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
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
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
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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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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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
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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