Routing

A router making a choice

Static Routes

“For packets with destinations in this network, send them to this router.”

!
ip route   192.168.100.0/24      10.10.10.5
!                ^ destination        ^ router
!

• Receive a packet (dst: 192.168.100.8)
• Is there a matching entry in the routing table? (show ip route, or show ip route 192.168.100.8)
• Yes! 192.168.100.0/24 via 10.10.10.5
• Forward that packet to the MAC address of 10.10.10.5
• .. but how did the router decide “Yes!”?

Binary logic briefing

• Binary operators include AND, OR, NAND, NOR, XOR…
• Use them on binary values
• Examples with 1 bit:
  • 1 AND 1 == 1
  • 1 AND 0 == 0
  • 0 AND 0 == 0
  • 1 OR 1 == 1
  • 1 OR 0 == 1
  • 0 OR 0 == 0
• Two bits:
  • 11 AND 10 == 10
  • 01 AND 00 == 00
  • 11 OR 10 == 11
  • 01 OR 00 == 01

More binary logic examples

• Three bits:
  • 101 AND 110 == 100 (5 AND 6 == 4)
  • 101 AND 101 == 101 (5 AND 5 == 5)
  • 001 AND 101 == 001 (1 AND 5 == 1)
  • 101 OR 110 == 111 (5 OR 6 == 7)
  • 101 OR 101 == 101 (5 OR 5 == 5)
  • 001 OR 101 == 101 (1 OR 5 == 5)
• Four bits:
  • 1111 AND 1110 == 1110 (15 AND 14 == 14)
  • 1101 AND 0001 == 0001 (13 AND 1 == 1)
  • 1111 OR 1110 == 1111 (15 OR 14 == 15)
  • 1101 OR 0001 == 1101 (13 OR 1 == 13)
• .. but we really only care about AND at the moment.

ANDing for Quantisation with 8 bits

• Watch how ANDing these values with the value 252 (11111100 in binary) quantises to multiples of 4:
  • 00000000 AND 11111100 == 00000000 (0 AND 252 == 0)
  • 00000001 AND 11111100 == 00000000 (1 AND 252 == 0)
  • 00000010 AND 11111100 == 00000000 (2 AND 252 == 0)
  • 00000011 AND 11111100 == 00000000 (3 AND 252 == 0)
• 0 through 3 when ANDed with 252 all came out as 0.
  • 00000100 AND 11111100 == 00000100 (4 AND 252 == 4)
  • 00000101 AND 11111100 == 00000100 (5 AND 252 == 4)
  • 00000110 AND 11111100 == 00000100 (6 AND 252 == 4)
  • 00000111 AND 11111100 == 00000100 (7 AND 252 == 4)
• 4 through 7, when ANDed with 252, all came out as 4!

More ANDing for fun and quantisation

• Further..
  • 00001000 AND 11111100 == 00001000 (8 AND 252 == 8)
  • 00001001 AND 11111100 == 00001000 (9 AND 252 == 8)
  • 00001010 AND 11111100 == 00001000 (10 AND 252 == 8)
  • 00001011 AND 11111100 == 00001000 (11 AND 252 == 8)
• And these all resulted in 8 ^
• The numbers are being rounded down to the nearest multiple of 4.
  • (side note mega tip: 4 is also the value of the right-most “on” bit in the network mask…)
• This works all the way up to 11111111 (255)
  • 11111100 AND 11111100 == 11111100 (252 AND 252 == 252)
  • 11111101 AND 11111100 == 11111100 (253 AND 252 == 252)
  • 11111110 AND 11111100 == 11111100 (254 AND 252 == 252)
  • 11111111 AND 11111100 == 11111100 (255 AND 252 == 252)

ANDing and IPv4

• You can also do this with a whole IP address!
• It’s just 32 bits instead of 8
• the resulting quantised value is called the Network Address.
  • 11000000.10101000.00000000.00000101 (192.168.0.5)
  • 11111111.11111111.11111111.11111100 (255.255.255.252) (aka /30)
  • AND
  • 11000000.10101000.00000000.00000100 (192.168.0.4)
  • ==> the network address of 192.168.0.5/30 is 192.168.0.4
  • ==> 192.168.0.5 is inside the 192.168.0.4/30 network

HOWTO select a route: 192.168.100.1

A packet arrives for forwarding with destination 192.168.100.1. Which route will be chosen?

r7#show ip route
...
Gateway of last resort is not set

 C        10.0.0.0/24 is directly connected, Ethernet1
 S        192.168.100.0/32 [1/0] via 10.0.0.5, Ethernet1
 S        192.168.100.0/30 [1/0] via 10.0.0.4, Ethernet1
 S        192.168.100.0/27 [1/0] via 10.0.0.3, Ethernet1
 S        192.168.100.0/24 [1/0] via 10.0.0.2, Ethernet1

r7#

• The most specific entry will take priority.
• This is also called the longest match.
• WAHT DOES IT MEAN

Static route 192.168.100.0/32

• Our destination address, converted to binary:

•   packet's dst:     192  .   168  .   100  .   1
                   11000000.10101000.01100100.00000001

• The first static route is 192.168.100.0 with a 32 bit netmask (255.255.255.255). Does it match?

• 
   S        192.168.100.0/32 [1/0] via 10.0.0.5, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.1  =>  11000000.10101000.01100100.00000001     |
                        bitwise AND to find network          |
            /32  =>  11111111.11111111.11111111.11111111     |
                                   equals                    |
                     11000000.10101000.01100100.00000001   <-+

• The result of ANDing the destination IP with the /32 netmask is different to the network of this static route.

• Therefore, the destination IP 192.168.100.1 is not inside the 192.168.100.0/32 network.

Static route 192.168.100.0/30

• 30 bit netmask (255.255.255.252):

•  S        192.168.100.0/30 [1/0] via 10.0.0.4, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.1  =>  11000000.10101000.01100100.00000001     |
                        bitwise AND to find network          |
            /30  =>  11111111.11111111.11111111.11111100     |
                                   equals                    |
                     11000000.10101000.01100100.00000000   <-+

• The result of ANDing the destination IP with the /30 netmask is the same as the network of the static route.
• This route is a valid candidate!
• The LENGTH of this match is 30 bits.

Static route 192.168.100.0/27

• 27 bit netmask (255.255.255.224):

•  S        192.168.100.0/27 [1/0] via 10.0.0.3, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.1  =>  11000000.10101000.01100100.00000001     |
                        bitwise AND to find network          |
            /27  =>  11111111.11111111.11111111.11100000     |
                                   equals                    |
                     11000000.10101000.01100100.00000000   <-+

• The result of ANDing the destination IP with the /27 netmask is the same as the network of the static route.
• This route is a valid candidate!
• The LENGTH of this match is 27 bits.

Static route 192.168.100.0/24

24 bits of netmask (255.255.255.0):

•  S        192.168.100.0/24 [1/0] via 10.0.0.2, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.1  =>  11000000.10101000.01100100.00000001     |
                        bitwise AND to find network          |
            /24  =>  11111111.11111111.11111111.00000000     |
                                   equals                    |
                     11000000.10101000.01100100.00000000   <-+

• The result of ANDing the destination IP with the /24 netmask is the same as the network of the static route.
• This route is a valid candidate!
• The LENGTH of this match is 24 bits.

The route is chosen.

• The three candidate matches for the destination 192.168.100.1 were:
  • 192.168.100.0/30 (30-bit match)
  • 192.168.100.0/27 (27-bit match)
  • 192.168.100.0/24 (24-bit match)
• longest match wins, so,
• S 192.168.100.0/30 [1/0] via 10.0.0.4, Ethernet1 is the chosen route

HOWTO select a route: 192.168.100.25

A packet arrives for forwarding with destination 192.168.100.25. Which route will be chosen?

r7#show ip route
...
Gateway of last resort is not set

 C        10.0.0.0/24 is directly connected, Ethernet1
 S        192.168.100.0/32 [1/0] via 10.0.0.5, Ethernet1
 S        192.168.100.0/30 [1/0] via 10.0.0.4, Ethernet1
 S        192.168.100.0/27 [1/0] via 10.0.0.3, Ethernet1
 S        192.168.100.0/24 [1/0] via 10.0.0.2, Ethernet1

r7#

• The most specific entry will take priority.
• AKA the longest match.

Static route 192.168.100.0/32

• Our destination address, converted to binary:

•   packet's dst:     192  .   168  .   100  .   25
                   11000000.10101000.01100100.00011001

• The first static route is 192.168.100.0 with a 32 bit netmask (255.255.255.255). Does it match?

• 
   S        192.168.100.0/32 [1/0] via 10.0.0.5, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.25 =>  11000000.10101000.01100100.00011001     |
                        bitwise AND to find network          |
            /32  =>  11111111.11111111.11111111.11111111     |
                                   equals                    |
                     11000000.10101000.01100100.00011001   <-+

• The result of ANDing the destination IP with the /32 netmask is different to the network of this static route.

• Therefore, the destination IP 192.168.100.25 is not inside the 192.168.100.0/32 network.

Static route 192.168.100.0/30

• 30 bit netmask (255.255.255.252):

•  S        192.168.100.0/30 [1/0] via 10.0.0.4, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.25 =>  11000000.10101000.01100100.00011001     |
                        bitwise AND to find network          |
            /30  =>  11111111.11111111.11111111.11111100     |
                                   equals                    |
                     11000000.10101000.01100100.00011000   <-+

• The result of ANDing the destination IP with the /30 netmask is different to the network of the static route.
• Therefore, the destination IP 192.168.100.25 is not inside the 192.168.100.0/30 network.

Static route 192.168.100.0/27

• 27 bit netmask (255.255.255.224):

•  S        192.168.100.0/27 [1/0] via 10.0.0.3, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.25 =>  11000000.10101000.01100100.00011001     |
                        bitwise AND to find network          |
            /27  =>  11111111.11111111.11111111.11100000     |
                                   equals                    |
                     11000000.10101000.01100100.00000000   <-+

• The result of ANDing the destination IP with the /27 netmask is the same as the network of the static route.
• This route is a valid candidate!
• The LENGTH of this match is 27 bits.

Static route 192.168.100.0/24

24 bits of netmask (255.255.255.0):

•  S        192.168.100.0/24 [1/0] via 10.0.0.2, Ethernet1
  
  192.168.100.0  =>  11000000.10101000.01100100.00000000   <-+
                                                             |
  192.168.100.25 =>  11000000.10101000.01100100.00011001     |
                        bitwise AND to find network          |
            /24  =>  11111111.11111111.11111111.00000000     |
                                   equals                    |
                     11000000.10101000.01100100.00000000   <-+

• The result of ANDing the destination IP with the /24 netmask is the same as the network of the static route.
• This route is a valid candidate!
• The LENGTH of this match is 24 bits.

The route is chosen.

• The two candidate matches for the destination 192.168.100.25 were:
  • 192.168.100.0/27 (27-bit match)
  • 192.168.100.0/24 (24-bit match)
• longest match wins, so,
• S 192.168.100.0/27 [1/0] via 10.0.0.3, Ethernet1 is the chosen route

Summary

• the router uses the network mask to decide which network a given IP address is in
• … by ANDing the IP address with the network mask
• the result is the network address.
• the router selects a route from the routing table to send packets
• it matches packets to routes by checking …
• … if the destination IP of the packet
• … is within the network defined by the route.

Questions!