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

(1)

Basis Bilangan 2

Basis bilangan hanya ada dua nilai 0 dan 1

10101 21 1010 10 10100 20 1001 9 10011 19 1000 8 10010 18 111 7 10001 17 110 6 10000 16 101 5 1111 15 100 4 1110 14 11 3 1101 13 10 2 1100 12 1 1 1011 11 0 0 BINER DESIMAL BINER DESIMAL

(2)

10110

₂

(1 x 2

4

_{= 16)}

+ (0 x 2

3

_{= 0)}

+ (1 x 2

2

_{= 4)}

+ (1 x 2

1

_{= 2)}

+ (0 x 2

0

_{= 0) =}

22

10

=………..

2

22/2 = 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 IP

10111111.11111110.11111101.00001101

32 bit angka biner

Penulisan mengunakan notasi titik, tiap 8 bit dijadikan

angka desimal

10111111.11111110.11111101.00001101

191.254.253.13

(3)

Cara mudah menghitung

1 1 1 1 1 1 1 1

Bit bernilai kecil Bit bernilai besar

1 2 4 8 16 32 64 128 Desimal 1 x 20 1 x 21 1 x 22 1 x 23 1 x 24 1 x 25 1 x 26 1 x 27 Perhitu ngan 1 1 1 1 1 1 1 1 Biner

Contoh

10111111.11111110.11111101.00001101 =

191.254.253.13 Dasar perhitungan :

191 = 1011111 = 128 + 0 + 32 + 16 + 8 + 4 + 2 + 1

254 = 1111110 = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 0

253 = 1111101 = 128 + 64 + 32 + 16 + 8 + 4 + 0 + 1

13 = 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 jaringan

karena sebagai identifikasi antara host ke host

– cation is represented by an address

Dasar-dasar Alamat pada

TCP/IP

Dasar-dasar Alamat pada

TCP/IP

172.18.0.2 172.18.0.1 172.17.0.2 172.17.0.1 172.16.0.2 172.16.0.1 SA DA HDR DATA 10.13.0.0 192.168.1.0 10.13.0.1 192.168.1.1

(5)

Pengalamatan IP

255 ₂₅₅

255

Desimal dengan titik

Maksimal

Network

Host

128 64 32 16 8 4 2 1

11111111 11111111

1010110000010000

01111010 11001100

Biner

32 Bits

172

16

122

204 Contoh

Decimal

Contoh

Biner

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 IP

Network

Network HostHost HostHost HostHost Network

Network NetworkNetwork HostHost HostHost Network

Network NetworkNetwork NetworkNetwork HostHost

(6)

Kelas Alamat IP

1

kelas A:

Bits: 0NNNNNNN

0NNNNNNN HostHost HostHost HostHost

8 9 16 17 24 25 32 Range (1-126) 1

Kelas B:

Bits: 10NNNNNN

10NNNNNN NetworkNetwork HostHost HostHost

8 9 16 17 24 25 32 Range (128-191) 1

Kelas C:

Bits: 110NNNNN

110NNNNN NetworkNetwork NetworkNetwork HostHost

8 9 16 17 24 25 32 Range (192-223) 1

kelas D:

Bits: 1110MMMM

1110MMMM Multicast GroupMulticast Group Multicast GroupMulticast GroupMulticast GroupMulticast Group

8 9 16 17 24 25 32 Range (224-239)

Alamat Host

172.16.2.2 172.16.3.10 172.16.12.12 10.1.1.1 10.250.8.11 10.180.30.118 E1 172.16 12 12 Network 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 alamat

pada perusahaan, pelanggan pribadi, contoh :

192.168.10.32/28

• Notasi

slash (/) dalam

pemisah

untuk

menuliskan panjang bit alamat jaringan

(8)

11111111

Kententuan yan digunakan

untuk alamat Host

Kententuan yan digunakan

untuk 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 11111111

11111111 11111110

...

00000000 00000011

11111101

1

2

3 65534

65535

65536

–

...

2 65534

N

2

N

_{– 2}

_{= 2}

16

_{– 2 = 65534}

IP 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 ke

jaringan dalam sub jaringan

Keuntungan

Dapat membagi sub jaringan ke jaringan yang

lebih kecil

Mengurangi Broadcast traffic

Keanaman

(10)

Rumusan

Jumlah Jaringan – 2

x

_-2

dimana X = nilai bit

Jumlah Host – 2

y

_-2

dimana y = jumlah bit untuk host

Block Size = Total number of Address

Block Size = 256-Mask

Subnetting

Classful 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 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 Subnet

172.16.1.0 172.16.2.0

172.16.3.0

(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 172.16 Network Network Interface 172.16.0.0 172.16.0.0 E0 E1

New Routing Table

2 160 Host . . 172.16.3.1 E1

Ke Subnet Addressing

172.16.2.200 172.16.2.2 172.16.2.160 172.16.2.1 172.16.3.5 172.16.3.100 172.16.3.150 172.16.3.1 E0 E1 172.16 2 160 Network Host . . Network Interface 172.16.2.0 172.16.3.0 E0 E1

New Routing Table

(13)

Subnet Mask

172

16

0

255

0

255

0

IP Address Default Subnet Mask 8-Bit Subnet Mask Network Host Network Host

Network Subnet Host

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

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

11111111 11111111 00000000 00000000

Nilai 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 Mask

172.16.2.160 172.16.2.160 255.255.0.0 255.255.0.0

Network

Number

• 8 bit Network digunakan untuk sub net

Subnet Mask dalam Subnet

16

Network

Host

172.16.2.160 172.16.2.160 255.255.255.0 255.255.255.0 172 2 0 10101100 11111111 10101100 00010000 11111111 00010000 11111111 00000010 10100000 00000000 00000000 00000010

Subnet

Network

Number

128 192 224 240 248 252 254 255

(15)

Subnet Mask with Subnets

(cont.)

Subnet Mask with Subnets

(cont.)

Network

Host

172.16.2.160 172.16.2.160 255.255.255.192 255.255.255.192 10101100 11111111 10101100 00010000 11111111 00010000 11111111 00000010 10100000 11000000 10000000 00000010

Subnet

• Network number extended by ten bits

16

172 2 128

Network

Number

128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255

Subnet 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

(16)

Subnet 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 Addresses

172.16.1.0 172.16.2.0 172.16.3.0 172.16.4.0 172.16.3.255 (Directed Broadcast) 255.255.255.255 (Local Network Broadcast)

X

172.16.255.255 (All Subnets Broadcast)

(17)

Addressing 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 5 6 7 8 9 16 172 2 160 IP Host Address: 172.16.2.121 Subnet 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: 10101100 11111111 Subnet: 10101100 00010000 00010000 11111111 00000010 00000010 11111111 01111001 00000000 00000000

Class B Subnet Example

Broadcast:

(18)

Subnet Planning

Other Subnets 192.168.5.16 192.168.5.32 192.168.5.48 20 Subnets 5 Hosts per Subnet Class C Address:

192.168.5.0 20 Subnets 5 Hosts per Subnet Class C Address:

192.168.5.0

11111000 IP Host Address: 192.168.5.121

Subnet Mask: 255.255.255.248

Network Subnet Host 192.168.5.121: 11000000 11111111 Subnet: 11000000 10101000 10101000 11111111 00000101 00000101 11111111 01111001 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:

Network Network

(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 10.63 Broadcast 10.62 LHID 10.33 FHID 10.64 10.32 Subnets

(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 10.7 Broadcast 10.6 LHID 10.5 FHID 10.8 10.4 Subnets

(21)

Exercise

?

/30

?

/29

?

Mask

?

/28

?

/27

?

/26

Host

Subnets

Exercise

2

64

252 /30

6

32

248 /29

240

224

192 Mask

14

16 /28

30

8 /27

62

4 /26

Host

Subnets

(22)

Exam Question

• Find Subnet and Broadcast address

– 192.168.0.100/27

Exercise

192.168.10.54 /29

Mask ?

Subnet ?

Broadcast ?

(23)

Exercise

192.168.10.130 /28

Mask ?

Subnet ?

Broadcast ?

Exercise

192.168.10.193 /30

Mask ?

Subnet ?

Broadcast ?

(24)

Exercise

192.168.1.100 /26

Mask ?

Subnet ?

Broadcast ?

Exercise

192.168.20.158 /27

Mask ?

Subnet ?

Broadcast ?

(25)

Class B

172.16.0.0 /19

Subnets ?

Hosts ?

Block Size ?

Class B

172.16.0.0 /19

Subnets 2

3

_{-2 = 6}

Hosts 2

13

_{-2 = 8190}

Block Size 256-224 = 32

95.255 95.254 64.1 64.0 31.255 31.254 0.1 0.0 127.255 63.255 Broadcast 127.254 63.254 LHID 96.1 32.1 FHID 96.0 32.0 Subnets

(26)

Class B

172.16.0.0 /27

Subnets ?

Hosts ?

Block Size ?

Class B

172.16.0.0 /27

Subnets 2

11

_{-2 = 2046}

Hosts 2

5

_{-2 = 30}

Block Size 256-224 = 32

(27)

Class B

172.16.0.0 /23

Subnets ?

Hosts ?

Block Size ?

Class B

172.16.0.0 /23

Subnets 2

7

_{-2 = 126}

Hosts 2

9

_{-2 = 510}

Block Size 256-254 = 2

(28)

Class B

172.16.0.0 /24

Subnets ?

Hosts ?

Block Size ?

Class B

172.16.0.0 /24

Subnets 2

8

_{-2 = 254}

Hosts 2

8

_{-2 = 254}

Block Size 256-255 = 1

(29)

Class B

172.16.0.0 /25

Subnets ?

Hosts ?

Block Size ?

Class B

172.16.0.0 /25

Subnets 2

9

_{-2 = 510}

Hosts 2

7

_{-2 = 126}

Block Size 256-128 = 128

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 0.255 Broadcast 0.254 LHID 0.129 FHID 0.128 Subnets

(30)

Find out Subnet and Broadcast

Address

• 172.16.85.30/20

Find out Subnet and Broadcast

Address

(31)

Find out Subnet and Broadcast

Address

• 172.30.101.62/23

Find out Subnet and Broadcast

Address

(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

10.0.0.0 /10

Subnets ?

Hosts ?

Block Size ?

Class A

10.0.0.0 /10 Subnets 22_{-2 = 2} Hosts 222_{-2 = 4194302} Block 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 10.254.255.255 10.127.255.255 Broadcast 10.254.255.254 10.127.255.254 LHID 10.192.0.1 10.64.0.1 FHID 10.192 10.64 Subnets

(34)

Class A

10.0.0.0 /18

Subnets ?

Hosts ?

Block Size ?

Class A

10.0.0.0 /18 Subnets 210_{-2 = 1022} Hosts 214_{-2 = 16382} Block 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 10.0.254.255 10.0.127.255 Broadcast 10.0.254.254 10.0.127.254 LHID 10.0.192.1 10.0.64.1 FHID 10.0.192.0 10.0.64.0 Subnets

(35)

Broadcast 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 201.222.10.56 201.222.10.63

Subnet Mask

15.16.193.6 255.255.248.0 A 15.16.192.0 15.16.199.255

128.16.32.13 255.255.255.252 B 128.16.32.12 128.16.32.15 153.50.6.27 255.255.255.128 B 153.50.6.0

(36)

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

¾

VLSM allows us to use one class C address to

design a networking scheme to meet the

following requirements:

¾

Bangalore

60 Hosts

¾

Mumbai

28 Hosts

¾

Sydney

12 Hosts

¾

Singapore

12 Hosts

¾

WAN 1

2 Hosts

¾

WAN 2

2 Hosts

¾

WAN 3

2 Hosts

(37)

Networking Requirements

Bangalore 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 Scheme

Mumbai 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 2 2 192.168.10.128/30 192.168.10.136/30 192.168.10.132/30

(38)

VLSM Exercise

2 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 2 35

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

(40)

Summarization

Supernetting

Network Subnet 172.16.12.0 11000000 11111111 10101000 11111111

00001100

11111111

255.255.255.0 Network Network 00000000 00000000 16 8 4 2 1 172.16.13.0 11000000 10101000

00001101

00000000 172.16.14.0 11000000 10101000

00001110

00000000 172.16.15.0 11000000 10101000

00001111

00000000

(41)

Supernetting

Network Subnet 172.16.12.0 11000000 11111111 10101000 11111111

00001100

11111100

255.255.252.0 Network Network 00000000 00000000 16 8 4 2 1 172.16.13.0 11000000 10101000

00001101

00000000 172.16.14.0 11000000 10101000

00001110

00000000 172.16.15.0 11000000 10101000

00001111

00000000 172.16.12.0/24 172.16.13.0/24 172.16.14.0/24 172.16.15.0/24 172.16.12.0/22

Supernetting Question

17_2.1 .7.0_/2 4 17_2.1 .6.0_/2 4 17 2.1 .5._0/2 4 172_.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. 172.1.4.0/24172.1.5.0/24172.1.6.0/24172.1.7.0/24 B. 172.1.0.0/22 C. 172.1.4.0/25172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24 D. 172.1.0.0/21 E. 172.1.4.0/22