VLSM Question

qsubqsub Member Posts: 303
I just finished reading the section in Todd Lammle's CCNA book and cisco's website in regards to VLSM...

It seems quite forward, can someone tell me if I'm missing something or the point of it...

Let's say you have..

-2 routers
-2 computers on each router

VLSM, is pretty much putting different subnet masks on each interface of the router? and the max block id's you can use is 256.

This is what I'm understanding from the book.. am I doing something completely wrong, or is that pretty much what it is?
World Cup 2006 - Zidane - Never Forget.

Comments

  • JiggsawwJiggsaww Member Posts: 195
    seem to me if u have different subnets on every interface of each router, they won't be able to communicate :).......VLSM is using a subnet mask that tightly fits the number of hosts on a given network.....for example 25 hosts max with a class C address....rather than use 255.255.255.0 which is default that allows 254 host........u can use 255.255.255.224 which allows for 6 networks with 30 hosts each.........hope this helps some!
  • keenonkeenon Member Posts: 1,922 ■■■■□□□□□□
    the better way is to think of it as breaking a class C for example into as many as usable address blocks (subnets) accoding to how many hosts are on a given interface. depending on this the subnet mask would change, some networks would end up with a few more addresses depending on the block size


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  • qsubqsub Member Posts: 303
    Well pretend it's 30 hosts on each subnet.

    (Class C)
    router 1 <-> hosts = /29
    *.1 - .30

    router 2 <-> hosts = /29
    *.31 - .60

    router 1 <-> router 2 = /30
    *.60 - .62

    Is this correct?
    I could add more subnet's like /24, /25, /26 and so on as long as it does not pass 256 in total for the block ID's? I did already have a look at the first link you posted. When I look at it, all I really see is a ton of networks with different subnet masks combined togeather to make different host addressing..
    World Cup 2006 - Zidane - Never Forget.
  • forbeslforbesl Member Posts: 454
    spfdz wrote:
    Well pretend it's 30 hosts on each subnet.

    (Class C)
    router 1 <-> hosts = /29
    *.1 - .30

    router 2 <-> hosts = /29
    *.31 - .60

    router 1 <-> router 2 = /30
    *.60 - .62

    Is this correct?
    I could add more subnet's like /24, /25, /26 and so on as long as it does not pass 256 in total for the block ID's? I did already have a look at the first link you posted. When I look at it, all I really see is a ton of networks with different subnet masks combined togeather to make different host addressing..
    /29 will not provide you with 30 hosts; it provides you with six.
    A /27 will provide you with 30 hosts.

    Let's break it down like this using CIDR notation:

    /24 =
    .1 - .254

    /25 = 126 hosts/subnet
    .1 - .126 (.0 network, .127 broadcast)
    .129 - .254 (.128 network, .255 broadcast)

    /26 = 62 hosts/subnet
    .1 - .62 (.0 network, .63 broadcast)
    .65 - .126 (.64 network, .127 broadcast)
    .129 - .190 (.128 network, .191 broadcast)
    .193 - .254 (.192 network, .255 broadcast)

    /27 = 30 hosts/subnet
    .1 - .30 (.0 network, .31 broadcast)
    .33 - .62 (.32 network, .63 broadcast)
    .65 - .94 (.64 network, .95 broadcast)
    .97 - .126 (.96 network, .127 broadcast)
    .etc, etc up to .254

    /28 = 14 hosts/subnet
    .1 - .14 (.0 network, .15 broadcast)
    .17 - .30 (.16 network, .31 broadcast)
    .33 - .46 (.32 network, .47 broadcast)
    .etc up to .254

    /29 = 6 hosts/subnet
    .1 - .6 (.0 network, .7 broadcast)
    .9 - . 14 (.8 network, .15 broadcast)
    .17 - .22 (.16 network, .23 broadcast)
    .etc up to .254

    /30 = 2 hosts/subnet
    .1 - .2 (.0 network, .3 broadcast)
    .5 - .6 (.4 network, .7 broadcast)
    .9 - .10 (.8 network, .11 broadcast)
    .etc up to .254

    The prefix length (/24, /25, and so on), put simply is the number of "1's" bits in the subnet mask. Look at the subnet mask 255.255.255.248 in binary:
    11111111.11111111.11111111.11111000
    Count the number of "1's" bits and you get 29
    Prefix length = /29 (also know as CIDR notation)

    Let's say you had a router with 3 interfaces (1 serial for WAN link, 2 ethernet for LAN connections). You could use 192.168.1.0/30 (two hosts) for the serial interface, 192.168.1.8/29 (six hosts) for one ethernet interface, and 192.168.1.16/29 (six hosts) for the other ethernet interface.
    That would be considered using VLSM.

    RIP v1 and IGRP do NOT support VLSM. Those are neat little protocols to play with in a lab and for learning CCNA, but not for the "real deal".

    Here's a Cisco link:
    http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a00800a67f5.shtml#cidr
  • qsubqsub Member Posts: 303
    My mistake about the /29.

    I was thinking, how much bits are required to count to 30. which is 5 (32bits)

    I just borrowed five bits, instead of leaving five host bits (borrowing bits)

    In regards to VLSM..yeah it seems pretty straight forward.

    Thanks!
    World Cup 2006 - Zidane - Never Forget.
  • dmafteidmaftei Member Posts: 83 ■■□□□□□□□□
    Jiggsaww wrote:
    seem to me if u have different subnets on every interface of each router, they won't be able to communicate
    Of course they will be able to communicate, that's what routers do: connect different subnets. Or maybe I misunderstand what you want to say...
    BSEE, MSCS
    www.maftei.net
  • forbeslforbesl Member Posts: 454
    Dan, I see you got your own website now.
  • mwillmwill Member Posts: 51 ■■□□□□□□□□
    Hi spfdz,

    When I took the CCNA, I did have a question regarding VLSM but it was VERY easy. Let me explain to you how the question was phrased.

    I got a diagram with about 5 different open slots and about 7 possiable answers. Keep in mind this was a drag and drop question with VLSM. Dont let that freak you out, its still easy.

    It will point toward a WAN connection and the others were toward the LAN connection. Drag the correct answer into the slots which are being pointed at. Now the answers are ip address with a CIDR beside them such as. 192.168.1.5/30.

    If your not familar with CIDR see below if so just skip this part.

    This means, 30 bits have been used (hence the /30) so the subnet mask has to be 255.255.255.252. If that doesnt make sense then add up the octects. 255-8 bits, 255-8bits, 255-8bits 8+8+8=24 then add the other used bits. you know you already used 24 bits which is 255.255.255 so you have a CIDR of 30, so you need to borrow 6 more bits to make that 30. so borrow 6 more from the last octect. 11111100 = 252 from binary. So subnetmask 255.255.255.252 TADA! Reason you need to know this subnet mask, is because you need to figure out your usable host addresses and subnets in order to answer the problems. "They try to fool you by putting in the correct CIDR \" but put in the wrong IP like putting in a broadcast or subnet instead.

    It will say something like
    > 12 hosts,
    >25 hosts,
    >WAN connection with two possiable hosts.

    Ok this is very easy check this out. All you need to do is match the ip addresses to the LAN or WAN connection needed.

    So for a subnet mask where you need 12 hosts, and dont waste any addresses! Subnetmask: 255.255.255.240 would be one. With 240 subnet mask you would have 14 hosts, actually 16 but you drop 2, for broadcast and host. So again you have 14 possiable hosts and you only waste 2 addresses this is the best fit for that! the CIDR is \28 11111111.11111111.11111111.11110000

    25 hosts use subnetmask of: 255.255.255.224 this gives you 30 hosts! CIDR \27 11111111.11111111.11111111.11100000

    WAN connection: Use subnetmask of 255.255.255.252 this gives you 2 possiable hosts. Actually 4, 256-252=4, but remember you loose 2.
    CIDR \30 11111111.11111111.11111111.11111100

    easy way to remember the CIDR. Just remember that 255.255.255 = 24 then add up the last reminding bits in your head. /27 = 24 + 3 is 27. so 11100000 is 224 so subnet mask, 255.255.255.224.

    -So my point, figure out the subnetmask by using the /CIDR then then make sure that subnet mask matches the number of hosts that is requred for the problem. Also make sure that ip address is not a broadcast or subnet! and thats it!
    more examples of CIDR if needed for anyone.
    /20

    255.255 16 bits, need 4 more. so 1111000. 255.255.240.0

    Alright dude, so if you had any questions regarding CIDR I hope that helps, i know your question wasnt near CIDR but you need to know CIDR inorder to pass the VLSM questions. IF you did know how to work CIDR, then, i hope i gave you some insight regarding the VLSM.
    Marcus Williamson
  • dmafteidmaftei Member Posts: 83 ■■□□□□□□□□
    forbesl wrote:
    Dan, I see you got your own website now.
    Perpetually under construction, though...
    BSEE, MSCS
    www.maftei.net
  • JiggsawwJiggsaww Member Posts: 195
    dmaftei wrote:
    Jiggsaww wrote:
    seem to me if u have different subnets on every interface of each router, they won't be able to communicate
    Of course they will be able to communicate, that's what routers do: connect different subnets. Or maybe I misunderstand what you want to say...

    nah wah i meant iz if u have one interface like S1 connected to S1 on another router but both have different subnet mask.....will it communicate :)
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