Subnetting Fun!

alphawavealphawave 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 icon_mad.gif) and I still don't feel like I have the knoledge to pass.

I would greatly apreciate any attempt to enlighten me here.

Comments

  • suren327suren327 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

    network - range - broadcast
    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
  • osiris69osiris69 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 icon_mad.gif) 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
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