Network Address and Subnetting
A small office has a network with IP address 192.168.10.0 and subnet mask 255.255.255.192. They need to connect 50 devices. The network administrator must determine if this subnet mask provides enough addresses, and if not, propose a suitable subnet mask. Also, they need to find the network address, broadcast address, and usable host range for the first subnet.
1. How many usable host addresses does the subnet mask 255.255.255.192 provide? Show your reasoning.
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The mask 255.255.255.192 in binary is 11111111.11111111.11111111.11000000. The last octet has 2 host bits (since 192 = 11000000, the host part is the last 6 bits? Wait: 255.255.255.192 means the first 26 bits are network, last 6 bits are host. Actually 192 = 11000000, so the subnet mask has 24 + 2 = 26 network bits. Therefore host bits = 32 - 26 = 6. Number of addresses = 2^6 = 64. Usable hosts = 64 - 2 = 62. So it provides 62 usable addresses.
2. Is this subnet mask sufficient for 50 devices? If yes, explain. If no, suggest a mask that would work.
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Yes, 62 usable addresses are sufficient for 50 devices. The mask 255.255.255.192 works. If more devices were needed, a mask like 255.255.255.128 (which gives 126 usable hosts) could be used, but it's not necessary here.
3. For the subnet 192.168.10.0/26, find the network address, broadcast address, and the range of usable host addresses.
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Network address: 192.168.10.0 (since the first 26 bits are network, and the host bits are all zero). Broadcast address: set all host bits to 1: last octet = 00111111 in binary = 63, so 192.168.10.63. Usable host range: from 192.168.10.1 to 192.168.10.62.