IAP presentation-1.pptx
- 1. INTERNET ARCHITECTURE AND PROTOCOL PRESENTATION Submitted To: Mam SARA Submitted by: Group#5(SEC B) MAEDA QAISAR (19011556-078) ALISHBA SHAHBAZ (19011556-043) FAIZA (19011556-176)
- 2. ROUTING PROTOCOLS Routing protocols are the set of rules used by the routers to communicate between source & destination. They do not move the information source to destination only update the routing table. Each protocol has its own algorithm to choose the best path.
- 3. METRICES BY ROUTING PROTOCOLS • Number of network layer devices along with the path (hop count) • Bandwidth • Delay • Load
- 4. fgh TYPES OF ROUTING PROTOCOLS Static Routing Protocol Dynamic Routing Protocol Dynamic Routing Protocol Distance Vector Link State RIP RIPV 2 RIPV 1 EIGRP OSPF
- 5. STATIC ROUTING PROTOCOLS Static routing ,when an administrator manually assigns the path from source to destination network.This is feasible in small networks, but not in large networks.
- 6. ADVANTAGESAND DISADVANTAGES OF STATIC ROUTING PROTOCOL Advantages: • No overhead on router CPU. • No bandwidth usage between links. • Security (only administrator add routes.) Disadvantages: • All link will be down on a link failure. • Not practical on large networks. • Administrator must update all routes.
- 7. DEFAULT ROUTING PROTOCOLS Default Route is the network route used by a router when there is no other known route exists for a given IP datagram’s destination address. All the IP datagrams with unknown destination address are sent to the default route.
- 8. ADVANTAGESAND DISADVANTAGESOF DEFAULT ROUTING PROTOCOL Advantages:- • No overhead on router CPU. • No bandwidth usage between links. • Security (only administrator add routes.) Disadvantages:- • All link will be down on a link failure. • Not practical on large networks. • Administrator must update all routes.
- 9. DYNAMIC ROUTING PROTOCOLS Dynamic routing is the process in which routing tables are automatically updates by routing table of each neighbor. Dynamically discover & maintains routes. Calculate routes
- 10. ADVANTAGESAND DISADVANTAGES OF DYNAMIC ROUTING PROTOCOL Advantages • Less work in maintaining the configuration when adding & deleting networks. • Protocols automatically react to the topology changes. • Configuration is less-prone. Disadvantages • Routers resource are used. • More administrator knowledge is required for configuration
- 11. Dynamic routing protocol Distance vector Link state
- 12. DISTANCEVECTOR ROUTING PROTOCOL Distance vector routing protocols use distance to determine the best path to a remote network. The distance is usually the number of hops (routers) to the destination network. Distance vector protocols send complete routing table to each neighbor (a neighbor is directly connected router that runs the same routing protocol) RIPV1 and RIPV2 are examples of distance vector routing protocols.
- 13. RIP V1 It allows routers to exchange their routing tables at a predefined interval. It is a distance-vector routing protocol which employs the hop count as a routing metric. It transmitted updates in every 30 seconds. CHARACTERISTICS: • Uses hop count metric • Supports 15 hop-count limit • AD value is 120. • Supports classful networks.
- 14. RIP V2 • It is a Extended version of RIP routing protocol. • Maximum hop count is 15. • Supports small network o Supports classless network. • SupportsVLSM/CIDR. • Supports Auto-Summarization. • Route updates after 30 sec. • It supports Key-authentication.
- 15. LINK STATE PROTOCOL Link state routing protocols are the second type of dynamic routing protocols.They have the same basic purpose as distance vector protocols, to find a best path to a destination, but use different methods to do so. Unlike distance vector protocols, link state protocols don't advertise the entire routing table. Instead, they advertise information about a network topology (directly connected links, neighboring routers...), so that in the end all routers running a link state protocol have the same topology database. OSPF and EIGRP are examples of link state routing protocols.
- 16. LINK STATE PROTOCOLALGORITHM Initialization N = {A} // A is a root node. for all nodes v if v adjacent to A then D(v) = c(A,v) else D(v) = infinity loop find w not in N such that D(w) is a minimum. Add w to N Update D(v) for all v adjacent to w and not in N: D(v) = min(D(v) , D(w) + c(w,v)) Until all nodes in N
- 18. A E D C B F 1 3 2 2 5 1 1 2 5 Step 1: The first step is an initialization step. The currently known least cost path from A to its directly attached neighbors, B, C, D are 2,5,1 respectively. The cost from A to B is set to 2, from A to D is set to 1 and from A to C is set to 5. The cost from A to E and F are set to infinity as they are not directly linked to A. Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ LINK STATE PROTOCOL ALGORITHM
- 19. LINK STATE PROTOCOLALGORITHM Step 2: In the above table, we observe that vertex D contains the least cost path in step 1. Therefore, it is added in N. Now, we need to determine a least-cost path through D vertex. a) Calculating shortest path from A to B v = B, w = D D(B) = min( D(B) , D(D) + c(D,B) ) = min( 2, 1+2)> = min( 2, 3) The minimum value is 2. Therefore, the currently shortest path from A to B is 2. b) Calculating shortest path from A to C v = C, w = D D(B) = min( D(C) , D(D) + c(D,C) ) = min( 5, 1+3) = min( 5, 4) The minimum value is 4. Therefore, the currently shortest path from A to C is 4.
- 20. LINK STATE PROTOCOLALGORITHM c) Calculating shortest path from A to E v = E, w = D D(B) = min( D(E) , D(D) + c(D,E) ) = min( ∞, 1+1) = min(∞, 2) The minimum value is 2. Therefore, the currently shortest path from A to E is 2. Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ 2 AD 2,A 4,D 2,D ∞
- 21. LINK STATE PROTOCOLALGORITHM Step 3: In the above table, we observe that both E and B have the least cost path in step 2. Let's consider the E vertex. Now, we determine the least cost path of remaining vertices through E. a) Calculating the shortest path from A to B. v = B, w = E D(B) = min( D(B) , D(E) + c(E,B) ) = min( 2 , 2+ ∞ ) = min( 2, ∞) The minimum value is 2. Therefore, the currently shortest path from A to B is 2. b) Calculating shortest path from A to C v = C, w = E D(B) = min( D(C) , D(E) + c(E,C) ) = min( 4 , 2+1 ) = min( 4,3) The minimum value is 3. Therefore, the currently shortest path from A to C is 3.
- 22. LINK STATE PROTOCOLALGORITHM c) Calculating shortest path from A to E v = F, w = E D(B) = min( D(F) , D(E) + c(E,F) ) = min( ∞ , 2+2 ) = min(∞ ,4) The minimum value is 4. Therefore, the currently shortest path from A to F is 4. Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ 2 AD 2,A 4,D 2,D ∞ 3 ADE 2,A 3,E 4E
- 23. LINK STATE PROTOCOL ALGORITHM Step 4: In the above table, we observe that B vertex has the least cost path in step 3. Therefore, it is added in N. Now, we determine the least cost path of remaining vertices through B. a) Calculating the shortest path from A to C. v = C, w = B D(B) = min( D(C) , D(B) + c(B,C) ) = min( 3 , 2+3 ) = min( 3,5) The minimum value is 3. Therefore, the curre ntly shortest path from A to C is 3. b) Calculating shortest path from A to F v = F, w = B D(B) = min( D(F) , D(B) + c(B,F) ) = min( 4, ∞) = min(4, ∞) The minimum value is 4. Therefore, the currently shorte st path from A to F is 4. Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ 2 AD 2,A 4,D 2,D ∞ 3 ADE 2,A 3,E 4E 4 ADEB 3,E 4,E
- 24. LINK STATE PROTOCOL ALGORITHM Step 5: In the above table, we observe that C vertex has the least cost path in step 4. Therefore, it is added in N. Now, we determine the least cost path of remaining vertices through C. a) Calculating the shortest path from A to F. v = F, w = C D(B) = min( D(F) , D(C) + c(C,F) ) = min( 4, 3+5) = min(4,8) The minimum value is 4. Therefore, the currently shortest path from A to F is 4. Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ 2 AD 2,A 4,D 2,D ∞ 3 ADE 2,A 3,E 4E 4 ADEB 3,E 4,E 5 ADEBC 4,E
- 25. LINK STATE PROTOCOLALGORITHM Step A N B D(B),P(B) C D(C),P(C) D D(D),P(D) E D(E),P(E) F D(F),P(F) 1 A 2,A 5,A 1,A ∞ ∞ 2 AD 2,A 4,D 2,D ∞ 3 ADE 2,A 3,E 4E 4 ADEB 3,E 4,E 5 ADEBC 4,E 6 ADEBCF
- 27. EIGRP It’s supports the features both distance vector & link state protocol. It is a cisco proprietary protocol. By default, bandwidth & delay are the activated metrics. CHARACTERISTICS: - • Uses DUAL algorithm. • Supports classless network • SupportsVLSM/CIDR. • It supports trigger updates.
- 28. OSPF The large network can be broken into the small areas so the router in one area know less topology and they don’t have information about other areas routers. Creating OSPF areas result in smaller database which reduce the memory consumption and processing. OSPF maintains a two layer hierarchy consisting of: • Backbone area (area 0) • Off backbone area (areas1- 65,535) CHARACTERISTICS: • AD value is 110. • Supports classless network. • SupportsVLSM/CIDR & has unlimited hop counts • Supports hierarchical network. • Route propagation using multicasting.
- 29. Questions