Basis Bilangan Basis bilangan hanya ada dua nilai 0 dan DESIMAL 0 3 4 5 6 7 8 9 0 BINER 0 0 00 0 0 000 00 00 DESIMAL 3 4 5 6 7 8 9 0 BINER 0 00 0 0 0000 000 000 00 000 00
00 ( x 4 = 6) + (0 x 3 = 0) + ( x = 4) + ( x = ) + (0 x 0 = 0) = 0 =.. / = sisa 0 /= 5 sisa 5/ = sisa / = sisa 0 tidak bisa dibagi lagi Hasil 00 Untuk Alamat IP 0.0.0.00000 3 bit angka biner Penulisan mengunakan notasi titik, tiap 8 bit dijadikan angka desimal 0.0.0.00000 9.54.53.3
Cara mudah menghitung Bit bernilai besar Bit bernilai kecil Biner Perhitu ngan x 7 x 6 x 5 x 4 x 3 x x x 0 Desimal 8 64 3 6 8 4 Contoh 0.0.0.00000 = 9.54.53.3 Dasar perhitungan : 9 = 0 = 8 + 0 + 3 + 6 + 8 + 4 + + 54 = 0 = 8 + 64 + 3 + 6 + 8 + 4 + + 0 53 = 0 = 8 + 64 + 3 + 6 + 8 + 4 + 0 + 3 = 00000 = 0 + 0 + 0 + 0 + 8 + 4 + 0 + 3
Pengalamatan IP Versi IPv 4 3 bit, dibagi 4 oktet Ditulis dengan angka desimal dengan notasi titik Dasar-dasar Alamat pada TCP/IP 7.8.0. 7.6.0. 7.8.0. 7.6.0. HDR SA DA DATA 0.3.0.0 9.68..0 0.3.0. 7.7.0. 7.7.0. 9.68.. 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 55 55 55 55 8 9 6 7 4 5 3 8 64 3 684 8 64 3 684 8 64 3 684 8 64 3 684 Contoh Decimal Contoh Biner 7 6 04 00000000000 000 0000 Kelas IP 8 Bits 8 Bits 8 Bits 8 Bits Class A: Class B: Class C: Class D: Class E: Multicast Research 5
Bits: kelas A: Bits: Kelas B: Bits: Kelas C: Bits: kelas D: Kelas Alamat IP 8 9 6 7 4 5 3 0NNNNNNN Range (-6) 8 9 6 7 4 5 3 0NNNNNN Range (8-9) 8 9 6 7 4 5 3 0NNNNN Range (9-3) 8 9 6 7 45 3 0MMMM Range (4-39) Multicast Group Multicast Group Multicast Group 7.6.. 7.6.3.0 7.6.. Alamat 0.6.4. E E0 7.6.. 0... 0.50.8. 0.80.30.8 Routing Table 7.6.. Interface 7.6.0.0 0.0.0.0 E0 E 6
Classless Inter-Domain Routing (CIDR) Suatu dasar cara yang dipakai ISPs (Internet Service Providers) untuk mengalokasikan alamat pada perusahaan, pelanggan pribadi, contoh : 9.68.0.3/8 Notasi slash (/) dalam pemisah untuk menuliskan panjang bit alamat jaringan CIDR Values 7
Kententuan yan digunakan untuk alamat 7 6 0 0 6 5 4 3 09 0000 0000000 00000000 00000000 00000000 0000000 00000000 000000... 8 76 5 43... 0 0 N = 6 = 65534 N... 3 65534 65535 65536 65534 IP Address Classes Exercise Address Class 0... 8.63..00 0..5.64 9.6.4. 30.3.64.6 56.4.0.0 8
IP Address Classes Exercise Answers Address Class 0... A 0.0.0.0 0... 8.63..00 B 8.63.0.0 0.0..00 0..5.64 C 0..5.0 0.0.0.64 9.6.4. C 9.6.4.0 0.0.0. 30.3.64.6 B 30.3.0.0 0.0.64.6 56.4.0.0 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
Jumlah Jaringan x - dimana X = nilai bit Rumusan Jumlah y - dimana y = jumlah bit untuk host Block Size = Total number of Address Block Size = 56-Mask Subnetting Classful IP Addressing SNM are a set of 55 s and 0 s. In Binary it s contiguous s and 0 s. SNM cannot be any value as it won t follow the rule of contiguous s and 0 s. Possible subnet mask values 0 8 9 4 40 48 5 54 55 0
Pengalamatan diluar Subnet 7.6.0. 7.6.0. 7.6.0.3 7.6.55.53 7.6.55.54... 7.6.0.0 7.6.0.0 Pengalamatan dalam Subnet 7.6.3.0 7.6.4.0 7.6..0 7.6..0 7.6.0.0
Belum di bagi ke Subnet Addressing 7.6..00 7.6.. 7.6.3. E E0 7.6.. 7.6.3.5 7.6.3.00 7.6..60 7.6.3.50 7.6.. 60 New Routing Table Interface 7.6.0.0 E0 7.6.0.0 E Ke Subnet Addressing 7.6..00 7.6.. 7.6..60 7.6.3. E E0 7.6.. 7.6.3.5 7.6.3.00 7.6.3.50 New Routing Table 7.6.. 60 Interface Subnet 7.6..0 7.6.3.0 E0 E
Subnet Mask IP Address Default Subnet Mask 8-Bit Subnet Mask 7 6 0 0 55 55 0 0 00000000 00000000 Juga bisa dutulis /6, 6 =adalah panjang bit dalam mask Subnet 55 55 55 0 Juga bisa ditulis /4, 4= adalah panjang bit dalam mask Nilai desimal dengan pola bit 8 64 3 6 8 4 0 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 = 8 0 0 0 0 0 0 = 9 0 0 0 0 0 = 4 0 0 0 0 = 40 0 0 0 = 48 0 0 = 5 0 = 54 = 55 3
Net Mask 7.6..60 0000 0000000 0000000 000000 55.55.0.0 00000000 00000000 Number 0000 0000000 00000000 00000000 7 6 0 0 Tidak menggunakan subnet Subnet Mask dalam Subnet Subnet 7.6..60 0000 0000000 0000000 000000 55.55.55.0 00000000 0000 0000000 0000000 00000000 8 9 4 40 48 5 54 55 Number 7 6 0 8 bit digunakan untuk sub net 4
Subnet Mask with Subnets (cont.) Subnet 7.6..60 0000 0000000 0000000 000000 55.55.55.9 000000 0000 0000000 0000000 0000000 8 9 4 40 48 5 54 55 8 9 4 40 48 5 54 55 Number 7 6 8 number extended by ten bits Subnet Mask Exercise Address Subnet Mask Class Subnet 7.6..0 0.6.4.0 0.30.36. 55.55.55.0 55.55.40.0 55.55.55.0 5
Subnet Mask Exercise Answers Address Subnet Mask Class Subnet 7.6..0 55.55.55.0 B 7.6..0 0.6.4.0 55.55.40.0 A 0.6.6.0 0.30.36. 55.55.55.0 A 0.30.36.0 Broadcast Addresses 7.6.3.0 7.6.4.0 7.6..0 7.6.3.55 (Directed Broadcast) 55.55.55.55 X (Local Broadcast) 7.6.55.55 (All Subnets Broadcast) 7.6..0 6
Addressing Summary Example 7 6 60 3 7.6..60 0000 0000000 0000000 000000 55.55.55.9 9 8 7.6..8 0000 0000000 0000000 000000 0000000 Mask Subnet 4 7.6..9 7.6..9 0000 0000000 0000000 0 0000 0000000 0000000 000000 Broadcast 5 First 6 7.6..90 0000 0000000 0000000 00 Last 7 Class B Subnet Example IP Address: 7.6.. Subnet Mask: 55.55.55.0 Subnet 7.6..: 0000 0000000 0000000 000 55.55.55.0: 00000000 Subnet: 0000 0000000 0000000 00000000 Broadcast: 0000 0000000 0000000 Subnet Address = 7.6..0 Addresses = 7.6.. 7.6..54 Broadcast Address = 7.6..55 Eight Bits of Subnetting 7
Subnet Planning 0 Subnets 5 s per Subnet Class C Address: 9.68.5.0 Other Subnets 9.68.5.6 9.68.5.3 9.68.5.48 Class C Subnet Planning Example IP Address: 9.68.5. Subnet Mask: 55.55.55.48 Subnet 9.68.5.: 000000 55.55.55.48: 00000 000000 000 000 Subnet: 000000 00000 000000 0000 Broadcast: 000000 00000 000000 0 Subnet Address = 9.68.5.0 Addresses = 9.68.5. 9.68.5.6 Broadcast Address = 9.68.5.7 Five Bits of Subnetting 8
9.68.0.0 /7 Exercise? SNM? Block Size?- Subnets Exercise /7? SNM 4? Block Size = 56-4 = 3?- Subnets Subnets 0.0 0.3 0.64 FHID 0. 0.33 LHID 0.30 0.6 Broadcast 0.3 0.63 9
9.68.0.0 /30 Exercise? SNM? Block Size?- Subnets Exercise /30? SNM 5? Block Size = 56-5 = 4?- Subnets Subnets 0.0 0.4 0.8 FHID 0. 0.5 LHID 0. 0.6 Broadcast 0.3 0.7 0
Exercise /6 /7 /8 /9 /30 Mask????? Subnets?????????? Exercise /6 /7 /8 /9 /30 Mask 9 4 40 48 5 Subnets 4 8 6 3 64 6 30 4 6
Exam Question Find Subnet and Broadcast address 9.68.0.00/7 9.68.0.54 /9 Mask? Subnet? Broadcast? Exercise
9.68.0.30 /8 Mask? Subnet? Broadcast? Exercise 9.68.0.93 /30 Mask? Subnet? Broadcast? Exercise 3
9.68..00 /6 Mask? Subnet? Broadcast? Exercise 9.68.0.58 /7 Mask? Subnet? Broadcast? Exercise 4
Class B 7.6.0.0 /9 Subnets? s? Block Size? Class B 7.6.0.0 /9 Subnets 3 - = 6 s 3 - = 890 Block Size 56-4 = 3 Subnets 0.0 3.0 64.0 96.0 FHID 0. 3. 64. 96. LHID 3.54 63.54 95.54 7.54 Broadcast 3.55 63.55 95.55 7.55 5
Class B 7.6.0.0 /7 Subnets? s? Block Size? Class B 7.6.0.0 /7 Subnets - = 046 s 5 - = 30 Block Size 56-4 = 3 Subnets 0.0 0.3 0.64 0.96 FHID 0. 0.33 0.65 0.97 LHID 0.30 0.6 0.94 0.6 Broadcast 0.3 0.63 0.95 0.7 6
Class B 7.6.0.0 /3 Subnets? s? Block Size? 7.6.0.0 /3 Subnets 7 - = 6 s 9 - = 50 Block Size 56-54 = Class B Subnets 0.0.0 4.0 6.0 FHID 0.. 4. 6. LHID.54 3.54 5.54 7.54 Broadcast.55 3.55 5.55 7.55 7
Class B 7.6.0.0 /4 Subnets? s? Block Size? 7.6.0.0 /4 Subnets 8 - = 54 s 8 - = 54 Block Size 56-55 = Class B Subnets 0.0.0.0 3.0 FHID 0... 3. LHID 0.54.54.54 3.54 Broadcast 0.55.55.55 3.55 8
Class B 7.6.0.0 /5 Subnets? s? Block Size? 7.6.0.0 /5 Subnets 9 - = 50 s 7 - = 6 Block Size 56-8 = 8 Class B Subnets 0.0 0.8.0.8.0.8 FHID 0. 0.9..9..9 LHID 0.6 0.54.6.54.6.54 Broadcast 0.7 0.55.7.55.7.55 9
Find out Subnet and Broadcast Address 7.6.85.30/0 Find out Subnet and Broadcast Address 7.6.85.30/9 30
Find out Subnet and Broadcast Address 7.30.0.6/3 Find out Subnet and Broadcast Address 7.0.0.80/4 3
Exercise Find out the mask which gives 00 subnets for class B Exercise Find out the Mask which gives 00 hosts for Class B 3
Class A 0.0.0.0 /0 Subnets? s? Block Size? Class A 0.0.0.0 /0 Subnets - = s - = 49430 Block Size 56-9 = 64 Subnets 0.0 0.64 0.8 0.9 FHID 0.0.0. 0.64.0. 0.8.0. 0.9.0. LHID 0.63.55.54 0.7.55.54 0.9.55.54 0.54.55.54 Broadcast 0.63.55.55 0.7.55.55 0.9.55.55 0.54.55.55 33
Class A 0.0.0.0 /8 Subnets? s? Block Size? Class A 0.0.0.0 /8 Subnets 0 - = 0 s 4 - = 638 Block Size 56-9 = 64 Subnets 0.0.0.0 0.0.64.0 0.0.8.0 0.0.9.0 FHID 0.0.0. 0.0.64. 0.0.8. 0.0.9. LHID 0.0.63.54 0.0.7.54 0.0.9.54 0.0.54.54 Broadcast 0.0.63.55 0.0.7.55 0.0.9.55 0.0.54.55 34
Broadcast Addresses Exercise Address Subnet Mask Class Subnet Broadcast 0..0.60 55.55.55.48 5.6.93.6 55.55.48.0 8.6.3.3 55.55.55.5 53.50.6.7 55.55.55.8 Broadcast Addresses Exercise Answers Address Subnet Mask Class Subnet Broadcast 0..0.60 55.55.55.48 C 0..0.56 0..0.63 5.6.93.6 55.55.48.0 A 5.6.9.0 5.6.99.55 8.6.3.3 55.55.55.5 B 8.6.3. 8.6.3.5 53.50.6.7 55.55.55.8 B 53.50.6.0 53.50.6.7 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 60 s Mumbai 8 s Sydney s Singapore s WAN s WAN s WAN 3 s 36
ing Requirements Bangalore 60 WAN 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 9.68.0.64/7 8 WAN 9.68.0.9 and 30 WAN 9.98.0.33 and 34 9.68.0.8/30 9.68.0.3/30 WAN 9.98.0.37 and 38 9.68.0.36/30 Bangalore 9.68.0.0/6 60 Sydney 9.68.0.96/8 Singapore 9.68.0./8 37
VLSM Exercise 40 5 9.68..0 VLSM Exercise 9.68..64/6 9.68..8/30 9.68..6/8 40 9.68..4/30 9.68../30 9.68..0 5 9.68..3/7 38
VLSM Exercise 8 5 35 5 9.68..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 6 8 4 Subnet 7.6..0 000000 00000000000 00000000 7.6.3.0 000000 0000000000 00000000 7.6.4.0 000000 0000000000 00000000 7.6.5.0 000000 000000000 00000000 55.55.55.0 00000000 40
Supernetting 6 8 4 Subnet 7.6..0 000000 00000000000 00000000 7.6.3.0 000000 0000000000 00000000 7.6.4.0 000000 0000000000 00000000 7.6.5.0 000000 000000000 00000000 55.55.5.0 00 00000000 7.6..0/4 7.6.3.0/4 7.6.4.0/4 7.6.5.0/4 7.6..0/ Supernetting Question 7..4.8/5 7..4.8/5 7..5.0/4 7..7.0/4 7..6.0/4 What is the most efficient summarization that TK can use to advertise its networks to TK? A. 7..4.0/47..5.0/47..6.0/47..7.0/4 B. 7..0.0/ C. 7..4.0/57..4.8/57..5.0/47..6.0/47..7.0/4 D. 7..0.0/ E. 7..4.0/ 4