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Ch 3: Underlying Technologies (remainder)

Ch 3: Underlying Technologies (remainder). Lecture #5. Reminder - Exam 1 on Tuesday. CONNECTING DEVICES. Repeater. Operates at the physical layer – layer 1 Receives the signal and regenerates the signal in it’s original pattern. A repeater forwards every bit; it has no filtering capability.

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Ch 3: Underlying Technologies (remainder)

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  1. Ch 3: Underlying Technologies (remainder) Lecture #5 Reminder - Exam 1 on Tuesday Lecture

  2. CONNECTING DEVICES Lecture

  3. Repeater Operates at the physical layer – layer 1 Receives the signal and regenerates the signal in it’s original pattern A repeater forwards every bit; it has no filtering capability Is there a difference between a regen or repeater and an amp ?? Lecture

  4. Repeaters d For the architecture above, will a signal ever traverse through more than 2 repeaters ? Lecture

  5. Hubs Hub – multi-port repeater Typically used to create a physical star topology Also used to create multiple levels of hierarchy For bus technology type networks, hubs can be used to increase the collision domain Lecture

  6. Bridge • Operates at both the physical and data link layers • At layer 1, it regenerates the signal. At layer 2, it checks the Tx/Rx physical address (using a bridge table) • Example Below: • If packet arrives to bridge-interface #1 for either of the 71….. stations, the packet is dropped because the 71…. Stations will see the packet • If packet arrives to bridge-interface #2 for either of the 71….. stations, the packet is forwarded to bridge-interface #1 With such an approach, the “bridged” network segments will acted as a single larger network What is a “smart” bridge ?? Lecture

  7. Routers • Is a 3-layer device: (1) at layer 1, regen the signals, (2) at layer 2, check physical address and (3) at layer 3, check network addresses • Routers are internetworking devices • Routers contain a physical and logical/IP address for it’s interfaces (repeaters/bridges don’t) • Routers only act on the packets needing to pass through • Routers change the physical address of the packets needing to pass through (repeaters/bridges don’t change physical addresses) Show example where a decision is needed d d Lecture

  8. Routing example • Routers can change the physical address of a packet • Example: as a packet flow from LAN 1 to LAN 2 • In LAN 1, the source address is the Tx’s address and the destination address is the Router’s interface address • In LAN 2, the source address is the Router’s interface address and the destination address is the Rx’s address LAN 1 LAN2 Lecture

  9. You are a High Priced Network Consultant Marketing Dept Engineering Dept (Super Computer) Manufacturing Dept (Robots) d They want all departments to communicate with one another;you want the network to maintain top performance – which design would you recommend ? Which devices would you recommend for empty circles ? – the least cost solution is the best solution Lecture

  10. Note: Lectures for Exam 1 just ended. We now start lectures for Exam 2 Lecture

  11. Chapters 4 & 5 Addressing Lecture

  12. IP ADDRESSING (Ch 4) Lecture

  13. IP Addresses • Internetworking Protocol (IP) of the Network Layer is responsible for uniquely identifying all devices and connections on the Internet • The unique identifier is called an IP address • IP address consist of 32 bits (for version 4) • Keep in mind that, if a single device had multiple connections to the Internet, you would need an IP address for each connection • Address space is 232 = 4,294,967,296 32-bit addresses • In theoretical terms, 4,294,967,296 connections can be made to the Internet (not really true in real life) Lecture

  14. 0111 0101 1001 0101 0001 1101 1110 1010 75 95 1D EA 75951DEA IP Addresses • The IP Address has 3 notations: Binary, Dotted-decimal and Hexadecimal • Binary: 4 Octets: 01110101 10010101 00011101 11101010 • Dotted-Decimal (or dot notation): • For Dotted-Decimal, each number can range from 0 to 255 • Hexadecimal: Lecture

  15. EXAMPLES Change the following IP address from binary notation to dotted-decimal notation: 10000001 00001011 00001011 11101111 Solution 129.11.11.239 Change the following IP address from dotted-decimal notation to binary notation: 111.56.45.78 Solution 01101111 00111000 00101101 01001110 Find the error, if any, in the following IP address: 111.56.045.78 Solution There are no leading zeroes in dotted-decimal notation (045). Change the following IP addresses from binary notation to hexadecimal notation: 10000001 00001011 00001011 11101111 810B0BEF16 Solution Lecture

  16. IP Addresses: Classful Addressing • When IP addressing was first started, it used a concept called “classful addressing”. A newer concept called “classless addressing” is slowly replacing it though. • Regarding “classful addressing”, the address space is divided into five classes: A, B, C, D and E. Lecture

  17. Finding the class in binary notation Finding the class in decimal notation Lecture

  18. EXAMPLES Solution Find the class of the address: 00000001 00001011 00001011 11101111 The first bit is 0. This is a class A address. Solution Find the class of the address: 11000001 10000011 00011011 11111111 The first 2 bits are 1; the third bit is 0. This is a class C address. Solution Find the class of the address: 227.12.14.87 The first byte is 227 (between 224 and 239); the class is D. Lecture

  19. Netid and hostid • A, B and C class-addresses are divided into network id and host id • For Class A, Netid=1 byte, Hostid = 3 bytes • For Class B, Netid=2 bytes, Hostid = 2 bytes • For Class C, Netid=3 bytes, Hostid = 1 byte Lecture

  20. Blocks in class A • Class A has 128 blocks or network ids • First byte is the same (netid), the remaining 3 bytes can change (hostids) • Network id 0 (first), Net id 127 (last) and Net id 10 are reserved – leaving 125 ids to be assigned to organizations/companies • Each block contains 16,777,216 addresses – this block should be used by large organizations. How many Host can be addressed ???? • The first address in the block is called the “network address” – defines the network of the organization • Example • Netid 73 is assigned • Last address is reserved • Recall: routers have addressees Lecture

  21. Blocks in class B • Class B is divided into 16,384 blocks (65,536 addresses each) • 16 blocks are reserved • First 2 bytes are the same (netid), the remaining 2 bytes can change (hostids) • For example, Network id 128.0 covers addresses 128.0.0.0 to 128.0.255.255 • Network id 191.225 is the last netid for this block • Example • Netid 180.8 is assigned • Last address is reserved • Recall: routers have addresses Lecture

  22. Blocks in class C • Class C is divided into 2,097,152 blocks (256 addresses each) • 256 blocks are reserved • First 3 bytes are the same (netid), the remaining 1 byte can change (hostids) • For example, Network id 192.0.0 covers addresses 192.0.0.0 to 192.0.0.255 Lecture

  23. Class D addresses are used for multicasting; there is only one block in this class. Class E addresses are reservedfor special purposes; most of the block is wasted. Lecture

  24. Network Addresses The network address is the first address. The network address defines the network to the rest of the Internet. Given the network address, we can find the class of the address, the block, and the range of the addresses in the block Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. 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. Solution Given the network address 132.21.0.0, find the class, the block, and the range of the addresses. The class is B because the first byte is between 128 and 191. The block has a netid of 132.21. The addresses range from 132.21.0.0 to 132.21.255.255. Solution Given the network address 220.34.76.0, find the class, the block, and the range of the addresses. The class is C because the first byte is between 192 and 223. The block has a netid of 220.34.76. The addresses range from 220.34.76.0 to 220.34.76.255. Solution Lecture

  25. REMINDER: Go Over Converting Binary to Decimal and Vice Versa Lecture

  26. Converting Number Systems - Review Base-10 The decimal number system is based on power of the base 10. For example, for the number 1259, the 9 is in the 10^0 column - 1s column the 5 is in the 10^1 column - 10s column the 2 is in the 10^2 column - 100s column the 1 is in the 10^3 column - 1000s column 1259 is 9 X 1 = 9 + 5 X 10 = 50 + 2 X 100 = 200 + 1 X 1000 = 1000 ----- 1259 Base-2 (Binary) The Binary number system uses the same mechanism and concept however, the base is 2 versus 10  The place values for binary are based on powers of the base 2: … 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 128 64 32 16 8 4 2 1 Lecture

  27. Converting Number Systems - Review So, the binary number 10110011 can be converted to a decimal number 1 X 1 = 1 (right most bit or position) 1 X 2 = 2 0 X 4 = 0 0 X 8 = 0 1 X 16 = 16 1 X 32 = 32 0 X 64 = 0 1 X 128 = 128 (left most bit or position) ------ 179 in decimal To convert from decimal to binary requires a different method called the division/remainder method. The idea is to repeatedly divide the decimal number and resulting quotients by 2. The answer will be the remainders. Example: convert 155 to binary (Start from the top and work down) 155/2 Q = 77, R = 1 (Start) 77/2 Q = 38, R = 1 38/2 Q = 19, R = 0 19/2 Q = 9, R = 1 9/2 Q = 4, R = 1 4/2 Q = 2, R = 0 2/2 Q = 1, R = 0 1/2 Q = 0, R = 1 (Stop) Answer is 10011011. Be careful to place the digits in the correct order. Lecture

  28. Converting Number Systems - Review Check the answer (10011011) : 1 X 1 = 1 1 X 2 = 2 0 X 4 = 0 1 X 8 = 8 1 X 16 = 16 0 X 32 = 0 0 X 64 = 0 1 X 128 = 128 ----- 155 Base-16 (Hex) The hexadecimal number system is based 16, and uses the same mechanisms and conversion routines we have already examined. The place values for hexadecimal are based on powers of the base 16 The digits for 10-15 are the letters A - F (A is 10, …….., F is 15) …….. 16^3 16^2 16^1 16^0 4096 256 16 1 The binary number 10110011 can be converted to hexadecimal by grouping bits into groups of 4 bits: 1011 0011 B 3 Lecture

  29. Mask • Given the network address, we can easily determine the block and range of addresses • Suppose given the IP address, can we determine the network address (beginning of the block) ? • To route packets to the correct network, a router must extract the network address from the destination IP address • For example, given 134.45.78.2, we know this is a class B, therefore 134.45 is the netid and 134.45.0.0 is the network address (starting address of the block) • How would we EXTRACT the network address from the IP address? We would use a MASK. A mask is a 32-bit binary number that gives the first address in the block (the network address) when bitwise ANDed with an address in the block. Lecture

  30. AND operation • If bit is ANDed with 1, it’s preserved • If bit is ANDed with 0, it’s changed to a 0. • There are 3 default masks: one for each class. The default masks preserve the netid when ANDed with the addresses • Class A Default Mask: 255.0.0.0 • Class B Default Mask: 255.255.0.0 • Class C Default Mask: 255.255.255.0 • A simple way to determine the netid for un-subnetted cases: (1) if mask byte is 255, retain corresponding byte of the address, (2) if mask byte is 0, set corresponding address byte to 0. Lecture

  31. Examples Given the address 23.56.7.91 and the default class A mask, find the beginning address (network address). 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. Solution Given the address 132.6.17.85 and the default class B mask, find the beginning address (network address). 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. Solution Given the address 201.180.56.5 and the class C default mask, find the beginning address (network address). 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. Solution Lecture

  32. 5-bit Address Space Illustration No Netid case 32 addresses/block Number of blocks: 1 Address range per block: 0 to 31 Netids: N/A Network Addresses : 00000 Broadcast Addresses: 11111 Lecture

  33. 5-bit Address Space Illustration 1-bit Netid case 16 addresses/block Number of blocks: 2 Address range per block: 0 to 15 Netids: 0, 1 Network Addresses : 00000, 10000 Broadcast Addresses: 01111, 11111 Lecture

  34. 5-bit Address Space Illustration 2-bit Netid Case 8 addresses/block Number of blocks: 4 Address range per block: 0 to 7 Netids: 00, 01, 10, 11 Network Addresses : 00000, 01000, 10000, 11000 Broadcast Addresses: 00111, 01111, 10111, 11111 Lecture

  35. 5-bit Address Space Illustration 3-bit Netid Case 4 addresses/block Number of blocks: 8 Address range per block: 0 to 3 Netids: 000, 001, 010, 011, 100, 101, 110, 111 Network Addresses : 00000, 00100, 01000, 01100 10000, 10100, 11000, 11100 Broadcast Addresses: 00011, 00111, 01011, 01111 10011, 10111, 11011, 11111 Lecture

  36. Mixing 3-bit & 2-bit Cases (think of the 32-bit case) 4 addresses/block and 8 addresses/block Number of blocks: 6 Address range per block: 0 to 3 and 0 to 7 Netids: 000, 001, 010, 011, 10, 11 Network Addresses : 00000, 00100, 01000, 01100 10000, 11000 Broadcast Addresses: 00011, 00111, 01011, 01111 10111, 11111 Lecture

  37. Multihomed devices • As we mentioned that, any device with one or more connections to the Internet will need an IP address for EACH connection – such devices are called “multihomed” devices. • A Router could be a multihomed device Lecture

  38. Example of direct broadcast address Router sending to all hosts on a network If the hostid is all 1’s, it’s called a “broadcast address” and the router use it to send a packet to all host in a specific network. In this case, hosts 20, 64, 126 and etc. will receive the packet from the router Example of limited broadcast address Host sending to all other hosts on a network If the hostid and netid are all 1’s, it’s called a “limited broadcast address”. If the host wants to send a packet to all host in a specific network, it would use this address. The router would block this address so that data stays contained within a specific network. Lecture

  39. Example of this host on this address IPless Host sending message to bootstrap server An address of all 0’s is used during bootstrap time if the host doesn’t know it’s IP address. The un-named host sends an all 0 source address and limited broadcast (all 1’s) destination address to the bootstrap server. Example of specific host on this network Host sending to some other specific host on a network An address with a netid of all 0’s is used by a host or router to send another host with in the same network a message. Lecture

  40. Example of loopback address • The IP address with the 1st byte equal to 127 is used for the loop back address. • Loopback address is used to test software on a machine – the packet never leaves the machine – it returns to the protocol software • Example: a “ping” command can send a packet with a loopback address as the destination address to see if the IP software is capable of receiving and processing a packet. Lecture

  41. Sample internet Ethernet ATM Token Ring Ethernet With your new found knowledge, think about Project 2 Lecture

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