This is my first time on the forum and i couldnt really find a student area anywhere so if this is the wrong section i apologize. If this is not a place where students can ask for help and you know of another place that would be best suited then i would appreciate if you could take the time to leave some links.
I have currently just started my first year of Networking in College around January, we are doing "Introduction To Networks CCNA LV.5" i believe. It is alll quite basic stuff but i am really struggling with subnetting. We have just started doing our skills assessment and one of the first tasks is to create an IP scheme for a basic network and apply it in Cisco.
I have been watching tutorials on youtube on what to do and i already have the basics down. The part i am struggling with is creating a scheme when given the number of hosts. I have been studying all day i think i have grasped it but would like someone with knowledge on the subject to look over my document and check to see if what i put down is correct. If you also have some time to help me where i am going wrong that would be great.
I will attach the document to this post. It is Part 3: Develop the IPv4 Addressing Scheme i am having some trouble with
I am going to continue to study tonight and hopefully i will eventually get to grips with it.
Thanks for any help in advance guys and again i apologize if this is the wrong place to come for this sort of thing
Looking at the topology, you can visualize their plan. First, on Subnet A they mention 'two host', one is the PC and the other is router (AKA default gateway). The second, Subnet B, is the switch which guess access to multiple users and the router (VLAN/default gateway). The important think to never forget is those two reserved addresses, the network address and the broadcast address. Based on these facts, let's break it down:
- First Memorize 1 2 4 8 16 32 64 128 256. It's not hard, start with 1 and double till you get to 256. You will have paper in your CCNA or other lab, right that down first thing before you start your test (it's your friend, I promise)
- Subnet A Needs four IPs, 2 Host (router and PC) and 2 Reserved (broadcast and network); so, you need four IPs, AKA four 'host bits' if doing this in binary. Remember in step 1 I told you to write down those numbers. Now you need them. Find the number that is the equal or closest bigger number to the number of IPs you need. You lucked out, four is an exact match. Here is a shortcut, since we don't have time to teach binary in this forum, if the number of 'host bits' (host + 2 reserved) are smaller than 256, simply subtract from 256. In this case you need 4 host bits, so 256 - four = 252. Your subnet mask will be 255.255.255.252. That is a short cut that only works for figuring out the subnet /IP mask when there are less than 256 host + reserved/host bits. This in no way will prepare you, this will just help you check your work. YOU NEED TO UNDERSTAND THESE BINARY CONVERSIONS.
- Subnet B 23 Host, which will allow for 22 PCs plus your router/gateway, and need to add the 2 reserved. That equals 25. Now you have to go back to step one to find a number that is equal are bigger than 25. 25 isn't there, so 32 is the next larger number. 256 - 32 = 224, so your subnet / IP mask is 255.255.255.224. Again, please don't forget this only works for network with out 254 host (+2 reserved for 256 host bits).
For everything else, you need to go learn binary. There are a ton of good YouTube videos, but just not enough time or space here to convey what is one of the hardest items for newbies to grasp. Hopefully I have given you another tool to quickly find simple subnet mask values and check your work later derived from the binary method.
Hmmm...... I am guessing they have given you packet tracer since this is net academy?
I have been at this a VERY long time, and i can tell you now the 256 is not vailed unless we are speaking of ranges... each octet actually looks like this 128, 64, 32, 16, 8, 4, 2, 1 with each bit representing that number and you add them up to figure our your mask with the last bit being the multiyear... basicly 1 will equal the subnetwork bits used with 0 as your host bits. so in this case for the all are set to 1, besides 2 and 1 which means two bits being 2^2= your 4 bits -2 = 2bits for hosts (network devices). the first range of addresses will have the mask 255.255.255.252 or 11111100 in binary.
1, 1, 1, 1, 1, 1,0, 0
Part one has given you this ip address and mask 192.168.25.0 255.255.255.0 (/24 is represented as this mask)
This means that you have a range of 1 - 254 one what the above guy was talking about from 0 - 255
Where did this range come from...?
well when figuring out Classless Inter domain Routing, you first need to figure our were your bitts are and in what octet.
/24 means the subnet is using 24 bits out of the possible 32 ( one octet = 8 bits, you have 4 octets in an address so 4x8 = 32), this leaves you with a remaining 8. Now that means the last bit that the network sides has browed will be on the 1 of that respective side in that octet..
so your ranges will be as follows So 2^8=256 - 2 = 254 possible hosts.
0. 0(1 - 254 for hosts) 255
1.0 (1 - 254 for hosts) 255
and so on,
Looking at part 3 you simply need to calculate how many hosts you need according to the requirement, since you are looking at 2, they are referring to connections between network devices such as routers or switches. Now you can’t have 2 subnets alone as you need broadcast address and Subnetwork address, so 2^2 will give you 4 addresses for subnet a with 2^5-2 then giving you 30 addresses for subnet B using the above method in the .25.1- .25.254 range.
(.25.0SNW) 25.1 - .25.2 (bc add = 25.3bc) <-----used for your devices
next calculation for hosts 32 is closer to 23 so 30 host using calculation above so..
0 - 31 <---- taken as you are using the first 4 for other devices, and its best practice to avoid other subnet ranges in use or you will confuse your routes. there for you will need to use the next available range with 32 being the multiplier since its the last bit borrowed for subnetwork.
so.. NW BC
32 (33 - 62 usable)- 63 with subnet mask of 255.255.255.224 since the bits will only go up to 32 (128, 64, 32) for that sub network. meanins the mask will look like 11100000 in binary.
well i hope i have given you food for thought. But if no then look up CBT nuggets, as they have some awesome tutorials!
I know this is sometimes difficult to visualize, but when counting from 0 - 255, there are 256 IP possibilities (0 counts in the world of IP addressing, although often the network address, since pure/unsubnetted class C mask (/24)is the most commonly deployed). 256 does not represent one of the BITs in the per octet 8-bit byte. It represents the total of all bits - 256; so as KK aluded to, not used in binary, but an important value to understand if you want to be able to figure out the subnet mask in less than 10 seconds. Of course as KK and I are aware, and I mentioned, that only nets 254 usable since two are always reserved in IPv4 addressing. So in using the shortcut for getting quick calculations of subnets needing 256 or less TOTAL IPs, including the reserved two, that works. I reality, most seasoned network engineers are not pulling out binary to figure out an ACL wildcard mask or network (they can quickly subtract the last octet from 256 to know IPs in the subnet), and if you are pulling out a subnet calculator or binary, you probably are not seasoned yet - still in admin vs engineer stages; also, since many low-end integrated NICs in printers and IP phones only have CAM table (storage for MAC addresses) with room for 256, 512, or 1024 entries, it is quite rare and often not advised to deploy subnets larger than 256 or 512. Can you and will it work in production to do so, yes. Will you be more exposed to the unexplainable VoIP phone weirdisms (especially when not-using Cisco) and unresponsive or locking up printers, likely.
So now after digressing and back to (256) 128 64 32 16 8 4 2 1:
- All 0's = 256 host bits in the last octet (after the last dot)
- you have to go in order turning on bit on at a time after that; so, the first workable bit (working left to right) is 128: 256 - 128 =128; so, you could get 128 IPs - 2 reserved - 126 host
- The next bit you could turn on is 64, but remember you had to have 128 turned on to go to the next; so, 128 + 64 = 192 (255.255.255.192) and 256 - 192 = 64 so you could have 64 IPs - 2 reserved for 62 host IPs
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