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Difficulty: Very hardIPv4 Addressing and Subnetting

A network administrator is allocating subnets sequentially from the 10.10.16.0/2010.10.16.0/20 parent block using Variable Length Subnet Masking (VLSM), assigning the largest required blocks first. The administrator has already completed subnets for three departments:

- Department A requires 500500 usable hosts
- Department B requires 250250 usable hosts
- Department C requires 120120 usable hosts

Immediately following these allocations, the administrator provisions a new subnet for Department D, which requires at least 6060 usable hosts. What is the broadcast address of Department D's newly provisioned subnet?

  1. 10.10.19.19110.10.19.191Answer
  2. B
    10.10.19.19010.10.19.190
  3. C
    10.10.19.25510.10.19.255
  4. D
    10.10.19.12710.10.19.127

Answer

The broadcast address of Department D's newly provisioned subnet is 10.10.19.19110.10.19.191.
To find the broadcast address of Department D's subnet, subnets must be allocated sequentially in order of decreasing size starting from 10.10.16.0/2010.10.16.0/20:
1. Department A (500500 hosts) requires a /23/23 block (512512 addresses), covering 10.10.16.010.10.16.0 – 10.10.17.25510.10.17.255.
2. Department B (250250 hosts) requires a /24/24 block (256256 addresses), covering 10.10.18.010.10.18.0 – 10.10.18.25510.10.18.255.
3. Department C (120120 hosts) requires a /25/25 block (128128 addresses), covering 10.10.19.010.10.19.0 – 10.10.19.12710.10.19.127.
4. Department D (6060 hosts) requires a /26/26 block (6464 addresses), starting at 10.10.19.12810.10.19.128 and ending at 10.10.19.19110.10.19.191. Thus, the broadcast address for Department D is 10.10.19.19110.10.19.191.

Step-by-Step Solution

1
Determine prefix length and address range for Department A (500 usable hosts)
Prefix length: /23/23 (29−2=510≥5002^9 - 2 = 510 \ge 500). Range: 10.10.16.010.10.16.0 to 10.10.17.25510.10.17.255. Next available IP: 10.10.18.010.10.18.0.
A /23/23 subnet provides 512 total addresses (510 usable), satisfying the 500 host requirement.
2
Determine prefix length and address range for Department B (250 usable hosts)
Prefix length: /24/24 (28−2=254≥2502^8 - 2 = 254 \ge 250). Range: 10.10.18.010.10.18.0 to 10.10.18.25510.10.18.255. Next available IP: 10.10.19.010.10.19.0.
A /24/24 subnet provides 256 total addresses (254 usable), satisfying the 250 host requirement.
3
Determine prefix length and address range for Department C (120 usable hosts)
Prefix length: /25/25 (27−2=126≥1202^7 - 2 = 126 \ge 120). Range: 10.10.19.010.10.19.0 to 10.10.19.12710.10.19.127. Next available IP: 10.10.19.12810.10.19.128.
A /25/25 subnet provides 128 total addresses (126 usable), satisfying the 120 host requirement.
4
Calculate the subnet boundaries and broadcast address for Department D (60 usable hosts)
Subnet block: 10.10.19.128/2610.10.19.128/26. Network ID: 10.10.19.12810.10.19.128, Usable host range: 10.10.19.12910.10.19.129 – 10.10.19.19010.10.19.190, Broadcast address: 10.10.19.19110.10.19.191.
Department D requires 6060 hosts, so 26−2=62≥602^6 - 2 = 62 \ge 60 requires a /26/26 prefix (block size of 64). Adding 64 to the starting address 10.10.19.12810.10.19.128 gives the next network boundary 10.10.19.19210.10.19.192, making 10.10.19.19110.10.19.191 the broadcast address.

Key Concept

Variable Length Subnet Masking (VLSM) Allocation and Boundary Calculation
Estimated Time:3m 0s
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Topics

IPv4 Addressing and Subnetting

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