Subnetting & CIDR Fakultas Rekayasa Industri Institut Teknologi Telkom
Soal 1 Diketahui IP Address 172.128.127.24 dengan netmask 255.255.255.240. tentukanlah network address dengan broadcast address yang digunakan oleh IP address tersebut.
Soal 2 Diketahui IP address 172.128.127.24 dengan netmask 255.255.240.0. tentukan Broacast dan Networknya dan ada berapa jumlah subnetwork yang dapat dibentuk dari kombinasi IP Address dan netmask tersebut!
Soal 3 Dua buah network yaitu 131.0.0.0/8 dan 131.7.0.0/16 akan di aggregate menjadi satu. Tentukan netmask hasil supernetting untuk kedua network tsb Network Address dan Broadcast Addressnya
Soal Tugas Dikumpul 9 Des 2011 pukul 08.30-09.00 WIB di C203 Sebuah perusahan IT bernama PT. Majuterus hendak membangun jaringan internet yang terdiri atas 4 buah divisi. Divisi marketing, divisi produk, divisi IT dan divisi keungan. Total IP address yang telah disiapkan untuk divisi marketing 58 komputer, divisi produk 28, divisi IT 13 komputer, divisi keuangan 5 komputer. IP yang digunakan adalah IP kelas C yaitu 192.168.2.0. bagaimana konfigurasi IP untuk masing-masing divisi.
Address Translation IPv4 Network IPv4 Header Packet ` IPv6 Network IPv4 Network IPv6 Network IPv6 Header Packet IPv6 Header Packet IPv4 Header Packet IPv4 Header Packet 6
Subnetting & Supernetting
Mengapa SubNetting? SubNetting adalah proses membagi sebuah network menjadi beberapa Sub-network. Sebagai contoh, dalam sebuah jaringan lokal yang menggunakan alamat kelas B 172.16.0.0 terdapat 65.534 host address. Efisiensi pengelolaan jaringan dapat ditingkatkan dengan cara melakukan subnetting terhadap network tersebut.
Mengapa SubNetting (Cont.) Alasan-alasan perlunya dibentuk subnetting antara lain : - Memudahkan pengelolaan jaringan. - Mereduksi traffic yang disebabkan oleh broadcast maupun benturan (collision). - Membantu pengembangan jaringan ke jarak geografis yang lebih jauh (LAN ke MAN).
Subnetting The increasing number of host connected to the internet Restrictions on the network size In subnetting, a network is divided into smaller subnetworks with each subnet having its own subnet address In supernetting, a organization can combine several class C to create a large range of addresses. In other word, several networks are combined to create a supernetwork 10
Without subnetting 11
With Subnettiing 12
Masking 13
Applying bit-wise-and operation to achieve masking 14
Ilustrasi sebuah Network tanpa Subnet
SubNetting Pembentukan subnet dilakukan dengan cara mengambil beberapa bit pada bagian HostId untuk dijadikan SubnetId. Contoh: Source: www.tcpipguide.com
Subnet Mask Source: www.tcpipguide.com
Subnet Mask (Cont.) Dalam contoh di atas, sebuah jaringan kelas B dengan Network-Id : 154.71.0.0. Subnet Mask dalam bentuk desimal adalah: 255.255.248.0 Dengan demikian 5 bit pertama pada octet ke 3 adalah Subnet-Id, sedangkan sisa bit adalah Host-Id.
Default Subnet-Mask
Konversi Subnet-Mask 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
Menentukan SubNet-Id Source: www.tcpipguide.com
Menentukan Subnet-Id Router menentukan sebuah IP address merupakan anggota dari subnet tertentu melalui proses masking seperti dalam gambar di atas. IP address: 154.71.150.42 dioperasikan AND dengan subnetmask. Didapat Subnet-Id: 18. Sedangkan IP address dari subnet tersebut adalah: 154.71.144.0.
IP Address dari Subnet Determining the Subnet ID of an IP Address Through Subnet Masking Component Octet 1 Octet 2 Octet 3 Octet 4 IP Address Subnet Mask Result of AND Masking 10011010 (154) 11111111 (255) 10011010 (154) 01000111 (71) 11111111 (255) 01000111 (71) 10010110 (150) 11111000 (248) 10010000 (144) Dengan CIDR, dapat dituliskan sebagai: 154.71.150.42/21. 00101010 (42) 00000000 (0) 00000000 (0)
Contoh Kasus 1 Sebuah jaringan dengan network-id: 192.16.9.0 akan dibagi ke dalam 3 buah subnet. Tentukan IP address untuk setiap subnet. No IP 192.16.9.0 adalah Kelas C, dengan host-id berada pada 8 bit terakhir. Karena itu, subnet-id harus berada pada 8 bit terakhir.
Penyelesaian Kasus 1 Kebutuhan 3 subnet berarti membutuhkan sebanyak 3 bit. Karena itu subnet-mask ditentukan: 11111111.11111111.11111111.11100000 255. 255. 255. 224
Penyelesaian Kasus 1 Kombinasi subnet: 000, 001, 010, 011, 100, 101, 110, 111. Karena itu 3 bit pertama dialokasikan untuk subnet. 192.16.9.b b b b b b b b subnet
Penyelesaian Kasus 1: Subnet Host Decimal 000 00000-11111 0-31 001 00000 11111 32 63 010 00000 11111 64 95 011 00000 11111 96-127 100 00000 11111 128-159 101 00000 11111 160 191 110 00000 11111 192 223 111 00000-11111 224-255
Kesimpulan Kasus 1 Jumlah subnet yang terbentuk ada 2 3 =8. Tetapi subnet 000 dan 111 tidak dapat digunakan. Karena itu jumlah subnet yang dapat digunakan adalah: (2 3-2=6). Jumlah host yang terbentuk untuk masing-masing subnet 2 5 =32. Sedang host yang dapat digunakan sebanyak 2 5-2=30. Host-Id: 00000 dan 11111 tidak dapat digunakan.
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Addressing example The example given in the curriculum shows subnetting without VLSM using 172.16.0.0/22. (172.16.0.0 172.16.3.255) They produce 4 subnets each with 510 addresses. This is impossible. It will be corrected. You can do it if you start with 172.16.0.0/21 (172.16.0.0 30
Addressing example no VLSM 172.16.0.0/21 31
What we have and need Given IP address 172.16.0.0/21 That s 172.16.0.0 to 172.16.7.255 4 subnets needed: Student LAN has 481 hosts Instructor LAN has 69 hosts Administrator LAN has 23 hosts WAN has 2 hosts 32
Without VLSM same size subnets Biggest subnet has 481 hosts. Formula for hosts is 2 n 2 n = 9 gives 510 hosts (n = 8 gives only 254) So 9 host bits needed. That means 32 9 = 23 network bits /23 or subnet mask 255.255.254.0 33
Network addresses /23 so subnet mask in binary is 11111111 11111111 11111110.00000000 Octet 3 is the interesting one. Value of last network bit in octet 3 is 2 So network numbers go up in 2s 172.16.0.0 172.16.2.0 172.16.4.0 172.16.6.0 34
Subnet with no VLSM Network Subnet address Host range Student 172.16.0.0/23 172.16.0.1-172.16.1.254 Instructor 172.16.2.0/23 172.16.2.1-172.16.3.254 Admin 172.16.4.0/23 172.16.4.1-172.16.5.254 WAN 172.16.6.0/23 172.16.6.1-172.16.7.254 Broadcast address 172.16.1.255 172.16.3.255 172.16.5.255 172.16.7.255 35
Addressing example with VLSM 172.16.0.0/22 is OK 36
What we have and need Given IP address 172.16.0.0/22 That s 172.16.0.0 to 172.16.3.255 4 subnets needed: Student LAN has 481 hosts Instructor LAN has 69 hosts Administrator LAN has 23 hosts WAN has 2 hosts 37
With VLSM Student subnet has 481 hosts. Formula for hosts is 2 n 2 n = 9 gives 510 hosts (n = 8 gives only 254) So 9 host bits needed. That means 32 9 = 23 network bits /23 or subnet mask 255.255.254.0 Network address 172.16.0.0 Broadcast address 172.16.1.255 38
With VLSM Instructor subnet has 69 hosts. Formula for hosts is 2 n 2 n = 7 gives 126 hosts (n = 6 gives only 62) So 7 host bits needed. That means 32 7 = 25 network bits /25 or subnet mask 255.255.255.128 Network address 172.16.2.0 Broadcast address 172.16.2.127 39
With VLSM Admin subnet has 23 hosts. Formula for hosts is 2 n 2 n = 5 gives 30 hosts (n = 4 gives only 14) So 5 host bits needed. That means 32 5 = 27 network bits /27 or subnet mask 255.255.255.224 Network address 172.16.2.128 Broadcast address 172.16.2.159 40
With VLSM WAN subnet has 2 hosts. Formula for hosts is 2 n 2 n = 2 gives 2 hosts So 2 host bits needed. That means 32 2 = 30 network bits /30 or subnet mask 255.255.255.252 Network address 172.16.2.160 Broadcast address 172.16.2.163 41
Visually with VLSM 172.16.0.0 172.16.1.0 172.16.2.0 172.16.3.0 Student Instructor Admin WAN 42
Case 2. Given 192.168.1.0/24 43
Subnet 192.168.1.0/24 2 subnets with 28 hosts each (largest) 5 host bits 2 5-2 = 30 would be just enough But allow for expansion: 6 host bits give 62 Network Subnet address Host range Broadcast address Network bits 32-6 = 26 B 192.168.1.0/26 192.168.1.1 so /26 or subnet mask - 192.168.1.62 255.255.255.192 E 192.168.1.64/26 192.168.1.65-192.168.1.126 192.168.1.63 192.168.1.127 44
Subnet 192.168.1.1/24 1 subnets with 14 hosts 4 host bits 2 4-2 = 14 would be just enough But allow for expansion: 5 host bits give 30 Network bits 32-5 = 27 Network Subnet address Host range Broadcast address so /27 or subnet mask 255.255.255.224 A 192.168.1.128/27 192.168.1.129-0-127 range 192.168.1.158 already used 192.168.1.159 45
Subnet 192.168.1.1/24 1 subnets with 7 hosts 4 host bits 2 4-2 = 14 is enough Network bits 32-4 = 28 so /28 or subnet mask 255.255.255.240 0-159 range already used Network Subnet address Host range Broadcast address D 192.168.1.160/28 192.168.1.161-192.168.1.174 192.168.1.175 46
Subnet 192.168.1.1/24 1 subnets with 2 hosts 2 host bits 2 2-2 = 2 is enough Network bits 32-2 = 30 so /30 or subnet mask 255.255.255.252 0-175 range already used Network Subnet address Host range Broadcast address C 192.168.1.176/30 192.168.1.177-192.168.1.178 192.168.1.179 47
Subnet plan with VLSM Networ k Subnet address Host range Broadcast address B 192.168.1.0/26 192.168.1.1-192.168.1.62 E 192.168.1.64/26 192.168.1.65-192.168.1.126 A 192.168.1.128/27 192.168.1.129-192.168.1.158 D 192.168.1.160/28 192.168.1.161-192.168.1.174 C 192.168.1.176/30 192.168.1.177-192.168.1.178 192.168.1.63 192.168.1.127 192.168.1.159 192.168.1.175 192.168.1.179 48
One octet available Visual B E A D C 49
Subnetting class A A class A address: Is made of a one-byte netid and a three-byte hostid Can have one single physical network with up to 16.777.214 (2 24-2) If we want more physical networks, we can divide this one range into several smaller ranges 50
Example : A organization with a class A needs a least 1000 subnetworks. Find the subnet mask and configuration of each network Solution: We need at least 1002 subnet to allow the all-1s and all-0s subnetids This means that the minimum number of bits to be allocated for subnetting should be 10 (2 9 < 1,002< 10 10 ) Fourteen bits are left to define the hostid 51
Without subnetting masking 255.0.0.0 In binarry : 11111111 00000000 00000000 0000000 Net id Host id With subnetting Masking 255.255.192.0 In binarry : 11111111 11111111 11000000 00000000 Net id SubNet id Host id 52
Theres is 1024 subnets Each subnet can have 16,384 hosts/computer 53
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Subnetting class B A class B: Is made A two byte netid and two-byte hostid Can have one single physical network and up to 65,534 hosts on the network. If we wan more physical network, we can divide this one big range into several smaller ranges. 55
Example An organization with a class B address needs at least 12 subnetwork. Find subnet mask and configuration of each subnetwork Solution: There is a need for at least 14 subnetworks, 12 as specified plus 2 reserve as special address. This means that the minimum number of bits should be 4 (2 3 < 14 < 2 4 ) 56
Without subnetting masking 255.0.0.0 In binarry : 11111111 11111111 00000000 0000000 Net id Host id With subnetting Masking 255.255.240.0 In binarry : 11111111 11111111 11110000 00000000 Net id SubNet id Host id 57
Each subnet 4094 hosts/computers 58
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Subnetting class C A class C address: is made of a three byte netid and one-byte hostid Can have one single physical network and up to 254 (2 8 2) host on that network If we want more physical network, we can divide this one range into several smaller range. 60
Example: An organization with a class C needs at least five subnetworks. Find the subnet mask and configuration of each subnetwork Solution: There is a need for at least seven subnetworks, five specifief and two reserved a special address. This means that minimum number of bits should be 3 ( 2 2 < 7 < 2 3 ) 61
Without subnetting masking 255.0.0.0 In binarry : 11111111 11111111 11111111 0000000 Net id Host id With subnetting Masking 255.255.240.0 In binarry : 11111111 11111111 11111111 11100000 Net id Host id SubNet id 62
There is 8 subnet Each subnet can have 32 hosts 63
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Supernetting Depend on the need of an organization One or more classes c can be jointed to make one supernetwork Example: an organization that needs 1000 address can be granted four class c addresses. The organization can then use these address in one supernetwork, in four network, or in more then four networks. 65
Supernetting 66
Supernet Mask Can be assigned to a block of class C network address, if the number of net. Address is a power of two Default mask for a class C address 255.255.255.0 If some of the 1s are changed to 0s, we can have a mask for a group of class C Supernet mask is the reverseof the subnet mask 67
The beginning address can be X.Y.32.0, but it can not be X.Y.33.0 68
Example: With supernet 255.255.252.0, we can have four class C combined into one supernetwork If we choose the first address to be X.Y.32.0, the other three addres X.Y.33.0, X.Y.34.0 and X.Y.35.0 If the router recieves a packet, it applies the supernet mask to the destination address and compare the result to the lowest address. If the result and the lowest address are the same, the packet belong to the supernet Suppose a packet arrives with destination address X.Y.33.4. After applying the mask, the result is X.Y.32.0 (the lowest address), the packet belong to the supernet 69
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CIDR Classless Interdomain Routing 72