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An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 addresses). The ISP
needs to distribute these addresses to three groups of customers as follows:
a. The first group has 64 customers; each needs 256 addresses.
b. The second group has 128 customers; each needs 128 addresses.
c. 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

solution:

1. 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:
2nd Customer:
190.100.0.0/24
190.100.1.0/24
190.100.0.255/24
190.100.1.255/24
64th Customer: 190.100.63.0/24 190.100.63.255/24
Total =64 X 256 =16,384
2. Group2
For this group, each customer needs 128 addresses. This means that 7 (10g2 128) 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
2nd Customer: 190.100.64.128/25
190.100.64.127/25
190.100.64.255/25
128th Customer: 190.100.127.128/25 190.100.127.255/25
Total =128 X 128 = 16,384
3. Group3
For this group, each customer needs 64 addresses. This means that 6 (logz 64) bits are needed
to each host. The prefix length is then 32 - 6 =26. The addresses are
SECTION 19.1 IPv4 ADDRESSES 563
1st Customer:
2nd Customer:
190.100.128.0/26
190.100.128.64/26
190.100.128.63/26
190.100.128.127/26
128th Customer: 190.100.159.192/26 190.100.159.255/26
Total =128 X 64 =8192
Number of granted addresses to the ISP: 65,536
Number of allocated addresses by the ISP: 40,960
Number of available addresses: 24,576

In this question I am not able to understand how they divide the groups 

closed as a duplicate of: Variable Lengths Subnets
asked in Computer Networks by (171 points)
closed by | 18 views


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