Subnetting IP 4 ICNDCCNA Prep What is subnetting

Subnetting IP 4 ICND/CCNA Prep

What is subnetting and why do I care? • Basically, subnetting is the process of taking one network and dividing it into smaller networks. For example, if we have a 172. 16. 0. 0/16 network, this gives us 65, 536 (2^16) possible hosts on the network. But, we could divide this up into two networks with 32, 768 hosts each. More on this in a bit… • For your ICND and CCNA exams, you need to be able to do this in your head, very quickly. I assure you that while you may not get exam questions that ask you to directly subnet at network (like we did above), you will be given information about a network and have to figure out what’s wrong with it – and that will require that you can subnet quickly.

Baby Steps: Counting in Binary. • The first thing you need to do is be able to count in binary and know your powers of two (up to 2^8). See, an IP address looks like this: 172. 16. 22. 44 But in binary it looks like this: 10101100. 00010000. 000101100 • So how did we get there? The first thing is to look at these octets (each portion of the IP separated by a “. ”) as individual numbers. Don’t worry, we’ll put it all together later. First though, our powers of 2: 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128

More on the powers of 2 • So right now we know: 2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 • The first section of our IP address 172. 16. 22. 44 is “ 172”. Our binary representation of that was 10101100. Let’s tear that apart: Powers of 2: 172 in Binary: Result: Sum of Results: 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 1 0 1 1 0 0 128 0 32 0 8 4 0 0 128+32+8+4 = 172

More on the powers of 2 • If you’re like me, you’re already sick of this, but you need to know it, so let’s tear down the rest of that address (172. 16. 22. 44). On the previous slide, we did 172 – now let’s do the rest: 16 Powers of 2: 16 in Binary: Result: 22 Powers of 2: 16 in Binary: Result: 44 Powers of 2: 16 in Binary: Result: 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 0 1 0 0 0 0 16 0 0 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 0 1 0 1 1 0 0 16 0 4 2 0 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 0 0 1 1 0 0 32 0 8 4 0 0 Result 16+4+2 22 Result 32+8+4 44

Recap on powers of 2 • Powers of two are key in subnetting. The good news is, you only need to know the first 8 (2^0 through 2^7), and they are 1, 2, 4, 8, 16, 32, 64, and 128 • Now, a bit of practice of what we’ve done so far: • Powers of 2: • 2^4 = 16 = 2^ 2^0 = • 128 = 2^ 32 = 2^ 2^6 = Convert to Binary Convert to an IP Address 172. 16. 1. 1 = 10. 177 = 224. 0. 0. 5 = 1. 199. 63. 17 = 11000000. 00111111. 00000010. 11111101 = 00000011. 01100010. 10000001. 10101011 = 00100001. 0101. 1010. 0000 = 11100101. 0011. 1111. 00010000 =

Back to subnetting • Ok, we can count by twos now – good game. Now that those basics are out of the way, let’s see how we can take a network address – which could have been assigned to us from a network administrator or an ISP – and divide into multiple, smaller networks, a. k. a sub-networks, a. k. a. “subnet it. ” • Example: We’re given the classful (c-class) network address of 192. 168. 10. 0/24 and need to subnet it. First, we need to recognize that we can’t make the “network” that has already been given to us (by whomever) bigger, we can only make it smaller. That means that the /24 (known as CIDR notation), which defines the network portion of the address that was assigned to you, is “fixed” when looking to the left. That /24 means the first 24 bits (192. 168. 10) are unchangeable. However, we have the last 8 bits that we can play with. I know that’s wordy – for now, just accept that the 192. 168. 10 part won’t be changed, ever…

Subnetting • We have our 192. 168. 10. 0/24 network. We know the 192. 168. 10 part is locked. We also know that this is denoted by the /24, meaning the first 24 bits (the first 3 octects – remember, in binary those are groups of eight 1 s and 0 s and 3 x 8=24) will never change. But now we’re the network admin, and we want to dish out network addresses to two junior admins. How do we do this? Simple, we steal 1 more bit, this turns our /24 into a /25. Remember, the first 24 (to the left) are locked, but we can play with the stuff to the right…. • …. let’s look at this in binary: • 192. 168. 10. 0/24 = 11000000. 10101000. 00001010. 0000 First 24 bits, locked Last 8 bits. We can play with these.

Subnetting • From out last slide: • 192. 168. 10. 0/24 = 11000000. 10101000. 00001010. 0000 First 24 bits, locked Last 8 bits. We can play with these. • Let’s turn this into two networks. To do that, we’re going to steal 1 bit from the last 8 bits. This gives us 2 networks (2^1 – told you powers of 2 were important) • The one bit we’re stealing from the last octet does three things: • 1) It means we’re using a /25 mask now (24 + the 1 we’re stealing) • 2) It defines our network boundary. Remember your powers of two – the far left bit we’re stealing is in the 2^7 th spot – which is 128. That means our networks are 192. 168. 10. 0 (if our 25 th bit is a 0) and 192. 168. 10. 128 (if our 25 th bit is a 1) • 3) It defines our subnet mask. More on this later, but a /25 gives us a mask of 255. 128

Subnetting • Remember, our possible network addresses are now 192. 168. 10. 0 (if our 25 th bit is a 0) and 192. 168. 10. 128 (if our 25 th bit is a 1): Decimal: 192. 168. 10. 0 Binary: 192. 168. 10. 128 11000000. 10101000. 00001010. 0000 11000000. 10101000. 00001010. 10000000 First 24 bits - locked 25 th bit that we stole To make 2 networks Available for Hosts

Subnetting • Given: Decimal: 192. 168. 10. 0 Binary: 192. 168. 10. 128 11000000. 10101000. 00001010. 0000 11000000. 10101000. 00001010. 10000000 First 24 bits - locked 25 th bit that we stole To make 2 networks Available for Hosts Network Definitions: Network Address Host Range Broadcast Address Network 1 192. 168. 10. 0 -126 192. 168. 10. 127 Network 2 192. 168. 10. 128 192. 168. 10. 129 -254 192. 168. 10. 255

A Step Further • Using our original network, 192. 168. 10. 0/24, what do we do if we need 3 networks? Remember back on slide 9 how we “stole” 1 bit to give us two /25 networks (we stole 1, and 2^1 is 2 …so we have two networks)? We’re going to do the same thing. If we steal 2 bits, we have 2^2 available networks – so 4. We need 3, so this works and has the added benefit of giving us one more for growth. • Look back at slide 9 – we stole 1 bit, which could have a value of 1 or 0, making our networks 192. 168. 10. 0 or 192. 168. 10. 128 (because the value of the far left bit in any octet is 128), now we’re borrowing 2 bits, which looks like this: • 11000000. 10101000. 00001010. 0000 First 24 bits - locked 25 th and 26 th bit that we stole To make 2 networks Available for Hosts

A Step Further • 11000000. 10101000. 00001010. 0000 First 24 bits - locked 25 th and 26 th bit that we stole To make 2 networks Available for Hosts • So, when we borrowed just 1 bit, the network boundaries were in 128 bit blocks (2^7). When we borrow 2 bits, the blocks are 64 (2^6). We have four networks, because each of our two borrowed bits can be either a 1 or 0 – this gives us the combinations of 00, 01, 10, and 11 – which makes the binary above look like one of these 4 combinations (with their decimal equivalents: 11000000. 10101000. 00001010. 0000 192. 168. 10. 0 11000000. 10101000. 00001010. 01000000 192. 168. 10. 64 11000000. 10101000. 00001010. 10000000 192. 168. 10. 128 11000000. 10101000. 00001010. 110000000 192. 168. 10. 192

A Step Further • So, given this information from the last slide: 11000000. 10101000. 00001010. 0000 192. 168. 10. 0 11000000. 10101000. 00001010. 01000000 192. 168. 10. 64 11000000. 10101000. 00001010. 10000000 192. 168. 10. 128 11000000. 10101000. 00001010. 110000000 192. 168. 10. 192 We have the networks: Network 1 Network Address 192. 168. 10. 0 Host Range 192. 168. 10. 0 -62 Broadcast 192. 168. 10. 63 Network 2 192. 168. 10. 64 -126 192. 168. 10. 127 Network 3 192. 168. 10. 128 192. 168. 10. 129 -190 192. 168. 10. 191 Network 4 192. 168. 10. 193 -254 192. 168. 10. 255

A Class B Example • Everything we’ve done so far as been with a class C address. They’re easier to learn on. Let’s take a look at this class be address and subnet the heck out of it: • 172. 32. 0. 0/22 Whoa, wait up, what’s this /22 stuff you’re giving me? No worries, brah, I got you covered. First, convert to binary: 10101100. 00100000000 First 22 bits - locked Last 10 bits – we can play with Let’s say we want to turn this into 4 networks. Easy enough by now, right? We just steal the last two bits of the 3 rd octet. Work it out on paper. Those last two placeholders in the 3 rd octet have a value of 2 and 1 (2^1, and 2^0). Remember, each bit can have a value of 1 or 0. So, our potential networks are: Network Address Host Range Broadcast Network 1 172. 16. 0. 0 -254 172. 16. 0. 255 Network 2 172. 16. 1. 0 -254 172. 16. 1. 255 Network 3 172. 16. 2. 0 -254 172. 16. 2. 255 Network 4 172. 16. 3. 0 -254 172. 16. 3. 255 Note: With your new /24 networks, your subnet mask is now 255. 0 - /22 was 255. 252. 0

A Harder Class B Example • Your network assignment from your ISP was: 172. 64. 0. 0/12 10101100. 0100000000 First 12 bits - locked Last 20 bits – we can play with You need 7 networks. 2^3 is 8, so that works just fine. So we’re going to steal 3 bits to play with. Now, in binary, we have something that looks like this: 10101100. 0100000000 First 12 bits - locked Last 17 bits for hosts Next 3 bits, for subnetting Let’s continue this on the next page…

A Harder Class B Example • 10101100. 0100000000 First 12 bits - locked Last 17 bits for hosts Next 3 bits, for subnetting • What are our networks in this example? The rightmost bit in the 3 bits that we’re subnetting has a value of 2 – in that second octet, that zero is in the 2^1 place – which makes our second network look like this (the first one is the original network we were given): 10101100. 01000010. 00000000 In decimal, this is 172. 64. 0. 0

A Harder Class B Example • 10101100. 0100000000 First 12 bits - locked Last 17 bits for hosts Next 3 bits, for subnetting The other networks are 172. 66. 0. 0, 172. 68. 0. 0, 172. 70. 0. 0, 172. 74. 0. 0, 172. 76. 0. 0. These are our 8 networks (again, remember the first one is still 172. 62. 0. 0). Here’s what that looks like 172. 62. 0. 0 172. 64. 0. 0 172. 66. 0. 0 172. 68. 0. 0 172. 70. 0. 0 172. 74. 0. 0 172. 76. 0. 0 10101100. 0100000000 10101100. 01000010. 00000000 10101100. 01000110. 00000000 10101100. 01001000. 00000000 10101100. 01001010. 00000000 10101100. 01001100. 00000000 10101100. 01001110. 00000000

Summary and Tips • Subnetting is important. At very least you’ll need it to pass your CCENT/CCNA exam(s). If you ever setup a new network, office, or branch office in the real world, you’ll need it then as well. • Know your powers of 2 up to 2^7 • Be able to convert from binary to decimal and back. You should know that 11100000 is 224, and that 177 is 10110001 • Practice. A lot. Consult the internet and search for subnetting practice. There’s tons of material out there that can keep you sharp with this stuff. • If you’re given an IP address and subnet mask, you should be able to quickly calculate the network ID. If you have a network ID and netmask, you should be able to list the broadcast address and range of usable IP addresses. It shouldn’t take you more than 10 seconds to do this. • You should practice to the point where you’re so sick of subnetting that by the time you walk into the exam, that you never want to see it again.