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Routing within an Organization Subnets are a subset of the entire network Networks can be divided into subnets Subnets can be divided into subnets Each subnet is treated as a separate network Problem: Organizations have multiple networks which are independently managed Solution 1: Allocate a separate network address for each network ▪ Difficult to manage ▪ From the outside of the organization, each network must be addressable. Solution 2: Add another level of hierarchy to the IP addressing structure University Network Engineering School Medical School Library Subnetting A subnet can be divided into subnets If the first layer of subnetting used 2 bits for the subnet portion, a second layer can be used to subnet within each subnet ▪ Using more bits from the host portion of the address Layer 1 Layer 2 Subnet 1 Subnet 2a Network 1 Subnet 2 Subnet 2b Subnet 3 Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. Solution The class is A because the first byte is between 0 and 127. The block has a netid of 17. The addresses range from 17.0.0.0 to 17.255.255.255. Given the network address 132.21.0.0, find the class, the block, and the range of addresses. Solution The class is B, the block is 132.21, and the range is 132.21.0.0 to 132.21.255.255 Given the network address 220.34.76.0, find the class, the block, and the range of addresses Solution The class is C, the block is 220.34.76, and the range of addresses is 220.34.76.0 to 220.34.76.255 Given an address from a block of addresses, we can find the network address by AND-ing with a mask. Default masks The network address is the beginning address of each block. It can be found by applying the default mask to any of the addresses in the block (including itself). It retains the netid of the block and sets the hostid to zero. Given the address 23.56.7.91, find the beginning address (network address). Solution The default mask is 255.0.0.0, which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is 23.0.0.0. Given the address 132.6.17.85, find the beginning address (network address). Solution The default mask is 255.255.0.0, which means that the first 2 bytes are preserved and the other 2 bytes are set to 0s. The network address is 132.6.0.0 Given the address 201.180.56.5, find the beginning address (network address). Solution The default mask is 255.255.255.0, which means that the first 3 bytes are preserved and the last byte is set to 0. The network address is 201.180.56.0 Class A (24 bits for hosts) 224 - 2* = 16,777,214 maximum hosts Class B (16 bits for hosts) 216 - 2* = 65,534 maximum hosts Class C (8 bits for hosts) 28 - 2* = 254 maximum hosts * Subtracting the network and broadcast reserved address An IP address such as 176.10.0.0 that has all binary 0s in the host bit positions is reserved for the network address. An IP address such as 176.10.255.255 that has all binary 1s in the host bit positions is reserved for the broadcast address. Is the act of “borrowing” bits from the host portion to create smaller networks (called subnetworks) Minimum bits that can be borrowed is 2. Subnetting is used to reduce the number of broadcast domains Communication between these subnetworks is achieved through a router 19 Determines which part of an IP address is the network field and which part is the host field Follow these steps to determine the subnet mask: 1. Express the subnetwork IP address in binary form. 2. Replace the network and subnet portion of the address with all 1’s. 3. Replace the host portion of the address with all 0’s. 4. Convert the binary expression back to dotted-decimal notation. Subnet mask in decimal = 255.255.240.0 Imagine that you have a class "B” network. This time however, instead of borrowing all eight bits of the third octet, only seven bits are borrowed to create subnetworks. Using binary representation, in this example, the subnet mask would be 11111111.11111111.11111110.00000000. Therefore, 255.255.255.0 can no longer be used as the subnet mask. Class B address with 8 bits borrowed for the subnet 130.5.2.144 (8 bits borrowed for subnetting) routes to subnet 130.5.2.0 rather than just to network 130.5.0.0. Class C address 197.15.22.131 with a subnet mask of 255.255.255.224 (3 bits borrowed) 11000101 00001111 00010110 100 Network Field SN 00011 Host Field The address 197.15.22.131 would be on the subnet 197.15.22.128. The number of lost IP addresses with a Class C network depends on the number of bits borrowed for subnetting. How do we determine how many bits to “borrow” for a subnet? Determine the number of sub networks required Work from the MOST significant (LHS) bits of the first octet after the network number and calculate the number of bits needed to create the required number of subnetworks 31 Example: You are given a class B address and you are required to create 1000 subnetworks. By working from the LHS of the (first octet after the network number) 3rd octet, calculate the number of bits to equal or slightly exceed 1000. (ie 2x = > 1000) This would equate to 210 or 1024-2 networks Hence you will need to borrow 10 bits from the host portion to create 1000 subnetworks 32 Example: The subnetmask in this instance would be 255.255.255.192 Natural Class B netmask 11111111 11111111 10 Bits Borrowed (subnetmask) 11111111 6 bits left for hosts 11 000000 How many host per network can you obtain from this addressing scheme? 33 How do we determine how many bits to “borrow” for a subnet given the number of hosts required? Determine the number of hosts required Work from the LEAST significant (RHS) bits of the last octet and calculate the number of bits needed to create the required number of subnetworks. 34 Example: You are given a class B address and you require 1000 nodes per subnet By working from the RHS (last octet) of the 4th octet, calculate the number of bits to equal or slightly exceed 1000. (ie 2x = > 1000) This would equate to 210 or 1024-2 networks Hence you will need to borrow 6 bits from the host portion to create subnetworks with 1000 hosts each 35 Example: The subnetmask in this instance would be 255.255.252.0 6 Bits Borrowed 10 bits required Natural Class B netmask (subnetmask) for hosts 11111111 11111111 111111 00 00000000 How many subnetworks per network can you obtain from this addressing scheme? 36 Given an IP address, you will usually be given a net/subnetmask If you are given the mask Subtract the mask from 256 This is known as the multiplier The first number in each multiplier value is the network number The broadcast address is the next multiplier value subtract 1 37 E.g. given the IP address 192.168.0.100 with the subnet mask of 255.255.255.240 Or 192.168.0.100/28 What is the network number What is the broadcast address What are the valid IP hosts for the subnet 38 Subtract 256 from the netmask 256 - 240 = 16 This is the multiplier ie the networks are in steps of 16 (16,32,48,64,96,112 etc) The IP address 192.168.0.100 is in the range of |100 / 16| which is the 6 th subnetwork The network address is 16* 6 = 96 (01100000b) The Broadcast address is 96 + 16 - 1 => 192.168.0.111 (01101111b) ie (next multiplier – 1) 39 Network Portion IP address 192.168.0.100 Netmask 255.255.255.240 SN Host portion 11000000 10101000 00000000 0110 0100 11111111 11111111 11111111 1111 0000 Network Address 11000000 10101000 00000000 192.168.0.96 Broadcast Address 11000000 10101000 00000000 192.168.0.111 0110 0000 0110 1111 40 Valid ranges are 192.168.0.97 to 192.168.0.110 Number of allowable hosts 97 to 110 (incl) = 14 or [24]16 - 2 = 14 ▪ Remember you cannot use the first address (network address) and the last address (broadcast address) in the range The number of allowable networks [24]16 - 2 = 14 ( ie 4 bits used. If a class B address with the last bit subnet, then add another 8 bits to give you 212 –2 allowable subnet) 41 What if the IP range goes over 2 octets Use the same principal Remember octets with all 0’s are considered “boring” and will be assigned the mask of 0 You will then have to locate the position in the address with both 1’s and 0’s (interesting byte) and use the same algorithm Similarly all 1’s are also considered boring and will be given the mask of 255 (e.g. subnetting the last byte of a class B address) 42 Example Organization address 131.181.112.0/22 Netmask expanded : 255.255.252.0 Last byte is “not interesting” hence we set it to “0” for network and “1” for broadcast The third byte is “interesting” 256 – 252 = 4 (multiplier) Networks are in increments of 4 steps 112/4 = 28 (the 28th subnetwork). Since there is no remainder, it is the beginning of the network address. 43 Assignable addresses in Network address : 131.181.112.0 this subnetwork Netmask : 131.181.112.1 – 255.255.252.0 131.181.115.254 Broadcast : Remember, the 1st and 112 + 4 – 1 = 115 last addresses cannot be =>131.181.115.255 used (Network and Number of valid hosts : broadcast). 210 –2 = 1024 – 2 = 1022 hosts Number of subnetworks available for this network 26 – 2 = 64 –2 =62 subnetworks 44 Let's assume that a device on another network with an IP address of 197.15.22.44 wants to send data to another device attached to Cisco's network with an IP address of 131.108.2.2. The data is sent out over the Internet until it reaches the router that is attached to Cisco's network. The router's job is to determine which one of Cisco's subnetworks the data should be routed to. 45 46 when the router performs this ”AND-ing" operation, the host portion falls through. 47 The router looks at what is left which is the network number including the subnetwork. 48 The router then looks in its routing table and tries to match the network number including the subnet with an interface. 49 Imagine that you have a class "B” network with the network number 172.16.0.0. After assessing the needs of his network, the network administrator has decided to borrow eight bits in order to create subnetworks. When eight bits are borrowed to create subnets, the subnet mask is 255.255.255.0. Someone outside the network sends data to the IP address 172.16.2.120. 50 51 52 What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0? Solution We apply the AND operation on the address and the subnet mask. Address ➡ 11001000 00101101 00100010 00111000 Subnet Mask ➡ 11111111 11111111 11110000 00000000 Subnetwork Address ➡ 11001000 00101101 00100000 00000000. or 200.45.32.0 53 You are given an IP address for a host 172.168.35.10/20 What is/are the Subnet address? Broadcast address? The number of useable hosts available for this subnet? The number of useable subnets available for this network? The assignable address range for this subnet? 54 Your organisation has been assigned a class B IP address of 130.10.0.0 You require about 2000 subnetworks Work out the Subnet mask required for this subnet The network and broadcast addresses for the first 5 useable subnets The number of hosts for each subnet The assignable address range of the first 5 useable subnets 55