Upload
ahnan-prastito
View
189
Download
4
Embed Size (px)
Citation preview
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
15
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
16
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
18
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
21
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
192.168.10.130 /28Mask ?Subnet ?Broadcast ?
Exercise
192.168.10.193 /30Mask ?Subnet ?Broadcast ?
24
Exercise
192.168.1.100 /26Mask ?Subnet ?Broadcast ?
Exercise
192.168.20.158 /27Mask ?Subnet ?Broadcast ?
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
26
Class B
172.16.0.0 /27Subnets ?Hosts ?Block Size ?
Class B172.16.0.0 /27Subnets 211 -2 = 2046Hosts 25 -2 = 30Block Size 256-224 = 32
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
172.16.0.0 /23Subnets ?Hosts ?Block Size ?
Class B172.16.0.0 /23Subnets 27 -2 = 126Hosts 29 -2 = 510Block Size 256-254 = 2
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
28
Class B
172.16.0.0 /24Subnets ?Hosts ?Block Size ?
Class B172.16.0.0 /24Subnets 28 -2 = 254Hosts 28 -2 = 254Block Size 256-255 = 1
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
29
Class B
172.16.0.0 /25Subnets ?Hosts ?Block Size ?
Class B172.16.0.0 /25Subnets 29 -2 = 510Hosts 27 -2 = 126Block 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
0.255Broadcast
0.254LHID
0.129FHID
0.128Subnets
30
Find out Subnet and Broadcast Address
• 172.16.85.30/20
Find out Subnet and Broadcast Address
• 172.16.85.30/29
31
Find out Subnet and Broadcast Address
• 172.30.101.62/23
Find out Subnet and Broadcast Address
• 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
10.0.0.0 /10Subnets ?Hosts ?Block Size ?
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
34
Class A
10.0.0.0 /18Subnets ?Hosts ?Block Size ?
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
35
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
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
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
38
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
40
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
41
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