Given the base network address 192.168.10.0/24, perform Variable Length Subnet Masking (VLSM) to accommodate the following host requirements…

VLSM: serve the largest requirement first, each from the smallest block that fits (block size = 2^host bits, usable = block − 2).

A — 25 hosts → needs 32 addresses (30 usable): /27

• Network 192.168.10.0/27, mask 255.255.255.224
• Usable 192.168.10.1 – .30, broadcast .31

B — 22 hosts → /27 (30 usable)

• Network 192.168.10.32/27, mask 255.255.255.224
• Usable 192.168.10.33 – .62, broadcast .63

C — 8 hosts → a /29 gives only 6 usable, so /28 (14 usable)

• Network 192.168.10.64/28, mask 255.255.255.240
• Usable 192.168.10.65 – .78, broadcast .79

D — 2 hosts → /30 (2 usable)

• Network 192.168.10.80/30, mask 255.255.255.252
• Usable 192.168.10.81 – .82, broadcast .83

E — 2 hosts → /30

• Network 192.168.10.84/30, mask 255.255.255.252
• Usable 192.168.10.85 – .86, broadcast .87

192.168.10.88 – 192.168.10.255 is left free for future subnets.

— rhs