Subnet Allocation from a /20 into Half (/21) and Quarter (/22)

We start with the available block:

• CIDR block: 245.248.128.0/20
  A /20 means the first 20 bits are fixed, leaving $32-20=12$ host bits, so the block size is $2^{12}=4096$ addresses.

The ISP wants:

• Half of the /20 to organization A (\Rightarrow 4096/2 = 2048) addresses
• Quarter of the /20 to organization B (\Rightarrow 4096/4 = 1024) addresses
• The remaining half (another 2048) kept by the ISP

In CIDR, halving a block corresponds to increasing the prefix length by 1:

• /20 split into two equal halves: /21 and /21
  And quarter corresponds to splitting into four equal parts:
• /20 into quarters: /22 blocks

So we need:

• A gets a /21 block
• B gets a /22 block (one quarter)
• Remaining is whatever is left (also consistent with other /21 or /22 portions)

Key facts:

• Prefix length determines block size
• Subnetting uses contiguous boundaries
• Contiguous subnet is required for valid allocations

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Visualizing the /20 range

For 245.248.128.0/20:

• The third octet spans from 128 to 143 inclusive (16 values).
• That means the block covers:
  • 245.248.128.0 through 245.248.143.255

A /21 splits at the 21st bit, i.e., in the third octet by chunks of 8:

• First /21: 245.248.128.0/21 covers third octets 128–135
• Second /21: 245.248.136.0/21 covers third octets 136–143

A /22 splits further: chunk size in the third octet is 4:

• /22 blocks inside the first /21:
  • 245.248.128.0/22 → third octets 128–131
  • 245.248.132.0/22 → third octets 132–135
• /22 blocks inside the second /21:
  • 245.248.136.0/22 → third octets 136–139
  • 245.248.140.0/22 → third octets 140–143

Thus, any valid quarter allocation must be exactly one of the /22 blocks above.

Now evaluate each option.

Option (i)

A = 245.248.136.0/21 (valid half)
B = 245.248.128.0/22 (this /22 is inside 245.248.128.0/21, not inside A)

This can still be valid only if B is meant to be a quarter of the whole /20 while A is the other half. Since A already claims the entire second /21 (136–143), the remaining portion is exactly the first /21 (128–135), which consists of two /22 quarters: 128.0/22 and 132.0/22.
So B = 245.248.128.0/22 is indeed one quarter of the original /20 and does not overlap A.

✅ Option (i) is valid.

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Option (ii)

A = 245.248.128.0/21 (valid half)
B = 245.248.128.0/22 (this is inside A)

But if B is allocated to the same half that A already fully occupies, then the ISP’s “half to A and quarter to B” constraint implies non-overlapping allocations (A is a half-block; B is a quarter-block carved out from the remaining space, otherwise you don’t uniquely allocate the intended half to A without overlap).
In standard subnet allocation problems like this, the intended meaning is: A gets one whole /21 block, B gets one whole /22 block from the remaining addresses, i.e., no overlap.

B overlapping A contradicts that intent.

❌ Option (ii) is not valid.

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Option (iii)

A = 245.248.132.0/22 (but A must be /21 for half!)

A is given a /22, which is only 1024 addresses, not 2048.

❌ Option (iii) is not valid.

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Option (iv)

A = 245.248.136.0/22 (but A must be /21 for half!)

Again, A is /22 not /21.

❌ Option (iv) is not valid.

Conclusion

The only option where:

• A is a /21 (half of /20), and
• B is a /22 (quarter of /20) within the remaining addresses (no overlap with A)

is:

(i) 245.248.136.0/21 and 245.248.128.0/22