Badiyanto, S.Kom., M.Kom STMIK AKAKOM Yogyakarta Dasar Perhitungan Basis Bilangan (Biner) Basis bilangan hanya ada dua nilai 0 dan 1 DESIMAL BINER DESIMAL BINER 0 0 11 1011 1 1 1 1100 10 13 1101 3 11 14 1110 4 100 15 1111 5 101 16 10000 6 110 17 10001 7 111 18 10010 8 1000 19 10011 9 1001 0 10100 10 1010 1 10101 1
Contoh 10110 (1 x 4 = 16) + (0 x 3 = 0) + (1 x = 4) + (1 x 1 = ) + (0 x 0 = 0) = 10=.. 10 =.. / = 11 sisa 0 11/= 5 sisa 1 5/ = sisa 1 / = 1 sisa 0 1 tidak bisa dibagi lagi Hasil 10110 Untuk Alamat IP 10111111.11111110.11111101.00001101 3 bit angka biner Penulisan mengunakan notasi titik, tiap 8 bit dijadikan angka desimal 10111111.11111110.11111101.00001101 191.54.53.13
Cara mudah menghitung 1 1 1 1 1 1 1 1 Bit bernilai besar Bit bernilai kecil Biner 1 1 1 1 1 1 1 1 Perhitu ngan 1 x 7 1 x 6 1 x 5 1 x 4 1 x 3 1 x 1 x 1 1 x 0 Desimal 18 64 3 16 8 4 1 Contoh 10111111.11111110.11111101.00001101 = 191.54.53.13 Dasar perhitungan : 191 = 1011111 = 18 + 0 + 3 + 16 + 8 + 4 + + 1 54 = 1111110 = 18 + 64 + 3 + 16 + 8 + 4 + + 0 53 = 1111101 = 18 + 64 + 3 + 16 + 8 + 4 + 0 + 1 13 = 00001101 = 0 + 0 + 0 + 0 + 8 + 4 + 0 + 1 3
Pengalamatan IP Versi IPv 4 3 bit, dibagi 4 oktet Ditulis dengan angka desimal dengan notasi titik Dasar-dasar Alamat pada TCP/IP 17.18.0.1 17.16.0.1 17.18.0. 17.16.0. HDR SA DA DATA 10.13.0.0 19.168.1.0 10.13.0.1 17.17.0.1 17.17.0. 19.168.1.1 Penggunaan alamat harus unik dalam satu jaringan karena sebagai identifikasi antara host ke host cation is represented by an address 4
Pengalamatan IP Desimal dengan titik Maksimal Biner 3 Bits Host 55 55 55 55 1 8 9 16 17 4 5 3 Contoh Decimal Contoh Biner 18 64 3 16 84 1 18 64 3 16 84 1 18 64 3 16 84 18 64 3 16 84 17 16 1 04 10101100 00010000 01111010 11001100 1 1 Kelas IP 8 Bits 8 Bits 8 Bits 8 Bits Class A: Class B: Class C: Host Host Host Host Host Host Class D: Multicast Class E: Research 5
Kelas Alamat IP Bits: kelas A: Bits: Kelas B: Bits: Kelas C: Bits: kelas D: 1 1 8 9 16 17 4 5 3 0NNNNNNN Host Host Host Range (1-16) 8 9 16 17 4 5 3 10NNNNNN Host Host Range (18-191) 1 8 9 16 17 4 5 3 110NNNNN Host Range (19-3) 1 8 9 16 17 4 5 3 1110MMMM Multicast Group Multicast Group Multicast Group Range (4-39) Alamat Host 17.16.. 17.16.3.10 E0 17.16..1 10.6.4. E1 10.1.1.1 10.50.8.11 17.16.1.1 10.180.30.118 Routing Table 17.16. 1. 1 Interface Host 17.16.0.0 10.0.0.0 E0 E1 6
Classless Inter-Domain Routing (CIDR) Suatu dasar cara yang dipakai ISPs (Internet Service Providers) untuk mengalokasikan alamat pada perusahaan, pelanggan pribadi, contoh : 19.168.10.3/8 Notasi slash (/) dalam pemisah untuk menuliskan panjang bit alamat jaringan CIDR Values 7
Kententuan yan digunakan untuk alamat Host 17 16 0 0 10101100 00010000 00000000 00000000 00000000 00000001 00000000 00000011 16 15 14 13 1 11 10 9 Host 8 76 5 43 1 11111101 11111110... N = 16 = 65534... N 1 3... 65534 65535 65536 65534 IP Address Classes Exercise Address Class Host 10..1.1 18.63..100 01..5.64 19.6.141. 130.113.64.16 56.41.01.10 8
IP Address Classes Exercise Answers Address Class Host 10..1.1 A 10.0.0.0 0..1.1 18.63..100 B 18.63.0.0 0.0..100 01..5.64 C 01..5.0 0.0.0.64 19.6.141. C 19.6.141.0 0.0.0. 130.113.64.16 B 130.113.0.0 0.0.64.16 56.41.01.10 Nonexistent Subnetting Subnetting adalah logika pembagian ke jaringan dalam sub jaringan Keuntungan Dapat membagi sub jaringan ke jaringan yang lebih kecil Mengurangi Broadcast traffic Keanaman Memudahkan mengelola 9
Rumusan Jumlah Jaringan x - dimana X = nilai bit Jumlah Host y - dimana y = jumlah bit untuk host Block Size = Total number of Address Block Size = 56-Mask Subnetting Disebut Classful jika alamat Sub Net Mask 55 dan 0. Dalam Binary hanya bersebelahan 1 dan 0 Mengihitungnya mudah hanya ada 1 dan 0 Kemungkinan nilai subnet mask 0 18 19 4 40 48 5 54 55 10
Pengalamatan diluar Subnet 17.16.0.1 17.16.0. 17.16.0.3 17.16.55.53 17.16.55.54... 17.16.0.0 17.16.0.0 Pengalamatan dalam Subnet 17.16.3.0 17.16.4.0 17.16.1.0 17.16..0 17.16.0.0 11
Belum di bagi ke Subnet Addressing 17.16..00 17.16.. E0 17.16..1 17.16.3.1 E1 17.16.3.5 17.16.3.100 17.16..160 17.16.3.150 17.16.. 160 Host New Routing Table Interface 17.16.0.0 E0 17.16.0.0 E1 Ke Subnet Addressing 17.16..00 17.16.. E0 17.16..1 17.16.3.1 E1 17.16.3.5 17.16.3.100 17.16..160 17.16.3.150 New Routing Table 17.16.. 160 Interface Subnet Host 17.16..0 17.16.3.0 E0 E1 1
Subnet Mask IP Address Host 17 16 0 0 Default Subnet Mask 8-Bit Subnet Mask Host 55 55 0 0 00000000 00000000 Juga bisa dutulis /16, 16 =adalah panjang bit 1 dalam mask Subnet Host 55 55 55 0 Juga bisa ditulis /4, 4= adalah panjang bit 1 dalam mask Nilai desimal dengan pola bit 18 64 3 16 8 4 1 0 0 0 0 0 0 0 0 = 0 1 0 0 0 0 0 0 0 = 18 1 1 0 0 0 0 0 0 = 19 1 1 1 0 0 0 0 0 = 4 1 1 1 1 0 0 0 0 = 40 1 1 1 1 1 0 0 0 = 48 1 1 1 1 1 1 0 0 = 5 1 1 1 1 1 1 1 0 = 54 1 1 1 1 1 1 1 1 = 55 13
18 19 4 40 48 5 54 55 017-05-17 Menghitung IP Host 17.16..160 10101100 00010000 00000010 10100000 55.55.0.0 Number 00000000 00000000 10101100 00010000 00000000 00000000 17 16 0 0 AND Tidak menggunakan subnet Subnet Mask dalam Subnet Subnet Host 17.16.. 160 10101100 00010000 00000010 10100000 55.55.55.0 00000000 10101100 00010000 00000010 00000000 Number 17 16 0 8 bit digunakan untuk sub net 14
Subnet Mask with Subnets (cont.) Subnet Host 17.16..160 10101100 00010000 00000010 10100000 55.55.55.19 11000000 AND 10101100 00010000 00000010 10000000 Number 17 16 18 Subnet Mask Exercise Address Subnet Mask Class Subnet 17.16..10 10.6.4.0 10.30.36.1 55.55.55.0 55.55.40.0 55.55.55.0 15
Subnet Mask Exercise Answers Address Subnet Mask Class Subnet 17.16..10 55.55.55.0 B 17.16..0 10.6.4.0 55.55.40.0 A 10.6.16.0 10.30.36.1 55.55.55.0 A 10.30.36.0 Broadcast Addresses 17.16.3.0 17.16.4.0 17.16.1.0 17.16.3.55 (Directed Broadcast) 55.55.55.55 (Local Broadcast) 17.16.55.55 (All Subnets Broadcast) X 17.16..0 16
Mencari IP network, broadcast 17 16 160 17.16..160 10101100 00010000 00000010 10100000 Host 55.55.55.19 11000000 Mask 17.16..18 10101100 00010000 00000010 10000000 Subnet 17.16..191 17.16..19 17.16..190 10101100 00010000 00000010 10111111 10101100 00010000 00000010 10000001 10101100 00010000 00000010 10111110 Broadcast First Last Pemetaan jaringan Alokasi ip : 17.16..0 Mask : 55.55.55.0 Di pecah 4 jaringan Jaringan 1. : 17.16..0 mask : 55.55.55.19 broadcast : 17.16..63 Jaringan. : 17.16..64 mask : 55.55.55.19 broadcast : 17.16..17 Jaringan 3. : 17.16..18 mask : 55.55.55.19 broadcast : 17.16..191 Jaringan 4. : 17.16..19 mask : 55.55.55.19 broadcast : 17.16..55 17
Class B Subnet Example IP Host Address: 17.16..11 Subnet Mask: 55.55.55.0 Subnet Host 17.16..11: 10101100 00010000 00000010 01111001 55.55.55.0: 00000000 Subnet: 10101100 00010000 00000010 00000000 Broadcast: 10101100 00010000 00000010 Subnet Address = 17.16..0 Host Addresses = 17.16..1 17.16..54 Broadcast Address = 17.16..55 Eight Bits of Subnetting Subnet Planning 0 Subnets 5 Hosts per Subnet Class C Address: 19.168.5.0 Other Subnets 19.168.5.16/ 19.168.5.3 19.168.5.48 18
Class C Subnet Planning Example IP Host Address: 19.168.5.11 Subnet Mask: 55.55.55.48 Subnet Host 19.168.5.11: 11000000 10101000 00000101 01111001 55.55.55.48: 11111000 Subnet: 11000000 10101000 00000101 01111000 Broadcast: 11000000 10101000 00000101 01111111 Subnet Address = 19.168.5.10 Host Addresses = 19.168.5.11 19.168.5.16 Broadcast Address = 19.168.5.17 Five Bits of Subnetting Exercise 19.168.10.0 /7? SNM? Block Size?- Subnets 19
Exercise /7? SNM 4? Block Size = 56-4 = 3?- Subnets Subnets 10.0 10.3 10.64 First Host ID 10.1 10.33 Last Host ID 10.30 10.6 Broadcast 10.31 10.63 Exercise 19.168.10.0 /30? SNM? Block Size?- Subnets 0
Exercise /30? SNM 5? Block Size = 56-5 = 4?- Subnets Subnets 10.0 10.4 10.8 FHID 10.1 10.5 LHID 10. 10.6 Broadcast 10.3 10.7 Exercise Mask Subnets Host /6??? /7??? /8??? /9??? /30??? 1
Exercise Mask Subnets Host /6 19 4 6 /7 4 8 30 /8 40 16 14 /9 48 3 6 /30 5 64 Exam Question Find Subnet and Broadcast address 19.168.0.100/7
Exercise 19.168.10.54 /9 Mask? Subnet? Broadcast? Exercise 19.168.10.130 /8 Mask? Subnet? Broadcast? 3
Exercise 19.168.10.193 /30 Mask? Subnet? Broadcast? Exercise 19.168.1.100 /6 Mask? Subnet? Broadcast? 4
Exercise 19.168.0.158 /7 Mask? Subnet? Broadcast? Class B 17.16.0.0 /19 Subnets? Hosts? Block Size? 5
Class B 17.16.0.0 /19 Subnets 3 - = 6 Hosts 13 - = 8190 Block Size 56-4 = 3 Subnets 0.0 3.0 64.0 96.0 FHID 0.1 3.1 64.1 96.1 LHID 31.54 63.54 95.54 17.54 Broadcast 31.55 63.55 95.55 17.55 Class B 17.16.0.0 /7 Subnets? Hosts? Block Size? 6
Class B 17.16.0.0 /7 Subnets 11 - = 046 Hosts 5 - = 30 Block Size 56-4 = 3 Subnets 17.16.0.0 17.16.0.3 17.16.0.64 17.16.0.96 FHID 17.16.0.1 17.16.0.33 17.16.0.65 17.16.0.97 LHID 17.16.0.30 17.16.0.6 17.16.0.94 17.16.0.16 Broadcast 17.16.0.31 17.16.0.63 17.16.0.95 17.16.0.17 Class B 17.16.0.0 /3 Subnets? Hosts? Block Size? 7
Class B 17.16.0.0 /3 Subnets 7 - = 16 Hosts 9 - = 510 Block Size 56-54 = Subnets 0.0.0 4.0 6.0 FHID 0.1.1 4.1 6.1 LHID 1.54 3.54 5.54 7.54 Broadcast 1.55 3.55 5.55 7.55 Class B 17.16.0.0 /4 Subnets? Hosts? Block Size? 8
Class B 17.16.0.0 /4 Subnets 8 - = 54 Hosts 8 - = 54 Block Size 56-55 = 1 Subnets 0.0 1.0.0 3.0 FHID 0.1 1.1.1 3.1 LHID 0.54 1.54.54 3.54 Broadcast 0.55 1.55.55 3.55 Class B 17.16.0.0 /5 Subnets? Hosts? Block Size? 9
Class B 17.16.0.0 /5 Subnets 9 - = 510 Hosts 7 - = 16 Block Size 56-18 = 18 Subnets 0.0 0.18 1.0 1.18.0.18 FHID 0.1 0.19 1.1 1.19.1.19 LHID 0.16 0.54 1.16 1.54.16.54 Broadcast 0.17 0.55 1.17 1.55.17.55 Find out Subnet and Broadcast Address 17.16.85.30/9 30
Find out Subnet and Broadcast Address 17.30.101.6/3 Find out Subnet and Broadcast Address 17.0.10.80/4 31
Exercise Find out the mask which gives 100 subnets for class B Exercise Find out the Mask which gives 100 hosts for Class B 3
Class A 10.0.0.0 /10 Subnets? Hosts? Block Size? Class A 10.0.0.0 /10 Subnets - = Hosts - = 419430 Block Size 56-19 = 64 Subnets 10.0 10.64 10.18 10.19 FHID 10.0.0.1 10.64.0.1 10.18.0.1 10.19.0.1 LHID 10.63.55.54 10.17.55.54 10.191.55.54 10.54.55.54 Broadcast 10.63.55.55 10.191.55.55 10.54.55.55 33
Class A 10.0.0.0 /18 Subnets? Hosts? Block Size? Class A 10.0.0.0 /18 Subnets 10 - = 10 Hosts 14 - = 1638 Block Size 56-19 = 64 Subnets 10.0.0.0 10.0.64.0 10.0.18.0 10.0.19.0 FHID 10.0.0.1 10.0.64.1 10.0.18.1 10.0.19.1 LHID 10.0.63.54 10.0.17.54 10.0.191.54 10.0.54.54 Broadcast 10.0.63.55 10.0.17.55 10.0.191.55 10.0.54.55 34
Broadcast Addresses Exercise Address Subnet Mask Class Subnet Broadcast 01..10.60 55.55.55.48 15.16.193.6 55.55.48.0 18.16.3.13 55.55.55.5 153.50.6.7 55.55.55.18 Broadcast Addresses Exercise Answers Address Subnet Mask Class Subnet Broadcast 01..10.60 55.55.55.48 C 01..10.56 01..10.63 15.16.193.6 55.55.48.0 A 15.16.19.0 15.16.199.55 18.16.3.13 55.55.55.5 B 18.16.3.1 18.16.3.15 153.50.6.7 55.55.55.18 B 153.50.6.0 153.50.6.17 35
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 Mumbai Sydney Singapore WAN 1 WAN WAN 3 60 Hosts 8 Hosts 1 Hosts 1 Hosts Hosts Hosts Hosts 36
ing Requirements Bangalore 60 WAN 1 WAN WAN 3 Mumbai 60 Sydney 60 Singapore 60 In the example above, a /6 was used to provide the 60 addresses for Bangalore and the other LANs. There are no addresses left for WAN links ing Scheme Mumbai 19.168.10.64/7 8 WAN 19.168.10.19 and 130 WAN 19.198.10.133 and 134 19.168.10.18/30 19.168.10.13/30 WAN 19.198.10.137 and 138 19.168.10.136/30 60 1 1 Bangalore 19.168.10.0/6 Sydney 19.168.10.96/8 Singapore 19.168.10.11/8 37
VLSM Exercise 40 1 5 19.168.1.0 VLSM Exercise 19.168.1.64/6 19.168.1.8/30 40 19.168.1.16/8 1 19.168.1.4/30 19.168.1.1/30 5 19.168.1.3/7 19.168.1.0 38
VLSM Exercise 8 5 35 15 19.168.1.0 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 39
Summarization Supernetting 16 8 4 1 Subnet 17.16.1.0 11000000 10101000 00001100 00000000 17.16.13.0 11000000 10101000 00001101 00000000 17.16.14.0 11000000 10101000 00001110 00000000 17.16.15.0 11000000 10101000 00001111 00000000 55.55.55.0 00000000 40
Supernetting 16 8 4 1 Subnet 17.16.1.0 11000000 10101000 00001100 00000000 17.16.13.0 11000000 10101000 00001101 00000000 17.16.14.0 11000000 10101000 00001110 00000000 17.16.15.0 11000000 10101000 00001111 00000000 55.55.5.0 11111100 00000000 17.16.1.0/4 17.16.13.0/4 17.16.14.0/4 17.16.15.0/4 17.16.1.0/ Supernetting Question What is the most efficient summarization that TK1 can use to advertise its networks to TK? A. 17.1.4.0/417.1.5.0/417.1.6.0/417.1.7.0/4 B. 17.1.0.0/ C. 17.1.4.0/517.1.4.18/517.1.5.0/417.1.6.0/417.1.7.0/4 D. 17.1.0.0/1 E. 17.1.4.0/ 41