EASILY DESIGN SUBNETS
Wednesday, October 29, 2008
this is what i liked in this tutorial.. designing subnets was never been easy for me and also very much confusing .. thanks to the author iam now satisfied with my designing subnets knowledge
192.168.1.0
4 subnets required
each subnet needs at least 10 hosts
firstly for its a class c addr ..so for our subnet mask it should have three octets full ... as class c is 255.255.255.x
so
fill it out as 255.255.255.x
now we need 4 subnets ... so tick till we meet our subnet requirements in the table
now as marked two bits under subnets ...you also need to mark two bits under subnets in the upper colomn
hence in this way this can be determined
as it stopped at 192..our subnet mask is ends with 192..there fore our subnet mask is 255.255.255.192
now as you tick two bits under upper subnets coloum...there are 6 bits remaining unticked right ..(unticked 224 240 248 252 254 255 total 6)
tick them under hosts
look the table and come back ... we are left at 64 ..as we need to subtract two for broadcast and subnets .. there would be 62
solve simply
the crucial one
bits---------128--------64--------32------16-------8----4-----2------1
subnets
128-------*
192-------*
224
240
248
252
254
255
powers of 2 ------- subnets------------------hosts (-2)
2--------------------*------------------------------------*
4--------------------*-------------------------------------*
8----------------------------------------------------------*
16----------------------------------------------------------*
32----------------------------------------------------------*
64------------------------------------------------------------*
128
256
512
1024
192.168.1.0
4 subnets required
each subnet needs at least 10 hosts
firstly for its a class c addr ..so for our subnet mask it should have three octets full ... as class c is 255.255.255.x
so
fill it out as 255.255.255.x
now we need 4 subnets ... so tick till we meet our subnet requirements in the table
now as marked two bits under subnets ...you also need to mark two bits under subnets in the upper colomn
hence in this way this can be determined
as it stopped at 192..our subnet mask is ends with 192..there fore our subnet mask is 255.255.255.192
now as you tick two bits under upper subnets coloum...there are 6 bits remaining unticked right ..(unticked 224 240 248 252 254 255 total 6)
tick them under hosts
look the table and come back ... we are left at 64 ..as we need to subtract two for broadcast and subnets .. there would be 62
solve simply
the crucial one
bits---------128--------64--------32------16-------8----4-----2------1
subnets
128-------*
192-------*
224
240
248
252
254
255
powers of 2 ------- subnets------------------hosts (-2)
2--------------------*------------------------------------*
4--------------------*-------------------------------------*
8----------------------------------------------------------*
16----------------------------------------------------------*
32----------------------------------------------------------*
64------------------------------------------------------------*
128
256
512
1024