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welcome back, my name is David Bombal CCIE 1123
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and in this section, we're going to look at Variable Length Subnet Mask or VLSM
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and route summarization.
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VLSM allows us to better use IP addresses
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by allowing for Variable Length Subnet Mask on a single network.
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so what we're going to cover in this section is firstly VLSM
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once again Variable Length Subnet Mask, we're going to look at CIDR
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or classless Inter-Domain Routing
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I want to explain summarization and show you
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how you can summarize multiple routes into fewer or single a route.
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we're going to look at routing choices
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and how routers choose one route over another
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not just based on administrative distance
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but also based on the legnth of the prefix match
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and then lastly we're going to look at issues regarding discontiguous networks.
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so Variable Length Subnet Mask
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allow us to have varying or variable mask throughout our network
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so for example, a classful network of 10
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will have a mask of /8 with the first octet
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10 is the network portion of the address
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and the last 3 octets, in other words, the last 24 bits
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is the host portion of the address.
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that classful network was subnetted
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we could, for instance, have a subnet of 10.1.1.0/24
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where 10.1.1 is the network portion of the address
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in other words, the most siginificant 24 bits
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or most significant 3 octets is the netwok portion and the last octet
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or least significant portion of the address is the host portion.
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we could also subnet a 10 network down to 10.1.0.0/16
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where half of the address is the network portion
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and half of the address is the host portion.
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please refer to the ICND 1 part of this course
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or my subnetting explained e-book, for lots of detail on subnetting.
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CIDR or Classless Inter-Domain Routing was introduced in 1993
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to replace the prior addressing architecture of classful networks.
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classful networks were not scalable
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classful networking introduce 3 classes of addresses
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for allocation of IP addresses to hosts and networking devices
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we had A B and C networks.
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so for instance 10.0.0.0/8 is an example of a classful A network.
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172.16.0.0/16 is an example of a class B network
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and 192.168.1.0/24 is an example of a class C network.
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the problem with this method is that you were forced to use
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for instances a /16 mask which gave you approximately 65000 host addresses
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or you could use a class C address like this
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which only gave you 254 host addresses.
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this scheme would have resulted
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in the quick extortion of IP addresses and inflexibility.
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so CIDR replaced this which is based on Variable Length Subnet Mask
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were masks can vary on arbitrary lengths
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so they allow for different size subnets
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so rather than being forced to use for instance 10.0.0.0/8
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you can also use 10.0.0.0/16 or 24 or 30 depending on the requirements
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for the number of subnets or hosts in your network
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or you can use for instance 10.1.1.0/27 or 10.1.2.0/26 and so forth and so on.
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so you could take a single class A address
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and subdivide it into multiple subnets based on your requirements.
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it's also important to note that in CIDR
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we don't just advertise a network, we advertise a routing prefix
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in other words, we would advertise 192.168.1.0/24 not just 192.168.1.0
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CIDR also allows for the summarization of addresses
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which I'm going to explain in more detail in a moment.
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but this is were we take multiple subnets
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and put them into a super net or summary network
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which will reduce routing table sizes.
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Variable Length Subnet Mask have many advantages
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including the better utilization of IP addresses and subnets.
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on a WAN link as an example only 2 IP addresses are required.
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1 IP address for 1 end of the link
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and another IP address for the other end of the link
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thus a /30 mask would be better in say a /16 or /24 mask.
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a /30 mask only support 2 hosts on that subnet
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and thats all that's required in the point-to-point scenario like this.
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a /24 mask for instance would support 254 hosts
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and that would be over kill on this specific link.
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so VLSM gives assist the administrator
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the ability to better utilize IP addresses and subnets.
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where we subnet down subnets
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to support the required number of IP addresses on that specific segment.
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so as an example, here we have subnets
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that are using a /27 mask, on the WAN links we are using /30 mask
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and for the instance at the headquarters we are using a /16 mask.
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it also allows us to implement better summarization
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and scalability of networks as rather than the entire world
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being forced to use a specific subnet mask
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we are able to use varying mask in our own networks
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and advertise summaries to other companies or to the Internet.
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so heres a simple network to explain the problem
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when summarization is not used.
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on the left hand side we have networks 10.1.1.0/24
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up to 10.1.2.200.0/24
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on the right hand side we have networks 10.2.1.0/24
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up to 10.2.200.0/24 now summarization is now implemented
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and let's say you're running a routeing protocol like RIP
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which advertises the entire routing table every 30 seconds
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you're going to have scenario like this
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with the router on the left-hand side
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advertises 200 routes every 30 seconds and the router on the router side
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also advertises 200 routes every 30 seconds.
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RIP will continually sends out its entire routing table
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every 30 seconds even though there are no changes.
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so 400 routes are advertised across this WAN link
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which is very inefficient, those routing updates
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consume a lot of bandwidth and don’t accomplish much
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because they will continuously be advertise
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even though there are no changes
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so it makes more sense to summarize the routes
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so on the left-hand side, we have 10.1.1.0
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all the way up to 10.1.200.0
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So these are subnets of the 10.1.0.0 network
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so we could summarize those 200 routes into a single route
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and on the left hand side 10.2.1.0 up to 10.2.200.0
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could be summarize as 10.2.0.0/16
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so rather than advertising 400 routes a single route from the left-hand side
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is advertised to the left-hand side.
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and a single route from right-hand side is advertised to the left-hand side
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With summarization what we are have looking for
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is most significant bits that are equal, in other words
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starting from the left-and side and working towards the right-hand side
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we look for bits that are the same
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so in the first octet all the subnets contain 10
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so that’s common, in the second octet all the subnets contain 1
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so that's common but in the 3rd octet the values vary from 1 - 200
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so it’s not as easy to see what’s in common
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so we could just summarize it as 10.1.0.00/16
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knowing that 10.1 is common throughout
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and the router over here will summarize all of those routes
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and only send out 1 advertisement
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the router on the left-hand side can still get to all of this subnets
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because they are subnets of this network
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so even though we've summarized routes
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we still have full connectivity throughout the network
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here’s some more complicated example
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assume that we have networks 172.16.32.0/24
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all the way up to 172.16.63.0/24
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now rather than this router advertising all of those networks
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can we summarize this network into 1 or at least a few summaries
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now to work this out you start from the most significant bits
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in other words, you start from the left-hand side
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and you look for what is common throughout all of this networks
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so in the first octet 172 is common, so we know that’s common
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16 is also common, so we know does far that the 1st 2 octets are common
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in the third octet, however, the numbers are changing
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so we have 32, 33, 34, 35, all the way up to 63
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so it's not as easy to visualize or see what’s common here
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so what we're going to do is we can convert the third octet
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into binary to look for common bits, so for example 32 = 0010 0000
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now obviously there’s no gap halfway through an octet
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I’ve just put the space here to try and make it easier to read
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so an octet consists of 8 binary values, 32 would look at follows
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now as soon as you convert into binary
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you should convert the remaining bits into binary as well
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now in this example, you don’t need to do that
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because you can see that the last octet contains just a 0
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but I’ve done it here for completeness, the second network is 172.16.33.0
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and 33 in binary looks as follows, 34 looks as follows, 35 looks as follows
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and I wouldn’t recommend that you convert all the addresses into binary
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I would just do say the first 3 or 4
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and then the last one to see what’s in common.
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in the real world, however, you may come across far more complicated examples
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and you may need to convert all the addresses into binary
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but at the CCNA level that may not be necessary.
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So once we’ve done that we can look for bits that are in common
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so the first binary 0 in the 3rd octet is common throughout
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the second binary 0 is also common throughout, the 3rd binary bit is a 1
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and that is common throughout, however, in the 4th binary bit position
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the values change as you can see here the first few networks have 0's
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but the last network has a 1 in the fourth binary bit position
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so we can draw a line to the right of all the bits that are common
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so as you can see here to the left of the line in the 3rd octet
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we have 001 in binary
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the remaining bits in binary are not common in other words bits vary.
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Now Cisco does not support discontiguous subnet masks
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so you can see these bits are not in common and these this bits are in common
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there has to be a contiguous grouping of common bits
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starting from the left-hand side moving from the right-hand side
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and as soon as there are bits that are not common
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you have to draw a line to make a differentiation between common bits
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and bits that are not common
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so we know now that the first octet is common, the second octet is common
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so 172.16 in the third octet, the first 3 binary bits are common
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so 001 now any bits that are not in common are just set to 0
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so notice, we’ve populate the remaining portion of the address with binary 0's
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and then we convert the binary back to decimal
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so hopefully remember this from the ICND 1 course.
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So the first octet is 172, the second octet is 16
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the third octet if you convert this to decimal is equal to 32
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and the last octet converted to decimal is 0.
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So the summary address is 172.16.32.0
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The last step is to work out the bits that are in common
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now the first octet is in common and an octet is 8 bits
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the second octet is in common and that’s an additional 8 bits
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so thus far we have 8 bits + 8 bits in other words, 16 bits that are in common
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3 bits in the third octet are in common, so 8+8+3 will give you 19 bits in common
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thus the summary address for this subnets will be 172.16.32.0/19
196
00:13:10,000 --> 00:13:14,000 align:middle line:84%
it’s a simple as that to work out summarization
197
00:13:14,000 --> 00:13:16,000 align:middle line:84%
and I’m going to show you trick in a moment
198
00:13:16,000 --> 00:13:19,000 align:middle line:84%
that allows you to work this out in the few seconds
199
00:13:19,000 --> 00:13:29,000 align:middle line:84%
here’s another example, we have subnets 172.16.64.0/24 up to 172.16.127.0/24
200
00:13:29,000 --> 00:13:36,000 align:middle line:84%
can these subnets be summarized into a single subnet or fewer subnets?
201
00:13:36,000 --> 00:13:40,000 align:middle line:84%
so using the same process we start from the left-hand side
202
00:13:40,000 --> 00:13:42,000 align:middle line:84%
and we look for bits that are in common
203
00:13:42,000 --> 00:13:47,000 align:middle line:84%
in the first octet, we have 172 throughout all of those subnets
204
00:13:47,000 --> 00:13:51,000 align:middle line:84%
so that’s common, in the second octet we have 16
205
00:13:51,000 --> 00:13:53,000 align:middle line:84%
so that’s common throughout
206
00:13:53,000 --> 00:13:58,000 align:middle line:84%
So we know that the first 2 octets 172.16 are common throughout
207
00:13:58,000 --> 00:14:01,000 align:middle line:84%
but in the third octet the values are changing
208
00:14:01,000 --> 00:14:06,000 align:middle line:84%
we’ve got 64, 65, 66, 67 all the way up to 127
209
00:14:06,000 --> 00:14:10,000 align:middle line:84%
so that third octet we're going to convert to binary
210
00:14:10,000 --> 00:14:13,000 align:middle line:84%
to be able to better see what’s in common
211
00:14:13,000 --> 00:14:20,000 align:middle line:84%
so converting 64 into binary will give you 00100 0000
212
00:14:20,000 --> 00:14:26,000 align:middle line:84%
65 will look as follows, 66 as this, 67 as that
213
00:14:26,000 --> 00:14:30,000 align:middle line:84%
and you could convert a third octet of all of the subnets until you get a 127
214
00:14:30,000 --> 00:14:32,000 align:middle line:84%
which looks as follows
215
00:14:32,000 --> 00:14:38,000 align:middle line:84%
the last octet is a 0 in decimal which looks as follows in binary
216
00:14:38,000 --> 00:14:40,000 align:middle line:84%
so now what we need to do is we need to look for the common bits
217
00:14:40,000 --> 00:14:44,000 align:middle line:84%
so the common bits are once again 172.16
218
00:14:44,000 --> 00:14:50,000 align:middle line:84%
and then the first binary 0 is in common throughout the subnets
219
00:14:50,000 --> 00:14:55,000 align:middle line:84%
the second binary bit which is set to 1 is common throughout
220
00:14:55,000 --> 00:15:00,000 align:middle line:84%
but the third binary bit is not common throughout notice there’s 0
221
00:15:00,000 --> 00:15:02,000 align:middle line:84%
and then here there’s a binary 1
222
00:15:02,000 --> 00:15:05,000 align:middle line:84%
so we can draw a line after the second binary bit
223
00:15:05,000 --> 00:15:09,000 align:middle line:84%
to denote that everything to the left of the line is in common
224
00:15:09,000 --> 00:15:12,000 align:middle line:84%
and everything to the right of the line is not in common
225
00:15:12,000 --> 00:15:16,000 align:middle line:84%
so all of this is in common and all of this is not in common.
226
00:15:16,000 --> 00:15:23,000 align:middle line:84%
therefore 172.16 in decimal, the first 2 octets are in common
227
00:15:23,000 --> 00:15:26,000 align:middle line:84%
and the first 2 binary bits of the third octet are in common
228
00:15:26,000 --> 00:15:30,000 align:middle line:84%
the remaining binary bits need to be set to 0's
229
00:15:30,000 --> 00:15:34,000 align:middle line:84%
so filling the remaining binary bits is to 0 will look as follows
230
00:15:34,000 --> 00:15:36,000 align:middle line:84%
and then converting the binary back to decimal
231
00:15:36,000 --> 00:15:41,000 align:middle line:84%
will get 172 in the first octet, 16 in the second octet
232
00:15:41,000 --> 00:15:47,000 align:middle line:84%
this is 64 in decimal, so the third octet is 64
233
00:15:47,000 --> 00:15:52,000 align:middle line:84%
and the last octet binary 0's is equal to 0 in decimal
234
00:15:52,000 --> 00:16:01,000 align:middle line:84%
so the address is 172.16.64.0 that is the summary network for all of these subnets.
235
00:16:01,000 --> 00:16:04,000 align:middle line:84%
the last step is to count the number of bits in common
236
00:16:04,000 --> 00:16:08,000 align:middle line:84%
the first octet is 8 bits, the second octet is 8 bits
237
00:16:08,000 --> 00:16:12,000 align:middle line:84%
so that’s 16 bits in total followed by another 2 binary bits
238
00:16:12,000 --> 00:16:15,000 align:middle line:84%
which gives you 18 bits in common
239
00:16:15,000 --> 00:16:19,000 align:middle line:84%
so the first 18 bits are in common throughout all of this subnets
240
00:16:19,000 --> 00:16:25,000 align:middle line:84%
so our final answer will be 172.16.64.0/18
241
00:16:25,000 --> 00:16:30,000 align:middle line:84%
that is the summary network for all of the listed subnets.
28656
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