Scenario 1 (Use the network 221.153.17.0 for the following questions.) We need to subnet this network...

Scenario 1 (Use the network 221.153.17.0 for the following questions.)

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?
How many hosts would each network be able to support?
What will the subnet mask be?
Give the network, IP range, and broadcast address for the first 3 networks:
Network 1:    _________________             Broadcast: _________________

Host IP Range:         _________________ to _________________

Network 2:    _________________             Broadcast: _________________

Host IP Range:         _________________ to _________________

Network 3:    _________________             Broadcast: _________________

Host IP Range:         _________________ to _________________

Solutions

Expert Solution

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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venereology answered 1 year ago

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