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  1. Junior Member Registered Member
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    #476
    kindly explain how to subnet if 1000 hosts are required. ip adress is 150.150.0.0/16. please explain. im doing it for your method but its not done
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  3. Junior Member Registered Member
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    #477
    can someone help with this:


    1) You’ve been given the Subnet ID 192.168.1.0/24, and are asked to divide this into 5 subnets of equal size. What are the network addresses, broadcast addresses, and ranges for each of these subnets?

    2) You are given a network of 10.50.24.0/21, which contains 2,048 addresses. What subnet mask should you use to divide this into four subnets of 512 addresses each?

    3) You are given the network 172.16.14.0/23, and asked to create subnets to meet the following requirements:

    a. 1 LAN that can accommodate 30 hosts
    b. 1 LAN that can accommodate 50 hosts
    c. 2 LANs that can accommodate 100 hosts each
    d. 2 Point-to-point networks
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  4. Member
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    #478
    For mani123:

    So you need 1000 hosts, and you're starting with 150.150.0.0/16. This is a pretty common type of question where one will say you need X # of networks, or Y # of hosts. Since this one is asking for hosts, we'll approach it that way.

    The first thing I'd do is start counting in powers of two until you reach the requirement. That will tell you how many bits you need for your hosts. Be careful, though, because when you're counting for hosts, you must subtract two for your broadcast and network IDs. So 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. That's 10 bits. 1024 - 2 is 1022, which is still enough to support our 1000 needed.

    So we now know we need 10 bits to support our host requirement. We're working with 150.150.0.0/16, which means we have only 16 bits to play with. We need 10 of those 16 for our hosts, so that leaves 6 bits for our network.

    Add those 6 bits to the /16, and you get a /22. So that's what our subnets will use to support 1000 hosts each.
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  5. Member
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    #479
    Brada130:

    Question 1 seems a little tricky, but remember you're only sizing your networks based on powers of two. So you start with your original, 192.168.1.0/24. That's one network. Start counting your powers of 2 for bits to determine how many networks you can make with the 8 remaining bits which also satisfy the number of subnets required. You need 5 subnets, but counting our bits gives us 2, 4, 8.

    So the only way to divide your original network into 5 equal parts is really to divide it into 8 equal parts and only use 5 of them. It took us 3 bits to get to that 8, so add that to your /24 for a /27, or 255.255.255.224 for the mask. That leaves 32 numbers for each network, so your networks would be 192.168.1.0 - 31, 192.168.1.32 - 63, 192.168.1.64 - 95, 192.168.1.96 - 127, and 192.168.1.128 - 159. There's still 3 other networks left in the original /24, but we're not using those because we only needed 5.

    Question 2 is pretty simple. You have a /21, which means you have 11 bits left to work with. Just start counting powers of 2 until you hit your requirement. 2, 4, 8, 16, 32, 64, 128, 256, 512. So that's 9 bits we need for our hosts. We had 11 to work with, we need 9 for hosts, so we have 2 left for the subnet. Add that 2 to your original /21, so you'll use a /23. /23 is 255.255.254.0.

    Question 3, you should always start out accommodating your largest networks first. So first you need networks for 100 hosts. Count out your bits, remembering for hosts you need to subtract 2. 2, 4, 8, 16, 32, 64, 128. 7 bits. You start with a /23, that leaves 9 bits to play with. Our first two networks will need 7 bits for the hosts, leaving 2 bits for the network. Add those 2 to the /23 for a /25, which is a mask of 225.255.255.128.

    So, we take 172.16.14.0/25 as our first subnet, and 172.16.14.128 /25 as the second. That satisfies part c.

    Now, we need a 50 host network. Count the bits. 2, 4, 8, 16, 32, 64. 6 bits. Now we're working with the /25 we left off from with part c, not the original /23. We only have 7 bits to play with, we need 6 for our hosts, so we'll add that remaining one to the /25 for a /26. We're also starting where the last network ended, which was 172.16.14.255.

    So, this 50 host network will begin 172.16.15.0/26. That satisfies part b.

    Now for the 30 host network. Count the bits. 2, 4, 8, 16, 32. 5 bits is just enough, because 32-2 is 30. We're starting with the previous /26, which leaves 6 bits total, we need 5 for our hosts, so add the remaining one to the /26 for the network and that gives /27. 255.255.255.224.

    Starting where we left off would be 172.16.15.64/27. That leaves 172.16.15.96 /27 as the start of our first point to point. We need 2 hosts on our point-to-point subnets. 2, 4. 2 bits, and 4-2 is 2, so that's exactly enough. We had 5 bits to play with, we need 2 bits for our hosts, so that leaves 3 remaining network bits, which is a /30 (mask 255.255.255.252) when we add those three to our /27.

    So, our first point to point network will be 172.16.15.96 /30, and that would make the second one 172.16.15.100 /30.

    Technically, this is a little sloppy, because you'll end up with a lot of leftover IP space, and ideally you'd chunk it out better with future needs in mind, but this still gets the job done within the given network.
    Last edited by albinorhino187; 10-23-2017 at 05:59 PM.
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  6. Junior Member
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    #480
    Check out this free video course :

    https://www.udemy.com/ipv4-and-ipv6-...nd-subnetting/
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  7. Junior Member Registered Member
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    #481
    Signed up just to say thank you for the advice LordFlasHeart. Never been able to subnet, find host ranges etc before and after about an hour of using your technique I can practically do it in my head.

    Thanks again
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  8. Junior Member Registered Member
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    #482
    Hey all. Hoping someone can help clear up what I sometimes get a little iffy on. With the following question What subnet does 10.34.67.234/12 sit on? The block size 16, so when I go to do my ranges, I just want to make sure the max is correct, that being the 3rd and 4th octet

    10.0.0.0---10.15.254.255
    10.16.0.0---10.31.254.255
    10.32.0.0-----10.47.254.255
    10.48.0.0......etc

    The answer is the 10.32.0.0 subnet. For that 3rd subnet, I put the max range of 10.47.254.255 but i don't exactly know why the third octet is 254. Is it correct that the subnet ID is 10.32.0.0 and the broadcast is 10.47.255.255? If so, I know that you cannot use the subnet ID and broadcast, but what is the "rule" that says the max third octet is 254? If I'm correct here, I'm only correct because I have seen it enough times but I don't why I'am right. Hopefully that makes sense.
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  9. Junior Member Registered Member
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    #483
    Hi all, thanks for this awesome technique. It really helped me.
    However, I stumbled upon a problem which I don't get it:

    10.101.99.17 -- 255.255.254.0

    So, we are in 3rd octet, right? 256-254 = 2, 2^2 = 4.

    Incrementing third octet:
    10.101.0.0
    10.101.4.0
    ...
    10.101.92.0
    10.101.96.0
    10.101.100.0

    Therefore, 10.101.99.17 falls between 100 and 96 one, so it must be 10.101.96.0 the Network Address?
    But it is wrong? Calculators say it must be 10.101.98.0?

    Btw, my broadcast address (next NA .100, therefore 100-1: 99, 10.101.99.255 is correct)
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  10. Junior Member Registered Member
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    #484
    Hi everyone,

    @therealone, you are getting it wrong.

    Raising to the power is something which you have to do when you have the netmask given in the CIDR notation (here it would be /23).

    Then it's one bit missing to 24, so you raise 2^1 = 2 and you know the networks will change in the increments of 2.

    See - you have n bits in the octet (the bits being zeros in the netmask) available for the addresses for your network. How many binary numbers can you compose using those n bits? You have n positions which can take 2 possible values: 0 or 1. So the result is: 2 times 2 times 2 etc. repeated n times, in other words, 2^n. And this is the reason why we increment by 2^n to get to the address of the next network. This is the reason why it works - from the mathematical perspective, explained in such a way that, I hope, everyone is able to understand it.

    When you have the netmask given in the 255.255.254.0 notation, then it's enough to do the subtraction.

    256-254 = 2.

    If someone wants a mathematical explanation - I will add it at the end of this post.

    So you have the networks:
    10.101.0.0
    10.101.2.0
    10.101.4.0
    and so on, the decimal numbers representing the third octet will be the multiples of 2, so simply all the even numbers from 0 to 254.

    For your address - 10.101.99.17 - the network will therefore range from 10.101.98.0 to 10.101.99.25 - of course, those are the network and the broadcast address and the available addresses are from 10.101.98.1 to 10.101.99.254.


    I promised an explanation, why the subtraction method works.

    Let's think of an example, such one which will be clearer than the one discussed before.

    A common netmask for point-to-point links: /30 or 255.255.255.252. Let's focus on the last octet. From the CIDR notation you easily see that the binary representation of the last octet is 11111100.

    In binary you easily see all the possible network addresses possible to create with this mask (let's separate the network part and the host part with a space for clarity):
    000000 00
    000001 00
    000010 00
    000011 00
    and so on, up to:
    111111 00

    They will be, generally, incremented by binary 100. Which is decimal 4.

    How to easily get this 100 from the netmask?

    When we subtract 11111100 from 11111111, we get simply 11. This is what you have in the host part of the broadcast address (of course if it wasn't the last octet, this would be a whole subnetwork containing not only the broadcast address, but also plenty of addresses available for our hosts - but it doesn't change anything for us, the last address in this subnet would be anyway the broadcast).

    To move to the first address to the next network it's enough to add 1 to the broadcast address. We get the 100 by adding 1 to 11.

    And this works for any netmask.

    In decimal... our netmask has 252 in the octet which interests us. We subtract it from 255 (11111111) and add 1. Which is equivalent to subtracting from 256. 256-252 = 4, so we know that we increment by 4 to move to the next network address, that addresses separated by at least 4 will be certainly in different networks and that we have 4 addresses available in the network (two of which are the network address and the broadcast address, so actually we have 2 addresses available for the hosts, which is ideal for point-to-point links - by the way, actually it's possible to have a /31 netmask and use the "network" and "broadcast" addresses as hosts addresses on a point-to-point link, but it's out of the scope for the CCNA and if they ask if it's possible, even though they shouldn't, the correct answer will be most likely that it isn't possible).
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