VLSM Help urgently required

exsquaddieexsquaddie Member Posts: 3 ■□□□□□□□□□
in CCNA & CCENT
Hi all. I am cramming for my CCNA exam and currently struggling with VLSM and overlapping.

How do I pick which networks to start with, those with the most hosts, or those with less or just randomly!??!!?!?!? And can anyone explain the overlapping??

I am reading both the Todd Lammle and Wendell Odom books, but it's just not sinking in yet. icon_sad.gificon_sad.gificon_sad.gif
If it don't work, hit with a hammer;-)
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  • nicklauscombsnicklauscombs Member Posts: 885
    Start with the network that needs the most hosts then work your way down, when creating the scheme overlapping is a bad thing as you don't want hosts to have the same ip addresses

    you should pick an example and try to solve it and post the question and your answer on here and we can help you understand it better
    WIP: IPS exam
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  • exsquaddieexsquaddie Member Posts: 3 ■□□□□□□□□□
    Right o,

    Taking your advice I tried the following exercise:

    Using IP address 204.15.5.0 /24 provide VLSM for the following networks.

    Net A - 14 Hosts
    Net B - 28 Hosts
    Net C - 2 Hosts
    Net D - 7 Hosts
    Net E - 28 Hosts

    I came up with:

    Net B:
    /27 mask
    204.15.5.0 - Subnet
    204.15.5.1 - 1st host
    204.15.5.30 - Last host
    204.15.5.31 - Broadcast address

    Net E:
    /27 mask
    204.15.5.32 - Subnet
    204.15.5.33 - 1st host
    204.15.5.62 - Last host
    204.15.5.63 - Broadcast address

    Net A:
    /28 mask
    204.15.5.64 - Subnet
    204.15.5.65 - 1st host
    204.15.5.78 - Last host
    204.15.5.79 - Broadcast address

    Net D:

    /28 mask
    204.15.5.80 - Subnet
    204.15.5.81 - 1st host
    204.15.5.94 - Last host
    204.15.5.95 - Broadcast address

    Net C:
    /30 mask
    204.15.5.96 - Subnet
    204.15.5.97 - 1st host
    204.15.5.98 - Last host
    204.15.5.99 - Broadcast address

    To avoid overlap, I just used the subnet number, that would logically be next if I was still using the same block number.

    How does it look???
    If it don't work, hit with a hammer;-)
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  • tech-airmantech-airman Member Posts: 953
    exsquaddie wrote:
    Hi all. I am cramming for my CCNA exam and currently struggling with VLSM and overlapping.

    How do I pick which networks to start with, those with the most hosts, or those with less or just randomly!??!!?!?!? And can anyone explain the overlapping??

    I am reading both the Todd Lammle and Wendell Odom books, but it's just not sinking in yet. icon_sad.gificon_sad.gificon_sad.gif

    exsquaddie,

    The first step is to look at the list of networks and then sort the networks from largest to smallest. For example:
    1. Network A - 32 hosts
    2. Network B - 8 hosts
    3. Network C - 64 hosts

    when sorted becomes:
    1. Network C - 64 hosts
    2. Network A - 32 hosts
    3. Network B - 8 hosts

    So to answer your direct questions:
    1. Q: "..., those with the most hosts,?"
      [list=1:94dfd19aed]
    2. A: Yes.
    [*]Q: "...,those with less?"
    1. A: No.
    [*]Q: "...just randomly?"
    1. A: No.
    [/list:o:94dfd19aed]

    Ok, so you're asking about overlapping. Firstly, each IP address is supposed to be unique for a given network, even if you use private IP addressing. So the problem of overlapping is when there are duplicate IP addresses because of overlapping subnets. For example:

    192.168.1.0/24
    192.168.0.0/23

    For the network 192.168.1.0/24, the values of IP address range from 192.168.1.1 through 192.168.1.255. In a linear view....
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    For the network 192.168.0.0/23, the values of IP address range from 192.168.0.1 through 192.168.1.255. In a linear view...
    1. 192.168.0.1
    2. 192.168.0.2
    3. 192.168.0.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    So the problem of overlapping is that two subnets have IP addresses in common, in this case:
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    The problem with overlapping from a router's perspective is if it receives a packet destined for say 192.168.1.3, does the router send the packet to the network 192.168.1.0/24 or 192.168.0.0/23? That's why it is important to avoid the problem of overlapping, especially with VLSM subnetting by being careful where one subnet ends and another subnet begins without overlap.
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  • Silver BulletSilver Bullet Member Posts: 676 ■■■□□□□□□□
    tech-airman wrote:
    exsquaddie wrote:
    Hi all. I am cramming for my CCNA exam and currently struggling with VLSM and overlapping.

    How do I pick which networks to start with, those with the most hosts, or those with less or just randomly!??!!?!?!? And can anyone explain the overlapping??

    I am reading both the Todd Lammle and Wendell Odom books, but it's just not sinking in yet. icon_sad.gificon_sad.gificon_sad.gif

    exsquaddie,

    The first step is to look at the list of networks and then sort the networks from largest to smallest. For example:
    1. Network A - 32 hosts
    2. Network B - 8 hosts
    3. Network C - 64 hosts

    when sorted becomes:
    1. Network C - 64 hosts
    2. Network A - 32 hosts
    3. Network B - 8 hosts

    So to answer your direct questions:
    1. Q: "..., those with the most hosts,?"
      [list=1:9d0e1f8d19]
    2. A: Yes.
    [*]Q: "...,those with less?"
    1. A: No.
    [*]Q: "...just randomly?"
    1. A: No.
    [/list:o:9d0e1f8d19]

    Ok, so you're asking about overlapping. Firstly, each IP address is supposed to be unique for a given network, even if you use private IP addressing. So the problem of overlapping is when there are duplicate IP addresses because of overlapping subnets. For example:

    192.168.1.0/24
    192.168.0.0/23

    For the network 192.168.1.0/24, the values of IP address range from 192.168.1.1 through 192.168.1.255. In a linear view....
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    For the network 192.168.0.0/23, the values of IP address range from 192.168.0.1 through 192.168.1.255. In a linear view...
    1. 192.168.0.1
    2. 192.168.0.2
    3. 192.168.0.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    So the problem of overlapping is that two subnets have IP addresses in common, in this case:
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    The problem with overlapping from a router's perspective is if it receives a packet destined for say 192.168.1.3, does the router send the packet to the network 192.168.1.0/24 or 192.168.0.0/23? That's why it is important to avoid the problem of overlapping, especially with VLSM subnetting by being careful where one subnet ends and another subnet begins without overlap.

    What you just posted was debated in another forum. Using /23 in a 192.168.1.0 network would indicate that you are subnetting a classful C address by borrowing from the network bits to accommodate for more hosts. Can't you only borrow from the host bits in subnetting unless you are summarizing routes?
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  • tech-airmantech-airman Member Posts: 953
    Silver Bullet wrote:
    tech-airman wrote:
    exsquaddie wrote:
    Hi all. I am cramming for my CCNA exam and currently struggling with VLSM and overlapping.

    How do I pick which networks to start with, those with the most hosts, or those with less or just randomly!??!!?!?!? And can anyone explain the overlapping??

    I am reading both the Todd Lammle and Wendell Odom books, but it's just not sinking in yet. icon_sad.gificon_sad.gificon_sad.gif

    exsquaddie,

    The first step is to look at the list of networks and then sort the networks from largest to smallest. For example:
    1. Network A - 32 hosts
    2. Network B - 8 hosts
    3. Network C - 64 hosts

    when sorted becomes:
    1. Network C - 64 hosts
    2. Network A - 32 hosts
    3. Network B - 8 hosts

    So to answer your direct questions:
    1. Q: "..., those with the most hosts,?"
      [list=1:ad3e9c07f8]
    2. A: Yes.
    [*]Q: "...,those with less?"
    1. A: No.
    [*]Q: "...just randomly?"
    1. A: No.
    [/list:o:ad3e9c07f8]

    Ok, so you're asking about overlapping. Firstly, each IP address is supposed to be unique for a given network, even if you use private IP addressing. So the problem of overlapping is when there are duplicate IP addresses because of overlapping subnets. For example:

    192.168.1.0/24
    192.168.0.0/23

    For the network 192.168.1.0/24, the values of IP address range from 192.168.1.1 through 192.168.1.255. In a linear view....
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    For the network 192.168.0.0/23, the values of IP address range from 192.168.0.1 through 192.168.1.255. In a linear view...
    1. 192.168.0.1
    2. 192.168.0.2
    3. 192.168.0.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    So the problem of overlapping is that two subnets have IP addresses in common, in this case:
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    The problem with overlapping from a router's perspective is if it receives a packet destined for say 192.168.1.3, does the router send the packet to the network 192.168.1.0/24 or 192.168.0.0/23? That's why it is important to avoid the problem of overlapping, especially with VLSM subnetting by being careful where one subnet ends and another subnet begins without overlap.

    What you just posted was debated in another forum. Using /23 in a 192.168.1.0 network would indicate that you are subnetting a classful C address by borrowing from the network bits to accommodate for more hosts. Can't you only borrow from the host bits in subnetting unless you are summarizing routes?

    Silver Bullet,

    According to RFC 1918, it states...
    RFC 1918 wrote:
    3. Private Address Space

    The Internet Assigned Numbers Authority (IANA) has reserved the
    following three blocks of the IP address space for private internets:

    10.0.0.0 - 10.255.255.255 (10/8 prefix)
    172.16.0.0 - 172.31.255.255 (172.16/12 prefix)
    192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

    So the network 192.168.0.0/23 can be a variably subnetted subnet of the 192.168.0.0/16 network.

    Source:
    1. Request for Comments: 1918 - Address Allocation for Private Internets - http://tools.ietf.org/html/rfc1918
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  • gojericho0gojericho0 Member Posts: 1,059 ■■■□□□□□□□
    tech-airman wrote:
    Silver Bullet wrote:
    tech-airman wrote:
    exsquaddie wrote:
    Hi all. I am cramming for my CCNA exam and currently struggling with VLSM and overlapping.

    How do I pick which networks to start with, those with the most hosts, or those with less or just randomly!??!!?!?!? And can anyone explain the overlapping??

    I am reading both the Todd Lammle and Wendell Odom books, but it's just not sinking in yet. icon_sad.gificon_sad.gificon_sad.gif

    exsquaddie,

    The first step is to look at the list of networks and then sort the networks from largest to smallest. For example:
    1. Network A - 32 hosts
    2. Network B - 8 hosts
    3. Network C - 64 hosts

    when sorted becomes:
    1. Network C - 64 hosts
    2. Network A - 32 hosts
    3. Network B - 8 hosts

    So to answer your direct questions:
    1. Q: "..., those with the most hosts,?"
      [list=1:0d8f3c2e83]
    2. A: Yes.
    [*]Q: "...,those with less?"
    1. A: No.
    [*]Q: "...just randomly?"
    1. A: No.
    [/list:o:0d8f3c2e83]

    Ok, so you're asking about overlapping. Firstly, each IP address is supposed to be unique for a given network, even if you use private IP addressing. So the problem of overlapping is when there are duplicate IP addresses because of overlapping subnets. For example:

    192.168.1.0/24
    192.168.0.0/23

    For the network 192.168.1.0/24, the values of IP address range from 192.168.1.1 through 192.168.1.255. In a linear view....
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    For the network 192.168.0.0/23, the values of IP address range from 192.168.0.1 through 192.168.1.255. In a linear view...
    1. 192.168.0.1
    2. 192.168.0.2
    3. 192.168.0.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    So the problem of overlapping is that two subnets have IP addresses in common, in this case:
    1. 192.168.1.1
    2. 192.168.1.2
    3. 192.168.1.3
    4. ...
    5. 192.168.1.253
    6. 192.168.1.254
    7. 192.168.1.255

    The problem with overlapping from a router's perspective is if it receives a packet destined for say 192.168.1.3, does the router send the packet to the network 192.168.1.0/24 or 192.168.0.0/23? That's why it is important to avoid the problem of overlapping, especially with VLSM subnetting by being careful where one subnet ends and another subnet begins without overlap.

    What you just posted was debated in another forum. Using /23 in a 192.168.1.0 network would indicate that you are subnetting a classful C address by borrowing from the network bits to accommodate for more hosts. Can't you only borrow from the host bits in subnetting unless you are summarizing routes?

    Silver Bullet,

    According to RFC 1918, it states...
    RFC 1918 wrote:
    3. Private Address Space

    The Internet Assigned Numbers Authority (IANA) has reserved the
    following three blocks of the IP address space for private internets:

    10.0.0.0 - 10.255.255.255 (10/8 prefix)
    172.16.0.0 - 172.31.255.255 (172.16/12 prefix)
    192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

    So the network 192.168.0.0/23 can be a variably subnetted subnet of the 192.168.0.0/16 network.

    Source:
    1. Request for Comments: 1918 - Address Allocation for Private Internets - http://tools.ietf.org/html/rfc1918

    It would also be a supernet of the classful C address :)
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  • Silver BulletSilver Bullet Member Posts: 676 ■■■□□□□□□□
    tech-airman wrote:
    Silver Bullet,

    According to RFC 1918, it states...
    RFC 1918 wrote:

    3. Private Address Space

    The Internet Assigned Numbers Authority (IANA) has reserved the
    following three blocks of the IP address space for private internets:

    10.0.0.0 - 10.255.255.255 (10/8 prefix)
    172.16.0.0 - 172.31.255.255 (172.16/12 prefix)
    192.168.0.0 - 192.168.255.255 (192.168/16 prefix)


    So the network 192.168.0.0/23 can be a variably subnetted subnet of the 192.168.0.0/16 network.

    Source:


    1. Request for Comments: 1918 - Address Allocation for Private Internets - http://tools.ietf.org/html/rfc1918
    From your source just below where you ended the quote it states
    We will refer to the first block as "24-bit block", the second as
    "20-bit block", and to the third as "16-bit" block. Note that (in
    pre-CIDR notation) the first block is nothing but a single class A
    network number, while the second block is a set of 16 contiguous
    class B network numbers, and third block is a set of 256 contiguous
    class C network numbers.
    The cidr notations are not defining a network range, rather, the cidr notations simply indicators of what was set aside to be used in a private address space. Subnetting is taking a classful address and borrowing bits from the host portion, not the network portion.

    Supernetting does allow for borrowing from the network portion of an address, but it's used for route summarization.
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  • tech-airmantech-airman Member Posts: 953
    Silver Bullet,
    Silver Bullet wrote:
    tech-airman wrote:
    Silver Bullet,

    According to RFC 1918, it states...
    RFC 1918 wrote:

    3. Private Address Space

    The Internet Assigned Numbers Authority (IANA) has reserved the
    following three blocks of the IP address space for private internets:


    10.0.0.0 - 10.255.255.255 (10/8 prefix)
    172.16.0.0 - 172.31.255.255 (172.16/12 prefix)
    192.168.0.0 - 192.168.255.255 (192.168/16 prefix)


    So the network 192.168.0.0/23 can be a variably subnetted subnet of the 192.168.0.0/16 network.

    Source:


    1. Request for Comments: 1918 - Address Allocation for Private Internets - http://tools.ietf.org/html/rfc1918
    From your source just below where you ended the quote it states
    We will refer to the first block as "24-bit block", the second as
    "20-bit block", and to the third as "16-bit" block. Note that (in
    pre-CIDR notation) the first block is nothing but a single class A
    network number, while the second block is a set of 16 contiguous
    class B network numbers, and third block is a set of 256 contiguous
    class C network numbers.
    The cidr notations are not defining a network range, rather, the cidr notations simply indicators of what was set aside to be used in a private address space. Subnetting is taking a classful address and borrowing bits from the host portion, not the network portion.

    Let's remember that a 32 bit subnet mask is:
    1. Binary = 11111111.11111111.11111111.11111111
    2. CIDR Notation = /32
    3. Decimal = 255.255.255.255

    With that said:
    1. For Class A = /32 - /8 = /24
    2. For Class B = /32 - /12 = /20
    3. For Class C = /32 - /16 = /16

    So the whole "24-bit block", "20-bit block", and "16-bit block" refers to the host part of the IP address and not the network part of the IP address. Therefore, /8 refers to a network mask of 255.0.0.0, /12 refers to a 255.240.0.0, and /16 refers to a network mask of 255.255.0.0. Therefore, I still maintain my stance that 192.168.0.0/23 is a subnet of the 192.168.0.0/16 network.
    Silver Bullet wrote:
    Supernetting does allow for borrowing from the network portion of an address, but it's used for route summarization.

    Since this thread was about VLSM and not Route Summarization, it was not appropriate to mention "Supernetting." While we're on that topic, yes, 192.168.0.0/23 would be the supernet for the 192.168.0.0/24 and the 192.168.1.0/24 subnets.
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  • Silver BulletSilver Bullet Member Posts: 676 ■■■□□□□□□□
    tech-airman wrote:
    Let's remember that a 32 bit subnet mask is:


    1. Binary = 11111111.11111111.11111111.11111111
    2. CIDR Notation = /32
    3. Decimal = 255.255.255.255



    With that said:


    1. For Class A = /32 - /8 = /24
    2. For Class B = /32 - /12 = /20
    3. For Class C = /32 - /16 = /16
    hmmm... No, actually, the A, B and C classes are defined as

    Class A = 0.0.0.0 - 127.255.255.255 Netmask = 255.0.0.0 CIDR = /8
    Class B = 128.0.0.0 - 191.255.255.255 Netmask = 255.255.0.0 CIDR = /16
    Class C = 192.0.0.0 - 223.255.255.255 Netmask = 255.255.255.0 CIDR = /24
    tech-airman wrote:
    Therefore, I still maintain my stance that 192.168.0.0/23 is a subnet of the 192.168.0.0/16 network.
    I personally hold nothing against you for maintaining your stance... just feel that those that are studying this stuff should have good information to go on.
    tech-airman wrote:
    Since this thread was about VLSM and not Route Summarization, it was not appropriate to mention "Supernetting."
    I know what the thread is about, but, I pointed out supernetting because another poster brought it up.
    tech-airman wrote:
    While we're on that topic, yes, 192.168.0.0/23 would be the supernet for the 192.168.0.0/24 and the 192.168.1.0/24 subnets
    Yes, that would be a supernet.
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  • dtlokeedtlokee Member Posts: 2,378 ■■■■□□□□□□
    When the mask is longer than the default classful mask it's a subnet, when it's equal to the default classful mask it's a "network" and when it's shorter than the default classful mask it's a "block of networks" or a "supernet" end of story.

    Since 192.168.x.x is a class C address its classful mask is 255.255.255.0 or /24. In this case anything /25 and greater is a subnet. Anything /23 and less is a "block of networks". So using 192.168.0.0/16 as a starting point you already have a "block of networks". It is in fact a block of 256 class C networks. Subnetting a block of networks does not necessarily indicate it's a subnet of a classful network.
    The only easy day was yesterday!
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  • PashPash Member Posts: 1,600 ■■■■■□□□□□
    dtlokee wrote:
    When the mask is longer than the default classful mask it's a subnet, when it's equal to the default classful mask it's a "network" and when it's shorter than the default classful mask it's a "block of networks" or a "supernet" end of story.

    Since 192.168.x.x is a class C address its classful mask is 255.255.255.0 or /24. In this case anything /25 and greater is a subnet. Anything /23 and less is a "block of networks". So using 192.168.0.0/16 as a starting point you already have a "block of networks". It is in fact a block of 256 class C networks. Subnetting a block of networks does not necessarily indicate it's a subnet of a classful network.

    You should do training :D
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
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  • exsquaddieexsquaddie Member Posts: 3 ■□□□□□□□□□
    Woah there folks. When I posted the original message it was because I having difficulty with where to start when using VLSM with a given IP address. I already understand the concept of subnetting, and just wanted some insight and guidance on the start point.

    I appreciate the time and effort you guys put into the replies, but would my answer for the example work?

    Thanks for help icon_cool.gif
    If it don't work, hit with a hammer;-)
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  • dynamikdynamik Banned Posts: 12,312 ■■■■■■■■■□
    Pash wrote:
    You should do training :D

    He runs his own training company ;)
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  • SVSV Member Posts: 166
    exsquaddie wrote:
    Woah there folks. When I posted the original message it was because I having difficulty with where to start when using VLSM with a given IP address. I already understand the concept of subnetting, and just wanted some insight and guidance on the start point.

    I appreciate the time and effort you guys put into the replies, but would my answer for the example work?

    Thanks for help icon_cool.gif

    I guess your calculation is correct.
    Life is a journey...
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  • PashPash Member Posts: 1,600 ■■■■■□□□□□
    dynamik wrote:
    Pash wrote:
    You should do training :D

    He runs his own training company ;)

    See my sarcasm was so well delivered that I even beat a seasoned forum vet like dynamik with no troubles.

    This is getting too easy!
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
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  • bertiebbertieb Member Posts: 1,031 ■■■■■■□□□□
    Pash wrote:
    dynamik wrote:
    Pash wrote:
    You should do training :D

    He runs his own training company ;)

    See my sarcasm was so well delivered that I even beat a seasoned forum vet like dynamik with no troubles.

    This is getting too easy!

    bowing.gif
    The trouble with quotes on the internet is that you can never tell if they are genuine - Abraham Lincoln
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  • dynamikdynamik Banned Posts: 12,312 ■■■■■■■■■□
    Aww icon_sad.gif

    To be fair, it can be difficult to identify someone's sarcasm when they only average 1.41 posts per day. Maybe you should get more involved ;)
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  • PashPash Member Posts: 1,600 ■■■■■□□□□□
    dynamik wrote:
    Aww icon_sad.gif

    To be fair, it can be difficult to identify someone's sarcasm when they only average 1.41 posts per day. Maybe you should get more involved ;)

    You have a point my friend, I have my S+ booked for 30th september, when ive passed that Ill be posting more!
    DevOps Engineer and Security Champion. https://blog.pash.by - I am trying to find my writing style, so please bear with me.
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