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Subnetting issues

lucky_67012001
Level 1
Level 1

Hello,

 I had this exam simulated question: 

1.If you take the network 192.168.1.0/24 and create four subnets from it, what are the new subnet IDs?

A. 192.168.1.128
B. 192.168.1.32
C. 192.168.1.96
D. 192.168.1.64
E. 192.168.1.192
F. 192.168.1.0
 Since i determined it was 4 subnets... i also determined that the increment is 32...but for some reason the answers "F,B,D,C" were wrong...so i was wondering why..or how exactly do you solve this ?
 I know this is a sort of a newb question, but i am kinda confused.. with 4=3 bits minimum to make the number..so i don`t understand.

1 Accepted Solution

Accepted Solutions

Simple way to calculate it -

1 bit = 2 to the power 1 = 2 subnets
2 bits = 2 to the power 2 = 4 subnets
3 bits = 2 to the power 3 = 8 subnets
etc.

So you need to use 2 bits from the last octet which gives 4 subnets ie. -

128  64  32  16 8 4 2 1
 0    0 
 0    1
 1    0
 1    1

which gives you the following subnets -

192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192

Jon

View solution in original post

5 Replies 5

Jon Marshall
Hall of Fame
Hall of Fame

The increment is 64.

You have calculated to get 4 you need 3 bits but you don't you only need 2 bits to give 4 different values which makes your subnet mask 255.255.255.192 ie. the first two bits in the last octet are being used.

Jon

Hello Jon,

 Can you explain it to me how it works exactly?

 I was always thought to calculate the subnet range based on how manny subnets i need.

 128 64 32 16 8 4 2 1

                          1 0 0

i have 4= 3

Class c = 11111111.11111111.11111111.11100000

                                                                  32

so shouldn`t the range be 192.168.1.0 -31

                                                         32- 63

                                                         64-95

                                                         96-127  ?

Simple way to calculate it -

1 bit = 2 to the power 1 = 2 subnets
2 bits = 2 to the power 2 = 4 subnets
3 bits = 2 to the power 3 = 8 subnets
etc.

So you need to use 2 bits from the last octet which gives 4 subnets ie. -

128  64  32  16 8 4 2 1
 0    0 
 0    1
 1    0
 1    1

which gives you the following subnets -

192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192

Jon

Thank you John, from now on i will use the 2 to the power of something to determine the needed bits to make a number.

No problem, glad to have helped.

Jon

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