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Difficulty: HardIPv4 Addressing and Subnetting

A network architect is subnetting the IP address block 10.112.0.0/1910.112.0.0/19 to provision dedicated VLANs for expanding branch offices. Each VLAN requires support for at least 127127 usable host IP addresses while conserving as much address space as possible. Subnets are allocated sequentially starting from the beginning of the block. What is the broadcast address of the 5th5^{\text{th}} subnet allocated?

  1. 10.112.4.25510.112.4.255Answer
  2. B
    10.112.2.12710.112.2.127
  3. C
    10.112.3.25510.112.3.255
  4. D
    10.112.5.25510.112.5.255

Answer

The broadcast address of the 5th allocated subnet is 10.112.4.25510.112.4.255.
To support 127127 usable hosts, the subnet must provide at least 127127 host IP addresses. Using the formula 2h−22^h - 2, h=7h=7 host bits yields 128−2=126128 - 2 = 126 usable hosts, which does not meet the requirement. Therefore, h=8h=8 host bits must be used (28−2=2542^8 - 2 = 254 usable hosts), corresponding to a /24/24 prefix with a block size of 256256 IP addresses per subnet. Counting sequentially from 10.112.0.010.112.0.0, the 5th subnet is 10.112.4.0/2410.112.4.0/24, which has a broadcast address of 10.112.4.25510.112.4.255.

Step-by-Step Solution

1
Determine the required number of host bits (hh).
h=8h = 8 host bits.
The usable host formula is 2h−2≥required hosts2^h - 2 \ge \text{required hosts}. For 127127 hosts, h=7h=7 gives 27−2=1262^7 - 2 = 126 usable hosts, which is insufficient. Selecting h=8h=8 gives 28−2=2542^8 - 2 = 254 usable hosts.
2
Calculate the subnet prefix and block size.
Prefix is /24/24 (32−8=2432 - 8 = 24), and block size is 28=2562^8 = 256 addresses per subnet.
Each /24/24 subnet increments the third octet by 11.
3
List the sequential /24/24 subnets starting from 10.112.0.0/1910.112.0.0/19.
Subnet 1: 10.112.0.0/2410.112.0.0/24
Subnet 2: 10.112.1.0/2410.112.1.0/24
Subnet 3: 10.112.2.0/2410.112.2.0/24
Subnet 4: 10.112.3.0/2410.112.3.0/24
Subnet 5: 10.112.4.0/2410.112.4.0/24
Subnets are assigned continuously starting from the network ID 10.112.0.010.112.0.0.
4
Identify the broadcast address for the 5th5^{\text{th}} subnet (10.112.4.0/2410.112.4.0/24).
Broadcast address is 10.112.4.25510.112.4.255.
The broadcast address is the last IP address in the subnet block immediately preceding the next subnet (10.112.5.010.112.5.0).

Key Concept

IPv4 Subnet Mask Calculation and Boundary Identification
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Topics

IPv4 Addressing and Subnetting

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