PERHITUNGAN ALAMAT IP

1

Basis Bilangan 2 Basis bilangan hanya ada dua nilai 0 dan 1

1010121101010101002010019100111910008100101811171000117110610000161015

11111510041110141131101131021100121110111100

BINERDESIMALBINERDESIMAL

2

101102

(1 x 24 = 16) + (0 x 23 = 0) + (1 x 22 = 4) + (1 x 21 = 2) + (0 x 20 = 0) =

22

2210=……………..222/2 = 11 sisa 0

11/2= 5 sisa 1

5/2 = 2 sisa 1

2/2 = 1 sisa 0

1 tidak bisa dibagi lagi

Hasil 10110

Untuk Alamat IP10111111.11111110.11111101.00001101

32 bit angka biner

Penulisan mengunakan notasi titik, tiap 8 bit dijadikanangka desimal

10111111.11111110.11111101.00001101

191.254.253.13191.254.253.13

3

Cara mudah menghitung

1 1 1 1 1 1 1 1Bit bernilai kecilBit bernilai besar

1248163264128Desimal

1 x 201 x 211 x 221 x 231 x 241 x 251 x 261 x 27Perhitungan

11111111Biner

Contoh

10111111.11111110.11111101.00001101 = 191.254.253.13

Dasar perhitungan :191 = 1011111 = 128 + 0 + 32 + 16 + 8 + 4 + 2 + 1254 = 1111110 = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 0253 = 1111101 = 128 + 64 + 32 + 16 + 8 + 4 + 0 + 113 = 00001101 = 0 + 0 + 0 + 0 + 8 + 4 + 0 + 1

4

Pengalamatan IP

• Versi IPv 4• 32 bit, dibagi 4 oktet• Ditulis dengan angka desimal dengan

notasi titik

– Penggunaan alamat harus unik dalam satu jaringankarena sebagai identifikasi antara host ke host

– cation is represented by an address

Dasar-dasar Alamat padaTCP/IP

Dasar-dasar Alamat padaTCP/IP

172.18.0.2

172.18.0.1

172.17.0.2172.17.0.1

172.16.0.2

172.16.0.1

SA DAHDR DATA10.13.0.0 192.168.1.0

10.13.0.1 192.168.1.1

5

Pengalamatan IPPengalamatan IP

255 255 255 255

DesimaldengantitikMaksimal

Network Host

128 64 32 16 8 4 2 1

11111111 11111111 11111111 11111111

1010110000010000 01111010 11001100

Biner

32 Bits

172 16 122 204ContohDecimalContohBiner

1 8 9 16 17 24 25 32

128 64 32 16 8 4 2 1

128 64 32 16 8 4 2 1

128 64 32 16 8 4 2 1

•Class A:

•Class B:

•Class C:

•Class D: Multicast

•Class E: Research

Kelas IPKelas IP

NetworkNetwork HostHost HostHost HostHost

NetworkNetwork NetworkNetwork HostHost HostHost

NetworkNetwork NetworkNetwork NetworkNetwork HostHost

8 Bits 8 Bits 8 Bits 8 Bits

6

Kelas Alamat IPKelas Alamat IP1

kelas A:Bits:

0NNNNNNN0NNNNNNN HostHost HostHost HostHost8 9 16 17 24 25 32

Range (1-126)1

Kelas B:Bits:

10NNNNNN10NNNNNN NetworkNetwork HostHost HostHost8 9 16 17 24 25 32

Range (128-191)1

Kelas C:Bits:

110NNNNN110NNNNN NetworkNetwork NetworkNetwork HostHost

8 9 16 17 24 25 32

Range (192-223)1

kelas D:Bits:

1110MMMM1110MMMM Multicast GroupMulticast Group Multicast GroupMulticast Group Multicast GroupMulticast Group

8 9 16 17 2425 32

Range (224-239)

Alamat HostAlamat Host172.16.2.2

172.16.3.10

172.16.12.12

10.1.1.1

10.250.8.11

10.180.30.118

E1

172.16 12 12Network Host

. . Network Interface

172.16.0.0

10.0.0.0

E0

E1

Routing Table

172.16.2.1

10.6.24.2

E0

7

Classless Inter-Domain Routing (CIDR)

• Suatu dasar cara yang dipakai ISPs (Internet Service Providers) untuk mengalokasikan alamatpada perusahaan, pelanggan pribadi, contoh : 192.168.10.32/28

• Notasi slash (/) dalam pemisah untukmenuliskan panjang bit alamat jaringan

CIDR Values

8

11111111

Kententuan yan digunakanuntuk alamat Host

Kententuan yan digunakanuntuk alamat Host

172 16 0 0

10101100 00010000 00000000 00000000

16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Network Host

00000000 00000001

11111111 1111111111111111 11111110

......

00000000 0000001111111101

123

655346553565536–

...

265534

N

2N – 2 = 216 – 2 = 65534

IP Address Classes ExerciseIP Address Classes Exercise

Address Class Network Host

10.2.1.1

128.63.2.100

201.222.5.64

192.6.141.2

130.113.64.16

256.241.201.10

9

IP Address Classes Exercise Answers

IP Address Classes Exercise Answers

Address Class Network Host

10.2.1.1

128.63.2.100

201.222.5.64

192.6.141.2

130.113.64.16

256.241.201.10

A

B

C

C

B

Nonexistent

10.0.0.0

128.63.0.0

201.222.5.0

192.6.141.0

130.113.0.0

0.2.1.1

0.0.2.100

0.0.0.64

0.0.0.2

0.0.64.16

Subnetting

Subnetting adalah logika pembagian kejaringan dalam sub jaringanKeuntungan

Dapat membagi sub jaringan ke jaringan yang lebih kecilMengurangi Broadcast trafficKeanamanMemudahkan mengelola

10

RumusanJumlah Jaringan – 2x-2dimana X = nilai bit

Jumlah Host – 2y-2dimana y = jumlah bit untuk host

Block Size = Total number of AddressBlock Size = 256-Mask

SubnettingClassful IP Addressing SNM are a set of 255’s and 0’s.In Binary it’s contiguous 1’s and 0’s.SNM cannot be any value as it won’t follow the rule of contiguous 1’s and 0’s.Possible subnet mask values– 0– 128– 192– 224– 240– 248– 252– 254– 255

11

• Network 172.16.0.0

172.16.0.0

Pengalamatan diluar SubnetPengalamatan diluar Subnet

172.16.0.1 172.16.0.2 172.16.0.3

…...

172.16.255.253 172.16.255.254

• Network 172.16.0.0

Pengalamatan dalam SubnetPengalamatan dalam Subnet

172.16.1.0 172.16.2.0

172.16.3.0

172.16.4.0

12

Belum di bagi ke Subnet Addressing

Belum di bagi ke Subnet Addressing

172.16.2.200

172.16.2.2

172.16.2.160

172.16.2.1

172.16.3.5

172.16.3.100

172.16.3.150

E0

172.16Network

Network Interface

172.16.0.0

172.16.0.0

E0

E1

New Routing Table2 160Host

. .

172.16.3.1E1

Ke Subnet AddressingKe Subnet Addressing172.16.2.200

172.16.2.2

172.16.2.160

172.16.2.1

172.16.3.5

172.16.3.100

172.16.3.150

172.16.3.1

E0E1

172.16 2 160Network Host

. . Network Interface

172.16.2.0

172.16.3.0

E0

E1

New Routing Table

Subnet

13

Subnet MaskSubnet Mask

172172 1616 00 00

255255 255255 00 00

255255 255255 255255 00

IPAddress

DefaultSubnet

Mask

8-BitSubnet

Mask

Network Host

Network Host

Network Subnet Host

• Juga bisa dutulis “/16,” 16 =adalah panjang bit 1 dalammask

• Juga bisa ditulis “/24,” 24= adalah panjang bit 1 dalammask

11111111 11111111 00000000 00000000

Nilai desimal dengan pola bitNilai desimal dengan pola bit

0 0 0 0 0 0 0 0 = 0

1 0 0 0 0 0 0 0 = 128

1 1 0 0 0 0 0 0 = 192

1 1 1 0 0 0 0 0 = 224

1 1 1 1 0 0 0 0 = 240

1 1 1 1 1 0 0 0 = 248

1 1 1 1 1 1 0 0 = 252

1 1 1 1 1 1 1 0 = 254

1 1 1 1 1 1 1 1 = 255

128 64 32 16 8 4 2 1

14

16

Network Host

172 0 0

10101100

11111111

10101100

00010000

11111111

00010000

00000000

00000000

10100000

00000000

00000000

•Tidak menggunakan subnet

00000010

Net MaskNet Mask

172.16.2.160172.16.2.160

255.255.0.0255.255.0.0

NetworkNumber

•8 bit Network digunakan untuk sub net

Subnet Mask dalam SubnetSubnet Mask dalam Subnet

16

Network Host

172.16.2.160172.16.2.160

255.255.255.0255.255.255.0

172 2 0

10101100

11111111

10101100

00010000

11111111

00010000

11111111

00000010

10100000

00000000

00000000

00000010

Subnet

NetworkNumber

128

192

224

240

248

252

254

255

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Subnet Mask with Subnets (cont.)

Subnet Mask with Subnets (cont.)

Network Host

172.16.2.160172.16.2.160

255.255.255.192255.255.255.192

10101100

11111111

10101100

00010000

11111111

00010000

11111111

00000010

10100000

11000000

10000000

00000010

Subnet

•Network number extended by ten bits

16172 2 128NetworkNumber

128

192

224

240

248

252

254

255

128

192

224

240

248

252

254

255

Subnet Mask ExerciseSubnet Mask Exercise

Address Subnet Mask Class Subnet

172.16.2.10

10.6.24.20

10.30.36.12

255.255.255.0

255.255.240.0

255.255.255.0

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Subnet Mask Exercise AnswersSubnet Mask Exercise Answers

Address Subnet Mask Class Subnet

172.16.2.10

10.6.24.20

10.30.36.12

255.255.255.0

255.255.240.0

255.255.255.0

B

A

A

172.16.2.0

10.6.16.0

10.30.36.0

Broadcast AddressesBroadcast Addresses

172.16.1.0172.16.2.0

172.16.3.0

172.16.4.0

172.16.3.255(Directed Broadcast)

255.255.255.255(Local Network Broadcast)XX

172.16.255.255(All Subnets Broadcast)

17

Addressing Summary ExampleAddressing Summary Example

10101100

11111111

10101100

00010000

11111111

00010000

11111111

00000010

10100000

11000000

10000000

00000010

10101100 00010000 00000010 10111111

10101100 00010000 00000010 10000001

10101100 00010000 00000010 10111110

Host

Mask

Subnet

Broadcast

Last

First

172.16.2.160

255.255.255.192

172.16.2.128

172.16.2.191

172.16.2.129

172.16.2.190

1

2

3

4

56

7

89

16172 2 160

IP Host Address: 172.16.2.121Subnet Mask: 255.255.255.0

• Subnet Address = 172.16.2.0• Host Addresses = 172.16.2.1–172.16.2.254• Broadcast Address = 172.16.2.255• Eight Bits of Subnetting

Network Subnet Host

10101100 00010000 00000010 11111111

172.16.2.121:255.255.255.0:

1010110011111111

Subnet: 10101100 00010000

0001000011111111

00000010

00000010

1111111101111001 00000000

00000000

Class B Subnet ExampleClass B Subnet Example

Broadcast:

Network

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Subnet PlanningSubnet Planning

Other Subnets

192.168.5.16

192.168.5.32 192.168.5.48

20 Subnets5 Hosts per SubnetClass C Address:192.168.5.0

20 Subnets5 Hosts per SubnetClass C Address:192.168.5.0

11111000

IP Host Address: 192.168.5.121Subnet Mask: 255.255.255.248

Network Subnet Host

192.168.5.121: 1100000011111111

Subnet: 11000000 10101000

1010100011111111

00000101

00000101

1111111101111001

01111000

255.255.255.248:

Class C Subnet Planning Example

Class C Subnet Planning Example

• Subnet Address = 192.168.5.120• Host Addresses = 192.168.5.121–192.168.5.126• Broadcast Address = 192.168.5.127• Five Bits of Subnetting

Broadcast:

NetworkNetwork

11000000 10101000 00000101 01111111

19

Exercise

• 192.168.10.0• /27

? – SNM? – Block Size?- Subnets

Exercise

• /27

? – SNM – 224? – Block Size = 256-224 = 32?- Subnets

10.31

10.30

10.1

10.0

10.63Broadcast

10.62LHID

10.33FHID

10.6410.32Subnets

20

Exercise

• 192.168.10.0• /30

? – SNM? – Block Size?- Subnets

Exercise

• /30

? – SNM – 252? – Block Size = 256-252 = 4?- Subnets

10.3

10.2

10.1

10.0

10.7Broadcast

10.6LHID

10.5FHID

10.810.4Subnets

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Exercise

???/30???/29

???Mask

??/28??/27??/26HostSubnets

Exercise

264252/30632248/29

240224192Mask

1416/28308/27624/26HostSubnets

22

Exam Question

• Find Subnet and Broadcast address– 192.168.0.100/27

Exercise

192.168.10.54 /29Mask ?Subnet ?Broadcast ?

23

Exercise


Exercise


24

Exercise


Exercise


25

Class B

172.16.0.0 /19Subnets ?Hosts ?Block Size ?

Class B172.16.0.0 /19Subnets 23 -2 = 6Hosts 213 -2 = 8190Block Size 256-224 = 32

95.255

95.254

64.1

64.0

31.255

31.254

0.1

0.0

127.25563.255Broadcast

127.25463.254LHID

96.132.1FHID

96.032.0Subnets

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Class B



0.95

0.94

0.65

0.64

0.31

0.30

0.1

0.0

0.1270.63Broadcast

0.1260.62LHID

0.970.33FHID

0.960.32Subnets

27

Class B



5.255

5.254

4.1

4.0

1.255

1.254

0.1

0.0

7.2553.255Broadcast

7.2543.254LHID

6.12.1FHID

6.02.0Subnets

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Class B



2.255

2.254

2.1

2.0

0.255

0.254

0.1

0.0

3.2551.255Broadcast

3.2541.254LHID

3.11.1FHID

3.01.0Subnets

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Class B



1.255

1.254

1.129

1.128

2.255

2.254

2.129

2.128

2.127

2.126

2.1

2.0

1.127

1.126

1.1

1.0

0.127

0.126

0.1

0.0

0.255Broadcast

0.254LHID

0.129FHID

0.128Subnets

30

Find out Subnet and Broadcast Address

• 172.16.85.30/20


• 172.16.85.30/29

31


• 172.30.101.62/23


• 172.20.210.80/24

32

Exercise

• Find out the mask which gives 100 subnets for class B

Exercise

• Find out the Mask which gives 100 hosts for Class B

33

Class A


Class A10.0.0.0 /10Subnets 22 -2 = 2Hosts 222 -2 = 4194302Block Size 256-192 = 64

10.191.255.255

10.191.255.254

10.128.0.1

10.128

10.63.255.255

10.63.255.254

10.0.0.1

10.0

10.254.255.25510.127.255.255Broadcast

10.254.255.25410.127.255.254LHID

10.192.0.110.64.0.1FHID

10.19210.64Subnets

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Class A


Class A10.0.0.0 /18Subnets 210 -2 = 1022Hosts 214 -2 = 16382Block Size 256-192 = 64

10.0.191.255

10.0.191.254

10.0.128.1

10.0.128.0

10.0.63.255

10.0.63.254

10.0.0.1

10.0.0.0

10.0.254.25510.0.127.255Broadcast

10.0.254.25410.0.127.254LHID

10.0.192.110.0.64.1FHID

10.0.192.010.0.64.0Subnets

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Broadcast Addresses ExerciseBroadcast Addresses Exercise

Address Class Subnet Broadcast

201.222.10.60 255.255.255.248

Subnet Mask

15.16.193.6 255.255.248.0

128.16.32.13 255.255.255.252

153.50.6.27 255.255.255.128

Broadcast Addresses Exercise Answers

Broadcast Addresses Exercise Answers

153.50.6.127

Address Class Subnet Broadcast

201.222.10.60 255.255.255.248 C 201.222.10.63201.222.10.56

Subnet Mask

15.16.193.6 255.255.248.0 A 15.16.199.25515.16.192.0

128.16.32.13 255.255.255.252 B 128.16.32.15128.16.32.12

153.50.6.27 255.255.255.128 B 153.50.6.0

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VLSM• VLSM is a method of designating a different subnet

mask for the same network number on different subnets

• Can use a long mask on networks with few hosts and a shorter mask on subnets with many hosts

• With VLSMs we can have different subnet masks for different subnets.

Variable Length Subnetting

VLSM allows us to use one class C address to design a networking scheme to meet the following requirements:

Bangalore 60 HostsMumbai 28 HostsSydney 12 HostsSingapore 12 HostsWAN 1 2 HostsWAN 2 2 HostsWAN 3 2 Hosts

37

Networking RequirementsBangalore 60

Mumbai 60 Sydney 60 Singapore 60

WAN 1 WAN 2

WAN 3

In the example above, a /26 was used to provide the 60 addresses for Bangalore and the other LANs. There are no addresses left for WAN links

Networking SchemeMumbai 192.168.10.64/27

Bangalore 192.168.10.0/26

Sydney 192.168.10.96/28

Singapore 192.168.10.112/28

WAN 192.168.10.129 and 130 WAN 192.198.10.133 and 134

WAN 192.198.10.137 and 138

60 12 12

28

2 22192.168.10.128/30

192.168.10.136/30

192.168.10.132/30

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VLSM Exercise2

2

2

40

25

12

192.168.1.0

VLSM Exercise

2 2

2

40

25

12

192.168.1.0

192.168.1.4/30

192.168.1.8/30

192.168.1.12/30

192.168.1.16/28

192.168.1.32/27

192.168.1.64/26

39

VLSM Exercise

2

2

8

15

5

192.168.1.0

2

235

Summarization• Summarization, also called route aggregation, allows

routing protocols to advertise many networks as one address.

• The purpose of this is to reduce the size of routing tables on routers to save memory

• Route summarization (also called route aggregation or supernetting) can reduce the number of routes that a router must maintain

• Route summarization is possible only when a proper addressing plan is in place

• Route summarization is most effective within a subnetted environment when the network addresses are in contiguous blocks

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Summarization

Supernetting

Network Subnet

172.16.12.0 11000000

11111111

10101000

11111111

00001100

11111111255.255.255.0

NetworkNetwork

00000000

0000000016 8 4 2 1

172.16.13.0 11000000 1010100000001101 00000000172.16.14.0 11000000 1010100000001110 00000000172.16.15.0 11000000 1010100000001111 00000000

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Supernetting

Network Subnet

172.16.12.0 11000000

11111111

10101000

11111111

00001100

11111100255.255.252.0

NetworkNetwork

00000000

0000000016 8 4 2 1

172.16.13.0 11000000 1010100000001101 00000000172.16.14.0 11000000 1010100000001110 00000000172.16.15.0 11000000 1010100000001111 00000000

172.16.12.0/24172.16.13.0/24172.16.14.0/24172.16.15.0/24

172.16.12.0/22

Supernetting Question

172.1.7.0/24

172.1.6.0/24

172.1.5.0/24

172.1.4.128/25

172.1.4.128/25

What is the most efficient summarization that TK1 can use to advertise its networks to TK2?

A. 172.1.4.0/24172.1.5.0/24172.1.6.0/24172.1.7.0/24B. 172.1.0.0/22C. 172.1.4.0/25172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24D. 172.1.0.0/21E. 172.1.4.0/22

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