Subnetting Fun!

Member Posts: 8 ■□□□□□□□□□
I am having a problem understanding some subnetting concepts.
Say your company has leased the network 138.16.0.0/21. You want to subdivide this network
into 10 evenly sized subnets. This would require extending the mask 4 bit places for a /25 or 255.255.255.128 subnetmask giving you 16 subnets. This would leave 7 bits to assigh to hosts. 2^7 = 128 - 2 for network and broadcast addresses gives 126. 126hots * 7subnets = 882total clients.

Is my logic sound in coming to these conclusions? How would the networks be divided up? I would guess like this:
138.16.0.0 - 138.16.0.127 /25
138.16.2.128 - 138.16.2.255 /25
138.16.3.0 - 138.16.3.127 /25
138.16.4.128 - 138.16.4.255 /25
ect...

Ip addressing was the reason I failed the 291 the first time (got a 676 ) and I still don't feel like I have the knoledge to pass.

I would greatly apreciate any attempt to enlighten me here.

• Member Posts: 44 ■■□□□□□□□□
u are correct with taking 4 bits for the network which leaves 7 bits for hosts..which gives u 16 subnets and 128-2=126 usable hosts

the networks should go like this

138.16.0.0 - 138.16.0.1-138.16.0.126 - 138.16.0.127
138.16.0.128 - 138.16.0.129-138.16.0.254 - 138.16.0.255
138.16.1.0 - 138.16.1.1-138.16.1.126 - 138.16.1.127
138.16.1.128 - 138.16.1.129-138.16.1.254 -138.16.1.255
and so on until

138.16.7.0 - 138.16.7.1-138.16.7.126 - 138.16.7.127
138.16.7.128 - 138.16.7.129-138.16.7.254 - 138.16.7.255

hope that helps
• Member Posts: 12 ■□□□□□□□□□
alphawave wrote: »
I am having a problem understanding some subnetting concepts.
Say your company has leased the network 138.16.0.0/21. You want to subdivide this network
into 10 evenly sized subnets. This would require extending the mask 4 bit places for a /25 or 255.255.255.128 subnetmask giving you 16 subnets. This would leave 7 bits to assigh to hosts. 2^7 = 128 - 2 for network and broadcast addresses gives 126. 126hots * 7subnets = 882total clients.

Is my logic sound in coming to these conclusions? How would the networks be divided up? I would guess like this:
138.16.0.0 - 138.16.0.127 /25
138.16.2.128 - 138.16.2.255 /25
138.16.3.0 - 138.16.3.127 /25
138.16.4.128 - 138.16.4.255 /25
ect...

Ip addressing was the reason I failed the 291 the first time (got a 676 ) and I still don't feel like I have the knoledge to pass.

I would greatly apreciate any attempt to enlighten me here.

Read Chapter 3 Sybex CCNA you will then fully understand subnetting