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An ISP is granted a block of addresses starting with 190.100.0.0/16 (65536) addresses the ISP needs to distribute these addresses to three groups of customers as follows:

1) The first group has 64 customers, each needs 256 addresses

2) The second group has 128 customers, each needs 128 bits.

3) The third group has 128 customers, each needs 64 addresses.

Design the sub blocks and find out how many addresses are still available after these allocations.

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Group 1:

For this group, each customer needs 256 addresses this means that 8(log2256) bits are needed to define each host the prefix length is then 32 – 8 = 24 the addresses are

1st customer 190.100.0.0/24 190.100.2.255/24
2nd customer 190.100.1.0/24 190.100.1.255/24
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64th customer 190.100.63.0/24 190.100.63.255/24

Total = 64 x 256 = 16384

Group 2:

For this group, each customer needs 128 addresses this means that 7(log2128) bits are needed to define each host the prefix length is then 32 – 7 = 25 the addresses are

1st customer 190.100.64.0/25 190.100.64.127/25
2nd customer 190.100.64.128/25 190.100.64.255/25
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128th customer 190.100.127.128/25 190.100.127.255/25

Total = 128 x 128 = 16384

Group 2:

For this group, each customer needs 128 addresses this means that 6(log264) bits are needed to define each host the prefix length is then 32 – 6 = 26 the addresses are

1st customer 190.100.128.0/26 190.100.128.63/26
2nd customer 190.100.128.64/26 190.100.128.127/26
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128th customer 190.100.159.192/26 190.100.159.255/26

Total = 128 x 64 = 8192

Number granted addresses to the ISP -> 65536

Number of allocated addresses by the ISP -> 40960

Number of available addresses -> 24576