Determine what the network ID subnet mask host address range

Determine what the network ID, subnet mask, host address range and broadcast address is for each of the subnets that you create from the IP address range given in the scenario....

Provided IP address range: 127.27.232.0/21

Building # 1 (850 hosts)

Building # 2 (600 hosts)

Building # 3 (360 hosts) *This also needs to be divided into 3 smaller subnets.

Solution

Answer:

1)

Network ID = 127.27.0.0

First Network address = 127.27.232.0

Broadcast address = 127.27.232.127

Subnet Mask : 255.255.252.0

Explaination :

See for 850 hosts : we need minimum 10 bits to be in host id [as 2^10 --> 1024]. So initially 21 masking is done and we reserve 10 bits for host so. Masking can be done in 22[32 - 10] ... so sm 255.255.252.0

Building # 2

Network id = 127.27.0.0

First address = 127.27.232.128

Broadcast Address = 127.27.232.191

Subnet mask = 255.255.252.0

Same as for building # 1

Building # 3

Network ID = 127.27.0.0

First IP = 127.27.232.192

Broadcast address = 127.27.232.255

Subnet Mask = 255.255.148.0

Explaination :

see for 360 hosts : we need minimum 9 bits to be in host id [as 2^9--> 512]... so initially 23 masking is done and we reserve 9 bits for host so.. masking can be done in 23[32 - 9] ... so sm 255.255.148.0