2. An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 a
ID: 3598024 • Letter: 2
Question
2. 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 i. The first group has 64 customers, each needs 256 addresses. ii. The second group has 128 customers, each needs 128 addresses. ii. The third group has 128 customers; each needs 64 addresses Design the sub-blocks, showing the first and last addresses each for the first and last customer in each block a. b. Find out how many addresses are still available after these allocations.Explanation / Answer
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: 190.100.0.0/24 190.100.0.255/24
2nd Customer: 190.100.1.0/24 190.100.1.255/24
.....................
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 (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
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 (log264) bits are needed to each host. The prefix length is then 32 - 6 =26. The addresses are
1stCustomer: 190.100.128.0/26 190.100.128.63/26
2nd Customer: 190.100.128.64/26 190.100.128.127/26
.........
128th Customer: 190.100.159.192/26 190.100.159.255/26
Total =128X 64 =8192
Number of granted addresses to the ISP: 65,536
Number ofallocated addresses by the ISP: 40,960
Number ofavailable addresses: 24,576
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