VLSM practice

Tricon7Tricon7 Posts: 238Inactive Imported Users
We're just now getting into VLSM. The teacher went over it briefly, and I admit it went right over my head. Is there an easy way to explain this and some easy practice somewhere? Thanks.

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  • georgemcgeorgemc Posts: 429Member
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • Tricon7Tricon7 Posts: 238Inactive Imported Users
    georgemc wrote:

    Yes, I've been to that site lots and lots of times. However, I would like a different site or practice suggestions other than that if one is available.
  • georgemcgeorgemc Posts: 429Member
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • Tricon7Tricon7 Posts: 238Inactive Imported Users
    Hmm - this is going to be a more difficult issue than I had anticipated. I have been unable to find any good explanation of VLSM anywhere on the Net. It's mainly general subnetting issues that barely touch on VLSM, much less give tutorials on it. I'm not ready for practice questions and such, since I don't even understand it. Maybe I should find a good book that has it?
  • malcyboodmalcybood Posts: 900Member
    http://www.tcpipguide.com/free/t_IPVariableLengthSubnetMaskingVLSM.htm

    3 pages that covers it

    You have the sybex book?
  • spicc7spicc7 Posts: 47Member ■■□□□□□□□□
    Tricon7,

    I found the following resource helpful for VLSM during my studies:

    http://i.i.com.com/cnwk.1d/i/tr/downloads/home/1587201585_chapter_2.pdf

    http://reviews.techrepublic.com/download.aspx?&kw=variable-length+subnet+masking+&docid=172915

    You may need to create a free account to access the article!
    __________________________________________
    CCNA, CCNA Security, MCSA, MCP, A+, Network+
  • kenjensen24kenjensen24 Posts: 5Member ■□□□□□□□□□
    Sybex has by far the best information for VLSM that I hav found........I am still tryig to get it down.........I have a hard time with it on the practice tests......
    God please let me pass!!!!
  • outstreamoutstream Posts: 18Member ■□□□□□□□□□
    Alright, here goes brief explanation of VLSM.

    Most of the times, the network diagrams that you have seen by far in the books, have the networks which are either subnetted using default mask (/24) or some other mask, but the same mask through out the network.

    For example, lets say you have two workstations A and B, which are connected with each other via switch and switch is connected to internet via router. Lets say ip addresses are as follows:

    workstation A IP: 10.1.1.1/24---Default Gateway: 192.168.0.1/24

    workstation B IP: 10.1.1.2/24---Default Gateway: 192.168.0.1/24

    switch IP: 10.1.2.1/24---Default Gateway: 192.168.0.1/24

    Router E0 IP: 192.168.0.1/24

    Do you see anything "constant" in the above mentioned ip addressing scheme? the submet mask is constant. all devices have addresses which are subnetted using /24, default mask.

    VLSM is something that allows you to use "different" or "variable" masks in one network. VLSM is needed to preserve precious ip addresses.

    consider the following scenario. You have two routers, Router_A and Router_B, and you have to connect these two routers via serial interfaces.

    Now, if you assign ip addresses using the subnet mask of /24, you will be wasting alot of ip address space. You need only one ip address for one serial interface. which means you will need two ip addresses to configure both the serial interfaces. /24 mask allows you to have 254 usable ip addresses. if you use two out of them, u`ll be wasting 252 ip addresses.

    so what you will have to do is, use /30 mask to configure serial interfaces of both routers so that ip address space can be saved. now lets say you finally decided to use /30 mask to configure both routers. all done flawlessly. but you assign ip addresses to your hosts or workstations using /24 subnetting scheme. now what you have done is this that you have used two different subnet masks in a network. /30 for routers and /24 for hosts.

    when you are doing something like this in your network, you are using VLSM :)
  • remyforbes777remyforbes777 Posts: 499Member
    You first have to understand subnetting to understand VLSM. Learn the different subnets and not just the classful ones such as /8, /16 or /24. Learn the /12 and /27 subnets. What i do is this. Say for instance I have a subnet of /18, I know that my classful subnet is /16 which is 255.255.0.0 so that means I would be taking two bits out of the third octet so 11111111.11111111.11000000.00000000. In the third octet two bits are used so thats 128 + 64 = 192. So your subnet mask is 255.255.192.0.
    Okay so now to find your subnets you subtract 192 from 256 ... 256-192 = 64. So there are your blocks (64) So now you have , if you are using subnet zero 0-63 for the first block 64-127 as the second block and 128-191.
    Does that explain it a little, if not then let me know and I will continue with this example.
  • Tricon7Tricon7 Posts: 238Inactive Imported Users
    You first have to understand subnetting to understand VLSM. Learn the different subnets and not just the classful ones such as /8, /16 or /24. Learn the /12 and /27 subnets. What i do is this. Say for instance I have a subnet of /18, I know that my classful subnet is /16 which is 255.255.0.0 so that means I would be taking two bits out of the third octet so 11111111.11111111.11000000.00000000. In the third octet two bits are used so thats 128 + 64 = 192. So your subnet mask is 255.255.192.0.
    Okay so now to find your subnets you subtract 192 from 256 ... 256-192 = 64. So there are your blocks (64) So now you have , if you are using subnet zero 0-63 for the first block 64-127 as the second block and 128-191.
    Does that explain it a little, if not then let me know and I will continue with this example.

    I followed you all the way to "blocks." You're saying that in the end you have 64 subnets to work with total?

    Here is the VLSM example my teacher is using for our mid-term:

    "Use the most efficient IP addressing scheme possible."

    Corporate network: 202.2.2.0

    Corporate site: 50 hosts
    Site 1: 48 hosts
    Site 2: 30 hosts
    Site 3: 23 hosts
    Site 4: 12 hosts
    Serial links: 4

    "Give the IP subnet addy, 1st usable host, last usable host, broadcast, and subnet mask for each."

    If I can do this, I should ace the VLSM test.
  • remyforbes777remyforbes777 Posts: 499Member
    No , since you are only using 2 bits from the third octet you only have 4 subnets 2^2 = 4.
    Thus giving you 0-63, 64-127, 128-191, 192-255.
  • georgemcgeorgemc Posts: 429Member
    Tricon7 wrote:
    Here is the VLSM example my teacher is using for our mid-term:

    "Use the most efficient IP addressing scheme possible."

    Corporate network: 202.2.2.0

    Corporate site: 50 hosts
    Site 1: 48 hosts
    Site 2: 30 hosts
    Site 3: 23 hosts
    Site 4: 12 hosts
    Serial links: 4

    "Give the IP subnet addy, 1st usable host, last usable host, broadcast, and subnet mask for each."


    Site 1: 202.2.2.0/26
    Subnet=.0, 1st host=.1, last host=.62, broadcast=.63, mask=255.255.255.192

    Site 2: 202.2.2.64/27
    Subnet=.64, 1st host=.65, last host=.94, broadcast=.95,
    mask=255.255.255.224

    Site 3: 202.2.2.96/27
    Subnet=.96, 1st host=.97, last host=.126, broadcast=.127,
    mask=255.255.255.224

    Site 4: 202.2.2.128/28
    Subnet=.128, 1st host=.129, last host=.142, broadcast=.143,
    mask=255.255.255.240

    Serial Link 1: 202.2.2.144/30
    Subnet=.144, 1st host=.145, last host=.146, broadcast=.147,
    mask=255.255.255.252

    Serial Link 2: 202.2.2.148/30
    Subnet=.148, 1st host=.149, last host=.150, broadcast=.151,
    mask=255.255.255.252

    Serial Link 3: 202.2.2.152/30
    Subnet=.152, 1st host=.153, last host=.154, broadcast=.155,
    mask=255.255.255.252

    Serial Link 4: 202.2.2.156/30
    Subnet=.156, 1st host=.157, last host=.158, broadcast=.159,
    mask=255.255.255.252

    This will leave one 62host subnet(.192 - .255), one 30 host subnet (.164 - .191), and one point-to-point link subnet (.160 - .163) available for future use. The larger subnet can of cours be subnet further if required.

    That should do it for your example... :D

    I would use a calculator to check my math if I were you... icon_rolleyes.gif There's always a possibility that I misplaced a bit or two. icon_eek.gif


    Edit: Oops icon_redface.gif , Looks like I forgot the corporate site, good thing we planned for future use.

    We can place them in the 202.2.2.192/26 subnet. You can figure out the rest ...
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • Tricon7Tricon7 Posts: 238Inactive Imported Users
    georgemc wrote:
    Tricon7 wrote:
    Here is the VLSM example my teacher is using for our mid-term:

    "Use the most efficient IP addressing scheme possible."

    Corporate network: 202.2.2.0

    Corporate site: 50 hosts
    Site 1: 48 hosts
    Site 2: 30 hosts
    Site 3: 23 hosts
    Site 4: 12 hosts
    Serial links: 4

    "Give the IP subnet addy, 1st usable host, last usable host, broadcast, and subnet mask for each."


    Site 1: 202.2.2.0/26
    Subnet=.0, 1st host=.1, last host=.62, broadcast=.63, mask=255.255.255.192

    Site 2: 202.2.2.64/27
    Subnet=.64, 1st host=.65, last host=.94, broadcast=.95,
    mask=255.255.255.224

    Site 3: 202.2.2.96/27
    Subnet=.96, 1st host=.97, last host=.126, broadcast=.127,
    mask=255.255.255.224

    Site 4: 202.2.2.128/28
    Subnet=.128, 1st host=.129, last host=.142, broadcast=.143,
    mask=255.255.255.240

    Serial Link 1: 202.2.2.144/30
    Subnet=.144, 1st host=.145, last host=.146, broadcast=.147,
    mask=255.255.255.252

    Serial Link 2: 202.2.2.148/30
    Subnet=.148, 1st host=.149, last host=.150, broadcast=.151,
    mask=255.255.255.252

    Serial Link 3: 202.2.2.152/30
    Subnet=.152, 1st host=.153, last host=.154, broadcast=.155,
    mask=255.255.255.252

    Serial Link 4: 202.2.2.156/30
    Subnet=.156, 1st host=.157, last host=.158, broadcast=.159,
    mask=255.255.255.252

    This will leave one 62host subnet(.192 - .255), one 30 host subnet (.164 - .191), and one point-to-point link subnet (.160 - .163) available for future use. The larger subnet can of cours be subnet further if required.

    That should do it for your example... :D

    I would use a calculator to check my math if I were you... icon_rolleyes.gif There's always a possibility that I misplaced a bit or two. icon_eek.gif


    Edit: Oops icon_redface.gif , Looks like I forgot the corporate site, good thing we planned for future use.

    We can place them in the 202.2.2.192/26 subnet. You can figure out the rest ...

    I believe that you have Site 1 for what the Corporate info should have been, I think is what you said. Corporate would start at .0, etc.

    I'm confused on what you have for Site 2 forward. My Corporate site needs 50 hosts, so I use 2^6, which is 64 bits borrowed, giving me /26 (and .64). I have that fine.

    For what you have for Site 2, I need 48 hosts, so I'm going to have to go up to 2^6 again, going up to .127 - not .95 which is what you have. I have Site 2 with the first usable host at .65, last usable host at .126, and a broadcast of .127.

    Did I miss something?

    And I'm having a hard time figuring out the easiest way to determine the CIDR for each site. Can someone assist with this, too? THANKS!
  • georgemcgeorgemc Posts: 429Member
    Tricon7 wrote:
    I believe that you have Site 1 for what the Corporate info should have been, I think is what you said. Corporate would start at .0, etc.

    OK, one of the reasons we use VLSM is flexibility. The corporate subnet does not need to be the first available. It could be any valid 64 bit subnet within the range.

    Work the problem out, pretending that you don't need to account for the corporate subnet, just sites 1-4 and the 4 serial links.

    When you're done with that(and you've already set up all of your equpment for the projected scheme), your boss walks in and tosses the requirement for another 50 host subnet (corporate) in your lap. Do you re-subnet your entire range and reconfigure all of your equipment? NO, or course not, you don't have the time or the necessity to do so. Just place the required 50 hosts in the next available subnet large enough to accept them (as in the example above.)
    Tricon7 wrote:
    I'm confused on what you have for Site 2 forward. My Corporate site needs 50 hosts, so I use 2^6, which is 64 bits borrowed, giving me /26 (and .64). I have that fine.

    For what you have for Site 2, I need 48 hosts, so I'm going to have to go up to 2^6 again, going up to .127 - not .95 which is what you have. I have Site 2 with the first usable host at .65, last usable host at .126, and a broadcast of .127.

    Did I miss something?
    Remember flexibility, your subnets don't have to go in the order presented, (corporate, site1, site 2, site 3, site4, serial links). As in the example, I ordered them site1, site2, site3, site4, serial links (x4), and then corporate.

    We can easily do it the way you're thinking (linearly), as below:

    Corporate: 202.2.2.0/26
    Subnet=.0, 1st host=.1, last host=.62, broadcast=.63, mask=255.255.255.192

    Site 1: 202.2.2.64/26
    Subnet=.64, 1st host=.65, last host=.126, broadcast=.127,
    mask=255.255.255.192

    Site 2: 202.2.2.96/27
    Subnet=.128, 1st host=.129, last host=.1158, broadcast=.159,
    mask=255.255.255.224

    Site 3: 202.2.2.128/27
    Subnet=.160, 1st host=.161, last host=.190, broadcast=.191,
    mask=255.255.255.224

    Site 4: 202.2.2.192/28
    Subnet=.192, 1st host=.193, last host=.206, broadcast=.207,
    mask=255.255.255.240

    Serial Link 1: 202.2.2.208/30
    Subnet=.208, 1st host=.209, last host=.210, broadcast=.211,
    mask=255.255.255.252

    Serial Link 2: 202.2.2.212/30
    Subnet=.212, 1st host=.213, last host=.214, broadcast=.215,
    mask=255.255.255.252

    Serial Link 3: 202.2.2.216/30
    Subnet=.216, 1st host=.217, last host=.218, broadcast=.219,
    mask=255.255.255.252

    Serial Link 4: 202.2.2.220/30
    Subnet=.220, 1st host=.221, last host=.222, broadcast=.223,
    mask=255.255.255.252


    This will leave one 30 host subnet(.224 - .255).
    Once again, I would use a calculator to check my math.
    georgemc wrote:
    This will leave one 62host subnet(.192 - .255), one 30 host subnet (.164 - .191), and one point-to-point link subnet (.160 - .163) available for future use. The larger subnet can of cours be subnet further if required.
    This should read:

    This will leave one 62 host subnet(.192 - .255) and one 30 host subnet (.160 - .191), available for future use. The subnets can of course be subnetted further if required.

    I must of been tired when I wrote that. :)
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • Tricon7Tricon7 Posts: 238Inactive Imported Users
    I have everything sorted now - however, I still am confused with my example on how best to determine the CIDRs on my sites, particularly sites 2,3,4, and serials. Can someone help? Thanks.
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