7/3/2015 Unit-3 : Network Layer 1 CS 1302 Computer Networks — Unit - 3 — — Network Layer — Text Book Behrouz.A. Forouzan, “Data communication and Networking”, Tata McGrawHill, 2004

Network Layer 7/3/20152Unit-3 : Network Layer

Position of network layer 7/3/20153Unit-3 : Network Layer

Network layer duties 7/3/20154Unit-3 : Network Layer

Host-to-Host Delivery: Internetworking, Addressing, and Routing 7/3/20155Unit-3 : Network Layer

19.1 Internetworks Need For Network Layer Internet As A Packet-Switched Network Internet As A Connectionless Network 7/3/20156Unit-3 : Network Layer

Figure 19.1 Internetwork 7/3/20157Unit-3 : Network Layer

Figure 19.2 Links in an internetwork 7/3/20158Unit-3 : Network Layer

Figure 19.3 Network layer in an internetwork 7/3/20159Unit-3 : Network Layer

Figure 19.4 Network layer at the source 7/3/201510Unit-3 : Network Layer

Figure 19.5 Network layer at a router 7/3/201511Unit-3 : Network Layer

Figure 19.6 Network layer at the destination 7/3/201512Unit-3 : Network Layer

Figure 19.7 Switching 7/3/201513Unit-3 : Network Layer

Figure 19.8 Datagram approach 7/3/201514Unit-3 : Network Layer

Switching at the network layer in the Internet is done using the datagram approach to packet switching. Note: 7/3/201515Unit-3 : Network Layer

Communication at the network layer in the Internet is connectionless. Note: 7/3/201516Unit-3 : Network Layer

19.2 Addressing Internet Address Classful Addressing Supernetting Subnetting Classless Addressing Dynamic Address Configuration Network Address Translation 7/3/201517Unit-3 : Network Layer

An IP address is a 32-bit address. Note: 7/3/201518Unit-3 : Network Layer

The IP addresses are unique and universal. Note: 7/3/201519Unit-3 : Network Layer

Figure 19.9 Dotted-decimal notation 7/3/201520Unit-3 : Network Layer

The binary, decimal, and hexadecimal number systems are reviewed in Appendix B. Note: 7/3/201521Unit-3 : Network Layer

Example 1 Change the following IP addresses from binary notation to dotted- decimal notation. a b Solution We replace each group of 8 bits with its equivalent decimal number (see Appendix B) and add dots for separation: a b /3/201522Unit-3 : Network Layer

Example 2 Change the following IP addresses from dotted-decimal notation to binary notation. a b Solution We replace each decimal number with its binary equivalent (see Appendix B): a b /3/201523Unit-3 : Network Layer

In classful addressing, the address space is divided into five classes: A, B, C, D, and E. Note: 7/3/201524Unit-3 : Network Layer

Figure Finding the class in binary notation 7/3/201525Unit-3 : Network Layer

Figure Finding the address class 7/3/201526Unit-3 : Network Layer

Example 3 Find the class of each address: 0 a b Solution See the procedure in Figure a.The first bit is 0; this is a class A address. b.The first 4 bits are 1s; this is a class E address. 7/3/201527Unit-3 : Network Layer

Figure Finding the class in decimal notation 7/3/201528Unit-3 : Network Layer

Example 4 Find the class of each address: a b c Solution a.The first byte is 227 (between 224 and 239); the class is D. b.The first byte is 252 (between 240 and 255); the class is E. c.The first byte is 134 (between 128 and 191); the class is B. 7/3/201529Unit-3 : Network Layer

Figure Netid and hostid 7/3/201530Unit-3 : Network Layer

Figure Blocks in class A 7/3/201531Unit-3 : Network Layer

Millions of class A addresses are wasted. Note: 7/3/201532Unit-3 : Network Layer

Figure Blocks in class B 7/3/201533Unit-3 : Network Layer

Many class B addresses are wasted. Note: 7/3/201534Unit-3 : Network Layer

The number of addresses in class C is smaller than the needs of most organizations. Note: 7/3/201535Unit-3 : Network Layer

Figure Blocks in class C 7/3/201536Unit-3 : Network Layer

Figure Network address 7/3/201537Unit-3 : Network Layer

In classful addressing, the network address is the one that is assigned to the organization. Note: 7/3/201538Unit-3 : Network Layer

Example 5 Given the address , find the network address. Solution The class is A. Only the first byte defines the netid. We can find the network address by replacing the hostid bytes ( ) with 0s. Therefore, the network address is /3/201539Unit-3 : Network Layer

Example 6 Given the address , find the network address. Solution The class is B. The first 2 bytes defines the netid. We can find the network address by replacing the hostid bytes (17.85) with 0s. Therefore, the network address is /3/201540Unit-3 : Network Layer

Example 7 Given the network address , find the class. Solution The class is A because the netid is only 1 byte. 7/3/201541Unit-3 : Network Layer

A network address is different from a netid. A network address has both netid and hostid, with 0s for the hostid. Note: 7/3/201542Unit-3 : Network Layer

Figure Sample internet 7/3/201543Unit-3 : Network Layer

IP addresses are designed with two levels of hierarchy. Note: 7/3/201544Unit-3 : Network Layer

Figure A network with two levels of hierarchy 7/3/201545Unit-3 : Network Layer

Figure A network with three levels of hierarchy (subnetted) 7/3/201546Unit-3 : Network Layer

Figure Addresses in a network with and without subnetting 7/3/201547Unit-3 : Network Layer

Figure Hierarchy concept in a telephone number 7/3/201548Unit-3 : Network Layer

Table 19.1 Default masks Class In Binary In Dotted- Decimal Using Slash A /8 B /16 C /24 7/3/201549Unit-3 : Network Layer

The network address can be found by applying the default mask to any address in the block (including itself). It retains the netid of the block and sets the hostid to 0s. Note: 7/3/201550Unit-3 : Network Layer

Example 8 A router outside the organization receives a packet with destination address Show how it finds the network address to route the packet. Solution The router follows three steps: 1.The router looks at the first byte of the address to find the class. It is class B. 2.The default mask for class B is The router ANDs this mask with the address to get The router looks in its routing table to find out how to route the packet to this destination. Later, we will see what happens if this destination does not exist. 7/3/201551Unit-3 : Network Layer

Figure Subnet mask 7/3/201552Unit-3 : Network Layer

Example 9 A router inside the organization receives the same packet with destination address Show how it finds the subnetwork address to route the packet. Solution The router follows three steps: 1.The router must know the mask. We assume it is /19, as shown in Figure The router applies the mask to the address, The subnet address is The router looks in its routing table to find how to route the packet to this destination. Later, we will see what happens if this destination does not exist. 7/3/201553Unit-3 : Network Layer

Figure DHCP transition diagram 7/3/201554Unit-3 : Network Layer

Table 19.2 Default masks Range Total to to to /3/201555Unit-3 : Network Layer

Figure NAT 7/3/201556Unit-3 : Network Layer

Figure Address translation 7/3/201557Unit-3 : Network Layer

Figure Translation 7/3/201558Unit-3 : Network Layer

Table 19.3 Five-column translation table Private Address Private Port External Address External Port Transport Protocol TCP TCP... 7/3/201559Unit-3 : Network Layer

19.3 Routing Routing Techniques Static Versus Dynamic Routing Routing Table for Classful Addressing Routing Table for Classless Addressing 7/3/201560Unit-3 : Network Layer

Figure Next-hop routing 7/3/201561Unit-3 : Network Layer

Figure Network-specific routing 7/3/201562Unit-3 : Network Layer

Figure Host-specific routing 7/3/201563Unit-3 : Network Layer

Figure Default routing 7/3/201564Unit-3 : Network Layer

Figure Classful addressing routing table 7/3/201565Unit-3 : Network Layer

Example 10 Using the table in Figure 19.32, the router receives a packet for destination For each row, the mask is applied to the destination address until a match with the destination address is found. In this example, the router sends the packet through interface m0 (host specific). 7/3/201566Unit-3 : Network Layer

Example 11 Using the table in Figure 19.32, the router receives a packet for destination For each row, the mask is applied to the destination address until a match with the next-hop address is found. In this example, the router sends the packet through interface m2 (network specific). 7/3/201567Unit-3 : Network Layer

Example 12 Using the table in Figure 19.32, the router receives a packet for destination For each row, the mask is applied to the destination address, but no match is found. In this example, the router sends the packet through the default interface m0. 7/3/201568Unit-3 : Network Layer

Routing Algorithms 1.Distance Vector Routing 2.Link State Routing 7/3/201569Unit-3 : Network Layer

Figure The Concept of Distance Vector Routing 7/3/201570Unit-3 : Network Layer

Figure Distance Vector Routing Table 7/3/201571Unit-3 : Network Layer

Figure Routing Table Distribution 7/3/201572Unit-3 : Network Layer

Figure Updating Routing Table for Router A 7/3/201573Unit-3 : Network Layer

Figure Final Routing Tables 7/3/201574Unit-3 : Network Layer

Figure Example /3/201575Unit-3 : Network Layer

Figure Concept of Link State Routing 7/3/201576Unit-3 : Network Layer

Figure Cost in Link State Routing 7/3/201577Unit-3 : Network Layer

Figure Link State Packet 7/3/201578Unit-3 : Network Layer

Figure Flooding of A’s LSP 7/3/201579Unit-3 : Network Layer

Figure Link State Database 7/3/201580Unit-3 : Network Layer

Figure Costs in the Dijkstra Algorithm 7/3/201581Unit-3 : Network Layer

Figure 21-30, Part I Shortest Path Calculation, Part I 7/3/201582Unit-3 : Network Layer

Figure 21-30, Part II Shortest Path Calculation, Part II 7/3/201583Unit-3 : Network Layer

Figure 21-30, Part III Shortest Path Calculation, Part III 7/3/201584Unit-3 : Network Layer

Figure 21-30, Part IV Shortest Path Calculation, Part IV 7/3/201585Unit-3 : Network Layer

Figure 21-30, Part V Shortest Path Calculation, Part V 7/3/201586Unit-3 : Network Layer

Figure 21-30, Part VI Shortest Path Calculation, Part VI 7/3/201587Unit-3 : Network Layer

Figure 21-31, Part VII Shortest Path Calculation, Part VII 7/3/201588Unit-3 : Network Layer

Figure 21-31, Part I Shortest Path Calculation, Part VIII 7/3/201589Unit-3 : Network Layer

Figure 21-31, Part II Shortest Path Calculation, Part IX 7/3/201590Unit-3 : Network Layer

Figure 21-31, Part III Shortest Path Calculation, Part X 7/3/201591Unit-3 : Network Layer

Figure 21-31, Part IV Shortest Path Calculation, Part XI 7/3/201592Unit-3 : Network Layer

Figure 21-31, Part V Shortest Path Calculation, Part XII 7/3/201593Unit-3 : Network Layer

Figure 21-31, Part VI Shortest Path Calculation, Part XIII 7/3/201594Unit-3 : Network Layer

Routing Table for Router A Figure /3/201595Unit-3 : Network Layer