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<title>Advanced Networking - Module 1 Chapter 8 - Subnetting IP Networks</title>
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<section>
<h1>Advanced Networking</h1>
<h2>Routing & Switching:</h2>
<h2>Introduction to Networks</h2>
<h3>Chapter 8: Subnetting IP Networks</h3>
<small><a href="http://hlcs.it">Hacklab Cosenza</a> / Centro di Ricerca su Tecnologia e Innovazione</small>
</section>
<section>
<h2>Why Subnetting?</h2>
<p><strong>Subnetting</strong> is the deepening of the network/host hierarcy into a network/<em>subnet(s)</em>/host one.</p>
<p><small>The terms <em>network</em> and <em>subnet</em> both identify computer networks. <em>Subnet</em> is just used to imply that we are talking about <strong>a part of some larger address space</strong>.</small></p>
<ul>
<li><strong>Optimizing address space usage</strong> by wasting less.</li>
<li>By minimizing the definition of "local", we can <strong>reduce broadcast</strong> traffic (smaller <em>broadcast domain</em>).</li>
<li>A <strong>better packet delivery</strong> can be achieved by providing more network targets to meet our routing requirements. This is called <em>network segmentation</em>.</li>
<li>Creating more targets against which to define our policies can also <strong>improve security</strong>.</li>
</ul>
</section>
<section>
<h2>Subnetting and Routing</h2>
<p>Subnetting means taking a <strong>single network address</strong> and partitioning it into <strong>multiple logical networks</strong>.</p>
<p>Subnets <strong>are</strong> networks, so as with any multiple networks <strong>you need routing</strong> to exchange traffic between subnets.</p>
<p>As in every network, for every subnet you need to configure:</p>
<ul>
<li>every host with <strong>network address, subnet mask and default gateway</strong> appropriate to the subnet they belong to.</li>
<li>the router interface that is part of the subnet with <strong>an IPv4 address that belongs to the subnet</strong> range.</li>
</ul>
</section>
<section>
<h2>Planning Subnetting</h2>
<p>What we'll learn about <strong>subnetting will apply to both public and private IPv4 address space</strong>. Subnetting public addresses is an ISP concern, we'll focus on subnetting private blocks.</p>
<p>So we start with private IP address ranges:</p>
<ul>
<li>10.0.0.0/8 (subnet mask 255.0.0.0)</li>
<li>172.16.0.0/12 (subnet mask 255.240.0.0)</li>
<li>192.168.0.0/16 (subnet mask 255.255.0.0)</li>
</ul>
<p>and then we ask ourselves these questions:</p>
<p><em>How many subnets? What belongs to each subnet? How many hosts for each? What will I need in the future?</em></p>
</section>
<section>
<h2>Subnetting = Borrowing Bits</h2>
<p>Network address = Network part + all-0s host part, distinguished thanks to the subnet mask (or prefix).</p>
<p><u>Subnets are created by <strong>taking one or more bits of the host part</strong> and making them "network bits"</u>.</p>
<p>The <strong>subnet mask is expanded to account for the bits <em>borrowed</em></strong> from the host portion (previously 0s in the mask) that are now part of the network portion (becoming 1s).</p>
<p><strong>The more bits are borrowed the more subnets are created</strong>, but with each additional subnet the available addresses decrease. Each borrowed bit <strong>doubles the subnets and halves the addresses</strong>.</p>
</section>
<section>
<section>
<h2>Subnetting 192.168.1.0/24</h2>
<h3>2 Subnets</h3>
<pre><code class="no-highlight"> 1 bit borrowed
▼
NNN.NNN.NNN. HHHHHHHH
Original Network Address 192.168. 1. 00000000 = 192.168.1.0/24
Original Subnet Mask 255.255.255. 00000000 = 255.255.255.0</code></pre>
<p>By <strong>borrowing 1 bit</strong> we can create 2 different network addresses, and thus 2 different subnets.</p>
<pre><code class="no-highlight"> NNN.NNN.NNN. NHHHHHHH
Subnet 0 Network Address 192.168. 1. 00000000 = 192.168.1.0/25
Subnet 1 Network Address 192.168. 1. 10000000 = 192.168.1.128/25
New Subnet Mask 255.255.255. 10000000 = 255.255.255.128</code></pre>
<ul>
<li>You always <strong>start borrowing from the MSB</strong> of the host portion.</li>
<li><strong>Subnet mask is the same for every generated subnet</strong>.</li>
</ul>
</section>
<section>
<h2>Subnetting 192.168.1.0/24</h2>
<h3>Available Ranges</h3>
<p>Before we assign addresses to devices in the subnet we must understand the available range:</p>
<ul>
<li><strong>Network Address</strong>: all 0s in the host part.</li>
<li><strong>First Host Address</strong>: all 0s except the last bit of the host part.</li>
<li><strong><em>n-th</em> Host Address</strong>: all other combinations.</li>
<li><strong>Last Host Address</strong>: all 1s except the last bit of the host part.</li>
<li><strong>Broadcast Address</strong>: all 1s in the host part.</li>
</ul>
</section>
<section>
<h2>Subnetting 192.168.1.0/24</h2>
<h3>192.168.1.0/25</h3>
<pre><code class="no-highlight"> NNN.NNN.NNN. N HHHHHHH
Subnet0 Network Address 192.168. 1. 0 0000000 = 192.168.1.0/25
Subnet0 1st Host Address 192.168. 1. 0 0000001 = 192.168.1.1
Subnet0 Last Host Address 192.168. 1. 0 1111110 = 192.168.1.126
Subnet0 Broadcast Address 192.168. 1. 0 1111111 = 192.168.1.127</code></pre>
<h3>192.168.1.128/25</h3>
<pre><code class="no-highlight"> NNN.NNN.NNN. N HHHHHHH
Subnet1 Network Address 192.168. 1. 1 0000000 = 192.168.1.128/25
Subnet1 1st Host Address 192.168. 1. 1 0000001 = 192.168.1.129
Subnet1 Last Host Address 192.168. 1. 1 1111110 = 192.168.1.254
Subnet1 Broadcast Address 192.168. 1. 1 1111111 = 192.168.1.255</code></pre>
</section>
</section>
<section>
<h2>Subnetting Formulas</h2>
<p>Where</p>
<ul>
<li><strong><em>n</em></strong> is the number of borrowed bits and</li>
<li><strong><em>h</em></strong> is the number of remaining bits in the host portion.</li>
</ul>
<ul>
<li><strong>To calculate the number of subnets</strong>: 2<sup>n</sup>.</li>
<li><strong>To calculate the number of hosts in a subnet</strong>: 2<sup>h</sup> - 2.</li>
</ul>
<p>For every network, <strong>the network address address and the broadcast address are not valid</strong> addresses to assign an host, so 2 addresses must be removed from the total.</p>
</section>
<section>
<section>
<h2>Subnetting 192.168.2.0/24</h2>
<h3>4 Subnets</h3>
<p><em>How many bits I need to borrow?</em> 2<sup>2</sup> = 4. So: 2 bits.</p>
<pre><code class="no-highlight"> 2 bit borrowed
▾▾
NNN.NNN.NNN. HH HHHHHH
Original Network Address 192.168. 2. 00 000000 = 192.168.2.0/24
Original Subnet Mask 255.255.255. 00 000000 = 255.255.255.0</code></pre>
<p>Each subnet will have 2<sup>6</sup> - 2 = 64 - 2 = 62 hosts.</p>
<pre><code class="no-highlight"> NNN.NNN.NNN. NN HHHHHH
Subnet0 Network Address 192.168. 2. 00 000000 = 192.168.1.0/26
Subnet1 Network Address 192.168. 2. 01 000000 = 192.168.1.64/26
Subnet2 Network Address 192.168. 2. 10 000000 = 192.168.1.128/26
Subnet3 Network Address 192.168. 2. 11 000000 = 192.168.1.192/26
New Subnet Mask 255.255.255. 11 000000 = 255.255.255.192</code></pre>
</section>
<section>
<h2>Subnetting 192.168.2.0/24</h2>
<h3>Address Ranges for 4 subnets</h3>
<pre><code class="no-highlight"> NNN.NNN.NNN. NN HHHHHH
Subnet0 Network Addr. 192.168. 2. 00 000000 = 192.168.2.0/26
Subnet0 1st Host Addr. 192.168. 2. 00 000001 = 192.168.2.1
Subnet0 Last Host Addr. 192.168. 2. 00 111110 = 192.168.2.62
Subnet0 Broadcast Addr. 192.168. 2. 00 111111 = 192.168.2.63
Subnet1 Network Addr. 192.168. 2. 01 000000 = 192.168.2.64/26
Subnet1 1st Host Addr. 192.168. 2. 01 000001 = 192.168.2.65
Subnet1 Last Host Addr. 192.168. 2. 01 111110 = 192.168.2.126
Subnet1 Broadcast Addr. 192.168. 2. 01 111111 = 192.168.2.127
Subnet2 Network Addr. 192.168. 2. 10 000000 = 192.168.2.128/26
Subnet2 1st Host Addr. 192.168. 2. 10 000001 = 192.168.2.129
Subnet2 Last Host Addr. 192.168. 2. 10 111110 = 192.168.2.190
Subnet2 Broadcast Addr. 192.168. 2. 00 111111 = 192.168.2.191
Subnet3 Network Addr. 192.168. 2. 11 000000 = 192.168.2.192/26
Subnet3 1st Host Addr. 192.168. 2. 11 000001 = 192.168.2.193
Subnet3 Last Host Addr. 192.168. 2. 11 111110 = 192.168.2.254
Subnet3 Broadcast Addr. 192.168. 2. 11 111111 = 192.168.2.255</code></pre>
</section>
</section>
<section>
<section>
<h2>Subnetting 192.168.3.0/24</h2>
<h3>5 Subnets</h3>
<img src="http://i.imgur.com/D2cTrKz.png" style="width: 450px;">
<p>If we need 5 subnets, we have to borrow 3 bits (2<sup>4</sup> = 8 > 5).</p>
<p>The remaing 5 bits in the host part allow for 2<sup>5</sup> - 2 = 30 hosts.</p>
</section>
<section>
<h2>Subnetting 192.168.3.0/24</h2>
<h3>Ranges</h3>
<pre><code class="no-highlight"> ▾▾▾
NNN.NNN.NNN. HHH HHHHH
Original Network Addr. 192.168. 3. 000 00000 = 192.168.3.0/24
Original Subnet Mask 255.255.255. 000 00000 = 255.255.255.0
NNN.NNN.NNN. NNN HHHHH
Subnet0 Network Addr. 192.168. 3. 000 00000 = 192.168.3.0/27
Subnet1 Network Addr. 192.168. 3. 001 00000 = 192.168.3.32/27
Subnet2 Network Addr. 192.168. 3. 010 00000 = 192.168.3.64/27
Subnet3 Network Addr. 192.168. 3. 011 00000 = 192.168.3.96/27
Subnet4 Network Addr. 192.168. 3. 100 00000 = 192.168.3.128/27
NNN.NNN.NNN. NNN HHHHH
Subnet0 Network Addr. 192.168. 3. 000 00000 = 192.168.3.0/27
Subnet0 1st Host Addr. 192.168. 3. 000 00001 = 192.168.3.1
Subnet0 Last Host Addr. 192.168. 3. 000 11110 = 192.168.3.30
Subnet0 Broadcast Addr. 192.168. 3. 000 11111 = 192.168.3.31
Subnet1 Network Addr. 192.168. 3. 001 00000 = 192.168.3.32/27
Subnet1 1st Host Addr. 192.168. 3. 001 00001 = 192.168.3.33
Subnet1 Last Host Addr. 192.168. 3. 001 11110 = 192.168.3.62
Subnet1 Broadcast Addr. 192.168. 3. 001 11111 = 192.168.3.63
Subnet2 Network Addr. 192.168. 3. 010 00000 = 192.168.3.64/27
Subnet2 1st Host Addr. 192.168. 3. 010 00001 = 192.168.3.65
Subnet2 Last Host Addr. 192.168. 3. 010 11110 = 192.168.3.94
Subnet2 Broadcast Addr. 192.168. 3. 010 11111 = 192.168.3.95
Subnet3 Network Addr. 192.168. 3. 011 00000 = 192.168.3.96/27
Subnet3 1st Host Addr. 192.168. 3. 011 00001 = 192.168.3.97
Subnet3 Last Host Addr. 192.168. 3. 011 11110 = 192.168.3.126
Subnet3 Broadcast Addr. 192.168. 3. 011 11111 = 192.168.3.127
Subnet4 Network Addr. 192.168. 3. 100 00000 = 192.168.3.128/27
Subnet4 1st Host Addr. 192.168. 3. 100 00001 = 192.168.3.129
Subnet4 Last Host Addr. 192.168. 3. 100 11110 = 192.168.3.158
Subnet4 Broadcast Addr. 192.168. 3. 100 11111 = 192.168.3.159</code></pre>
</section>
</section>
<section>
<section>
<h2>100 Subnets!</h2>
<p>If we need 100 subnets, we need to borrow <strong>at least 7 bits</strong> from the host part of our address block: 2<sup>7</sup> = 128 > 100.</p>
<p>If our address block is a /24 network, that would generate a /31 prefix and leave 1 bit in the host part, so 2<sup>1</sup>-2 = <u>0 hosts</u>!</p>
<p>This means we need more host bits to borrow from, and therefore <strong>a larger starting address block</strong>. For instance...</p>
</section>
<section>
<h2>100 Subnets?</h2>
<h3>172.20.0.0/16</h3>
<pre><code class="no-highlight"> ▾▾▾▾▾▾▾
NNN.NNN.HHHHHHH H.HHHHHHHH
Original Network Add. 172. 20.0000000 0.00000000 = 172.20.0.0/16
Original Subnet Mask 255.255.0000000 0.00000000 = 255.255.0.0
NNN.NNN.NNNNNNN H.HHHHHHHH
New Subnet Mask 255.255.1111111 0.00000000 = 255.255.254.0
Subnet0 Network Addr. 172. 20.0000000 0.00000000 = 172.20.0.0/23
Subnet1 Network Addr. 172. 20.0000001 0.00000000 = 172.20.2.0/23
Subnet2 Network Addr. 172. 20.0000010 0.00000000 = 172.20.4.0/23
Subnet3 Network Addr. 172. 20.0000011 0.00000000 = 172.20.6.0/23
[..............................CUT...............................]
Subnet126 Netw. Addr. 172. 20.1111110 0.00000000 = 172.20.252.0/23
Subnet127 Netw. Addr. 172. 20.1111111 0.00000000 = 172.20.254.0/23</code></pre>
<p>Subnetting is done in the 3rd octect, <strong>host bits share part of the 3rd</strong>, plus the whole 4th octect.</p>
<p>Total hosts per subnet: 2<sup>9</sup> - 2 = 512 - 2 = 510.</p>
</section>
</section>
<section>
<section>
<h2>5000+ Hosts?</h2>
<p>Let's say we need subnets capable of > 5000 hosts. We:</p>
<ul>
<li>need at least <strong>13 bits</strong> left in the host part, <strong>after</strong> the subnetting borrowings (2<sup>13</sup> - 2 = 8190 > 5000).</li>
<li>are left with <strong>just 3 bits to borrow</strong> for subnetting, in a /16 space. This amounts to just 2<sup>3</sup> = 8 subnets.</li>
</ul>
<p>This could very well not be enough for most needs. <strong>The solution is to start from a larger block</strong>, like 10.0.0.0/8.</p>
<p>In a "10/8", after reserving the last 13 bits for our 5000+ hosts, we are left with 11 bits to borrow for subnetting. That's 2<sup>11</sup> = 2048 subnets with a /19 prefix (255.255.224.0).</p>
</section>
<section>
<h2>5000+ Hosts?</h2>
<pre><code class="no-highlight"> ▾▾▾▾▾▾▾▾ ▾▾▾
NNN.HHHHHHHH.HHH HHHHH.HHHHHHHH
Original Net. Addr. 10.00000000.000 00000.00000000 = 10.0.0.0/8
Original Mask 255.00000000.000 00000.00000000 = 255.0.0.0
New Subnet Mask 255.11111111.111 00000.00000000 = 255.255.224.0
NNN.NNNNNNNN.NNN HHHHH.HHHHHHHH
Subnet0 Net. Addr. 10.00000000.000 00000.00000000 = 10.0.0.0/19
Subnet1 Net. Addr. 10.00000000.001 00000.00000000 = 10.0.32.0/19
Subnet2 Net. Addr. 10.00000000.010 00000.00000000 = 10.0.64.0/19
[..............................CUT...............................]
Subnet8 Net. Addr. 10.00000001.000 00000.00000000 = 10.1.0.0/19
[..............................CUT...............................]
Subnet2045 Net.Addr. 10.11111111.101 00000.00000000 = 10.255.160.0/19
Subnet2046 Net.Addr. 10.11111111.110 00000.00000000 = 10.255.192.0/19
Subnet2047 Net.Addr. 10.11111111.111 00000.00000000 = 10.255.224.0/19</code></pre>
</section>
</section>
<section>
<h2>Subnetting "Short Blanket"</h2>
<p>The <strong>key to basic subnetting</strong> is to balance number of subnets, and hosts for each, based on requirements.</p>
<ul>
<li>Start by determining the <strong>total number of hosts</strong> to serve. This will hint at the <strong> starting address space</strong> you need.</li>
<li>Analize how your network is organized to <strong>determine the number of subnetworks you need</strong>. Think about possibile <strong>future segmentations</strong>.</li>
<li>Determine the <strong>highest number of host you will have on a single subnet</strong>. Try to think about if and how much this number will <strong>grow in the foreseeable future</strong>.</li>
<li>Now you have <strong>the data you need to choose</strong> how many bits to borrow and how many host bits to "leave alone".</li>
</ul>
</section>
<section>
<h2>Variable Length Subnet Mask</h2>
<p><strong>With basic subnetting, every subnet has the same size</strong>. <u>When the number of hosts needed in each is very different</u>, that would lead to significant <strong>waste and unused addresses</strong>.</p>
<p>Subnets too large for the actual need will also <strong>limit growth</strong> by reducing the number of available subnets.</p>
<p>The solution to these scenarios is a <strong>subnetting method</strong> called <em>Variable Length Subnet Mask (VLSM)</em>.</p>
<p>It does what the name says: the number of bits borrowed (hence, the subnet mask) <strong>varies with each subnet</strong>.</p>
</section>
<section>
<h2>How does VLSM work?</h2>
<img src="http://i.imgur.com/8JdgAJz.png">
<ul>
<li>you still borrow bits to create subnets.</li>
<li>the subnetting formulas are still valid.</li>
</ul>
<p>The difference is <strong>it's not a single-pass process</strong> anymore:</p>
<ul>
<li>The network is subnetted, and <strong>subnets are subnetted again</strong>, creating differently-sized subnets.</li>
<li>The <strong>process is repeated</strong> as needed.</li>
</ul>
</section>
<section>
<section>
<h2>VLSM Scenario</h2>
<img src="http://i.imgur.com/QMjXJ6M.jpg" style="width: 550px;">
<p><u>Starting from a /24</u>: we need 9 subnets, so 4 bits to borrow. It means 16 subnets with a /28 mask, each capable of 14 hosts.</p>
<p>This means <u>12 wasted address on each PtP link</u>, 48 total!</p>
<p>And 14 hosts per subnet are a bit low, aren't they?</p>
</section>
<section>
<h2>VLSM Scenario</h2>
<p>Operating with <strong>traditional, basic</strong> subnetting:</p>
<pre><code class="no-highlight">Original Network Address 192.168.1.0/24
Original Subnet Mask 255.255.255.0
New Subnet Mask 255.255.255.240 = /28
Subnet 0 192.168.1.0/28
Subnet 1 192.168.1.16/28
Subnet 2 192.168.1.32/28
Subnet 3 192.168.1.48/28
[...]
Subnet 14 192.168.1.224/28
Subnet 15 192.168.1.240/28</code></pre>
<p>Wasteful. Let's take the first /28 subnet and subnet it again into <strong>4 /30 subnets</strong> of 2 hosts each: <u>0 wasted addresses</u>!:</p>
<ul>
<li>2 hosts needed meant 2 host bits (2<sup>2</sup> - 2 = 2 hosts)</li>
<li>this left us with 2 bits to borrow: 2<sup>2</sup> = 4 subnets.</li>
</ul>
</section>
<section>
<h2>VLSM Scenario</h2>
<h3>VLSM Applied</h3>
<pre><code class="no-highlight">Subnet0 - PtP #1 192.168.1.0/30 # subnetted from 192.168.1.0/28
Subnet1 - PtP #2 192.168.1.4/30 # subnetted from 192.168.1.0/28
Subnet2 - PtP #3 192.168.1.8/30 # subnetted from 192.168.1.0/28
Subnet3 - PtP #4 192.168.1.12/30 # subnetted from 192.168.1.0/28
Subnet4 192.168.1.16/28
Subnet5 192.168.1.32/28
Subnet6 192.168.1.48/28
[...]
Subnet12 192.168.1.144/28
Subnet13 192.168.1.160/27 # 30 hosts capable
Subnet14 192.168.1.192/27 # 30 hosts capable
Subnet15 192.168.1.224/27 # 30 hosts capable</code></pre>
<p>In addition to the perfect fit for the PtP links, VLSM also allows us to create <strong>more flexible subnetting</strong>.</p>
<p>In this example <u>we ditched the last 6 /28 subnets for 3 /27 subnets</u> in which we can host up to 30 devices.</p>
</section>
<section>
<h2>VLSM Charts</h2>
<img src="http://i.imgur.com/WgSJYKv.jpg" style="width: 600px;">
<p>When using VLSM:</p>
<ul>
<li>Check that address ranges <strong>do not overlap</strong>.</li>
<li>It's easier to proceed from larger subnet to smaller.</li>
</ul>
</section>
</section>
<section>
<section>
<h2>Addressing Plan</h2>
<p>We plan the addressing scheme for our network for the following purposes:</p>
<ul>
<li><strong>Preventing address duplication</strong>: by carefully documenting how the addresses are assigned we can prevent address collision.</li>
<li><strong>Access control</strong>: by knowing precisely which address is assigned to which host, we can better control the access to network resources based on our policies.</li>
<li><strong>Network monitoring</strong>: the network admin needs to be able to readily identify sources and destinations of the traffic while analyzing it for performance and security issues.</li>
</ul>
</section>
<section>
<h2>Assigning Addresses</h2>
<p>It is best to assign addresses to hosts based on a <strong>recognizable pattern</strong>, which will help the administrator and the documentation process.</p>
<ul>
<li><strong>Clients</strong> - DHCP is the common assignment method.</li>
<li><strong>Servers & peripherals</strong> - As they provide almost all network resources, they should have static addressing, with consistent numbering, making them easy to remember.</li>
<li><strong>Internet-accessible hosts</strong> - Both public and private static addressing. Public addresses are mostly associated to these hosts by performing NAT on the perimeter.</li>
</ul>
</section>
<section>
<h2>Assigning Addresses</h2>
<ul>
<li><strong>Intermediary Devices</strong> - If they are L3-capable, they should have a static, predictable address to allow management.</li>
<li><strong>Gateways</strong> - They belong to multiple networks, so they have an IP address assigned to each interface. Almost always it is the <u>highest or lowest address available on that network segment</u>.</li>
</ul>
</section>
</section>
<section>
<section>
<h2>IPv6 Subnetting</h2>
<p>IPv4 Subnetting is done to manage address scarcity.</p>
<p>Subnetting IPv6 is done for an <u>entirely different reason</u>: to support <strong>hierarchical and logical network design</strong>.</p>
<img src="http://i.imgur.com/LGVfYS4.png">
<p>An IPv6 (public) address block with a /48 (Global Routing) prefix has <strong>16 bits for subnet ID</strong> (65536 /64 subnets).</p>
<p>A dedicated Subnet ID field means there's <u>no need to borrow bits</u> from Interface ID (host portion).</p>
</section>
<section>
<h2>IPv6 Subnetting Example</h2>
<p>Subnets created by the Subnet ID are easy to identify and represent, it just works like an hexadecimal counter.</p>
<pre><code class="no-highlight">Address Block 2001:0DB8:ACAD::/48
Subnet 0 2001:0DB8:ACAD:0000::/64
Subnet 1 2001:0DB8:ACAD:0001::/64
Subnet 2 2001:0DB8:ACAD:0002::/64
Subnet 3 2001:0DB8:ACAD:0003::/64
Subnet 4 2001:0DB8:ACAD:0004::/64
Subnet 5 2001:0DB8:ACAD:0005::/64
Subnet 6 2001:0DB8:ACAD:0006::/64
Subnet 7 2001:0DB8:ACAD:0007::/64
Subnet 8 2001:0DB8:ACAD:0008::/64
Subnet 9 2001:0DB8:ACAD:0009::/64
Subnet 10 2001:0DB8:ACAD:000A::/64
[ . . . . . . . . . . . . . . . . . . . ]
Subnet 65533 2001:0DB8:ACAD:FFFD::/64
Subnet 65534 2001:0DB8:ACAD:FFFE::/64
Subnet 65535 2001:0DB8:ACAD:FFFF::/64</code></pre>
</section>
<section>
<h2>IPv6 Subnetting Considerations</h2>
<p>Subnetting IPv6 is not that much different from IPv4.</p>
<p>The leading practice is to receive at least a <strong>/48 prefix from an ISP</strong>. This leaves 2<sup>80</sup> bits to manipulate for subnetting.</p>
<p><a href="https://tools.ietf.org/html/rfc4291">RFC 4291</a> says the <strong>current smallest recommended prefix</strong> is /64. You can assign a /64 to PtP links and <em>not feel guilty</em>.</p>
<p>Using a prefix smaller than /64 is actually <u>strongly discouraged</u> because <strong>it breaks IPv6 features</strong> that rely on a 64-bit Interface ID, like SLAAC.</p>
</section>
<section>
<h2>IPv6 Subnetting Considerations</h2>
<p>Some ISPs insist to give customers a single /64 IPv6 subnet, so they can't use the Subnet ID field to do subnetting.</p>
<p>You can of course <strong>borrow bits from the Interface ID</strong> to have a deeper hierarchy. The best practice is to borrow <strong>4 bits at a time</strong> and used aligned netmasks (/68, /72, /76, etc.). Why?</p>
<p>A <strong>nibble</strong> is a 4 bit group, or <u>one hex digit</u>. By subnetting on <em>nibble boundaries</em> we create subnets <strong>just by adding 0x01</strong>.</p>
</section>
</section>
<section>
<h1>End of Lesson</h1>
</section>
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