IPv4 Subnetting and VLSM for ISP Allocations within 80.70.56.0/21

We are given an ISP block **keyword allocated to the ISP: 80.70.56.0/21. The ISP must allocate subnets to:

• two organizations each needing 500 addresses,
• two organizations each needing 250 addresses,
• three organizations each needing 50 addresses.

We will use keywordVLSM VLSM. The general rule for picking a prefix length $/p$ for $H$ required usable host addresses in IPv4 is:

• Total addresses in a subnet: $2^{(32-p)}$
• Usable hosts in a subnet: $2^{(32-p)}-2$ (excluding network and broadcast)

Then we compute the range (network IP to broadcast IP) and the unallocated remainder inside the original /21.

A small visual of the subnet math:

Two key terms:

• keywordNetwork address
• keywordBroadcast address

Note: The computations below use standard IPv4 host usability ($2^{n}-2$ for usable). If a problem’s convention differs (e.g., counting total addresses including network/broadcast), the chosen prefix lengths may change.

Part (i): Number and range of addresses in the ISP block (80.70.56.0/21)

1) Total number of addresses A /21 means $32-21=11$ host bits, so total addresses in the block are: $2^{11} = 2048$

So the ISP block contains 2048 total IP addresses.

2) Usable host addresses Usable IPv4 hosts are: $2048-2 = 2046$

So usable hosts in 80.70.56.0/21 are 2046.

3) Determine the block range A /21 subnet has a block size in the last octet of: $256 - 21\text{ mask shift} = 2^{(8- (21-16))} = 2^{(8-5)} = 2^3 = 8?$

Let’s do it cleanly by subnetting boundaries:

• Prefix /21 splits across the second and third octets: /21 means:
  • First 16 bits fixed (first two octets),
  • Next 5 bits (third octet) define subnet increments.
• In the third octet, subnet increment size is $2^{(8-5)} = 2^{3} = 8$. So the third-octet values covered by 80.70.56.0/21 run from:
• Start third octet = 56
• End third octet = 56 + 7 = 63

Thus:

• Network address: 80.70.56.0
• Broadcast address: 80.70.63.255
• Range: 80.70.56.0 – 80.70.63.255

Host requirement to prefix mapping (pre-work)

We need prefix sizes that can accommodate the usable hosts.

For 500-address organizations

Find smallest $/p$ such that: $2^{(32-p)}-2 \ge 500$ Try $2^{9}-2=512-2=510$ usable ⇒ works. $2^{9}$ total implies host bits = 9 ⇒ $32-p=9$ ⇒ $p=23$.

So each 500-address org gets: /23

• Subnet size total: $2^{9}=512$
• Usable hosts: $512-2=510$

For 250-address organizations

Need: $2^{(32-p)}-2 \ge 250$ Try $2^{8}-2=256-2=254$ usable ⇒ works. Host bits = 8 ⇒ $32-p=8$ ⇒ $p=24$.

So each 250-address org gets: /24

• Total: $2^{8}=256$
• Usable: $256-2=254$

For 50-address organizations

Need: Try $2^{6}-2=64-2=62$ usable ⇒ works. Host bits = 6 ⇒ $32-p=6$ ⇒ $p=26$.

So each 50-address org gets: /26

• Total: $2^{6}=64$
• Usable: $64-2=62$

Actual allocations (ranges)

We will allocate in this order: two /23, then two /24, then three /26, all within 80.70.56.0/21 (80.70.56.0 to 80.70.63.255).

A) Two organizations with 500 addresses each ⇒ two /23 subnets

A /23 starting at 80.70.56.0 covers:

• third octet: 56 to 57
• fourth octet: 0 to 255

So Org500-1:

• Range: 80.70.56.0 – 80.70.57.255
• Network/Broadcast: 80.70.56.0 / 80.70.57.255

Next available /23 begins at 80.70.58.0.

Org500-2:

• Range: 80.70.58.0 – 80.70.59.255
• Network/Broadcast: 80.70.58.0 / 80.70.59.255

After these, the next free address is 80.70.60.0.

---

B) Two organizations with 250 addresses each ⇒ two /24 subnets

Each /24 starting at 80.70.60.0:

Org250-1:

• Range: 80.70.60.0 – 80.70.60.255
• Network/Broadcast: 80.70.60.0 / 80.70.60.255

Next /24: 80.70.61.0

Org250-2:

• Range: 80.70.61.0 – 80.70.61.255
• Network/Broadcast: 80.70.61.0 / 80.70.61.255

Next free address becomes 80.70.62.0.

---

C) Three organizations with 50 addresses each ⇒ three /26 subnets

Within 80.70.62.0/24, a /26 splits as:

• /26 block size = 64 addresses
• fourth-octet ranges per /26: .0-.63, .64-.127, .128-.191, .192-.255

Org50-1:

• Range: 80.70.62.0 – 80.70.62.63

Org50-2:

• Range: 80.70.62.64 – 80.70.62.127

Org50-3:

• Range: 80.70.62.128 – 80.70.62.191

After three /26 blocks, the next free block is:

• 80.70.62.192 – 80.70.62.255 (still available inside the /24) and then everything in 80.70.63.0/24 (also still available).

Range of unallocated addresses

The original /21 range is:

• 80.70.56.0 – 80.70.63.255

Allocated up to:

• Highest allocated address = 80.70.62.191

So unallocated begins at 80.70.62.192 and continues to 80.70.63.255.

Therefore the unallocated range is:

• 80.70.62.192 – 80.70.63.255

If you want it split into convenient CIDR-aligned leftover blocks, the leftover space is:

1. From 80.70.62.192 – 80.70.62.255 (that is one /26)
2. Plus the entire 80.70.63.0 – 80.70.63.255 (one /24)

So leftover can be expressed as:

• 80.70.62.192/26
• 80.70.63.0/24