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Difficulty: Very hardIPv4 Addressing and Subnetting

An enterprise network architect is designing a Variable Length Subnet Mask (VLSM) address allocation plan starting at the base private IPv4 block 10.200.16.0/2010.200.16.0/20. Three internal subnets must be allocated sequentially in contiguous order from the lowest available boundary:

- Subnet A (WLAN Clients): Requires at least 1,0001,000 usable host IP addresses
- Subnet B (VoIP Phones): Requires at least 450450 usable host IP addresses
- Subnet C (Management): Requires at least 6060 usable host IP addresses

If Subnet A is allocated first, followed immediately by Subnet B, and then Subnet C, what is the valid usable host IPv4 address range for Subnet C?

  1. 10.200.22.1 to 10.200.22.62Answer
  2. B
    10.200.22.0 to 10.200.22.63
  3. C
    10.200.20.1 to 10.200.20.62
  4. D
    10.200.21.1 to 10.200.21.62

Answer

The valid usable host IPv4 address range for Subnet C is 10.200.22.1 to 10.200.22.62.
To calculate VLSM subnets sequentially from the base block 10.200.16.0/20:
1. Subnet A requires 1,000 usable host IPs. The smallest power of 2 minus 2 that satisfies this is 2^10 - 2 = 1,022 (10 host bits, /22 prefix). Allocated block: 10.200.16.0 to 10.200.19.255.
2. Subnet B starts at 10.200.20.0 and requires 450 usable host IPs. 2^9 - 2 = 510 (9 host bits, /23 prefix). Allocated block: 10.200.20.0 to 10.200.21.255.
3. Subnet C starts at 10.200.22.0 and requires 60 usable host IPs. 2^6 - 2 = 62 (6 host bits, /26 prefix). Network ID is 10.200.22.0 and broadcast is 10.200.22.63.
Thus, the usable host range for Subnet C is 10.200.22.1 through 10.200.22.62.

Step-by-Step Solution

1
Calculate host bits and prefix length for Subnet A (1,000 hosts).
Host bits needed: h=10h = 10 (210−2=1,022≥1,0002^{10} - 2 = 1,022 \ge 1,000). Prefix length: /22/22 (32−1032 - 10). Address block: 10.200.16.0/2210.200.16.0/22, spanning 10.200.16.010.200.16.0 through 10.200.19.25510.200.19.255.
Subnetting formula 2h−22^h - 2 requires 10 host bits to fit 1,000 usable host addresses.
2
Determine starting address and size for Subnet B (450 hosts).
Start address: 10.200.20.010.200.20.0. Host bits needed: h=9h = 9 (29−2=510≥4502^9 - 2 = 510 \ge 450). Prefix length: /23/23 (32−932 - 9). Address block: 10.200.20.0/2310.200.20.0/23, spanning 10.200.20.010.200.20.0 through 10.200.21.25510.200.21.255.
Sequential allocation places Subnet B immediately after Subnet A's broadcast boundary.
3
Determine starting address, size, and usable range for Subnet C (60 hosts).
Start address: 10.200.22.010.200.22.0. Host bits needed: h=6h = 6 (26−2=62≥602^6 - 2 = 62 \ge 60). Prefix length: /26/26 (32−632 - 6). Network ID: 10.200.22.0/2610.200.22.0/26. Broadcast address: 10.200.22.6310.200.22.63. Usable range: 10.200.22.110.200.22.1 to 10.200.22.6210.200.22.62.
Subnet C starts at 10.200.22.0/2610.200.22.0/26. Usable hosts range from Network ID +1+ 1 to Broadcast −1- 1.

Key Concept

Variable Length Subnet Masking (VLSM) host requirement sizing and sequential subnet boundary calculation.
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Topics

IPv4 Addressing and Subnetting

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