ip addressing
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333© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
IP Addressing
• Basic Addressing
• Working with Addresses
• Summarization & Subnets
• VLSM
• Working with VLSM Networks
• Classful Addressing
• Working with Classful Addressing
444© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1
• IP addresses are written in dotted decimal format.
• Four sections are separated by dots.
• Each section contains a number between 0 and 255.
Dots separate the sections
Each section contains a number between 0 and 255
555© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1
• Why is each section a number between 0 and 255?
• Computers operate in binary, humans operate in decimal.
• Computers treat IP addresses as a single large 32 digit binary number, but this is hard for people to do.
• So, we split them up into four smaller sections so we can remember and work with them better!
Dots separate the sections
Each section contains a number between 0 and 255
Why????
666© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1
• 32/4 == 8.
• 28 = 256.
• But, computers number starting at 0, so to make a space of 256 numbers, we number from 0 to 255.
00001010 00000001 00000001 000000018 8 8 8
32
Each 8 digit group represents a number between 0 and 255
777© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1• Each device on a network is
assigned an IP address.
• Each IP address has two fundamental parts:
• The network portion, which describes the physical wire the device is attached to.
• The host portion, which identifies the host on that wire.
• How can we tell the difference between the two sections?
00001010 00000001 00000001 00000001
Ne
two
rk
Ho
st
888© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1• The network mask shows us
where to split the network and host sections.
• Each place there is a 1 in the network mask, that binary digit belongs to the network portion of the address.
• Each place there is a 0 in the network mask, that binary digit belongs to the host portion of the address.
00001010 00000001 00000001 00000001
Ne
two
rk
Ho
st
255.255.255.0
11111111 11111111 11111111 00000000
999© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
10.1.1.1• An alternative set of
terminology is:
• The network portion of the address is called the prefix.
• The host portion of the address is called the host.
• The network mask is expressed as a prefix length, which is a count of the number of 1’s in the subnet mask.
00001010 00000001 00000001 00000001
Pre
fix
Ho
st
11111111 11111111 11111111 00000000
8 + 8 + 8 = 24
10.1.1.1/24
101010© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Basic Addressing
• The network address is the IP address with all 0’s in the host bits.
• The broadcast address is the IP address with all 1’s in the host bits.
• Packets sent to either address will be delivered to all the hosts connected to the wire.
10 1 1 0/2400001010 000000011 00000001 00000000
prefix host
these bits are 0, so this is the network address
10 1 1 255/2400001010 000000011 00000001 11111111
prefix host
these bits are 1, so this is the broadcast address
111111© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses
• Two of the most common questions you are going to face when dealing with IP addresses are:• What’s the network?
• What’s the host?
• How dow we figure this out?
192.168.100.80/26????
121212© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Hard Way)
• First, convert the IP address into binary. This is easier than it looks.
• Work with one octet at a time.
• Divide by two, farm out the remainder on the side.
• The bottom is the binary MSD, the top the binary LSD.
192
96 0
divide by 2
remainder
48 0
divide by 2
remainder
24 0
divide by 2
remainder
12 0
divide by 2
remainder
6 0
divide by 2
remainder
3 0
divide by 2
remainder
1 1
divide by 2
remainder
0 1
divide by 2
remainder Le
ftR
igh
t
131313© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Hard Way)
Write down the IP address.
11000000 10101000 01100100 01010000192 168 100 80
If you have a prefix length, just wrote down the number of 1’s. If you have a network mask, computer the binary as with the IP address.
11111111 11111111 11111111 110000008 +8 +8 +2 == 26
AND these two. 11000000 10101000 01100100 01000000
Convert back to dotted decimal. This is the network address.
192 168 100 64
141414© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Hard Way)
Write down the IP address.
11000000 10101000 01100100 01010000192 168 100 80
If you have a prefix length, just wrote down the number of 1’s. If you have a network mask, computer the binary as with the IP address.
11111111 11111111 11111111 110000008 +8 +8 +2 == 26
NOR these two. 00000000 00000000 00000000 00010000
Convert back to dotted decimal. This is the host address.
0 0 0 16
151515© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Hard Way)
• To convert from binary to decimal, use a simple chart.
• Add the number indicated for each 1 set in the binary number.
128 1 128
64 0 0
32 1 32
16 0 0
8 1 8
4 0 0
2 0 0
1 0 0
168
161616© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• First, if you are using a network mask, convert it to a prefix length.
• For each octet in the network mask that is 255, add 8 to the prefix length.
• For the one octet that isn’t 255, convert to binary and add the right number of bits--or use a chart!
255.255.255.1928 +8 +8 +2 == 26
192 == 11000000
171717© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• Take the prefix length and divide by 8.
• Take the resulting number, and ignore those octets out of the IP address--these are all part of the network address!
• We’re going to use the remainder to find the fourth octet of the network address.
26/8 == 3 (remainder 2)
192.168.100.80/26
These three octets are part of the network
The remainder tells us what the network address
in the fourth octet is
181818© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• Take the remainder, and find the corresponding “multiple” on the chart; in this case, 64.
• The largest multiple of 64 that will fit into 80 is 64, so the network is 64.
• Add the three octets we “set aside” earlier, and the network (prefix!) is 192.168.100.64/26.
• 80 - 64 == 16, so the host address is 16.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
64 x 1 == 6464 x 2 == 128
Remainder == 2
Network is 64!
80 - 64 == 16
192.168.100.64/26
16 Hosts!
191919© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• How many hosts are in this network? The remainder tells us there are 64 addresses, minus the network and broadcast addresses, so 62 hosts.
• To find the broadcast address, subtract 1 from the number of hosts, and add that number to the network address.
• The key is to work in octets, rather than trying to work with the entire IP address at once!
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
Remainder == 2
64 - 2 == 62 hosts
64 addresses
64 + (64 - 1) == 127
192.168.100.127 is the broadcast address
202020© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• What if the prefix length is less than 24?
• Take the prefix length and divide by 8.
• Take the resulting number, and ignore those octets out of the IP address--these are all part of the network address!
• We’re going to use the remainder to find the third octet of the network address.
22/8 == 2 (remainder 6)
192.168.100.80/22
These three octets are part of the network
The remainder tells us what the network address
in the third octet is
212121© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• Take the remainder, and find the corresponding “multiple” on the chart; in this case, 4.
• The largest multiple of 64 that will fit into 80 is 64, so the network is 64.
• Add the two octets we “set aside” earlier, and make any octets after the network 0’s (the fourth octet).
• The network (prefix!) is 192.168.100.0/22.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
4 x 25 == 1004 x 26 == 104
Remainder == 6
Third octet is 100!Set the fourth octet to 0.
192.168.100.0/22
222222© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• To find the number of hosts, take the number of octets set to 0, which is 1 in this case (the fourth octet), and multiply by 256.
• Next, take the number relating to the remainder from the chart, and multiple this by the number we just found above.
• Subtract two.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
4 x 256 == 10241024 – 2 == 1022 hosts
Remainder == 6
“0” octets == 11 x 256 == 256
232323© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Addresses (The Easy Way)
• The key is to work in octets, rather than trying to work with the entire IP address at once!
242424© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Summarization & Subnets
• A single network address (prefix!) represents a set of hosts attached to a wire.
• We can abstract this, and simply say that a prefix represents a set of reachable addresses.
• We can say that we’ve “summarized” information about the hosts attached to the physical wire by referring to the entire group as a single network.
10.1
.1.2
10.1
.1.4
10.1
.1.7
10.1
.1.8
10.1.1.0/26
252525© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Summarization & Subnets
• In effect, we’ve shortened the network part of the address (prefix!), and lengthened the host portion of the address, in effect describing more hosts (destinations) in a single address.
• If we can shorten the prefix length to describe multiple hosts with a single network address, why can’t we shorten the prefix length so a single network address describes two networks?
• We can! It’s called address summarization, or just summarization.
10.1.1.0/2610.1.1.64/26
10.1.1.2/3210.1.1.4/3210.1.1.7/3210.1.1.8/32
These host addresses are described by this network
10.1.1.0/25
These networks are described by this network
262626© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Summarization & Subnets
10.1.1.0 through 10.1.1.31.
00001010 00000001 00000001 0000000010 1 1 0
11111111 11111111 11111111 11000000
10.1.1.32 through 10.1.1.63.
00001010 00000001 00000001 0100000010 1 1 64
11111111 11111111 11111111 11000000
10.1.1.0 through 10.1.1.63, so it’s the same space!
00001010 00000001 00000001 0000000010 1 1 0
11111111 11111111 11111111 10000000
Changing the mask bit from 1 to 0, which shortens the prefix length, means the bit in the two networks that
distinguish them from one another are now considered host bits!
272727© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Summarization & Subnets
• A network which is a part of another network is called a subnet.
• There is another term, the supernet, but it’s definition depends on whether you are using VLSM subnetting, or calssful subnetting, so it will be defined in the next two sections.
10.1.1.0/2610.1.1.64/26
10.1.1.2/3210.1.1.4/3210.1.1.7/3210.1.1.8/32
These host addresses are subnets of this network
10.1.1.0/25
These networks are subnets of this network
282828© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
VLSM
• VLSM: Variable Length Subnet Masking
• It simply means that the entire IP address space is treated as one flat address space.
• Any prefix length is allowed in the network at any point.
10.1.1.0/2410.1.2.0/2510.1.2.128/2610.1.2.192/27
All of these are valid in the same network!
292929© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
VLSM
• At this point, you pretty much already know VLSM! You already know how to find the network address, broadcast address, and number of hosts in a network.
• Two other common problems in working with VLSM networks remain:• Building summary addresses from groups of networks.
We won’t cover this here (maybe later in routing).
• Building network addressing schemes from a given number of hosts and networks.
303030© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• You have 5 subnets with the following numbers of hosts on them: 58, 14, 29, 49, 3
• You are given the address space 10.1.1.0/24.
• Determine what subnets you could use to fit these hosts into it.
• How to solve this:
• Start with the chart!
• Order the networks from the largest to the smallest.
• Find the smallest number in the chart that will fit the number of the largest number of hosts + 2.
• Continue through each space needed until you either run out of space, or you finish.
313131© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• 58, 14, 29, 49, 3: reorder to 58, 49, 29, 14, 3. Start with 58.
• Smallest number larger than (58 + 2) is 64. 64 is 2 bits.
• 24 bits of prefix length in the address space given, add 2 for 26.
• First network is 10.1.1.0/26.
• The next network is 10.1.1.0 + 64, so we start the next “round” at 10.1.1.64.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
32 < (58 + 2) < 64
24 + 2 == 26
10.1.1.0/26 takes care of the first 58
hosts
Start the next block at 10.1.1.64
323232© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• Next block is 49 hosts.
• Smallest number larger than (49 + 2) is 64. 64 is 2 bits.
• 24 bits of prefix length in the address space given, add 2 for 26.
• We start this block at 10.1.1.64, so network is 10.1.1.64/26.
• The next network is 10.1.1.64 + 64, so we start the next “round” at 10.1.1.128.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
32 < (49 + 2) < 64
24 + 2 == 26
10.1.1.64/26 takes care of the next 49
hosts
Start the next block at 10.1.1.128
333333© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• Next block is 29 hosts.
• Smallest number larger than (29 + 2) is 32. 32 is 3 bits.
• 24 bits of prefix length in the address space given, add 3 for 27.
• We start this block at 10.1.1.128, so network is 10.1.1.128/27.
• The next network is 10.1.1.128 + 32, so we start the next “round” at 10.1.1.160.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
16 < (29 + 2) < 32
24 + 3 == 27
10.1.1.128/27 takes care of the
next 29 hosts
Start the next block at 10.1.1.160
343434© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• Next block is 14 hosts.
• Smallest number larger than (14 + 2) is 16. 16 is 4 bits (actually equal, but it still works!).
• 24 bits of prefix length in the address space given, add 4 for 28.
• We start this block at 10.1.1.160, so network is 10.1.1.160/27.
• The next network is 10.1.1.160 + 16, so we start the next “round” at 10.1.1.176.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
(14 + 2) == 16
24 + 4 == 28
10.1.1.160/28 takes care of the
next 14 hosts
Start the next block at 10.1.1.176
353535© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• Last block is 3 hosts.
• Smallest number larger than (3 + 2) is 8. 8 is 5 bits.
• 24 bits of prefix length in the address space given, add 5 for 29.
• We start this block at 10.1.1.176, so network is 10.1.1.176/29.
• This is the last block of hosts, so we’re done!
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
4 < (5 + 2) < 8
24 + 5 == 29
10.1.1.176/29 takes care of the
next 14 hosts
363636© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with VLSM Networks
• A subnet is any network which is “part of” a larger network space.
• A supernet is any network which covers a larger space than a given network, including the space covered by the network.
10.1.2.0/24
10.1.1.0/24
10.1.0.0/23
10.1.2.0/25
10.1.2.128/25
10.1.2.128/26
sub
net
s
sub
net
ssu
bn
et
sup
ern
etsu
per
net
sup
ern
et
373737© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Classful Addressing
• Classful subnetting is similar to VLSM, with two more rules:
• The IP address space is divided into “classes,” with each class having a specific “natural” prefix length. Each block of address space is called a “major net.”
• You cannot have more than one prefix length within a major net.
383838© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Classful Addressing
Network Class Beginning Digits in Binary
Natural Prefix Length
Range of Addresses
Example Major Networks
Class A 10XX 8 1.0.0.0/8 through 126.0.0.0/8
11.0.0.0/8100.0.0.0/8120.0.0.0/8
Class B 110X 16 128.0.0.0/16 through 191.0.0.0/16
130.1.0.0/16148.45.0.0/16190.100.0.0/16
Class C 1110 24 192.0.0.0/24 through 223.0.0.0/24
193.1.3.0/24193.1.4.0/24192.2.5.0/24
393939© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Classful Addressing
• It’s illegal to have multiple network masks within a single major network.
• There cannot be a mix of /24’s and /25’s in the 10.0.0.0/8 major network.
• There cannot be a mix of /25’s and /26’s in the 11.0.0.0/8 network.
10.1.1.0/2410.1.2.0/2410.1.3.0/2510.1.3.128/25
11.1.1.0/2511.1.1.128/26
two different prefix lengths in the same major network
404040© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• You can find the network address, broadcast address, and number of hosts as we described earlier.
• You can find the number of networks by subtracting the network mask from the natural mask, and then using the chart.
414141© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• 10.1.1.0/25 is in the 10.0.0.0 class A major network.
• The natural prefix length for a class A network is /8.
• Subtract the natural prefix length from the actual prefix length.
• Divide by 8, holding the remainder on the side.
10.1.1.0/2510.0.0.0/8 is class A
25 – 8 == 1717/8 == 2, 1 remaining
424242© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• Find the remainder in the power of two’s chart.
• Multiply the result, 256, and the number from the power of two’s chart.
• Subtract 2.
8 7 6 5 4 3 2 1
1 2 4 8 16 32 64 128
(256 x 2) x 128 == 65536
10.1.1.0/25
10.0.0.0/8 is class A
25 – 8 == 1717/8 == 2, 1 remaining
65536 – 2 == 65534 networks
434343© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• Subnet 0 • The network with
all the between the host and the natural major net set to 0.
• This only exists in classful addressing schemes.
10 0 0 0/2400001010 00000000 00000000 00000000
natural network
natural host
configured network these bits are 0, so this is subnet 0
10.0.0.0/16
10.0.1.0/16
172.31.0.0/24
172.31.1.0/24
192.168.100.0/25
Yes
No
Yes
No
Yes
444444© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• Broadcast Subnet• The network with
all the bits between the host and the natural major network set to 1.
• This only exists in calssful address schemes.
10 255 255 0/2400001010 11111111 11111111 00000000
natural network
natural host
configured network these bits are 1, so this is
the broadcast network
10.255.0.0/16
10.255.0.0/24
172.31.255.0/24
172.31.255.0/25
192.168.100.128/25
Yes
No
Yes
No
Yes
454545© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• You have 5 subnets with the following numbers of hosts on them: 58, 14, 29, 49, 3
• You are given the address space 10.1.0.0/22.
• Determine what subnets you could use to fit these hosts into it.
• How to solve this:
• Start with the chart!
• Find the largest set of hosts.
• Find the smallest number in the chart that will fit the number of the largest number of hosts + 2.
• Use that prefix length for all the subnets (remember you cannot have different subnet masks within the same major network).
464646© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Working with Classful Addressing
• A subnet is any prefix with a prefix length longer than the natural prefix length of the major network.
• A supernet is any prefix with a prefix length shorter than the natural prefix length of the major network.
172.18.1.0/24 Subnet
10.2.0.0/9 Subnet
172.34.0.0/15 Supernet
192.168.44.64/25 Subnet
192.168.44.0/23 Supernet
474747© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Private & Special Address Space
Address Space Range of Addresses
10.0.0.0/8 10.0.0.0 through 10.255.255.255
172.16.0.0/19 172.16.0.0 through 172.31.0.0
192.168.0.0/16 192.168.0.0 through 192.168.255.255
Network Class Beginning Digits in Binary Range of Addresses
Class D (Multicast)
11110x 224.0.0.0 through 239.255.255.255
Class E (Experimental)
11111x 240.0.0.0 through ....
484848© 2003, Cisco Systems, Inc. All rights reserved.RST-2002
Cisco IOS Show IP Route
2651A#sho ip route....
Gateway of last resort is not set
C 208.0.12.0/24 is directly connected, Serial0/2....S 208.1.10.0/24 [1/0] via 208.0.12.11.... 144.2.0.0/16 is variably subnetted, 2 subnets, 2 masksS 144.2.2.0/24 [1/0] via 208.0.12.11S 144.2.3.0/29 [1/0] via 208.0.12.11C 208.0.7.0/24 is directly connected, Serial0/0C 208.0.6.0/24 is directly connected, FastEthernet0/0C 208.0.0.0/24 is directly connected, FastEthernet0/1S 208.1.0.0/16 [1/0] via 208.0.12.11
two different prefix lengths under the
same major network
a supernet and natural mask in the same network address space
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