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Wednesday, October 31, 2012

IP Addressing


Acknowledgements
Note: This PowerPoint is based on version 2.0 of the Curriculum
        If I have not directly quoted Cisco Networking Academy material, then I have summarized it.  Therefore, the content of this PowerPoint Presentation is the exclusive property of Cisco Systems, Inc. and all rights that pertain to the actual curriculum apply. You may not copy, print, or otherwise use this material for any other purpose than viewing and taking notes. Other Cisco Certified Academy Instructors (CCAI) may use it for lecture preparations and classroom presentations in CNAP licensed classrooms only.
        In addition to the Cisco Networking Academy curriculum, I have relied heavily on Todd Lammle’s books and material. You can purchase his products at www.sybex.com.
        For those of you who have emailed me in the past with comments, questions, critiques, and criticism—Thank You!! I can be reached at allan1962@hotmail.com.                                 Created 12/2001
IP Addressing
*      IP Addressing is a logical addressing scheme at the Network Layer of the OSI Model.
*      Like all Network Layer addressing schemes (IPX, AppleTalk, DECnet, CLNS, etc.), IP addresses have two parts:
      Network—identifies the network or subnet
      Host—identifies the device on that network/subnet
*      An IP Address’ 32 bits are expressed in 4 octets (called dotted-decimal notation).
*      IP addresses are divided into five class types depending upon the value of bit positions in the first octet.
IP Address Classes
Class A: 1.0.0.0 to 127.0.0.0
Network
Host
Host
Host
1st Octet Bits:                      0          x         x          x           x           x
                                   (The 128 bit is off.)
Class B: 128.0.0.0 to 191.255.0.0
Network
Network
Host
Host
1st Octet Bits:                      1          0         x          x           x           x
                                 (The 128 bit is on and the 64 bit is off.)
  Class C: 192.0.0.0 to 223.255.255.0
Network
Network
Network
HOst
1st Octet Bits:                      1          1         0          x           x           x
                                (The 128 and 64 bits are on. The 32 bit is off.)
Reserved IP Address Classes
Multicasting
Class D: 224.0.0.0 to 239.0.0.0
1st Octet Bits:       1         1        1         0          x          x         x        x
                                  (The 128, 64, and 32 bit are on. The 16 bit is off.)
Experimental
Class E: 240.0.0.0 to 255.0.0.0
1st Octet Bits:       1         1        1         1          x          x         x        x
                                 (The 128, 64, 32, and 16 bit are all on.)
Private IP Addresses
*      Private IP Addresses cannot exist on the public Internet.
*      Your gateway router uses Name Address Translation (NAT) to give outbound packets a “legitimate” IP source address.
*      Private Addressing and NAT are discussed later.
Class A: 10.0.0.0
(Favored by large enterprises because of its flexibility)

Class B: 172.16.0.0 to 172.31.0.0
(In the 3rd Octet, the 128, 64, and 32 bit are off. The 16 bit is on.)

Class C: 192.168.0.0 to 192.168.255.0
(256 separate Class C Addresses)

Why Subnet?
*      Remember: we are usually dealing with a broadcast topology.
*      Can you imagine what the network traffic overhead would be like on a network with 254 hosts trying to discover each others MAC addresses?
*      Subnetting allows us to segment LANs into logical broadcast domains called subnets, thereby improving network performance.
Four Subnetting Steps
*      To correctly subnet a given network address into subnet addresses, ask yourself the following questions:
1.   How many bits do I need to borrow?
2.   What’s the subnet mask?
3.   What’s the “magic number” or multiplier?
4.   What are the first three subnetwork addresses?
*      Let’s look at each of these questions in detail
1. How many bits to borrow?
*      First, you need to know how many host bits you have to work with.
*      Second, you must know either how many subnets you need or how many hosts per subnet you need.
*      Finally, you need to figure out the number of bits to borrow.
How many host bits do I have to work with?
*      Depends on the class of your network address.
*      Class C: 8 host bits
*      Class B: 16 host bits
*      Class A: 24 host bits
*      Remember: you must borrow at least 2 bits for subnets and leave at least 2 bits for host addresses.
*      2 bits borrowed allows 22 - 2 = 2 subnets
*      Anyway, that’s how we learned it in our CCNA Curriculum. You will soon discover that subnet zero is actually available for your use.
How many subnets or hosts do I need?
*      A simple formula:
*      Host Bits = Bits Borrowed + Bits Left
*      HB = BB + BL
I need x subnets:                    2*BB      -         2>=   X
I need x hosts:                        2*BL      -         2>=   X
*      Remember: we need to subtract two hosts to provide for the subnetwork and broadcast addresses.

*      Class C Example: 210.93.45.0
      Design goals specify at least 5 subnets so how many bits do we borrow?
      How many bits in the host portion do we have to work with (HB)?
    Since it’s a Class C, we have 8 bits to work with.
      What’s the BB in our HB = BB + BL formula?
    8 = BB + BL
      2 to what power will give us at least 5 subnets?
    23 - 2 = 6 subnets
      How many bits are left for hosts?
    Since 8 = 3 + BL, then BL = 5
      So how many hosts can we assign to each subnet?
    25 - 2 = 30 hosts

*      Class B Example: 185.75.0.0
      Design goals specify no more than 126 hosts per subnet, so how many bits do we need to leave (BL)?
      How many bits in the host portion do we have to work with (HB)?
    Since it’s a Class B, we have 16 bits to work with.
      What’s the BL in our HB = BB + BL formula?
    16 = BB + BL
      2 to what power will give us 126 hosts per subnet?
    27 - 2 = 126 hosts
      How many bits are left for subnets?
    Since 16 = BB + 7, then BB = 9
      So how many subnets can we have?
    29 - 2 = 510 subnets
2. What’s the subnet mask?
*      We determine the subnet mask by adding up the decimal value of the bits we borrowed.
*      In the previous Class C example, we borrowed 3 bits. Below is the host octet showing the bits we borrowed and their decimal values.
       1       1         1          
      ---      ---      ---     ---      ---      ---      ---       ---
     128     64      32     16       8        4        2        1


 


We add up the decimal value of these bits and get 224.  That’s the last non-zero octet of our subnet mask.

So our subnet mask is 255.255.255.224
Remember: The subnet mask has all 1s in the network portion.
3. What’s the “magic number?”
*      To find the “magic number” or the multiplier we will use to determine the subnetwork addresses, we subtract the last non-zero octet from 256.
      Note: The “magic number” can also be found by determining the value of the last bit borrowed.
*      In our Class C example, our subnet mask was 255.255.255.224.  224 is our last non-zero octet.
*      Our magic number is 256 - 224 = 32
      Note: The last bit borrowed was the 32 bit.
Last Non-Zero Octet
*      Memorize this table.  You should be able to:
      Quickly calculate the last non-zero octet when given the number of bits borrowed or...
      Determine the number of bits borrowed when given the last non-zero octet
       Bit
    Borrowed
 Non Zone
  Octal
         1
      128
         2
      192
         3
      224
         4
      240
         5
      248
         6
      252
         7
      254
         8
      255
4. What are the subnets?
*      We now take our “magic number” and use it as a multiplier.
*      Our Class C address was 210.93.45.0.
*      We borrowed bits in the fourth octet, so that’s where our multiplier occurs.
      1st subnet:         210.93.45.32
      2nd subnet:        210.93.45.64
      3rd subnet:        210.93.45.96
      4th subnet: 210.93.45.128
      5th subnet: 210.93.45.160
      6th subnet: 210.93.45.192
Host & Broadcast Addresses
*      Now you can see why we subtract 2 when determining the number of host addresses.
      Let’s look at our 1st subnet: 210.93.45.32
      What is the total range of addresses up to our next subnet, 210.93.45.64?
    210.93.45.32 to 210.93.45.63 or 32 addresses
      .32 cannot be assigned to a host. Why?
    Because it is the subnet’s address.
      .63 cannot be assigned to a host. Why?
    Because it is the subnet’s broadcast address.
      So our host addresses are .33 - .62 or 30 host addresses--just like we figured out earlier.
Practice Your Subnetting!!
*      If you have not yet mastered subnetting, now is the time to do so.
      Semester 5’s curriculum assumes the ability to quickly subnet without pencil & paper! (much like the ability to add and subtract is assumed in Algebra)
      You will need to be able to evaluate an addressing scheme quickly just by looking at the address and subnet mask.
      Furthermore, Variable Length Subnet Masking (VLSM) becomes much easier if you’ve mastered subnetting.
      To practice, simply take any network address/design goal scenario and subnet it!! For example...
ü  192.168.1.0 with at least 30 subnets
ü  172.16.0.0 with at least 500 hosts per subnet
ü  10.0.0.0 with at least 2000 subnets

Depletion of IPv4
*      IP became ARPA’s protocol for host-to-host communications on January 1, 1982.
        “It is urgent that the implementation of IP/TCP be begun on all...ARPANET hosts as soon as possible and no later than 1 January 1982.” (RFC 801, p. 2)
*      The designers of IP could not foresee the explosive growth of the what they had come to call the Internet.
*      In 1981, they figured that a 32 bit address with more than 4 billion possible host addresses would never be exhausted.
*      However, ten years later they were scrambling to solve just that problem: address space depletion.
Solving the Depletion Crisis
*      In 1992, IETF had two main concerns:
      Class A is gone and Class B is almost gone
      Internet routing tables are huge!!
*      Therefore, over the next several years they came up with solutions:
      Route Summarization using CIDR Notation
      Variable Length Subnet Masking
      Private Addressing and NAT
      IP Unnumbered on WAN links
      IP version 6
*      VLSM will be discussed in the next section.
*      Private Addressing, IP Unnumbered, IPv6 will be discussed following VLSM.
CIDR Notation
*      Classless Interdomain Routing is a method of representing an IP address and its subnet mask with a network prefix and bitmask.
*      For example:  192.168.50.0/27
*      What do you think the 27 tells you?
      27 is the number of 1 bits in the subnet mask.  Therefore, 255.255.255.224
      Also, you know 192 is a Class C, so we borrowed 3 bits!! How do we know that?
   Default subnet mask for Class C is 255.255.255.0 or /24
      Finally, you know the magic number is 256 - 224 = 32, so the first useable subnet address is 192.168.50.32!!
*      Let’s see the power of CIDR notation.
202.151.37.0/26
*      Subnet mask?
      255.255.255.192
*      Bits borrowed?
      Class C so 2 bits borrowed
*      Magic Number?
      256 - 192 = 64
*      First useable subnet address?
      202.151.37.64
*      Third useable subnet address?
      64 + 64 + 64 = 192, so 202.151.37.192
198.53.67.0/30
*      Subnet mask?
      255.255.255.252
*      Bits borrowed?
      Class C so 6 bits borrowed
*      Magic Number?
      256 - 252 = 4
*      Third useable subnet address?
      4 + 4 + 4 = 12, so 198.53.67.12
*      Second subnet’s broadcast address?
      4 + 4 + 4 - 1 = 11, so 198.53.67.11
200.39.89.0/28
*      What kind of address is 200.39.89.0?
      Class C, so 4 bits borrowed
      Last non-zero octet is 240
      Magic number is 256 - 240 = 16
      32 is a multiple of 16 so 200.39.89.32 is a subnet address--the second subnet address!!
*      What’s the broadcast address of 200.39.89.32?
      32 + 16 -1 = 47, so 200.39.89.47
194.53.45.0/29
*      What kind of address is 194.53.45.26?
      Class C, so 5 bits borrowed
      Last non-zero octet is 248
      Magic number is 256 - 248 = 8
      Subnets are .8, .16, .24, .32, ect.
      So 194.53.45.26 belongs to the third subnet address (194.53.45.24) and is a host address.
*      What broadcast address would this host use to communicate with other devices on the same subnet?
      It belongs to .24 and the next is .32, so 1 less is .31 (194.53.45.31)
No Worksheet Needed!
*      After some practice, you should never need a subnetting worksheet again.
*      The only information you need is the IP address and the CIDR notation.
*      For example, the address 221.39.50.0/26
*      You can quickly determine that the first subnet address is 221.39.50.64.  How?
      Class C, 2 bits borrowed
      256 - 192 = 64, so 221.39.50.64
*      For the rest of the addresses, just do multiples of 64 (.64, .128, .192).
*      MEMORIZE THIS TABLE!!!



       Bit
    Borrowed
 Non Zone
  Octal
         1
      128
         2
      192
         3
      224
         4
      240
         5
      248
         6
      252
         7
      254
         8
      255
Practice On Your Own
*      Below are some practice problems.  Take out a sheet of paper and calculate...
      Bits borrowed
      Last non-zero octet
      Second subnet address and broadcast address
  1. 192.168.15.0/26
  2. 220.75.32.0/30
  3. 200.39.79.0/29
  4. 195.50.120.0/27
  5. 202.139.67.0/28
  6. Challenge: 132.59.0.0/19
  7. Challenge: 64.0.0.0/16

Route Summarization
*      Also known as Route Aggregation and Supernetting, Route Summarization is a method of representing multiple, contiguous subnets with one aggregated address.
*      Without route summarization, the routing tables of the Internet would’ve collapsed back in the mid 1990s.
      See a real routing table.
*      Route summarization benefits include...
      More efficient routing
      reduced CPU usage
      reduced memory requirements
Route Flapping
*      Route Flapping is the process of a route continuously going up and then down
      Can be caused by physical or data-link layer problems
*      Route Summarization effectively insulates upstream routers from continually recalculating their routing tables because of route flapping.
      The flapping network’s border router is summarizing and advertising all local networks as one route.
Route Summarization Example
*      Your enterprise has four Class C addresses:
      199.100.0.0/24
      199.100.1.0/24
      199.100.2.0/24
      199.100.3.0/24
*      Notice these addresses are contiguous.
*      With CIDR notation, we can represent all four addresses as 199.100.0.0/22. How?
*      Because all four addresses have the first 22 bits in common (called a prefix).
*      We can summarize these addresses because none of them have the 4 bit turned on in the 3rd octet.
*      Below is 199.100.0.0/22 worked out in binary.
*      Being able to work at the bit level is crucial when supernetting to summarize a range of addresses.
*      How does route summarization help reduce routing tables?
199.100.0.0
1100      0111
0110        0100
0000      00   00
0000     0000
199.100.1.0
1100      0111
0110        0100
0000      00   01
0000     0000
199.100.2.0
1100      0111
0110        0100
0000      00   10
0000     0000
199.100.3.0
1100      0111
0110        0100
0000      00   11
0000     0000
 Bitmask
1111      1111
1111        1111   
1111      11   00
0000     0000

Route Summarization Example
*      Your AS advertises a summarized route to your ISP.
*      The ISP, in turn, advertises a further summarized route to the Internet, thereby reducing the Internet’s routing table.

VLSM Overview
*      You may have noticed in your CCNA studies that addressing a WAN link is often a waste of host addresses.
*      VLSM allows you to subnet a subnet!
*      WAN links only need 2 addresses for hosts.
*      Therefore, using VLSM would yield a CIDR notation of /30 on WAN links.
*      In addition, with the ip subnet-zero command enabled by default on Cisco IOS 12.0 and higher, you can now use subnet zero.
Example
*      You have a small Class C network with 6 LANs & 30 hosts (192.168.1.0/27)
*      NO MORE ADDRESSES for WAN links!!
*      Solution: Use subnet zero and subnet it further:
      192.168.1.4/30
      192.168.1.8/30
      192.168.1.12/30
      192.168.1.16/30
      192.168.1.20/30
      192.168.1.24/30
      192.168.1.28/30
*      You now have enough addresses for 7 WAN links.
*      The graphic shows how you can have your 6 subnets with 30 hosts and still have subnets leftover for WAN links.
*      The hub router would then summarize all the subnets as 192.168.1.0/24
*      This simple demonstration of VLSM hides its true power... SCALABILITY!!
*      Let’s explore that power.

The Three-Layer Model
*      Remember our three layers from Ch. 1?
      Core, Distribution, and Access
*      With VLSM, route summarization and the appropriate routing protocol, we can scale our network making routing much more efficient.
*      Using the address 172.16.0.0, we could do the following, summarizing up to the Core Layer:
      All WAN links:
    172.16.0.4/30 through 172.16.0.248/30
      All Distribution routers and attached networks:
    172.16.1.0/24 through 172.16.255.0/24
      All Access routers and attached networks:
    172.16.1.32/27 through 172.16.255.32/27
VLSM & The Three Layers
VLSM Routing Protocols
*      Only the classless routing protocols shown in the table below support VLSM.
Classful
Classless
RIPv1
RIPv2
IGRP
EIGRP
EGP
OSPF
BGPv3
BGPv4


RIPv1 versus RIPv2
*      RIPv1...
      does not send subnet mask information
ü  the receiving router applies its subnet mask or the default
      broadcasts its updates
      does not support authentication
*      RIPv2...
      supports VLSM
      multicasts its updates
      supports authentication
      However, RIPv2 is still limited to 15 hops and only considers hops as its metric.
*      Configuring RIPv2...
      Router(config)#router rip
      Router(config-router)#version 2
VLSM Labs
*      This chapter has 3 labs for practicing VLSM.
*      Be sure you work them on your own before we do them together as a class
      Note: There are usually multiple correct solutions.
*      Mike Harris has developed an Excel spreadsheet tool to help you with VLSM. Download it here, add it to your Engineering Journal and copy it in your bound notebook.
      Mike’s spreadsheet is an excellent visual aid when designing a VLSM addressing scheme.
Private Addressing & NAT
*      As discussed earlier, private IP addresses cannot exist on the Internet.
*      Therefore, we use Name Address Translation (NAT) to dynamically give packets a real IP address.
      ISPs will only give you a limited number of real IP addresses (if any!). So NAT configuration also includes the ability to “overload” a real IP.
      The purpose of NAT overloading is to allow multiple local inside addresses to share a single global outside address.
      This is done by tracking source ports from the transport layer. As packets leave, not only do they get a real IP but are also tagged with a port number to identify the session (and host) as packets return from the destination.
      For more detail on NAT, review Semester 6’s Chapter 11 devoted to the subject. We will not configure NAT this semester.
IP Unnumbered
*      IP Unnumbered is used to conserve more space on WAN links.
      Serial interfaces “borrow” an IP address from another interface (typically a LAN interface)
*      Rules for using IP unnumbered:
      Only point-to-point serial interfaces
      Both sides must belong to the same major network with the same subnet mask or…
      Different major network with default subnet masks
*      Drawbacks to using IP unnumbered:
      Cannot ping the interface
      Cannot boot a network IOS image over interface
      Cannot use IP security
*      Configuring IP Unnumbered
Router(config)#interface s0
Router(config-if)#ip unnumbered e0



DHCP & Easy IP
*      Hosts configured to dynamically obtain their IP addresses will send a DHCP broadcast upon booting.
      The gateway router will respond either with an IP address or a DHCP router address.
*      Configuring DHCP (Be sure to do Interactive Lab 2.8.3)
Router(config)#ip dhcp excluded-address [address_range]
ü  Specifies a range of addresses to be excluded from the dhcp pool
Router(config)#ip dhcp pool [pool_name]
Router(dhcp-config)#network [network_address][subnet_mask]
ü  Defines the name of the dhcp pool and the address to be used to assign IPs
Router(dhcp-config)#default-router [router_address1]…[router_address8]
ü  Defines up to 8 routers from which the host can get IP addresses
*      Cisco’s Easy IP
      “Plug and Play” routing that allows a remote router to get a real IP address from the ISP
      Then the remote router uses DHCP/NAT to provide access to internal LAN clients.
Helper Addresses
*      DHCP uses BootP’s UDP port numbers 67 & 68 to broadcast for an IP addresses.
      Normally, routers will not forward UDP requests. This causes a problem if the local router is not the DHCP server.
      Therefore, we configure the host’s local router with a helper address to which it will forward UDP requests for services.
*      On the interface where hosts requesting services are located…
Router(config-if)#ip helper-address [server_address]
ü  Will forward the 8 UDP services below which includes DHCP
For UDP services not included in the 8, use the global command…
Router(config)#ip forward-protocol udp [port_number]
ü  For UDP services you want to exclude, use no in front of command
  UPD  Service Forwarded by Helper Command
           Service
           Port
              Service
Port
Time
             37
BOOTP/DHCP   client
68
TACAS
             49
TFTP
69
DNS
             53
NetBIOS  name
137
BOOTP/DHCPserver
             67
NetBIOS data gram
138
Internet Protocol, version 6
*      IPv4 will eventually perish even with…
      Subnetting (1985)
      VLSM (1987)
      CIDR (1993)
*      The proliferation of IP addressable devices will eventually exceed IPv4’s limit of 4 billion addresses.
*      IPv6 is a 128 bit address. But because of the success of NAT and private IPs, IPv4 will not go away for some time.
*      IPv6 will require network administrators to re-engineer their enterprises with new software and new hardware.
Expressing IPv6
*      IPv6 addresses are for interfaces and sets of interfaces, not nodes.
*      Its 128 bits are expressed in hexadecimal
      Leading zeros in each 16-bit value can be omitted
      16-bit values that are all zeros can be completely omitted and replaced with a double colon
ü  Fully expressed
     1080:0000:0000:0000:0008:200C:417A
ü  Omit leading zeros
     1080:0:0:0:8:800:200C:417A
ü  Omit 16-bit zeros
     1080::8:800:200C:417A
      Don’t yet know hex? It’s coming back to haunt you!!
Lab Notes
*      Lab 2.10.1: VLSM & IP Unnumbered
      You initially configure the network with VLSM and RIPv1 only to discover you do not have full connectivity
      Enter the command version 2 and you get convergence
      Also, you configure IP unnumbered and view the routing table
*      Lab 2.10.2: VLSM
      Three different VLSM scenarios you must solve by assigning every network an appropriate address for a limited pool of available addresses
*      Lab 2.10.2: DHCP & Helper Addresses
      Using two routers and two hosts, you get good practice at using DHCP to get your hosts’ an IP address
      You also use a helper address to allow a remote host to get an IP address



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