Subnetting help ...

christopherfestchristopherfest Posts: 1Member ■□□□□□□□□□
have a few questions if someone coul clarify them and then give me a good subnetting resource webiste :)

How many subnets and hosts per subnet can you get from the network
192.168.245.0 255.255.255.240?

and


Network ID: 121.0.0.0
Subnet into 5000 networks

What is the broadcast address of the #3 network?

Thanks :)

Comments

  • Dilbert65Dilbert65 Posts: 73Member ■■□□□□□□□□
    First it is very important to learn sub-netting with pen and paper only. You wont pass test/life without that knowledge. To check your answers here are 2 sites I recommend.

    TechExams.net - IP Subnet Calculator


    Online IP Subnet Calculator


    I can not stress this enough you have to know how to do this with just pen and paper. And yes it is a pain in the behind but you have to know it.
  • thehourmanthehourman Posts: 723Member
    I could be wrong.
    1. 16 subnets and 14 host/subnet

    2. 121.0.23.255

    Can someone please double check my answer? I want to practice subnetting as well.
    Studying:
    Working on CCNA: Security. Start date: 12.28.10
    Microsoft 70-640 - on hold (This is not taking me anywhere. I started this in October, and it is December now, I am still on page 221. WTH!)
    Reading:
    Network Warrior - Currently at Part II
    Reading IPv6 Essentials 2nd Edition - on hold
  • danb83danb83 Posts: 22Member ■□□□□□□□□□
    thehourman wrote: »
    I could be wrong.
    1. 16 subnets and 14 host/subnet

    2. 121.0.23.255

    Can someone please double check my answer? I want to practice subnetting as well.


    1. I also get 16 subnets, and 14 hosts per subnet.

    2. At first I got 121.0.31.255, but looking at your answer I realised I didnt begin with the 121.0.0.0 network, instead starting with the 121.0.8.0 network. I need to get out of the habit of doing that.
  • spartangtrspartangtr Posts: 111Member
    boswd1983 wrote: »
    1. I also get 16 subnets, and 14 hosts per subnet.

    I always see these Class C private address ranges using a /24 mask, can't remember if I've ever seen one that didn't. But I sat there and thought about it and isn't class C private 192.168.0.0/16?

    Meaning 14 hosts per subnet but 4096(2^12) subnets?
  • mensmens Posts: 69Member ■■□□□□□□□□
    spartangtr wrote: »
    I always see these Class C private address ranges using a /24 mask, can't remember if I've ever seen one that didn't. But I sat there and thought about it and isn't class C private 192.168.0.0/16?

    Meaning 14 hosts per subnet but 4096(2^12) subnets?

    class c is /24.
    x.x.x.240 is a /28
    which means 4 bits for the subnet and 4 bits for the hosts
    means 2^4 subnets and 2^4 -2 hosts.
  • spartangtrspartangtr Posts: 111Member
    mens wrote: »
    class c is /24.
    x.x.x.240 is a /28
    which means 4 bits for the subnet and 4 bits for the hosts
    means 2^4 subnets and 2^4 -2 hosts.

    Check the Odom ICND1 book. Or wikipedia. Or this.

    Class C Subnet Mask | 6 of 10
  • thehourmanthehourman Posts: 723Member
    boswd1983 wrote: »
    1. I also get 16 subnets, and 14 hosts per subnet.

    2. At first I got 121.0.31.255, but looking at your answer I realised I didnt begin with the 121.0.0.0 network, instead starting with the 121.0.8.0 network. I need to get out of the habit of doing that.
    Cool, so my answer is right then.
    Studying:
    Working on CCNA: Security. Start date: 12.28.10
    Microsoft 70-640 - on hold (This is not taking me anywhere. I started this in October, and it is December now, I am still on page 221. WTH!)
    Reading:
    Network Warrior - Currently at Part II
    Reading IPv6 Essentials 2nd Edition - on hold
  • PickerPicker Posts: 46Member ■■□□□□□□□□
    thehourman wrote: »
    Cool, so my answer is right then.

    So what will happen to the networks between 0-8 if you are saying 121.0.31.255 is right
  • thehourmanthehourman Posts: 723Member
    Picker wrote: »
    So what will happen to the networks between 0-8 if you are saying 121.0.31.255 is right
    So my answers are wrong?
    I just tried to answer the OP's 2nd question.

    I am still in the learning process.
    Studying:
    Working on CCNA: Security. Start date: 12.28.10
    Microsoft 70-640 - on hold (This is not taking me anywhere. I started this in October, and it is December now, I am still on page 221. WTH!)
    Reading:
    Network Warrior - Currently at Part II
    Reading IPv6 Essentials 2nd Edition - on hold
  • mensmens Posts: 69Member ■■□□□□□□□□
    spartangtr wrote: »
    Check the Odom ICND1 book. Or wikipedia. Or this.

    Class C Subnet Mask | 6 of 10

    Humm yea, good point. I never really thought about why the class c private range span across a /16.
  • danb83danb83 Posts: 22Member ■□□□□□□□□□
    thehourman wrote: »
    So my answers are wrong?
    I just tried to answer the OP's 2nd question.

    I am still in the learning process.

    I think its right:

    1st subnet - 121.0.0.0
    2nd subnet - 121.0.8.0
    3rd subnet - 121.0.16.0 (bcast = 121.0.23.255)
    4th subnet - 121.0.24.0
    etc
  • phoeneousphoeneous Go ping yourself... Posts: 2,333Member ■■■■■■■□□□
    5000 subnets = 2^13 or 8192.

    With that network address, this gives you 13 subnet bits (8 in second octet, 5 in the thrid octet) and an increment of 8 (3 host bits in 3rd octet). Mask is a /21 or 255.255.248.0.

    121.0.0.0 - 121.0.7.255
    121.0.8.0 - 121.0.15.255
    121.0.16.0 - 121.0.23.255 <--broadcast
    121.0.24.0 - and so on


    This is a great quick reference for doing quick subnet calculations:
    http://packetlife.net/media/library/images/tn_IPv4_Subnetting.pdf.jpg
  • itdaddyitdaddy Senior Member Posts: 2,088Member ■■■■□□□□□□
    5000 subnets = 2^12 or 4096. It cant be 2^13 which is 8192.


    phoeneous

    why do you say 2^13 is wrong? when 5000 can be taken from it
    and 2^12 is short by 1000 subnets. 2^12 doesnt meet the requirements?

    mask is /21 and 255.255.248.0 256248= 8 block size
    256-240 = 16 block size

    icon_confused.gif:

    it is 2^13 is part of the answer...why did you say that? explain.
    At least I dont get what you mean since it doesnt meet requirements..
    thanks
    yep I get
  • phoeneousphoeneous Go ping yourself... Posts: 2,333Member ■■■■■■■□□□
    itdaddy wrote: »
    5000 subnets = 2^12 or 4096. It cant be 2^13 which is 8192.


    phoeneous

    why do you say 2^13 is wrong? when 5000 can be taken from it
    and 2^12 is short by 1000 subnets. 2^12 doesnt meet the requirements?

    mask is /21 and 255.255.248.0 256248= 8 block size
    256-240 = 16 block size

    icon_confused.gif:

    it is 2^13 is part of the answer...why did you say that? explain.
    At least I dont get what you mean since it doesnt meet requirements..
    thanks
    yep I get

    Ooppss, fixed :)

    I typically like to short my subnets by 1000 hosts icon_thumright.gif
  • SysAdmin4066SysAdmin4066 Posts: 443Member
    Here's two practice resources I use daily.

    IP Subnet Practice

    Subnetting Quiz -- Steve Kehlet's Pages

    I do about 30 mins every morning, just to keep it fresh.
    In Progress: CCIE R&S Written Scheduled July 17th (Tentative)

    Next Up: CCIE R&S Lab
  • dummy123dummy123 Posts: 12Member ■□□□□□□□□□
    121.0.0.0
    5000 networks
    bcast of 3rd network

    no ip subnet zero:
    1. 121.00000000.00000|000.00000000 <- 13 host bits borrowed
    2. 121.00000000.00011|000.00000000 <- 3rd network (121.0.24.0)
    3. 121.00000000.00011|111.11111111 <- bcast address of 3rd network (121.0.31.255)

    ip subnet zero:
    1. 121.00000000.00000|000.00000000 <- 13 host bits borrowed
    2. 121.00000000.00010|000.00000000 <- 3rd network (121.0.16.0)
    3. 121.00000000.00010|111.11111111 <- bcast address of 3rd network (121.0.23.255)
  • tanixtanix Posts: 68Member ■■□□□□□□□□
    I will explain what helped me to understand it and maybe this may help you.

    First, you want to memorize some things. I know, I know, memorizing sucks, but... trust me, you want to do this.As I am going over this, watch for the patterns that evolve throughout everything we do.

    First, memorize your powers of 2.
    2^1=2
    2^2=4
    2^3=8
    2^4=16
    2^5=32
    2^6=64
    2^7=128
    2^8=256
    
    Memorize higher if you like, I'm just stopping here because it nicely deals with an octet.


    Now, memorize the relation between bit, decimal, and CIDR. I'm going to start from the last octet for a standard class C /24.

    Binary     Decimal   CIDR
    
    00000000 = 0         /24
    10000000 = 128       /25
    11000000 = 192       /26
    11100000 = 224       /27
    11110000 = 240       /28
    11111000 = 248       /29
    11111100 = 252       /30
    
    Now certainly you can go /23 and lower, but notice the binary and decimal repeat. So it is the same thing in every octet in its progression.

    Now, here is the trick. Notice the powers of 2? and then look at the list we just did? See something sticking out? That's right, block sizes.

    So if we memorize block size with this (and we also make sure we know how to count in 2's, 4's, 8's, 16's, etc...) we can blaze through these in seconds for pretty much any use we have.

    So the question was how many networks and how many hosts for:

    192.168.245.0
    255.255.255.240


    First, its a class C so our focus is the last octet. We know that a 240 decimal is 4 bits on, 4 bits off, and a /28. We also know that if we see a class C which is /24 and we recognize that 240 is a /28 we can simply subtract /28-24 = 4 bits on. Once you have these basics memorized you can come at it from multiple angles depending on the need. Just make sure you can recognize backwards and forwards between them. In a class C, 4 bits on = /28 = 240 or /28 = 240 = 4 bits on, etc..

    With that in hand, we only need to understand the basics of simple networking and how it relates to the subnet address and the broadcast address. For the number networks, we don't need to consider this unless we do not have "ip subnet zero" enabled which essentially is an old issue not common so much anymore, but basically it simply means that without it, you can't count the first and last subnet. It is on by default these days, so in most cases you don't have to worry about it and I think Cisco doesn't test on it anymore (someone correct me if I am wrong).

    Anyway, so 4 bits are being used for the subnets. We know since we memorized that 4 bits is 2^4=16 which is our block size and if you have memorized this, you know that a 240 (/2icon_cool.gif has a block size of 16. Notice that the number of networks is your block size and your bit multiple?

    So if a 240, or a block size of 16, or rather a /28 is 4 bits on, 4 bits off, then we know exactly what we have for both.

    We have 4 bits on for our network which is 16 subnets and we have 4 bits off available for our hosts. This is where identifying the subnets and hosts differ. You know that of those 4 bits 2^4=16 hosts, two of them have to be reserved to identify your subnet and your broadcast. So in this case, you have 16-2=14 hosts available.

    So, we have 16 networks and 14 hosts per network.

    If we start with each network and count in 16's (block size), we can identify the range of each network.
    192.168.245.0
    192.168.245.16
    192.168.245.32 
    192.168.245.48
    192.168.245.64
    192.168.245.80
    192.168.245.96
    192.168.245.112
    192.168.245.128
    192.168.245.144
    192.168.245.160
    192.168.245.176
    192.168.245.192
    192.168.245.208
    192.168.245.224
    192.168.245.240
    
    There we have it, 16 networks with each a block size of 16 and the host capacity of 14 each.

    If you have memorized multiples of 2, your CIDR notation, block size, decimal, and bits on and off for just that single octet backwards and forwards, you can snap these out extremely fast.

    It is the same for /8-/15 and /16-/23 as it is for /24-30 with the only difference being that you have 2 more bits to know for the class A and B networks which are easy to remember anyway. Point is, you can now subnet in your head through any address range and this will make doing VLSM's and route summarization all the more easier to understand and deal with. In fact, in many simple VLSM designs, you will also be able to do these in your head if you get good at it.

    Sorry if I was long winded, or if I was a bit over redundant in my mention, It is just that this approach really helped me and maybe, if I haven't blinded people with walls of text, it can help any of you having trouble.

    Also, I did this to make sure I remember it and am on track. If anyone sees any errors, by all means please correct me.
  • PickerPicker Posts: 46Member ■■□□□□□□□□
    tanix wrote: »
    I will explain what helped me to understand it and maybe this may help you.

    First, you want to memorize some things. I know, I know, memorizing sucks, but... trust me, you want to do this.As I am going over this, watch for the patterns that evolve throughout everything we do.

    First, memorize your powers of 2.
    2^1=2
    2^2=4
    2^3=8
    2^4=16
    2^5=32
    2^6=64
    2^7=128
    2^8=256
    
    Memorize higher if you like, I'm just stopping here because it nicely deals with an octet.


    Now, memorize the relation between bit, decimal, and CIDR. I'm going to start from the last octet for a standard class C /24.

    Binary     Decimal   CIDR
    
    00000000 = 0         /24
    10000000 = 128       /25
    11000000 = 192       /26
    11100000 = 224       /27
    11110000 = 240       /28
    11111000 = 248       /29
    11111100 = 252       /30
    
    Now certainly you can go /23 and lower, but notice the binary and decimal repeat. So it is the same thing in every octet in its progression.

    Now, here is the trick. Notice the powers of 2? and then look at the list we just did? See something sticking out? That's right, block sizes.

    So if we memorize block size with this (and we also make sure we know how to count in 2's, 4's, 8's, 16's, etc...) we can blaze through these in seconds for pretty much any use we have.

    So the question was how many networks and how many hosts for:

    192.168.245.0
    255.255.255.240


    First, its a class C so our focus is the last octet. We know that a 240 decimal is 4 bits on, 4 bits off, and a /28. We also know that if we see a class C which is /24 and we recognize that 240 is a /28 we can simply subtract /28-24 = 4 bits on. Once you have these basics memorized you can come at it from multiple angles depending on the need. Just make sure you can recognize backwards and forwards between them. In a class C, 4 bits on = /28 = 240 or /28 = 240 = 4 bits on, etc..

    With that in hand, we only need to understand the basics of simple networking and how it relates to the subnet address and the broadcast address. For the number networks, we don't need to consider this unless we do not have "ip subnet zero" enabled which essentially is an old issue not common so much anymore, but basically it simply means that without it, you can't count the first and last subnet. It is on by default these days, so in most cases you don't have to worry about it and I think Cisco doesn't test on it anymore (someone correct me if I am wrong).

    Anyway, so 4 bits are being used for the subnets. We know since we memorized that 4 bits is 2^4=16 which is our block size and if you have memorized this, you know that a 240 (/2icon_cool.gif has a block size of 16. Notice that the number of networks is your block size and your bit multiple?

    So if a 240, or a block size of 16, or rather a /28 is 4 bits on, 4 bits off, then we know exactly what we have for both.

    We have 4 bits on for our network which is 16 subnets and we have 4 bits off available for our hosts. This is where identifying the subnets and hosts differ. You know that of those 4 bits 2^4=16 hosts, two of them have to be reserved to identify your subnet and your broadcast. So in this case, you have 16-2=14 hosts available.

    So, we have 16 networks and 14 hosts per network.

    If we start with each network and count in 16's (block size), we can identify the range of each network.
    192.168.245.0
    192.168.245.16
    192.168.245.32 
    192.168.245.48
    192.168.245.64
    192.168.245.80
    192.168.245.96
    192.168.245.112
    192.168.245.128
    192.168.245.144
    192.168.245.160
    192.168.245.176
    192.168.245.192
    192.168.245.208
    192.168.245.224
    192.168.245.240
    
    There we have it, 16 networks with each a block size of 16 and the host capacity of 14 each.

    If you have memorized multiples of 2, your CIDR notation, block size, decimal, and bits on and off for just that single octet backwards and forwards, you can snap these out extremely fast.

    It is the same for /8-/15 and /16-/23 as it is for /24-30 with the only difference being that you have 2 more bits to know for the class A and B networks which are easy to remember anyway. Point is, you can now subnet in your head through any address range and this will make doing VLSM's and route summarization all the more easier to understand and deal with. In fact, in many simple VLSM designs, you will also be able to do these in your head if you get good at it.

    Sorry if I was long winded, or if I was a bit over redundant in my mention, It is just that this approach really helped me and maybe, if I haven't blinded people with walls of text, it can help any of you having trouble.

    Also, I did this to make sure I remember it and am on track. If anyone sees any errors, by all means please correct me.

    Nicely put. well done
  • itdaddyitdaddy Senior Member Posts: 2,088Member ■■■■□□□□□□
    sysadmin

    dude great stuff thank I agree doing about 30 minutes a day to keep fresh
    if not you kind of loose your mind tools super man I appreciate your extra stuff to add to my tool set icon_thumright.gif


    phoeneous

    hahaha icon_lol.gificon_lol.gif

    I was like maybe I am wrong? ahhahaha good catch dude
    I have done that too ;)


    Picker

    that is excatly what I do: here is my **** sheet to know only:

    http://subnettingquestions.com/ /* great for daily subnetting */

    network ID zero and the block size used

    1st usable host = network id + 1
    last usable host = bcast -1
    bdcast address = last host +1
    256-mask = block size

    HOST BITS
    32-mask/# = (host bits)
    2^H=H
    H-2=actual hosts

    SUBNET BITS
    mask/#-default mask =( subnet bits)

    2^5 = 32 memorize (evrything doubles)
    2^10=1024

    A /8 1-126 127 loopback 255.0.0.0
    B /16 128-191 255.255.0.0
    C/24 192-223 255.255.255.0
    D/32 224- multicast 255.255.255.255


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