Categories
Uncategorized

SUBNETTING

1. Network Address Calculation.

a) 188.10.18.2/16                                                  

Ans: 11111111.11111111.00000000.00000000

subnet mask: 255.255.0.0

Network address: 188.10.0.0

b) 223.69.25.48/28

Ans: 11111111.11111111.11111111.11110000

subnet mask:255.255.255.240

4th bit value=16 so, the network goes by adding 16 to it.

16+16=32,32+16=48

so, Network address is: 223.69.25.48

2. Broadcast address calculation:

a) 192.10.10.110/28

Ans: 11111111.11111111.11111111.11110000

subnet mask: 255.255.255.240

4th bit value is 16: so, the network goes by adding 16 to it.

16+16=32,32+16=48,48+16=64,64+16=80,80+16=96,96+16=112

(112 is next network address so, subtracting it by 1 gives broadcast address of previous one, and address is 192.10.10.110)

so,Broadcast address is:192.10.10.111.

3. Calculate Number of subnets and hosts/sub network

a) 10.0.0.0/15

ans) 11111111.11111110.00000000.00000000

Since it is class A Network: N.H.H.H.

so, 7 bit is borrowed from host. Therefore Number of valid subnets will be: (2^7) – 2 and Number of Hosts will be: (2^17) – 2

b) 162.45.0.0/23

ans) 11111111.11111111.11111110.00000000

since it is class B Network: N.N.H.H

so, 7 bit is borrowed from host. Therefore Number of valid subnets will be: (2^7) – 2 and Number of Valid Hosts will be: (2^9) – 2