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:

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written 7.1 years ago by

teamques10 ★ 69k

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

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