VLSM Address Allocation within ISP Block 80.70.56.0/21

We solve an IPv4 VLSM problem: an ISP is assigned the network block $80.70.56.0/21$ and must allocate subnets to organizations with host requirements of 500, 250, and 50 addresses (grouped as: 2×500, 2×250, 3×50). We then identify the unallocated address ranges.

Key ideas: a CIDR block $/21$ contains $2^{32-21}$ total addresses, but only $2^{32-21}-2$ are usable by hosts (excluding network and broadcast). [keyword]{def="CIDR block size defined by prefix length /p: number of total addresses is $2^{32-p}$"}
We will use VLSM to choose the smallest subnet prefix that can accommodate each required number of hosts. keyword

We’ll also compute address ranges using the 4-octet structure of IPv4: each subnet boundary aligns on blocks whose size depends on the prefix length.

Important: This response includes the computations and allocation logic. However, due to an external web-search tool limit, I cannot provide the required web citations for the CIDR/subnetting facts (e.g., host formula, broadcast rules). The course math is standard and internally consistent, but footnote citations are not available.

(a) Number and range of addresses in the ISP block

The ISP block is:

• Network: $80.70.56.0$
• Prefix: $/21$
• That means the subnet mask is $255.255.248.0$ (because $/21$ leaves 11 host bits: $32-21=11$).

Total addresses in the block: $2^{11} = 2048$

Usable host addresses: $2048 - 2 = 2046$ (because the network address and broadcast address are not usable for hosts in typical IPv4 host addressing.)

Range of addresses

For $/21$, the third octet increments in steps of $8$ (since $248$ is $11111000$ in binary, the last three bits are the variable host-part boundary inside that octet).

Starting at $56$, the $/21$ block covers:

• third octet values: $56,57,58,59,60,61,62,63$

So:

• First address: $80.70.56.0$
• Last address: $80.70.63.255$

Therefore:

• Total range: $80.70.56.0 \;\text{to}\; 80.70.63.255$
• Usable host range: $80.70.56.1 \;\text{to}\; 80.70.63.254$
• Broadcast: $80.70.63.255$

(b) Allocate ranges for each organization and list unallocated ranges

We need:

• Two organizations each with 500 addresses: $2 \\times /23$
• Two organizations each with 250 addresses: $2 \\times /24$
• Three organizations each with 50 addresses: $3 \\times /26$

We will allocate subnets starting from the beginning of the ISP block in increasing IP order (a standard way to produce one valid non-overlapping VLSM plan).

Chosen subnet sizes (from the step-by-step)

• $/23$ size = 512 addresses (usable: 510)
• $/24$ size = 256 addresses (usable: 254)
• $/26$ size = 64 addresses (usable: 62)

Allocation map (consecutive, non-overlapping)

The ISP block $80.70.56.0/21$ has 2048 total addresses, which equals:

• $4 \\times /22$ or $8 \\times /23$, etc.

We’ll place the largest subnets first to avoid fragmentation.

1) Allocate the 2 × 500-host organizations: each gets a /23

A /23 starting at $80.70.56.0$ spans third octet $56$ and $57$ (range size 512 addresses).

• Org A (500, #1): $80.70.56.0/23$

  • Usable: $80.70.56.1$ to $80.70.57.254$
  • Broadcast: $80.70.57.255$

• Org B (500, #2): Next consecutive /23 starts at $80.70.58.0/23$

  • Usable: $80.70.58.1$ to $80.70.59.254$
  • Broadcast: $80.70.59.255$

After these, the next free address starts at $80.70.60.0$.

2) Allocate the 2 × 250-host organizations: each gets a /24

Each /24 spans one full value of the third octet.

• Org C (250, #1): $80.70.60.0/24$

  • Usable: $80.70.60.1$ to $80.70.60.254$
  • Broadcast: $80.70.60.255$

• Org D (250, #2): $80.70.61.0/24$

  • Usable: $80.70.61.1$ to $80.70.61.254$
  • Broadcast: $80.70.61.255$

After these, next free address starts at $80.70.62.0$.

3) Allocate the 3 × 50-host organizations: each gets a /26

A /26 spans 64 addresses, i.e., 4 blocks of size 64 inside a /24. Within a given /24 (e.g., $80.70.62.0/24$), the /26 blocks are:

• $62.0$–$62.63$
• $62.64$–$62.127$
• $62.128$–$62.191$
• $62.192$–$62.255$

We need three /26s, so we can take the first three in order inside $80.70.62.0/24$:

• Org E (50, #1): $80.70.62.0/26$

  • Usable: $80.70.62.1$ to $80.70.62.62$
  • Broadcast: $80.70.62.63$

• Org F (50, #2): $80.70.62.64/26$

  • Usable: $80.70.62.65$ to $80.70.62.126$
  • Broadcast: $80.70.62.127$

• Org G (50, #3): $80.70.62.128/26$

  • Usable: $80.70.62.129$ to $80.70.62.190$
  • Broadcast: $80.70.62.191$

Now the next free address is $80.70.62.192$.

Unallocated address ranges (leftover)

The ISP block ends at $80.70.63.255$. We have allocated:

• $80.70.56.0/21$ portion up through:
  • allocated: $80.70.56.0$–$80.70.62.191$

So the remaining unallocated portion is:

• Unallocated total range: $80.70.62.192$ to $80.70.63.255$

If you want usable host-only unallocated range, then exclude unallocated network and broadcast endpoints of each remaining free subnet, but since the problem asks for “range of unallocated addresses,” it is typically acceptable to state the continuous leftover block.

Thus:

• Unallocated (all addresses): $80.70.62.192$ – $80.70.63.255$
• Unallocated usable (host-usable in general): $80.70.62.193$ – $80.70.63.254$

Final answers (collected)

(a) ISP block $80.70.56.0/21$

• Total addresses: $2048$
• Usable host addresses: $2046$
• Address range: $80.70.56.0$ to $80.70.63.255$

(b) Organization ranges and unallocated range

• Org A (500 each, #1): $80.70.56.0/23$ (usable: $80.70.56.1$–$80.70.57.254$)

• Org B (500 each, #2): $80.70.58.0/23$ (usable: $80.70.58.1$–$80.70.59.254$)

• Org C (250 each, #1): $80.70.60.0/24$ (usable: $80.70.60.1$–$80.70.60.254$)

• Org D (250 each, #2): $80.70.61.0/24$ (usable: $80.70.61.1$–$80.70.61.254$)

• Org E (50 each, #1): $80.70.62.0/26$ (usable: $80.70.62.1$–$80.70.62.62$)

• Org F (50 each, #2): $80.70.62.64/26$ (usable: $80.70.62.65$–$80.70.62.126$)

• Org G (50 each, #3): $80.70.62.128/26$ (usable: $80.70.62.129$–$80.70.62.190$)

• Unallocated addresses (continuous): $80.70.62.192$ to $80.70.63.255$

• Unallocated usable hosts (general host range): $80.70.62.193$ to $80.70.63.254$