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An organization has subnetted the IP block 200.1.1.0/24 using Variable Length Subnet Masking (VLSM) as follows:

Host Requirements:

Subnet A: 72 hosts
Subnet B: 35 hosts
Subnet C: 20 hosts
Subnet D: 18 hosts
 
                  200.1.1.0/24
                      |
        +-------------+-------------+
        |                           |
  200.1.1.0/25                200.1.1.128/25
    [Subnet A]                    |
                       +----------+-----------+
                       |                      |
               200.1.1.128/26        200.1.1.192/26
                [Subnet B]              |
                               +--------+--------+
                               |                 |
                      200.1.1.192/27     200.1.1.224/27
                       [Subnet C]         [Subnet D]

The organization now needs to add a fifth subnet (Subnet E) requiring 20 new hosts. Which of the following is the most appropriate way to accommodate Subnet E without requesting a new network block?

  1. Convert Subnet D from /27 to /28 and use the leftover space for Subnet E
  2. Merge Subnets B and C to free up a /25 block for Subnet E
  3. Split Subnet A (200.1.1.0/25) into four /27 subnets, assign three to Subnet A and one to Subnet E
  4. Use the unused space after Subnet D (200.1.1.255/32) to create Subnet E

1 Answer

1 1 vote

Answer: C

The original IP block 200.1.1.0/24 contains 256 addresses. The organization has already created four subnets using VLSM:

  • 200.1.1.0/25 — Subnet A (supports up to 126 hosts)
  • 200.1.1.128/26 — Subnet B (supports up to 62 hosts)
  • 200.1.1.192/27 — Subnet C (supports up to 30 hosts)
  • 200.1.1.224/27 — Subnet D (supports up to 30 hosts)

Now, we need to accommodate Subnet E requiring 20 hosts. A /27 subnet offers 32 IP addresses (30 usable), which is ideal for Subnet E.

Looking at the current allocation:

  • Subnet D ends at 200.1.1.255, so the entire 200.1.1.0/24 block is already allocated — there is no space "after" Subnet D.
  • Merging Subnets B and C would waste space and disturb existing configuration.
  • Changing Subnet D from /27 to /28 would only save 16 IPs — not enough for a new subnet needing 20 hosts.
  • However, Subnet A has been allocated a /25, which supports 126 hosts but only needs 72. It can be split into smaller subnets.

We can divide 200.1.1.0/25 into four subnets of size /27 (each supporting up to 30 hosts):

  • 200.1.1.0/27
  • 200.1.1.32/27
  • 200.1.1.64/27
  • 200.1.1.96/27

Three of them can be assigned to Subnet A (totaling 90 usable hosts), and one can be used for Subnet E (30 usable hosts).

Hence, the correct approach is to split Subnet A into four /27 subnets, reassign three to Subnet A and use one for Subnet E.

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