Examkin

Question

Difficulty: HardIPv4 Addressing and Subnetting

A network architect is implementing Variable Length Subnet Masking (VLSM) on the assigned address block 172.16.48.0/20172.16.48.0/20 to provision distinct functional zones in a datacenter. Zone A requires host capacity for at least 500 IP addresses, while Zone B requires host capacity for at least 250 IP addresses. Subnets are allocated sequentially starting from the lowest available network address of the block, with Zone A allocated first, followed immediately by Zone B. What is the broadcast address of the subnet allocated to Zone B?

  1. 172.16.50.255172.16.50.255Answer
  2. B
    172.16.50.254172.16.50.254
  3. C
    172.16.49.255172.16.49.255
  4. D
    172.16.51.255172.16.51.255

Answer

The broadcast address of the subnet allocated to Zone B is 172.16.50.255172.16.50.255.
To host at least 500 devices in Zone A, 9 host bits are needed (29−2=5102^9 - 2 = 510), resulting in a /23/23 subnet (172.16.48.0172.16.48.0 to 172.16.49.255172.16.49.255). The next available network address for Zone B is 172.16.50.0172.16.50.0. To host 250 devices in Zone B, 8 host bits are required (28−2=2542^8 - 2 = 254), resulting in a /24/24 subnet (172.16.50.0172.16.50.0 to 172.16.50.255172.16.50.255). The broadcast address of this subnet is the final address in its range, 172.16.50.255172.16.50.255.

Step-by-Step Solution

1
Determine prefix length and block size for Zone A (500 hosts required)
Host bits needed: 2h−2≥500  ⟹  h=92^h - 2 \ge 500 \implies h = 9 bits (29−2=5102^9 - 2 = 510 usable hosts). Subnet mask prefix is 32−9=/2332 - 9 = /23, which gives a block size of 29=5122^9 = 512 addresses.
Zone A requires at least 500 hosts, so a /24/24 (254 usable hosts) is insufficient and a /23/23 must be used.
2
Calculate the address boundary for Zone A
Zone A starts at 172.16.48.0/23172.16.48.0/23 and spans 172.16.48.0172.16.48.0 through 172.16.49.255172.16.49.255.
A /23/23 block starting at 172.16.48.0172.16.48.0 increments the third octet by 2 (48+2=5048 + 2 = 50).
3
Determine prefix length and block size for Zone B (250 hosts required)
Starting address for Zone B is 172.16.50.0172.16.50.0. Host bits needed: 2h−2≥250  ⟹  h=82^h - 2 \ge 250 \implies h = 8 bits (28−2=2542^8 - 2 = 254 usable hosts). Subnet mask prefix is 32−8=/2432 - 8 = /24, giving a block size of 28=2562^8 = 256 addresses.
250 hosts require 8 host bits, corresponding to a /24/24 subnet mask.
4
Determine the broadcast address of Zone B
The subnet range for Zone B is 172.16.50.0172.16.50.0 to 172.16.50.255172.16.50.255. The broadcast address is 172.16.50.255172.16.50.255.
The last address in an IP subnet block is reserved as the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) calculations, subnet block size determination, and broadcast address identification.
Rate this question

Topics

IPv4 Addressing and Subnetting

More questions from this topic