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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)
    245.248.136.0/24 and 245.248.132.0/21
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An Internet Service Provider (ISP) has the following chunk of CIDR-bas...
Since half of 4096 host addresses must be given to organization A, we can set 12th bit to 1 and include that bit into network part of organization A, so the valid allocation of addresses to A is 245.248.136.0/21
Now for organization B, 12th bit is set to ‘0’ but since we need only half of 2048 addresses,
13th bit can be set to ‘0’ and include that bit into network part of organization B so the valid allocation of addresses to B is 245.248.128.0/22
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Most Upvoted Answer
An Internet Service Provider (ISP) has the following chunk of CIDR-bas...
Allocation of IP addresses to A and B

To allocate IP addresses to Organization A and B, we need to divide the given CIDR block of addresses into two parts.

CIDR block: 245.248.128.0/20

Total number of IP addresses in the block: 2^12 = 4096

Half of the addresses should be allocated to A, which is 2048 addresses.
Quarter of the addresses should be allocated to B, which is 1024 addresses.

Option A: 245.248.136.0/21 and 245.248.128.0/22

- Organization A: 245.248.136.0/21
- Number of addresses: 2^11 = 2048 (half of the total addresses)
- Range of addresses: 245.248.136.0 – 245.248.143.255
- Organization B: 245.248.128.0/22
- Number of addresses: 2^10 = 1024 (quarter of the total addresses)
- Range of addresses: 245.248.128.0 – 245.248.131.255

Option B: 245.248.128.0/21 and 245.248.128.0/22

- Organization A: 245.248.128.0/21
- Number of addresses: 2^11 = 2048 (half of the total addresses)
- Range of addresses: 245.248.128.0 – 245.248.135.255
- Organization B: 245.248.128.0/22
- Number of addresses: 2^10 = 1024 (quarter of the total addresses)
- Range of addresses: 245.248.128.0 – 245.248.131.255

Option C: 245.248.132.0/22 and 245.248.132.0/21

- Organization A: 245.248.132.0/22
- Number of addresses: 2^10 = 1024 (quarter of the total addresses)
- Range of addresses: 245.248.132.0 – 245.248.135.255
- Organization B: 245.248.132.0/21
- Number of addresses: 2^11 = 2048 (half of the total addresses)
- Range of addresses: 245.248.132.0 – 245.248.143.255

Option D: 245.248.136.0/24 and 245.248.132.0/21

- Organization A: 245.248.136.0/24
- Number of addresses: 2^8 = 256
- Range of addresses: 245.248.136.0 – 245.248.136.255
- Organization B: 245.248.132.0/21
- Number of addresses: 2^11 = 2048 (half of the total addresses)
- Range of addresses: 245.248.132.0 – 245.248.143.255

Therefore, the valid allocation of addresses to A and B is option A.
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An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses availablewith 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 isa valid allocation of addresses to A and B?a)245.248.136.0/21 and 245.248.128.0/22b)245.248.128.0/21 and 245.248.128.0/22c)245.248.132.0/22 and 245.248.132.0/21d)245.248.136.0/24 and 245.248.132.0/21Correct answer is option 'A'. Can you explain this answer?
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