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An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536addresses). The ISP needs to distribute these addresses to three groups ofcustomers as follows:a. The first group has 64 customers; each needs 256addresses.b. The second group has 128 customers; each needs 128addresses.c. The third group has 128 customers; each needs 64addresses.Design the subblocks and find out how many addresses are still availableafter these allocations.

Question

An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536addresses). The ISP needs to distribute these addresses to three groups ofcustomers as follows:a. The first group has 64 customers; each needs 256addresses.b. The second group has 128 customers; each needs 128addresses.c. The third group has 128 customers; each needs 64addresses.Design the subblocks and find out how many addresses are still availableafter these allocations.

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Solution

Sure, let's break this down:

  1. The ISP has a block of addresses starting with 190.100.0.0/16. This means they have 2^(32-16) = 65,536 addresses.

  2. The first group has 64 customers, each needing 256 addresses. This means the first group needs 64*256 = 16,384 addresses. We can represent these addresses with a /24 prefix, because 2^(32-24) = 256.

  3. The second group has 128 customers, each needing 128 addresses. This means the second group needs 128*128 = 16,384 addresses. We can represent these addresses with a /25 prefix, because 2^(32-25) = 128.

  4. The third group has 128 customers, each needing 64 addresses. This means the third group needs 128*64 = 8,192 addresses. We can represent these addresses with a /26 prefix, because 2^(32-26) = 64.

  5. To find out how many addresses are still available, we subtract the total number of addresses used by the groups from the total number of addresses the ISP has: 65,536 - 16,384 - 16,384 - 8,192 = 24,576 addresses are still available.

So, the ISP can allocate the addresses as follows:

  • The first group: 190.100.0.0/24 to 190.100.63.0/24 (64 blocks of 256 addresses)
  • The second group: 190.100.64.0/25 to 190.100.191.0/25 (128 blocks of 128 addresses)
  • The third group: 190.100.192.0/26 to 190.100.255.0/26 (128 blocks of 64 addresses)

And there would still be 24,576 addresses available.

This problem has been solved

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