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Please determine the network 4D, host range, broadcast IP and the default-gateway ( 1 at

available host IP) for the following: a) 10.23 .33 .234 and MASK 255.255 .255 .224 b) 10.14 .64 .22 and MASK 255.255 .248 .0 c) 10.54.21.15 and MASK 255.252.0.0 d) 172.16,33.234 and MASK 255.255.255.224 e) 172.18.64.22 and MASK 255.255 .248 .0 f) 172.19 .21 .15 and MASK 255.255 .252 .0 g) 192.168 .33 .234 and MASK 255.255.255.224 h) 192.168 .64 .22 and MASK 255.255.255,248 i) 192.168.21.15 and MASK 255.255.255.252

1 Answer

5 votes

Final answer:

The question involves computing the network address, host range, broadcast IP, and default gateway for several IP addresses and subnet masks. The process includes subnet mask binary conversion, finding the network address through bitwise operations, and identifying the usable host IP range, including setting a default gateway.

Step-by-step explanation:

The student's question requires calculating the network address, host range, broadcast IP, and default gateway for a series of IP addresses and subnet masks. To obtain these, we must convert the subnet masks into binary to understand the network and host components of the address, determine the network address by performing a bitwise AND between the IP address and subnet mask, and then find the broadcast address and range of host addresses that can be used. Let's go through a couple of examples.

For IP address 10.23.33.234 with mask 255.255.255.224:

  • Network Address: 10.23.33.224
  • Host Range: 10.23.33.225 to 10.23.33.254
  • Broadcast IP: 10.23.33.255
  • Default-Gateway: Typically, the first available host IP would be 10.23.33.225

Similar steps are taken for each set of IP address and subnet mask to find the respective values. The host range is inclusive of the first and last IP address in the range post-network and pre-broadcast address. The default-gateway is usually the first or second usable IP in the host range.