Chapter 5-Subnetting-Supernetting and Classless Addressing

(1)

Chapter 5

Chapter 5 Subnetting/Supernetting

Subnetting/Supernetting

and Classless Addressing

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(3)

Outline

o o SubnettingSubnetting o o SupernernettingSupernernetting o

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(5)

SUBNETTING

5.1

(6)

(7)

Subnetting

o

o IP addresses are designed with two level of IP addresses are designed with two level of

hierarchy hierarchy n

n Two levels of hierarchy is not Two levels of hierarchy is not enoughenough o

o Solution:Solution: subnetting subnetting

n

n A network is divided into several smaller A network is divided into several smaller networks

networks

n

n Each smaller network is called aEach smaller network is called a subnetw subnetwork ork or or aa subnet

(8)

(9)

I

P addr

e

s

e

s

ar

e

de

s

i

gn

e

d wi

th

II P addr

P addr e

es

ss

se

es

s ar

ar e

e de

des

siign

gne

ed wi

d wi th

th

two l

e

ve

l

s

of h

i

e

r

ar

chy.

two l

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(11)

Figure 5-1

A

N

e

twor

k

wi

th

T

wo

L

e

v

e

l

s

of

H

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(13)

Figure 5-2

A

N

e

twor

k

wi

th

T

h

r

e

L

e

ve

l

s

of

H

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(15)

Su

bn

et

ti

ng

(

Co

nt

.)

o

o The subnetwThe subnetworksorks still still appear appear as a sas a singleingle

network to the rest of the Internet network to the rest of the Internet

o

o For example, a packet destined for hostFor example, a packet destined for host

141.14.192.2 still reaches router R1 141.14.192.2 still reaches router R1

o

o However, R1 knows the network 141.14 isHowever, R1 knows the network 141.14 is

physically divided into subnetworks physically divided into subnetworks

n

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(17)

Three Levels of

Hierarchy

o

o Three levelThree level

n

n Site, subnet, and hostSite, subnet, and host o

o The routing of an IP datagram now involvesThe routing of an IP datagram now involves

three step three step n

n Delivery to theDelivery to the site site

n

n Delivery to theDelivery to the subnetw subnetwork ork

n

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(19)

Figure 5-3

A

ddr

e

s

e

s

i

n

a N

e

twor

k

wi

th

and w

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Subnet Mask

o

o The network mask create the network addressThe network mask create the network address

o

o The subnet mask create the subnetwork The subnet mask create the subnetwork

address address

o

o Subnet Mask Subnet Mask

n

n Nonconti Noncontiguous: a mguous: a mixture of 0s aixture of 0s and 1snd 1s

o

o Out-of-dayOut-of-day

n

n ContiguoContiguous: a run of 1s us: a run of 1s followed by a run of 0sfollowed by a run of 0s

o

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(23)

Figure 5-5

D

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(25)

Finding the Subnet Address

o

o Given an IP address, we can find theGiven an IP address, we can find the subnet subnet

address

address in the same way as we fin the same way as we found theound the network address

network address n

n Apply the mask to the addressApply the mask to the address

o

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(27)

Straight Method

o

o Use binary notation for both the address andUse binary notation for both the address and

the mask the mask

o

o Then apply the AND operation to find theThen apply the AND operation to find the

subnet address subnet address

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E

Example xample 1 1

E

Example xample 1 1

W

Wha

hat i

t is t

s th

he s

e sub

ubne

netw

twor

ork

k ad

add

drres

ess i

s if t

f th

hee

destination address is 200.45.34.56 and the

subnet mask is

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(31)

Solution Solution Solution Solution

11001000 00101101 00100010 00111000

11111111 11111111 1111

0

0 11001000 00101101 0010

11001000 00101101 0010

0000

00000000

The

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(33)

Short-Cut Method

o

o If the byte in the mask is 255, copy thIf the byte in the mask is 255, copy the byte ine byte in

the address the address

o

o If the byte in the mask is 0, repIf the byte in the mask is 0, replace the byte inlace the byte in

the address with 0 the address with 0

o

o If the byte in the mask is neither 2If the byte in the mask is neither 255 nor 0,55 nor 0,

we write the mask and the address in binary we write the mask and the address in binary and apply the AND operation

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(35)

E

Example xample 2 2

E

Example xample 2 2

Wh

What

at is

is th

the s

e sub

ubne

netw

twor

ork

k ad

addr

dres

ess i

s if t

f th

hee

destination address is 19.30.80.5 and the

mask is 255.255.192.0?

Solution Solution Solution Solution

See Next Figure.

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(37)

Figure 5-6

E

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(39)

Default Mask and Subnet Mask

o

o The number of 1sThe number of 1s in a default mask isin a default mask is

perdetermined perdetermined

n

n 8, 16, or 248, 16, or 24

o

o But, in a subnet mask, the number of 1s isBut, in a subnet mask, the number of 1s is

more than the number of 1s in the more than the number of 1s in the corresponding default mask

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Figure 5-7 Figure 5-7

Co

mp

ar

i

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o

n of

a

D

e

f

a

u

l

t M

as

k and

a S

u

bn

e

t M

as

k

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Number of Subnetworks

o

o Found by counting the number of extra bitsFound by counting the number of extra bits

that are added to the default

that are added to the default mask in a subnetmask in a subnet mask

mask

o

o For example, in above figureFor example, in above figure

n

n The number of extra 1s is 3The number of extra 1s is 3

o

o The lThe lengength of sth of subneubnetidtid = 3= 3

n

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(45)

T

h

e

n

u

mbe

r

of

s

u

bne

ts

mu

s

t be

T

Th

he

e n

nu

umbe

mber

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of s

su

ubn

bne

ets

ts mu

mus

st be

t be

a po

we

r

of

2.

2. a po

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(47)

Number of Addresses per Subnet

o

o Found by counting the number of Found by counting the number of 0s0s in thein the

subnet mask subnet mask

o

o For example, in above figureFor example, in above figure

n

n The number of 0s is 13The number of 0s is 13

o

o The The lenlength gth of of hosthostidid = = 1313

n

n The number of addressed in each subnet is 2^13 =The number of addressed in each subnet is 2^13 = 8192

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(49)

Designing Subnets

o

o How a network managers design subnetsHow a network managers design subnets

n

n Deciding the number of subnetsDeciding the number of subnets

n

n Finding the subnet mask Finding the subnet mask

n

n Find the range of address in each subnetFind the range of address in each subnet

o

o Start with the first subnet and its first address is theStart with the first subnet and its first address is the

first address in the block first address in the block

o

o Add the number of addresses in each subnet minusAdd the number of addresses in each subnet minus

one to get the last address one to get the last address

o

o Add one to the last address in obtain step to obtain theAdd one to the last address in obtain step to obtain the

first address in the next subnet first address in the next subnet

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E

Example xample 3 3

E

Example xample 3 3

A company is granted the site address

201.70.64.0 (class C). The company needs

six subnets. Design the subnets.

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(53)

o

o The number of 1s in the defThe number of 1s in the default mask is 24ault mask is 24

(class C).

o

o The company needs six subnets.The company needs six subnets.

n

n This number 6 is not a power of 2.This number 6 is not a power of 2.

n

n The next number that is a power of 2 is The next number that is a power of 2 is 8 (23)8 (23) o

o We need 3 more 1s in the subnet mask.We need 3 more 1s in the subnet mask.

n

n The total number of 1s in the subnet mask is 27The total number of 1s in the subnet mask is 27 (24 + 3).

(24 + 3). n

n The tThe total notal numbumber of 0s is 5 (er of 0s is 5 (32 -32 - 27).27).

Solution

Solution Solution

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(55)

S

Soluolutition on (Co(Contintinunueed) d)

S

Soluolutition on (Co(Continti nunueed) d)

.

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(57)

S

The mask is

1111 11111111111 1111111 11111111 111111 111111111111 1111110000000000 or or 255.255.255.224 255.255.255.224

The number of subnets is 8. The number of subnets is 8.

The number of addresses in each subnet is 2 The number of addresses in each subnet is 255 (5 is the number of 0s) or 32.

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(59)

Figure 5-8

E

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(61)

E

Example xample 4 4

E

Example xample 4 4

A company is granted the site address

181.56.0.0 (class B). The company needs

1000 subnets. Design the subnets.

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(63)

o

o The number of 1s in the defThe number of 1s in the default mask is 16ault mask is 16

(class B).

o

o The company needs 1000 subnets.The company needs 1000 subnets.

n

n This number is not a power of 2.This number is not a power of 2.

n

n The next number that is a power of 2 is 1024 (210).The next number that is a power of 2 is 1024 (210). o

o We need 10 more 1s in the subnet mask.We need 10 more 1s in the subnet mask.

n

n The total number of 1s in the subnet mask is 26 (16The total number of 1s in the subnet mask is 26 (16 + 10).

+ 10). n

n The tThe total notal numbumber of 0s is 6 (er of 0s is 6 (32 -32 - 26).26).

Solution

Solution Solution

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(65)

o

o The mask isThe mask is

11 1111111111111 1 1111111111111111 1111111111111111 1111000000000000 or or 255.255 255.255.255.192..255.192. o

o The number of subnets is 1024.The number of subnets is 1024.

o

o The number of addresses in each subnet is 26The number of addresses in each subnet is 26

(6 is the number of 0s) or 64. (6 is the number of 0s) or 64. n

n See the Next FigureSee the Next Figure

S

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(67)

Figure 5-9

E

(68)

(69)

Variable Length Subnetting

o

o A site that is granted a class C address andA site that is granted a class C address and

have five subnets with the number of hosts have five subnets with the number of hosts n

n 60, 60, 60, 30, 3060, 60, 60, 30, 30 o

o Cannot use a subnet mask with only extra 2Cannot use a subnet mask with only extra 2

bits bits

n

n Allow only four subnetAllow only four subnet o

o Cannot use a subnet make with Cannot use a subnet make with extra 3 bitsextra 3 bits

n

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(71)

Var

iab

le Le

ngt

h Subn

ett

ing

(Co

nt.)

o

o SolutionSolution

n

n Variable length subnettingVariable length subnetting o

o Use two different masks, one applied afUse two different masks, one applied after theter the

other other n

n First use the mask with 26 1s (255.255.255.192)First use the mask with 26 1s (255.255.255.192) to divide the network into

to divide the network into four sub four subnet net

n

n Then, apply mask with 27 1s (255.255.255.224)Then, apply mask with 27 1s (255.255.255.224) to

to oneone of the subnet to divide it of the subnet to divide it intointo two smaller two smaller subnets

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(73)

Figure 5-10

V

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(75)

SUPERNETTING

5.2

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(77)

Supernetting

o

o Class A and B addresses are almost depleted.Class A and B addresses are almost depleted.

However, class C addresses are still available However, class C addresses are still available

o

o But, the size of class block, 256, is too smallBut, the size of class block, 256, is too small

o

o Solution:Solution: subnetting subnetting

n

n Combine several class C blocks to create a larger Combine several class C blocks to create a larger range of addresses

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(79)

Figure 5-11

A

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(81)

Make a Supernet

o

o RulesRules

n

n The number of blocks must beThe number of blocks must be a power of 2a power of 2 (1, 2,(1, 2, 4

4, , 88, , 1166, , .. .. ..))

n

n The blocks must beThe blocks must be contiguouscontiguous in the addressin the address space (no gaps between the blocks).

space (no gaps between the blocks).

n

n TheThe third bytethird byte of theof the first add first addressress in thein the supe

superblorblockck mumust be evenly divst be evenly divisiblisible bye by ththee nunumbmbeer r of blocks

of blocks..

o

o In other words, if the number of blocks isIn other words, if the number of blocks is N N , the third, the third

byte mu

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(83)

E

Example xample 5 5

E

Example xample 5 5

A company needs 600 addresses. Which of

the following set of class C blocks can be

used t

used to form

o form a su

a superne

pernett for th

for this co

is compan

mpany?

y?

198. 198.47.47.32.32.0 198.47.0 198.47.33.33.00 198.198.47.47.34.34.00 198.47. 198.47.32.032.0 198.47.198.47.42.0 42.0 198.47.198.47.52.0 52.0 198.47.198.47.62.062.0 198.47. 198.47.31.031.0 198.47.198.47.32.0 32.0 198.47.198.47.33.0 33.0 198.47.198.47.52.052.0 198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0 198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0

(84)

(85)

Solution

Solution Solution

1

1: No, there are only three blocks <> 2^n,: No, there are only three blocks <> 2^n,

2

2: No, the blocks are not contiguous.: No, the blocks are not contiguous.

3

3: No, 31 in the first block is not divisible by 4.: No, 31 in the first block is not divisible by 4.

4

(86)

(87)

Su

pe

rn

et

Ma

sk

o

o In original block of addresses, we knoIn original block of addresses, we know thew the

range of addresses fro

range of addresses from the first addressm the first address n

n Since Since the mthe mask is ask is perdefinedperdefined (default m(default mask)ask)

o

o In subnettingIn subnetting or supor supernetternetting, thing, the first ae first addressddress

alone cannot derive the range of addresses alone cannot derive the range of addresses n

n We need to know theWe need to know the mask mask , subnet mask or , subnet mask or supe

(88)

(89)

I

n

II n

n subnetting

subnetting

,,

,

we

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addr

e

s

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s

.

de

(90)

(91)

I

n

II n

n supernetting

supernetting

,,

,

we

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t

addr

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es

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of

t

h

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e supernet

supernet

and the

a

and th

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supernet

supernet mas

supernet

mas

mask

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k

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de

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(92)

(93)

Sup

er

ne

t

Ma

sk

(C

ont

.)

o

o A supernetA supernet mask mask is the is the reversreverse of a e of a subnetsubnet

mask mask n

n A subnet mask has more 1s than the default mask A subnet mask has more 1s than the default mask

n

(94)

(95)

Figure 5-12

Comp

ar

i

s

on of

S

u

bne

t, D

e

f

aul

t,

a

(96)

(97)

E

Example xample 6 6

E

Example xample 6 6

We n

We need t

eed to m

o make

ake a su

a super

pernet

netwo

work

rk out

out of 1

of 16

6 class

class C blo

C blocks.

cks. What i

What is the s

s the super

upernet

net mask

mask??

Solution

Solution Solution

We need 16 blocks. For 16 blocks we need to change four

1s to 0s in the default mask. So the mask is

11111111 11111111 1111 11111111 11111111 111100000000 0000000000000000 Or Or 255.255.240.0 255.255.240.0

(98)

(99)

E

Example xample 7 7

E

Example xample 7 7

A sup

A supernernetet has a fhas a firsirst adt addredress of ss of 205205.1.16.6.32.32.0 an0 and ad a super

supernetnet mamask of 255.2sk of 255.255.255.248.48.0. A router rece0. A router receives threives threee packets wi

packets with the follth the following desowing destination tination addressesaddresses::

205.16.37.44 205.16.37.44 205.16.42.56 205.16.42.56 205.17.33.76 205.17.33.76

Which packet belongs to the supernet? Which packet belongs to the supernet?

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(101)

Solution

Solution Solution We ap

We apply tply the suhe superpernetnet masmask to sek to see if we cae if we can findn find

the beginning address.

205.16.37.44 205.16.37.44 AND AND 255.255.248.0255.255.248.0 èè 205.16.32.0205.16.32.0 205.16.42.56 205.16.42.56 AND AND 255.255.248.0255.255.248.0 èè 205.16.40.0205.16.40.0 205.17.33.76 205.17.33.76 AND AND 255.255.248.0255.255.248.0 èè 205.17.32.0205.17.32.0

Only the first address belongs to this supernet.

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(103)

E

Example xample 8 8

E

Example xample 8 8

A sup

A supernernetet has has a firsa first addt addresress of 20s of 205.5.16.16.32.32.0 an0 and ad a supernet

supernet mask of mask of 255.255.255.255.248.0. H248.0. How manow many blocks y blocks are inare in this su

(104)

(105)

o

o ThThe sue superpernenett hahas 21 1ss 21 1s..

o

o The default mask has 24 1s.The default mask has 24 1s.

o

o Since the difference is 3Since the difference is 3

n

n There are 23 or There are 23 or 8 blocks in this supernet.8 blocks in this supernet. o

o The blocks are 205.16.32.0 to 205.16.39.0.The blocks are 205.16.32.0 to 205.16.39.0.

n

n The first address The first address is 205.16.32.0.is 205.16.32.0.

n

n The last address is 205.16.39.255.The last address is 205.16.39.255.

Solution

Solution Solution

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(107)

CLASSLESS

ADDRESSING

5.3

(108)

(109)

Classless Addressing

o

o Classfull addresClassfu addressing sing has has createcreated mad manyny

problems problems

o

o Solution: classless addressing Solution: classless addressing

o

o Idea:Idea: variable-length blocksvariable-length blocks

n

n The whole address space (2^32 The whole address space (2^32 addresses) isaddresses) is divided into blocks of different

(110)

(111)

Figure 5-13

V

(112)

(113)

Number of Addresses in a Block

o

o There is only one condition on the number of There is only one condition on the number of

addresses in a block addresses in a block n

n It It mmusust bt be a e a popowewer or of 2 f 2 (2(2, 4, 4, , 8,8, .. .. .).) o

o A household may be given a block of 2A household may be given a block of 2

addresses. addresses.

o

o A small business may be given 16 aA small business may be given 16 addresses.ddresses.

o

o A large organization may be given 1024A large organization may be given 1024

addresses addresses

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(115)

Beginning Addresses

o

o The beginning address mustThe beginning address must be evenly divisiblebe evenly divisible

by the number of addresses. by the number of addresses. n

n Ex: if a block contains 4 addresses, the beginningEx: if a block contains 4 addresses, the beginning address must be divisible by 4.

address must be divisible by 4. o

o If the block has less than 256 addresses, weIf the block has less than 256 addresses, we

need to check only the rightmost byte. need to check only the rightmost byte.

o

o If it has less than 65,536 addresses, we need toIf it has less than 65,536 addresses, we need to

check only the two rightmost bytes, and so on. check only the two rightmost bytes, and so on.

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(117)

E

Example xample 9 9

E

Example xample 9 9

Which of the following can be the beginning address of a Which of the following can be the beginning address of a block that

block that contains contains 16 addres16 addresses?ses? 205.16.37.32 205.16.37.32 190.16.42.44 190.16.42.44 17.17.33.80 17.17.33.80 123.45.24.52 123.45.24.52 Solution Solution Solution Solution

The address 205.16.37.32 is eligible because 32 is

divisible by 16. The address 17.17.33.80 is eligible

because 80 is divisible by 16.

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(119)

E

Examplxample e 10 10

E

Examplxampl e e 10 10

Which of the following can be the beginning address of a Which of the following can be the beginning address of a block that

block that contains contains 1024 ad1024 addresses?dresses? 205.16.37.32 205.16.37.32 190.16.42.0 190.16.42.0 17.17.32.0 17.17.32.0 123.45.24.52 123.45.24.52 Solution Solution Solution Solution

To be divisible by 1024, the rightmost byte of an

address should be 0 and the second rightmost byte

must be divisible by 4. Only the address 17.17.32.0

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(121)

Mask

o

o LikLike te the che classlassfulful addraddressessinging, s, subneubnettittingng , a, andnd

supernetting, in classless addressing, given supernetting, in classless addressing, given n

n First add First addressress

n

n The mask The mask

o

(122)

(123)

Slash Notation

o

o Attach the number of 1s in a mask Attach the number of 1s in a mask to the end of ato the end of a classless address

classless address o

o Also called CIDR (C Also called CIDR (C lassless I lassless I nter nter D Domainomain R Routing)outing) o

o The CIDR convey two ideasThe CIDR convey two ideas

n

n The address isThe address is classlessclassless n

n Routing is done usingRouting is done using inteinterdomrdomainain routrouting ing

o

o A mask and a slash followed by a number define theA mask and a slash followed by a number define the same thing

same thing

n

(124)

(125)

Figure 5-14

S

(126)

(127)

S

l

as

h

n

otati

on i

s

al

s

o c

al

l

e

d

S

Sllas

ash

h n

notati

otation i

on is

s al

als

so ca

o calllle

ed

d

CIDR

notation.

(128)

(129)

Prefix and Suffix

o

o PrefixPrefix

n

n Another name for the common part of the address rangeAnother name for the common part of the address range

o

o Prefix range: the length of the prefixPrefix range: the length of the prefix

n

n Equal toEqual to nn in the slash notationin the slash notation

o

o SuffixSuffix

n

n The varying part of the address rangeThe varying part of the address range

o

o Suffix range: the length of the suffixSuffix range: the length of the suffix

n

(130)

(131)

E

Exxample ample 11 11

E

A small organization is given a block with the A small organization is given a block with the beginning

beginning address address and and the the prefix prefix lengthlength 205.16.37.24/29

205.16.37.24/29 (in slash notation). What is the(in slash notation). What is the range of the block?

(132)

(133)

o

o The beginning address is 205.16.37.24.The beginning address is 205.16.37.24.

o

o To find the last address we keep the first 29To find the last address we keep the first 29

bits and change the last 3 bits to 1s.

n

n Beginning:Beginning: 11001111 00010000 00100101 0001111001111 00010000 00100101 00011000000 n

n Ending Ending :: 11001111 00010000 00100101 0001111001111 00010000 00100101 00011111111

o

o There are only 8 addresses in this block.There are only 8 addresses in this block.

Solution

Solution Solution

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E

We can find the range of addresses in Example 11 by We can find the range of addresses in Example 11 by another method. We can argue that the length of the another method. We can argue that the length of the suffix is 32

suffix is 32 −− 29 or 3. So there are 229 or 3. So there are 233 == 8 addresses in this8 addresses in this

block. If the

block. If the first address is 205.16.37.24, the last addressfirst address is 205.16.37.24, the last address is 205.16.37.31 (24

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Finding the Network Address

o

o We can derive the network address if we knowWe can derive the network address if we know

n

n One of the address in the block One of the address in the block

n

n TheThe prefix le prefix lengthngth, or a, or a mask mask , or the, or the suffix len suffix lengthgth o

o Solution 1Solution 1

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o Solution 2Solution 2

n

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E

Example 1xample 13 3

E

What is the network address if one of the addresses is What is the network address if one of the addresses is 167.199.170.82/27?

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o

o The prefix length is 27, which means that weThe prefix length is 27, which means that we

must keep the first 27 bits as it is and

must keep the first 27 bits as it is and changechange

the remaining bits (5) to 0s.

o

o The 5 bits affect only the last byte.The 5 bits affect only the last byte.

o

o The last byte is 01010010.The last byte is 01010010.

o

o Changing the last 5 bits to 0s, we getChanging the last 5 bits to 0s, we get

01000000 or 64.

o

o The network address is 167.199.170.64/27.The network address is 167.199.170.64/27.

Solution

Solution Solution

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Subnetting

o

o We can also We can also use subuse subnettinettingng with with classlclasslessess

addressing addressing n

n Just increase the prefix length to derive the subnetJust increase the prefix length to derive the subnet prefix lengt

prefix lengthh

o

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E

Exxample ample 14 14

E

An organization is granted the block An organization is granted the block 130.34.12.64/26. The organization needs to have 130.34.12.64/26. The organization needs to have four subnets. What are the subnet addresses and four subnets. What are the subnet addresses and the range of addresses for each subnet?

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o

o The suffix length is 6 (32-26).The suffix length is 6 (32-26). o

o This means the total number of addresses in theThis means the total number of addresses in the block is 6

block is 64 (2^6).4 (2^6).

o

o If we If we create four subnets, each subnet will have 16create four subnets, each subnet will have 16 addresses.

addresses.

o

o Let us first Let us first find the subnet prefix (subnet mask).find the subnet prefix (subnet mask). o

o We need four subnets, which means we need to addWe need four subnets, which means we need to add two more 1s to the site prefix.

two more 1s to the site prefix. Solution

Solution

Solution Solution

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S

The subnet prefix is then /28.

Subnet 1: 130.34.12.64/28 to 130.34.12.79/28. Subnet 1: 130.34.12.64/28 to 130.34.12.79/28. Subnet 2 : 130 Subnet 2 : 130.34.12.80/28 to 130.34.12.95/28..34.12.80/28 to 130.34.12.95/28. Subnet 3: 130.34.12.96/28 to 130.34.12.111/28. Subnet 3: 130.34.12.96/28 to 130.34.12.111/28. Subnet 4: 130.34.12.112/28 to 130.34.12.127/28. Subnet 4: 130.34.12.112/28 to 130.34.12.127/28.

See the Next Figure

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Figure 5-15

E

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E

An ISP is granted a block of addresses starting with An ISP is granted a block of addresses starting with 190.100.0.0/16. The ISP needs to distribute these 190.100.0.0/16. The ISP needs to distribute these addresses to three groups of customers as follows:

addresses to three groups of customers as follows:

1. The first group has 64 customers; each needs 256 addresses.

2. The second group has 128 customers; each needs 128 addresses.

3. The third group has 128 customers; each needs 64 addresses

3. The third group has 128 customers; each needs 64 addresses..

Design the

Design the subblocksubblockss and give and give the slasthe slash notatioh notation for eachn for each subblock. Find out how many addresses are still available subblock. Find out how many addresses are still available

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Solution Solution Solution Solution Group 1 Group 1

For this group, each customer needs 256 addresses.

This means the suffix length is 8 (2

This means the suffix length is 8 (288 == 256). The256). The

prefix length is then 32

prefix length is then 32 −− 88 == 24.24.

01: 190.100.0.0/24 01: 190.100.0.0/24 èè190.100.0.255/24190.100.0.255/24 02: 190.100.1.0/24 02: 190.100.1.0/24 èè190.100.1.255/24190.100.1.255/24 ……… ……….... 64: 190.100.63.0/24 64: 190.100.63.0/24èè190.100.63.255/24190.100.63.255/24 Total Total 6464 256256 16,38416,384

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S

Soluolutition on (Co(Continti nunueed) d) Group 2

Group 2

prefix length

prefix length is then 32is then 32 −− 77 == 25. The addresses are:25. The addresses are:

001: 190.100.64.0/25 001: 190.100.64.0/25 èè190.100.64.127/25190.100.64.127/25 002: 190.100.64.128/25 002: 190.100.64.128/25 èè190.100.64.255/25190.100.64.255/25 128: 190.100.127.128/25 128: 190.100.127.128/25 èè190.100.127.255/25190.100.127.255/25 Total Total == 128128 ×× 128128 == 16,38416,384

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S

Soluolutition on (Co(Continti nunueed) d) Group 3

Group 3

prefix length is then 32

prefix length is then 32 −− 66 == 26.26.

001 001:190.100.128.0/26:190.100.128.0/26 èè190.100.128.63/26190.100.128.63/26 002 002:190.100.128.64/26:190.100.128.64/26 èè190.100.128.127/26190.100.128.127/26 ……… ……… 128 128:190.100.159.192/26:190.100.159.192/26 èè190.100.159.255/26190.100.159.255/26

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S

Number of granted addresses: 65,536

Number of allocated addresses: 40,960

Number of available addresses: 24,576

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Other Issues

o

o SupernettingSupernetting

n

n TherThere is no need for superne is no need for supernettiettingng in classin classlessless addressing

addressing

n

n If an If an organizatiorganization become larger on become larger

o

o It may ask a larger block and return the original oneIt may ask a larger block and return the original one

o

o Migration: when the idea of classlessMigration: when the idea of classless

addressing come true addressing come true n

n Organization that own class A, B, or C mustOrganization that own class A, B, or C must

o

o Use slash notation (/8, /16, /24)Use slash notation (/8, /16, /24) o

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