In: Computer Science
Scenario 1 (Use the network 221.153.17.0 for the following questions.)
Network 1: _________________ Broadcast: _________________
Host IP Range: _________________ to _________________
Network 2: _________________ Broadcast: _________________
Host IP Range: _________________ to _________________
Network 3: _________________ Broadcast: _________________
Host IP Range: _________________ to _________________
Network Address = 221.153.17.0
Host bit = 8
Maximum number of network = 4
We need to subnet this network to support a minimum of 4 (four) networks. How many bits do we need to use from the last octet to support this?
Number of bit needed to use from the last octet = ceil (log2(4)) = 2 bits
How many hosts would each network be able to support?
Host bit in each network = (8 - 2) = 6 bits
What will the subnet mask be?
Number of hosts in each network = 2^6 - 2 = 62
Give the network, IP range, and broadcast address for the first 3 networks:
Network 1: 221.153.17.00000000 = 221.153.17.0
Broadcast = 221.153.17.00111111 = 221.153.17.63
Host IP range 221.153.17.00000001 to 221.153.17.00111110
Host IP range 221.153.17.1 to 221.153.17.62
Network 2: 221.153.17.01000000 = 221.153.17.64
Broadcast = 221.153.17.01111111 = 221.153.17.127
Host IP range 221.153.17.01000001 to 221.153.17.01111110
Host IP range 221.153.17.65 to 221.153.17.126
Network 1: 221.153.17.10000000 = 221.153.17.128
Broadcast = 221.153.17.10111111 = 221.153.17.191
Host IP range 221.153.17.10000001 to 221.153.17.10111110
Host IP range 221.153.17.129 to 221.153.17.190
Thank You....!!!!
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