Subnetting supernetting

Networking

1. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Subnetting/Supernetting
and
Classless Addressing

2. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
CONTENTS
•SUBNETTING
•SUPERNETTING
•CLASSLESS ADDRSSING

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SUBNETTING
5.1

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IP addresses are designed with
two levels of hierarchy.

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Figure 5-1
A network with two levels of
hierarchy (not subnetted)

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Figure 5-2
A network with three levels of
hierarchy (subnetted)

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Figure 5-3
Addresses in a network with
and without subnetting

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Figure 5-4
Hierarchy concept in a telephone number

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Figure 5-5
Default mask and subnet mask

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Finding the Subnet Address
Given an IP address, we can find the
subnet address by applying the mask to the
address. We can do this in two ways:
straight or short-cut.

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Straight Method
In the straight method, we use binary
notation for both the address and the
mask and then apply the AND operation
to find the subnet address.

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Example 1
Whatisthesubnetworkaddressifthe
destinationaddressis200.45.34.56andthe
subnetmaskis255.255.240.0?

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Solution
11001000 00101101 00100010 00111000
11111111 11111111 1111000000000000
11001000 00101101 0010000000000000
The subnetwork address is 200.45.32.0.

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Short-Cut Method
**If the byte in the mask is 255, copy
the byte in the address.
**If the byte in the mask is 0, replace
the byte in the address with 0.
**If the byte in the mask is neither 255
nor 0, we write the mask and the address
in binary and apply the AND operation.

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Example 2
Whatisthesubnetworkaddressifthe
destinationaddressis19.30.80.5andthe
maskis255.255.192.0?
Solution
See Figure 5.6

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Figure 5-6
Example 2

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Figure 5-7
Comparison of a default mask and
a subnet mask

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The number of subnets must be
a power of 2.

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Example 3
Acompanyisgrantedthesiteaddress
201.70.64.0(classC).Thecompanyneeds
sixsubnets.Designthesubnets.
Solution
Thenumberof1sinthedefault
maskis24(classC).

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Solution (Continued)
Thecompanyneedssixsubnets.Thisnumber
6isnotapowerof2.Thenextnumberthatis
apowerof2is8(2
3
).Weneed3more1sin
thesubnetmask.Thetotalnumberof1sin
thesubnetmaskis27(24+3).
Thetotalnumberof0sis5(32-27).The
maskis

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Solution (Continued)
11111111 11111111 1111111111100000
or
255.255.255.224
Thenumberofsubnetsis8.
Thenumberofaddressesineachsubnet
is2
5
(5isthenumberof0s)or32.
SeeFigure5.8

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Figure 5-8
Example 3

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Example 4
Acompanyisgrantedthesiteaddress
181.56.0.0(classB).Thecompanyneeds
1000subnets.Designthesubnets.
Solution
Thenumberof1sinthedefaultmaskis16
(classB).

24. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Solution (Continued)
Thecompanyneeds1000subnets.This
numberisnotapowerof2.Thenextnumber
thatisapowerof2is1024(2
10
).Weneed10
more1sinthesubnetmask.
Thetotalnumberof1sinthesubnetmaskis
26(16+10).
Thetotalnumberof0sis6(32-26).

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Solution (Continued)
Themaskis
11111111111111111111111111000000
or
255.255.255.192.
Thenumberofsubnetsis1024.
Thenumberofaddressesineachsubnetis2
6
(6isthenumberof0s)or64.
SeeFigure5.9

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Figure 5-9
Example 4

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Figure 5-10
Variable-length subnetting

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SUPERNETTING
5.2

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Figure 5-11
A supernetwork

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Rules:
**The number of blocks must be a power of 2 (1,
2, 4, 8, 16, ...).
**The blocks must be contiguous in the address
space (no gaps between the blocks).
**The third byte of the first address in the
superblock must be evenly divisible by the number
of blocks. In other words, if the number of blocks is
N, the third byte must be divisible by N.

31. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Example 5
Acompanyneeds600addresses.Whichof
thefollowingsetofclassCblockscanbe
usedtoformasupernetforthiscompany?
198.47.32.0198.47.33.0198.47.34.0
198.47.32.0198.47.42.0198.47.52.0198.47.62.0
198.47.31.0198.47.32.0198.47.33.0198.47.52.0
198.47.32.0198.47.33.0198.47.34.0198.47.35.0

32. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Solution
1:No,thereareonlythreeblocks.
2:No,theblocksarenotcontiguous.
3:No,31inthefirstblockisnotdivisibleby4.
4:Yes,allthreerequirementsarefulfilled.

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In subnetting,
we need the first address of the
subnet and the subnet mask to
define the range of addresses.

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In supernetting,
we need the first address of
the supernet
and the supernet mask to
define the range of addresses.

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Figure 5-12
Comparison of subnet, default,
and supernet masks

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Example 6
Weneedtomakeasupernetworkoutof16
classCblocks.Whatisthesupernetmask?
Solution
Weneed16blocks.For16blocksweneedto
changefour1sto0sinthedefaultmask.Sothe
maskis
11111111 11111111 1111000000000000
or
255.255.240.0

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Example 7
Asupernethasafirstaddressof205.16.32.0anda
supernetmaskof255.255.248.0.Arouterreceivesthree
packetswiththefollowingdestinationaddresses:
205.16.37.44
205.16.42.56
205.17.33.76
Whichpacketbelongstothesupernet?

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Solution
Weapplythesupernetmasktoseeifwecanfind
thebeginningaddress.
205.16.37.44AND255.255.248.0205.16.32.0
205.16.42.56AND255.255.248.0205.16.40.0
205.17.33.76AND255.255.248.0205.17.32.0
Onlythefirstaddressbelongstothissupernet.

39. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Example 8
Asupernethasafirstaddressof205.16.32.0anda
supernetmaskof255.255.248.0.Howmanyblocksarein
thissupernetandwhatistherangeofaddresses?
Solution
Thesupernethas211s.Thedefaultmaskhas24
1s.Sincethedifferenceis3,thereare2
3
or8
blocksinthissupernet.Theblocksare205.16.32.0
to205.16.39.0.Thefirstaddressis205.16.32.0.
Thelastaddressis205.16.39.255.

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CLASSLESS
ADDRESSING
5.3

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Figure 5-13
Variable-length blocks

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Number of Addresses in a Block
There is only one condition on the number
of addresses in a block; it must be a power
of 2 (2, 4, 8,...). A household may be given
a block of 2 addresses. A small business
may be given 16 addresses. A large
organization may be given 1024 addresses.

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Beginning Address
The beginning address must be evenly divisible
by the number of addresses. For example, if a
block contains 4 addresses, the beginning
address must be divisible by 4. If the block has
less than 256 addresses, we need to check only
the rightmost byte. If it has less than 65,536
addresses, we need to check only the two
rightmost bytes, and so on.

44. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Example 9
Whichofthefollowingcanbethebeginningaddressofa
blockthatcontains16addresses?
205.16.37.32
190.16.42.44
17.17.33.80
123.45.24.52
Solution
Theaddress205.16.37.32iseligiblebecause32is
divisibleby16.Theaddress17.17.33.80iseligible
because80isdivisibleby16.

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Example 10
Whichofthefollowingcanbethebeginningaddressofa
blockthatcontains1024addresses?
205.16.37.32
190.16.42.0
17.17.32.0
123.45.24.52
Solution
Tobedivisibleby1024,therightmostbyteofan
addressshouldbe0andthesecondrightmostbyte
mustbedivisibleby4.Onlytheaddress17.17.32.0
meetsthiscondition.

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Figure 5-14
Slash notation

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Slash notation is also called
CIDR
notation.

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Example 11
Asmallorganizationisgivenablockwiththebeginning
addressandtheprefixlength205.16.37.24/29(inslash
notation).Whatistherangeoftheblock?
Solution
Thebeginningaddressis205.16.37.24.Tofindthe
lastaddresswekeepthefirst29bitsandchangethe
last3bitsto1s.
Beginning:11001111 00010000 00100101 00011000
Ending : 11001111 00010000 00100101 00011111
There are only 8 addresses in this block.

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Example 12
WecanfindtherangeofaddressesinExample11by
anothermethod.Wecanarguethatthelengthofthesuffix
is32-29or3.Sothereare2
3
=8addressesinthisblock.
Ifthefirstaddressis205.16.37.24,thelastaddressis
205.16.37.31(24+7=31).

50. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
A block in classes A, B, and C
can easily be represented in slash
notation as
A.B.C.D/ n
where nis
either 8 (class A), 16 (class B), or
24 (class C).

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Example 13
Whatisthenetworkaddressifoneoftheaddressesis
167.199.170.82/27?
Solution
Theprefixlengthis27,whichmeansthatwemust
keepthefirst27bitsasisandchangetheremaining
bits(5)to0s.The5bitsaffectonlythelastbyte.
Thelastbyteis01010010.Changingthelast5bits
to0s,weget01000000or64.Thenetworkaddress
is167.199.170.64/27.

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Example 14
Anorganizationisgrantedtheblock130.34.12.64/26.
Theorganizationneedstohavefoursubnets.Whatarethe
subnetaddressesandtherangeofaddressesforeach
subnet?
Solution
Thesuffixlengthis6.Thismeansthetotalnumber
ofaddressesintheblockis64(2
6
).Ifwecreate
foursubnets,eachsubnetwillhave16addresses.

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Solution (Continued)
Letusfirstfindthesubnetprefix(subnetmask).
Weneedfoursubnets,whichmeansweneedtoadd
twomore1stothesiteprefix.Thesubnetprefixis
then/28.
Subnet1:130.34.12.64/28to130.34.12.79/28.
Subnet2:130.34.12.80/28to130.34.12.95/28.
Subnet3:130.34.12.96/28to130.34.12.111/28.
Subnet4:130.34.12.112/28to130.34.12.127/28.
SeeFigure5.15

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Figure 5-15
Example 14

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Example 15
AnISPisgrantedablockofaddressesstartingwith
190.100.0.0/16.TheISPneedstodistributethese
addressestothreegroupsofcustomersasfollows:
1.Thefirstgrouphas64customers;eachneeds256addresses.
2.Thesecondgrouphas128customers;eachneeds128addresses.
3.Thethirdgrouphas128customers;eachneeds64addresses.
Designthesubblocksandgivetheslashnotationforeach
subblock.Findouthowmanyaddressesarestillavailable
aftertheseallocations.

56. McGraw-Hill ©The McGraw-Hill Companies, Inc., 2000
Solution
Group1
Forthisgroup,eachcustomerneeds256addresses.
Thismeansthesuffixlengthis8(2
8
=256).The
prefixlengthisthen32-8=24.
01:190.100.0.0/24190.100.0.255/24
02:190.100.1.0/24190.100.1.255/24
………………………………… ..
64:190.100.63.0/24190.100.63.255/24
Total=64256=16,384

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Solution (Continued)
Group2
Forthisgroup,eachcustomerneeds128addresses.
Thismeansthesuffixlengthis7(2
7
=128).The
prefixlengthisthen32-7=25.Theaddresses
are:
001:190.100.64.0/25190.100.64.127/25
002:190.100.64.128/25190.100.64.255/25
003:190.100.127.128/25190.100.127.255/25
Total=128128=16,384

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Solution (Continued)
Group3
Forthisgroup,eachcustomerneeds64addresses.
Thismeansthesuffixlengthis6(2
6
=64).The
prefixlengthisthen32-6=26.
001:190.100.128.0/26190.100.128.63/26
002:190.100.128.64/26190.100.128.127/26
…………………………
128:190.100.159.192/26190.100.159.255/26
Total=12864=8,192

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Solution (Continued)
Numberofgrantedaddresses:65,536
Numberofallocatedaddresses:40,960
Numberofavailableaddresses:24,576