Wipro telecom training i pv4 concepts
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Here are the steps to divide the address space as per the requirement: 1. Net 1 (500 hosts) - 9 bits (512 addresses) /23 2. Net 2 (200 hosts) - 8 bits (256 addresses) /24 3. Net 3 (100 hosts) - 7 bits (128 addresses) /25 4. Net 4 (60 hosts) - 6 bits (64 addresses) /26 5. Net 5 (20 hosts) - 5 bits (32 addresses) /27 6. Net 6 (2 hosts) - 2 bits (4 addresses) /30 7. Net 7 (2 hosts) - 2 bits (4 addresses) /30
Wipro telecom training i pv4 concepts
- 1. RMK College of Engineering and Technology IP v 4 Department of Electronics and Communication Engineering
- 2. Internet Protocol Version 4 @ (IPv4) Prepared by Jai Ganesh S Asst.Professor - ECE
- 3. Training Contents • What is an IP? • Need for an IP? • Representation of an IP • Classes of an IP Address • Sub netting and its types
- 4. Introduction to IP Address • An IP address is a 32-bit address that identifies a connection to the Internet. • The IP addresses are universally unique. • The address space of IPv4 is 2^32 or 4,294,967,296.
- 5. Need for IP Address
- 7. How IP ‘s are used for Identification? Network Host Process Network Host Process www.google.com INDIA CALIFORNIA
- 8. Few Organized Numbering Systems +91 - 9790410977 COUNTRY CODE ACCESS CODE SERVICE PROVIDER CODE SUBSCRIBER NUMBER 1116 – 15 – 106 – 003 111615106003
- 9. Representation of IP Address Binary Representation Dotted Decimal Representation
- 10. IP Address Space • Initially 8 bits were allotted for Network ID and remaining 24 bits were allotted for Host ID • 8 bits 28 Networks 256 Networks • 24 bits 224 Hosts Per Network 16Million hosts per network • NOT SCALABLE
- 11. Class Full Addressing 1X 11X 111XX 1 1 . . . 0 0 . . . 231 231 232 address 231 address 10 10 . . . 11 11 . . . 230 230 230 address 229 address 110 110 . . . 111 111 . . . 229 229 1110 1110 . . . 1111 1111 . . . 228 228 CA CB CC CD CE 00000000 0 01111111 127 10000000 128 10111111 191 11000000 192 11011111 223 11100000 224 11101111 239 11110000 240 11111111 255
- 12. Finding Class of a Given IP
- 13. NID and HID Organization
- 14. Ok.. Now it’s your Time • How many Network ID’s and Host ID’s are present in each classes ???
- 15. Summary of all classes
- 16. Blocks in Class A
- 17. Blocks in Class B
- 18. Blocks in Class C
- 19. Simple Exercise Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. • Solution: • Class A because first byte lies between 0 – 127 • Net ID is 17.0.0.0 so address block ranges from (17.0.0.0 to 17.255.255.255)
- 20. Simple Exercise • Given the network address 132.21.0.0, find the class, the block, and the range of the addresses. • Solution: • The class is B because the first byte is between 128 and 191. • The addresses range from 132.21.0.0 to 132.21.255.255
- 21. Simple Exercise • Given the network address 220.34.76.0, find the class, the block, and the range of the addresses. • Solution: • The class is C because the first byte is between 192 and 223 • The addresses range from 220.34.76.0 to 220.34.76.255.
- 22. Not all IP Address are usable… • Network ID or Net ID – First IP of address block • Broad cast Address – Last IP of address block • Types of broadcast • Limited Broadcast • Directed Broadcast 11.0.0.0 LBA: 255.255.255.255 11.0.0.0 20.0.0.0 DBA: 20.255.255.255
- 23. Try to fill the Table IP Address Network ID Directed Broadcast ID Limited Broadcast ID 1.2.3.4 1.0.0.0 1.255.255.255 255.255.255.255 10.15.20.60 10.0.0.0 10.255.255.255 255.255.255.255 130.1.2.3 130.1.0.0 130.1.255.255 255.255.255.255 150.0.150.150 150.0.0.0 150.0.255.255 255.255.255.255 200.1.10.100 200.1.10.0 200.1.10.255 255.255.255.255 220.15.1.10 220.15.1.0 220.15.1.255 255.255.255.255 250.0.1.2 NOT POSSIBLE NOT POSSIBLE NOT POSSIBLE 300.1.2.3 INCORRECT IP ADDRESS
- 24. Subnets and Subnet Mask • What are subnets? • Dividing the large address space into smaller blocks of IP Address • Why do we go for subnets? • If the blocks of IP address are used as such millions of IP address are wasted. • Managing the Large block of IP Address is also difficult • What are the types of subnets? • Fixed Length Subnetting • Variable Length Subnetting • Subnet Mask • Will be discussed few slides later
- 25. Fixed Length Subnetting • Lets consider an Organization and the entire organization has different departments / Sections I II III IVECE EEE CSE MECH
- 26. Fixed Length – Example 1 • Divide the address block 200.1.2.0 into 2 subnets Given address block is Class C 24 bits for NID and 8 bits for HID 200.1.2.00000000 200.1.2.11111111 200.1.2.0 200.1.2.255 200.1.2.X0000000 X = 0 X = 1 Network ID : 200.1.2.0 Broadcast ID: 200.1.2.255 1 bit is required to make 2 parts Get Back
- 27. 200.1.2.X0000000 X = 0 X = 1 200.1.2.0_ _ _ _ _ _ _ Runs from 200.1.2.00000000 200.1.2.00000001 200.1.2.00000010 200.1.2.00000011 . . 200.1.2.01111111 200.1.2.1_ _ _ _ _ _ _ Runs from 200.1.2.10000000 200.1.2.10000001 200.1.2.10000010 200.1.2.10000011 . . 200.1.2.11111111 200.1.2.0 200.1.2.127 200.1.2.128 200.1.2.255 Network ID : 200.1.2.0 Broadcast ID: 200.1.2.127 Network ID : 200.1.2.128 Broadcast ID: 200.1.2.255 27 = 128 Address in Each Subnet Get Back
- 28. Fixed Length – Example 2 • Divide the address block 200.1.2.0 into 4 subnets Given address block is Class C 24 bits for NID and 8 bits for HID 200.1.2.00000000 200.1.2.11111111 200.1.2.0 200.1.2.255 200.1.2.XX000000 XX = 00 Network ID : 200.1.2.0 Broadcast ID: 200.1.2.255 XX = 01 XX = 10 XX = 11 2 bits are required to make 4 parts
- 29. 200.1.2.XX000000 XX = 00 XX = 01 XX = 10 XX = 11 200.1.2.00 _ _ _ _ _ _ Runs from 200.1.2.00000000 to 200.1.2.00111111 200.1.2.01 _ _ _ _ _ _ Runs from 200.1.2.01000000 to 200.1.2.01111111 200.1.2.11 _ _ _ _ _ _ Runs from 200.1.2.11000000 to 200.1.2.11111111 200.1.2.10 _ _ _ _ _ _ Runs from 200.1.2.10000000 to 200.1.2.10111111 200.1.2.0 Network ID 200.1.2.63 Broadcast ID 200.1.2.128 Network ID 200.1.2.191 Broadcast ID 200.1.2.192 Network ID 200.1.2.255 Broadcast ID 200.1.2.64 Network ID 200.1.2.127 Broadcast ID 26 = 64 Address in Each Subnet
- 30. Fixed Length – Exercise 1 Divide the address block 192.168.9.0 into 5 networks Given address block is Class C 24 bits for NID and 8 bits for HID 192.168.9.00000000 192.168.9.11111111 192.168.9.0 192.168.9.255 192.168.9.XXX00000 Network ID : 192.168.9.0 Broadcast ID: 192.168.9.255 3 bits are required to make 8 parts 000 001 010 011 100 101 110 111
- 31. NET 1 192.168.9.00000000 192.168.9.0 192.168.9.00011111 192.168.9.31 NET 2 192.168.9.00100000 192.168.9.32 192.168.9.00111111 192.168.9.63 NET 3 192.168.9.01000000 192.168.9.64 192.168.9.01011111 192.168.9.95 NET 4 192.168.9.01100000 192.168.9.96 192.168.9.01111111 192.168.9.127 NET 5 192.168.9.10000000 192.168.9.128 192.168.9.10011111 192.168.9.159 NET 6 192.168.9.10100000 192.168.9.160 192.168.9.10111111 192.168.9.191 NET 7 192.168.9.11000000 192.168.9.192 192.168.9.11011111 192.168.9.223 NET 8 192.168.9.11100000 192.168.9.224 192.168.9.11111111 192.168.9.255
- 32. Fixed Length – Exercise 2 Divide the address space 172.16.0.0 into 3 networks Given address block is Class B 16 bits for NID and 16 bits for HID 172.16.00000000.00000000 To 172.16.11111111.11111111 172.16.XX000000.00000000 XX = 00 Network ID : 172.16.0.0 Broadcast ID: 172.16.255.255 XX = 01 XX = 10 XX = 11 2 bits are required to make 4 parts
- 33. NET 1 172.16.00000000.00000000 172.16.0.0 172.16.00111111.11111111 172.16.63.255 NET 2 172.16.01000000.00000000 172.16.64.0 172.16.01111111.11111111 172.16.127.255 NET 3 172.16.10000000.00000000 172.16.128.0 172.16.10111111.11111111 172.16.191.255 NET 4 172.16.11000000.00000000 172.16.192.0 172.16.11111111.11111111 172.16.255.255
- 34. Fixed Length – Exercise 3 • Divide the address space 192.168.9.0 into suitable number of networks so that each network can handle at least 10 hosts In each subnet 2 address are not usable so total no of address required is 12 2? Gives 12 address This is not possible 24 Gives 16 address This is the nearest possible value 192.168.9.XXXX_ _ _ _
- 35. Subnet Mask
- 36. Subnet Mask • A Subnet mask is a 32-bit number that masks an IP address, and divides the IP address into network address and host address. • Subnet Mask is made by setting • network bits to all "1"s • host bits to all "0"s.
- 38. Representation of Subnet Mask • There are 3 types of representation • Binary Representation • Dotted Decimal Representation • /n Representation or CIDR Representation (n Defines #. Of bits allotted for Network)
- 39. Default Subnet Mask Class NID - HID Binary Class A NID – HID – HID – HID 11111111.00000000.00000000.00000000 Class B NID – NID – HID – HID 11111111.11111111.00000000.00000000 Class C NID – NID – NID – HID 11111111.11111111.11111111.00000000 Class Dotted Decimal /n Class A 255.0.0.0 /8 Class B 255.255.0.0 /16 Class C 255.255.255.0 /24
- 40. Need for Subnet Mask 1. Given any IP Address, Subnet mask can be used to identify the network ID
- 41. Need for Subnet Mask 2. To inform the router that subnetting is performed Lets go back to the Example problem 200.1.2.0 200.1.2.127 200.1.2.0 /25 200.1.2.127 /25 11111111.11111111.11111111.10000000 255 . 255 . 255 . 128
- 43. Variable Length Subnetting • There are certain drawbacks in fixed length methods • All subnets will have equal number of IP Address 64 64 6464 64 64 128 This scenario is called VLSM (Variable Length Subnet Masking)
- 44. Variable Length – Example 1 • Divide the given address space 172.16.0.0 / 16 into 7 networks whose requirement is given below • Net 1 500 hosts • Net 2 200 hosts • Net 3 100 hosts • Net 4 60 hosts • Net 5 20 hosts • Net 6 2 hosts • Net 7 2 hosts
- 45. 500 200 100 60 20 2 2 1. Arrange the required No. of hosts in Ascending Order 2. Allocate the No of bits required for Hosts in powers of 2 29 = 512 28 = 256 27 = 128 26 = 64 25 = 32 22 = 4 22 = 4 3. Write the given address space with required no of Host Bits 172.16.00000000.00000000 4. Count for required no of bits from right side ( ) and find the first and last address 172.16.00000001.11111111 5. Write the starting and ending address with subnet mask 172.16.0.0 / 23 172.16.1.255 / 23 6. Write the immediate next ip address for next subnet 172.16.00000010.00000000 7. Count the number of bits required for 2nd subnet 172.16.00000010.00000000 172.16.00000010.11111111 Write the first and last ip address with subnet mask 172.16.2.0 / 24 172.16.2.255 / 24 Repeat the procedure until last subnet 172.16.00000011.00000000 172.16.00000011.01111111 172.16.3.0 / 25 172.16.3.127 / 25 172.16.00000011.10000000 172.16.00000011.10111111 172.16.3.128 / 26 172.16.3.191 / 26 172.16.00000011.11000000 172.16.00000011.11011111 172.16.3.192 / 27 172.16.3.223 / 27 172.16.00000011.11100000 172.16.00000011.11100011 172.16.3.224 / 30 172.16.3.227 / 30 172.16.00000011.11100100 172.16.00000011.11100111 172.16.3.228 / 30 172.16.3.231 / 30
- 46. Variable Length – Example 2 Divide the address space 172.16.0.0 / 16 for 3 LANS as mentioned below Lan 1 should have 50 address Lan 2 should have 217 address Lan 3 should have 300 address
- 47. Solution NET # No of Bits Required Address space Address with subnet mask LAN 3 300 29 = 512 172.16.00000000.00000000 172.16.0.0 /23 172.16.00000001.11111111 172.16.1.255 /23 LAN 2 217 28 = 256 172.16.00000010.00000000 172.16.2.0 /24 172.16.00000010.11111111 172.16.2.255 /24 LAN 1 50 26 = 64 172.16.00000011.00000000 172.16.3.0 /26 172.16.00000011.00111111 172.16.3.63 /26
- 48. Mastering the Subnets Problem Divide the address space 172.16.0.0 / 16 for 3 LANS as mentioned below • Lan 1 should have 50 address • Lan 1.1 25 address • Lan 1.2 12 address • Lan 1.3 5 address • Lan 1.4 5 address • Lan 2 should have 217 address • Lan 2.1 70 address • Lan 2.2 30 address • Lan 2.3 16 address • Lan 2.4 10 address • Lan 3 should have 300 address • Lan 3.1 80 address • Lan 3.2 42 address • Lan 3.3 25 address • Lan 3.4 15 address • Lan 3.5 all remaining address
- 49. Solution net # Required Address no of bits Starting Address Ending Address net 3 300 address 2^9 = 512 172.16.0.0 /23 172.16.1.255 /23 3.1 80 address 2^7 = 128 172.16.0.0 /25 172.16.0.127 /25 3.2 42 address 2^6 = 64 172.16.0.128 / 26 172.16.0.191 /26 3.3 25 address 2^5 = 32 172.16.0.192 /27 172.16.0.223 /27 3.4 15 address 2^5 = 32 172.16.0.224 /27 172.16.0.255/27 3.5 remaining 172.16.1.0 172.16.1.255 net 2 217 address 2^8 = 256 172.16.2.0 /24 172.16.2.255 /24 2.1 70 address 2^7 = 128 172.16.2.0 /25 172.16.2.127 /25 2.2 30 address 2^5 = 32 172.16.2.128 / 27 172.16.2.159 / 27 2.3 16 address 2^5 = 32 172.16.2.160 /27 172.16.2.191 /27 2.4 10 address 2^4 = 16 172.16.2.192 /28 172.16.2.207 /28 remaining 172.16.2.208 172.16.2.255 net 1 50 address 2^6 = 64 172.16.3.0 /26 172.16.3.63 /26 1.1 25 address 2^5 = 32 172.16.3.0 /27 172.16.3.31 /27 1.2 12 address 2^4 = 16 172.16.3.32 /28 172.16.3.47 /28 1.3 5 address 2^3 = 8 172.16.3.48 /29 172.16.3.55 /29 1.4 5 address 2^3 = 8 172.16.3.56 /29 172.16.3.63 /29
- 50. Classless Inter Domain Routing (CIDR) CA CB CC CD CE 10 1110 110 1111 There are certain Drawbacks in class full addressing The least number of host Address that you can get is (2^8) 256 in class C.
- 51. Rules for CIDR Blocks • All IP address should be Contiguous • The size of the requested IP should be in power of 2 • Meaning – you cannot ask for 500 IP address. You can get only 512 IP Address as a block Taking this further will have their applications in Operating Systems So, I will leave it to your exercise