

1st customer => 10.80.64.0/25 to 10.80.64.127/25

2nd customer =>  10.80.64.128/25 to 10.80.64.255/25

3rd customer => 10.80.65.0/25 to 10.80.65.127/25

4th customer => 10.80.65.128/25 to 10.80.65.255/25

5th customer => 10.80.66.0/25 to 10.80.66.127/25

6th customer => 10.80.66.128/25 to 10.80.66.255/25

…………...

128th customer => 10.80.127.128/25 to 10.80.127.255/25

So, for 2nd group the ISP has allocated 128*128 = 16384 addresses

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For the 3rd group, each customer needs 32 addresses. So, 5 bits are needed to define each host. The prefix length is 32 – 5= 27.

Therefore the addresses of 3rd group are:

1st customer => 10.80.128.0/27 to 10.80.128.31/27

2nd customer => 10.80.128.32/27 to 10.80.128.63/27

3rd customer => 10.80.128.64/27 to 10.80.128.95/27

4th customer => 10.80.128.96/27 to 10.80.128.127/27

……………………

64th customer => 10.80.135.224/27 to 10.80.135.255/27

So, for 3rd group the ISP has allocated 64*32 = 2048 addresses.

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For the 4th group, each customer needs 32 addresses. So, 5 bits are needed to define each host. The prefix length is 32 – 5= 27.

Therefore the addresses of 4th group are:

1st customer => 10.80.136.0/27 to 10.80.136.31/27

2nd customer => 10.80.136.32/27 to 10.80.136.63/27

3rd customer => 10.80.136.64/27 to 10.80.136.95/27

4th customer => 10.80.136.96/27 to 10.80.136.127/27

……………………

64th customer => 10.80.143.224/27 to 10.80.143.255/27

So, for 4th group the ISP has allocated 64*32 = 2048 addresses.

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Hence the total no.of addresses available after allocating is 65536 – (16384+16384 +2048+2048) = 28672 addresses

edited by

3 Comments

Thank you  its very precise explanation . I got where i had a mistake .
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Sir, i think we should start the 3rd gp with
10.80.128.0/27 to 10.80.128.31/27 Please check it.Because there will be ovelapping of ip  in gp 2nd and 3rd.
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Now check @Aks9639

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