Computer network

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briefly define about the Routing, broadcasting and unicasting

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  • A Cpu cache is a cache used by the central processing unit of computer to reduce the average time to access memory. Cache is a smaller faster memory which store copies of data from the most frequently used main memory location.
  • Generally group mailing are used in multicasting.
  • 127.1.1.1 is the loop back address which is connect to the port. If the port are connect properly to the host then the packet are transferred.
  • Computer network

    1. 1. Computer Network ROUTING, BROADCASTING AND MULTICASTING
    2. 2. IP addresses  Every computer has a physical address which is called as MAC address(LAN Card address/hardware address) or NIC card address, this address is 48 bit address.  Communication in internetworking is not possible with the MAC address, because computer manufactured by different companies and they don’t follow the same format of generating the MAC address. General MAC address format 1. 00-10-FA-6E-38-4A OR 2. 50-1A-DE-90-42-DC
    3. 3. IP address  There will be a IANA(Internet Assigned number     authority)(www.iana.org). These are generally assigned a logical addressing system(called as ip address) and in system its store in Cache memory. A system in networking world is identify by ip address. The designer of internet protocol define as an ip address as a 32 bit number and this system is known as IPV4 (internet protocol version 4) is still we use today. However due to the enormous growth of internet and predicted depletion(decrease) of available addresses a new version IPV6 using 128 bit for the address was developed in 1995. Ipv4 address are 172.16.254.1 Ipv6 address are 2001:db8:0:1234:0:567:8:1
    4. 4. Classification of computer network by ip address  Classes are generally divided into two parts: 1. Net Id. (32 bit) 2. Host Id.  Classes are divided into five classes: 1. Class A- unicasting. 2. Class B- unicasting. 3. Class C- unicasting. 4. Class D-Multicasting. 5. Class E- Research work.
    5. 5.  Class A: 0 (2^7-2) (2^24-2) (8 bits) Net ID Host ID (24 bits) As octet value of class A is (0-127). There will be a special address (0.0.0.0) which is a default address or the DHCP client(Dynamic Host Configuration protocol). 127.x.y.z it is a Loop back address. Where x,y,z is from (0255). Number of network= 2^7-2= 128-2=126. So the range of class A is (1-126) and each network have host id (2^242). * Network Mask : 255.0.0.0
    6. 6.  Class B: 10 2^14 (2^16-2) (16 bits) Net ID Host ID (16 bits) The range of ip address is (128-191). There will be no special address. Number of networks = 2^14. * Network Mask: 255.255.0.0  Class C: 110 (2^21) (2^8-2) Net ID(24 bits) Host ID(8 bits) The range of class C is (192-223). There will be also no special address. * Network Mask: 255.255.255.0 (Default Mask)
    7. 7.  Class D: During unicasting we require Net id and Host Id where as during Multicasting we don’t require Net id and Host id. 1110 The Range is from(224-239).  Class E: This is for research area. 1111 The range is from (240-254), 16.257.11.13 is a wrong ip address because 257 is not a valid address.
    8. 8. Calculating the Net id and Host id with Network mask  63.14.19.67 calculate the net id? 63.14.19.67 63: 00111111 255.0.0.0 255:11111111 63.0.0.0 (Net id) :0011111 Net ID 63.0.0.0 Host ID 63.0.0.1 Host ID 63.0.0.2 Net ID 63.14.19.67 Host ID 63.255.255. 254 * 63.255.255.255 is the direct broadcast address.
    9. 9. Characteristic of loop back address  It is used to test if the ip address packets are coming from the other system or going out of the system.  Physical port are used to transfer the data. Loop back address will always used as a destination address. It is used for inter process communication with in the same system.  In class A 126 network have 2^24-2 host which approx 80,00,00 host if we use only 15000 then the large number of ip address are going to be waste. So this will be use in multinational company use.
    10. 10.  And in class B 2^14 network each network have (2^16-2) host 65,534 host here also the ip address are waste.  Class C have 2^21 network and each network have (2^8-2)= 254 hosts. So that in institution and small companies are used this because the ip address are not to be wasted over this.
    11. 11. Router  Router is a WAN device(where different LAN network are connected).  Connecting to different router with different model network we use GATEWAY. Suppose we use TCP/IP model network and X.25 model network. We use gateway to communicate with these two routers.
    12. 12. Router sending and receiving Data| 46.0.0.1 (SIP) | 130.12.0.1(DIP). In this the data is transmit to class A to class B.  Data| 46.0.0.1 (SIP) | 46.1.2.24(DIP). In this the data is send and receive to same class A this is called as a unicasting.  Data| 46.0.0.1 (SIP) | 255.255.255.255 (DIP). In this the DIP is the limited broadcast address. Which broadcast with in the network of Class A.  Data| 46.0.0.1 (SIP) | 130.1.255.255 (DIP). In this class A send the data to broadcast to all the network class B. 
    13. 13. Working of Ping?  When you start window-> goto command prompt and Run cmd.  This will be you see: C:>ping 11.1.2.3 Enter. Ping (packet internet grapher). Ping is a client to server utility that is provided by operating system. C:> ping 11.1.2.3 Enter TTL=2 MS RTT=4MS C:> ping 11.1.2.3 Enter Destination unreachable(DIP is not configure)
    14. 14.  TTL is time to line, Its going to indicate the no. of hops the packets are traverse before reaching the destination.  Distance from one router to another router is called 1 HOP.  RTT is Round Trip Time; Time taken for the data to reach the destination and coming back on acknowledgement to the centre.
    15. 15. How to transmit the data and calculate the time  Suppose if the size of the message is 1 KB. And     Bandwidth is 1 Mbps then the transmission time; TT= (1*10^3 ) B / (1*10^6) bits per sec = 8*10^-3 sec = 8 millisec. Propagation time= Distance/ Velocity. Acknowledgement size<< message size. TT ack= ack size/bandwidth. TT ack<< TT msg data.
    16. 16.  PT ack= distance/ velocity.  Total time= TT data + PT data + TT ack + PT ack. TT ack is negligible. Total time= TT data + 2*PT RTT(Round Trip Time) =2*PT Link Utilization of sender= TT/ (TT data + 2*PT)
    17. 17. Routing Algorithm  Static Algorithm: These are the algorithm which doesn’t take the network load as consideration. * Flooding is as a static algorithm.  Dynamic Algorithm: The algorithm that contain network load as consideration.  1. If the routing of packets are done with in the autonomous system, it is known as inter routing protocol; example : Distance vector routing algorithm, Link State routing algorithm.  2. If the routing of packets are done between autonomous system is known as inter domain routing algorithm. Example : Path vector routing algorithm.
    18. 18. Flooding B p p p flooding p A C p p p p EE D
    19. 19. Distance Vector Routing Algorithm New routing table Packet transfer to router A Vector table A B E D F C
    20. 20. Linked State Routing Algorithm 5 2 C B A 3 3 G 4 4 1 D 5 E 2 F
    21. 21. Shortest path tree from Router A A 3 5 2 C B G 4 1 D E 2 F
    22. 22. Conclusion  Computer network is the very large area. And generally we want a network which have a high speed of transmission and very less of error while communicating. Computer network wants speed, accuracy, adequacy and security. These are the all future aspects of computer network which we will have to consider.

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