Computer Networks Quiz | IP Addressing | Question 10
An Internet Service Provider(ISP) has the following chunk of CIDR-based IP addresses available with it:245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
(A) 245.248.136.0/21 and 245.248.128.0/22
(B) 245.248.128.0/21 and 245.248.128.0/22
(C) 245.248.132.0/22 and 245.248.132.0/21
(D) 245.248.136.0/22 and 245.248.132.0/21
Explanation: Since the routing prefix is 20, the ISP has 2^(32-20) or 2^12 addresses. Out of these 2^12 addresses, half (or 2^11) addresses have to be given to organization A and quarter (2^10) addresses have to be given to organization B. So routing prefix for organization A will be 21. For B, it will be 22. If we see all options given in the question, only options (A) and (B) are left as only these options have the same number of routing prefixes. Now we need to choose from option (A) and (B).
To assign addresses to organization A, ISP needs to take the first 20 bits from 245.248.128.0 and fix the 21st bit as 0 or 1. Similarly, ISP needs to fix 21st and 22nd bits for organization B. If we take a closer look at the options (A) and (B), we can see the 21st and 22nd bits for organization B are considered as 0 in both options. So the 21st bit of organization A must be 1. Now take the first 20 bits from 245.248.128.0 and 21st bit as 1, we get addresses for organization A as 245.248.136.0/21
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