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7-Lect_7 .pptxNetwork Layer. Addressing Subnetting Mask (default and subnet) Routing
- 1. Computer Networks LEC #2 Communicating Over the Network Computer Networks LEC #7 Network Layer (CONT.)
- 2. Lecture Outlines 2 Network Layer. Addressing Subnetting Mask (default and subnet) Routing Next Lecture
- 3. Remember 32 bit Binary Expressed in DOTTED DECIMAL NOTATION Divided into OCTETS
- 4. Subnet Mask Class First Octet Range Number of Network Bits Number of Host Bits Default Subnet Mask Number of Networks Number of Hosts per Network A 0-127 8 24 255.0.0.0 128 16,777,216 B 128-191 16 16 255.255.0.0 16,384 65,536 C 192-223 24 8 255.255.255.0 2,097,152 256 Each class has a default or "natural" subnet mask based on the default number of bits used for the network and host portion. Netid Hostid Hostid Hostid Netid Netid Hostid Hostid Netid Netid Netid Hostid Class A Class B Class C First Byte Second Byte Third Byte Fourth Byte
- 5. Network Prefixes How do you know the number of bits assigned to the network and the number of bits assigned to the host? Prefix Mask: The address is followed by a number that represents the number of bits (prefix length), beginning from the left, that apply to the network. A slash (/) is used to separate the address and the prefix length. 192.168.10.2/24 Means that the first 24 bits are the network portion. The last 8 bits are the host portion. Class Number of Network Bits Number of Host Bits Default Prefix Default Subnet Mask A 8 24 /8 255.0.0.0 B 16 16 /16 255.255.0.0 C 24 8 /24 255.255.255.0
- 6. Classful IPAddressing – Class C We know from the Class C default subnet mask (255.255.255.0): The first 24 bits are the network number and the last 8 bits are the host numbers. 11111111 01001101 00010100 11010010 255 77. 20. 210. • The last host address (all 1's) is reserved for the broadcast address. 00000000 01001101 00010100 11010010 0 77. 20. 210. • The first host address (all 0's) is reserved for the network address. The number of usable host addresses for the entire network is 28 – 2 = 254 11111110 01001101 00010100 11010010 254 77. 20. 210. 00000001 01001101 00010100 11010010 1 77. 20. 210. The range of available addresses is:
- 7. Example For example: In Class C : Address range: 192 - 223 Number of network bits: 24 Number of networks: 2^24 = 2,097,152 Number of host bits: 8 Number of hosts per network: 28 = 256 Number of Useable Hosts per network: 28 - 2 = 254 Default Subnet Mask: 255.255.255.0 or /24
- 8. Exemple
- 9. Exemple
- 10. Exemple
- 11. Resume
- 12. 12 Subnetting It divides the main network into several subnets Each one with its own subnet address. For example, if you want to take 5 bits from the Host field for subnetting and leaves 3 bits for defining hosts as shown in figure below. Having 5 bits available for defining subnets means that we can have up to 32 (2^5) different subnets. Let a class C network with 256 host addresses. One of these addresses is used to identify the network address and another one is used to identify the broadcast address on the network. Therefore, we are left with 254 addresses available for addressing hosts.
- 13. 13 Subnetting Suppose the IP address = 192.168.10.44 Default Mask = 255.255.255.0 How to calculate subnet using binary method? 1. Convert to binary 2. Calculate the Subnet Address To calculate the IP Address Subnet you need to perform a bit-wise AND operation (1 and 1 = 1, 1 and 0 or 0 and 1 = 0, 0 and 0 = 0) on the host IP address and subnet mask. The result is the subnet address.
- 14. Subnetting 3. Find Host Range
- 15. Subnetting 3. Find Host Range
- 16. Subnetting 4. Calculate the Total Number of Subnets and Hosts Per Subnet So number of subnets is = 2 ^# subnet bits Number of hosts in each subnet is = 2^ # host bits
- 17. Example 211.215.9.115/29 : 29 (24+5) means we will take 5 bits from the host part for the subnetting and leave 3 bits for the hosts in the subnet. So to find the first subnet Suppose the following IP 211.215.9.115/29 find first subnet address, first host in the subnet, Last host in the first subnet, broadcast address of the first subnet and the address of the second subnet. Answer 211.215.9. 115 = 01110011 subnetMask for the 5 bits is 11111000 Bit wise And operation 01110000 = 112 So, the first subnet is 211.215.9.112 First host in the subnet is 211.215.9.113 last host in the subnet is 211.215.9.118 Broadcast address is 211.215.9.119 Second subnet will start 211.215.9.120
- 18. Example Note that if you need for example 3 subnets only from your network, How to compute the number of bits that you need? So, convert number 3 to binary? 2 3 1 So, we need Two bits 1 1 11000110 10101000 11010100 00 000000 Netid (For network) Hostid Subnet 11000110 10101000 11010100 00000000 Netid (For network) Hostid
- 19. Routing To route the router, need to know: 1) Destination address 2) Different routes 3) Best route 4) Maintain and verify routing information
- 20. Routing Each router has a routing table to specify the different networks and best route to this networks To know the best route, router should perform routing process because it only know the direct connected devices. there are two different types of routing : Static and Dynamic Static uses the route that is entered manually by the admin Dynamic uses a network routing protocol adjusts automatically for topology or traffic changes
- 21. Static Route Configure unidirectional static routes to and from a sub network to allow communications to occur. For example, if we have packets on router A and need to travel to network 172.16.1.0 we should configure router A to decide the path to the destination address How? Destination address Subnet Mask Next Hop which means next router
- 22. Static Route How to decide more than route in the static way? The same command is used but with adding distance for each command to specify the priority of each route. The distance value ranged from 0 to 255 and default value is 1. The low value is the best route Distance value
- 23. Dynamic Routing Routing protocols are used to determine paths and maintain routing table such as RIP (Routing Information Protocol). Once the path is determined a router can route a routed protocol. Routing Protocols Interior Protocols Hybrid Routing EIGRP (Enhanced Interior Gateway Routing Protocol) Distance Vector RIP (Routing Information Protocol) Link State OSPF (Open Shortest Path First) Exterior Protocols BGP (Border Gateway Protocol)
- 24. Dynamic Routing – Distance Vector - RIP It uses hop count as a routing metric to find the best path between the source and the destination network. RIP uses port number 520. Hop count is the number of routers occurring in between the source and destination network. The path with the lowest hop count is considered as the best route to reach a network and therefore placed in the routing table. Routers periodically exchange the router information (default exchange time 30 second). if no update comes until 180 seconds, then the destination router considers the other router is out of service. Default distance is 120 (priority of the route). In case there are more than one route between two networks the RIP will prefers the route that has least hop.
- 25. 25 Dynamic Routing – RIP Example Each Router will send its Routing table to the routers in its neighbors. For example, Router C will send its table to Router B The 0 value in last column means number of hops to reach to the network because its direct connection so it’s equal 0. So, what is the content of the table at the end of round?
- 26. 26 So Router B will investigate the incoming table to register the missing information in its table for example routing table of router C have two networks 10.3.0.0 and 10.4.0.0 and in the table of router B there are two networks 10.2.0.0 and 10.3.0.0, So Router B will add network 10.4.0.1 in its table with port S1. Dynamic Routing – RIP Example