1. Teaching IP Subnetting

2.  I am trying to sell you something… ◦ I am trying to sell you a methodology for teaching subnetting ◦ I am trying to sell you on the “cheat sheet” method of figuring out subnetting problems ◦ If you like the method…GREAT! ◦ If you would like to use the book…GREAT!

3.  We are not here to teach! ◦ We are here to facilitate learning ◦ Do your best to ensure a good learning environment  Teach slowly and in small vignettes of information  Assume nothing - start at square one.

4.  Subnetting is a “new” language and like learning a new language, it takes time.  YOU must understand subnetting to effectively teach it.  I asked my class: ◦ What is it about IP addressing that makes it so hard for most people to understand?

5.  I think that the main problem most individuals have is the issue of translating an IP address to binary and back from it. It is hard for people to switch their thinking from one numeric system to another.  It may have been simple when it was first set up, but since the Internet grew faster and bigger than expected they needed to add subnetting and supernetting in order to accommodate the additional addresses needed. So understanding how and when to use subnetting and supernetting is what is difficult for me.  I think the hardest thing to grasp is, sub-netting, super-netting, moving to the right or left, or the different formulas in general. When siting in class going over everything, I'm right there with you, understanding how it works. I get home and work with it myself, I forget everything we went over.  I personally don't find IP addressing in and of itself hard to understand. Sub- netting however leaves me drooling like a moron. The IP address is the simple part of the equation.

6.  The biggest problem is figuring out what they were asking for. Number of Hosts number of networks, or number of subnets, add to the left or the right, are they usable or total number. The cheat sheet helps a lot if you know what they are asking for.  The hardest problem is the super netting and the subneting, when it is explained it looks easy, but when it is turned into story problems I just cant seem to know where to start.  I think IP addressing became more difficult over time with the addition of subnetting and supernetting. The basic concept was easy to understand, but when trying to figure out subnetting and supernetting it gets more confusing.  I know the hardest part for me to grasp was looking at the different numbers in the story problems and not thinking about them in binary but rather in decimal. Once I started changing everything into binary it was a lot easier for me to understand what was going on. Then it just got even easier when I started using the cheat sheet.

7.  Start simple!

8. Network addresses are binary representations of a special numbering system used to find devices on the internet Machines do not speak our language and do not know what decimal numbers (i.e. 1,2,3,4,5 etc) are Binary - - the language of computers I = on 0 = off Each part of a binary address (BIT) is a placeholder 128 64 32 16 8 4 2 1 I I 0 I 0 I I I Therefore: I I O I O I I I = 128 + 64 + 16 + 4 + 2 + 1 = 215 We sum the placeholders with a one and ignore the placeholders with a zero

9. What is a Network Address? 00000010 00010000 00001111 10000000 11111111 11111110 01100011 00000000 Cannot be done with 8 bits! Convert the following to binary using 8 bit positions: 2 16 15 128 255 254 99 0 300 Reference pg 9 IP Subnetting Made Easy

10.  Set the 1 st lesson aside – this is a new lesson  Do not overwhelm the student

11. Rules for IP addresses: 32 bits octets4 sections called octets Dotted decimal format Divided into a network portion and a host portion IP addresses range from 0 to 255 (128+64+32+16+8+4+2+1=255) Network addresses may look like this to us... 128.32.15.22... but they look like this to a computer: 10000000.0010000.00001111.00010110

12. Given the address of 128.32.15.22..... 128 64 32 16 8 4 2 1 I 0 0 0 0 0 0 0 128 64 32 16 8 4 2 1 0 0 0 I 0 I I 0 128 64 32 16 8 4 2 1 0 0 1 0 0 0 0 0 128 64 32 16 8 4 2 1 0 0 0 0 I I I I 128 32 15 22 128 + 0 = 128 32 + 0 = 32 8 + 4 + 2 + 1 = 15 16 + 4 + 2 = 22 This is why 128.32.15.22 = 10000000.00100000.00001111.00010110

13.  Set the 2 nd lesson aside – this is a new lesson – time for a mental stop  Again, do not overwhelm the student

14. 120.19.0.12 130.15.16.17 10.0.0.0 15.255.255.0 11.254.254.255 Note: Binary counting ALWAYS starts with a “0”, not a “1”. Also, counting like this does NOT apply to subnet masks 120.19.0.13 130.15.16.18 10.0.0.1 15.255.255.1 11.254.255.0 120.19.0.11 130.15.16.16 9.255.255.255 15.255.254.255 11.254.254.254

15. Network addresses consist of two parts Network address Host or node address Similar to an address for your home/business 128.32.15.22 Network Address Host Address 12050 Main Street Anytown, MI 48300 Regional Address Street Address Networks are like this; we have a few big cities with lots of homes and lots of small cities with few homes.

16. IP address 10.1.2.1 100.100.100.100 129.129.129.129 172.16.100.1 192.0.0.1 130.175.100.1 256.200.100.8 Class A - private A - public B - public B - private C - public B - public Illegal Address! What class (A/B/C) and type (public/private) are the following addresses?  Provide periodic topic reviews to encourage thinking

17. Subnet masks divide the network bits from the host bits Masks are used in conjunction with IP addresses ANDing – a binary adding process 1 + 1 = 1 1 + 0 = 0 0 + 1 = 0 0 + 0 = 0 Subnet masks are: Dotted decimal 4 octet 32 bit Complimentary to IP addresses, not identical Subnet masks are like duct tape

18. Classful masks A = 255.0.0.0 – or – IIIIIIII.00000000.00000000.00000000 B = 255.255.0.0– or – IIIIIIII.IIIIIIII.00000000.00000000 C = 255.255.255.0 – or – IIIIIIII.IIIIIIII. IIIIIIII.00000000 The “ones” indicate the network and the “zeros” indicate the host portion. Note that the IP address designates the class of address, the subnet mask DOES NOT! Given the IP address of 128.32.15.22 it’s classful mask would be????

19. Lets apply binary ANDing: 10000000.0010000.00001111.00010110 – 128.32.15.22 11111111.1111111.00000000.00000000 – 255.255.0.0 Lets look at it from the human perspective: 128.32.15.22 255.255.0.0 Lets look at it from the computer’s perspective: 10000000.0010000.00001111.00010110 11111111.11111111.00000000.00000000 10000000.00100000.00000000.00000000 = 128.32.0.0 128 32 0 0 Network Portion (ones) Host Portion (zeros) Everything in the host position MUST be a zero after ANDing

20. Class A: 10.0.0.0 255.0.0.0 10.0.0.1 – 10.255.255.255 Class B: 130.175.0.0 255.255.0.0 130.175.0.0 – 130.175.255.255 Class C: 200.201.202.0 255.255.255.0 200.201.202.0 – 200.201.202.255 Remember; it works like an odometer and you cannot change the “255”

21.  Yep..yet another mental stop  Might be a good time to take a break

22. How do we determine the number of possible combinations? Recall that binary rules say a bit can only be a “1” or a “0”. That leaves only 2 combinations per bit. 1 bit can equal either 1 or 0 2 bits can equal 00, 10, 01 or 00 3 bits can equal 000, 001, 010, 011, 100, 101, 110 or 111 This is how powers of “2” work: 1 bit = 2 1 or 2 x 1 = 2 combinations 2 bits = 2 2 or 2 x 2 = 4 combinations 3 bits = 2 3 or 2 x 2 x 2 = 8 combinations

23. Classful addressing does not work for all situations The solution? - subnetting! Moving the network/host boundary to the right of the “classful” boundary is called subnetting Now we give the student a real-world example!

24. “Classful Mask” 255.255.0.0 = 11111111. 11111111.00000000. 00000000 Network. Host Network. Host 128.32.15.22 = 10000000.0010000.00001111.00010110 – Class “B” Network 11111111. 11111111. 11111111. 11111100 Network. Host “Subnetted Mask” 11111111. 11111111.10000000. 00000000 Network. Host 11111111. 11111111.11100000. 00000000 Network. Host

25. 11111111. 11111111.11111111.11111100 = 30 Network bits – 2 Host bits Network. Host 2 30 = 1,073,741,824 2 2 = 4 11111111. 11111111.10000000. 00000000 = 17 Network bits – 15 Host bits Network. Host 2 17 = 131,072 2 15 = 32,768 11111111. 11111111.11100000. 00000000 = 19 Network bits – 13 Host bits Network. Host 2 19 = 524,288 2 13 = 8,192 Remember that each time you move a bit it either divides or multiplies by 2!

26. Problem Solved!

27. AKA aggregation Works only for contiguous subnets Scenario: 200.122.4.0/24 200.122.5.0/24 200.122.6.0/24 200.122.7.0/24 ADVERTISE Moving the network/host boundary to the left of the “classful” boundary is supernetting

28. 11001000. 01111010.00000100.00000000 = 200.122.4.0 11001000. 01111010.00000101.00000000 = 200.122.5.0 11001000. 01111010.00000110.00000000 = 200.122.6.0 11001000. 01111010.00000111.00000000 = 200.122.7.0 These bits are all alike.... Lets change the “like” bits to ones and apply ANDing... 11001000.01111010.00000100.00000000 - 200.122.4.0 11111111.11111111.11111100.00000000 - 255.255.252.0 Classful boundary 11001000.01111010.00000100.00000000 = 200.122.4.0 200 122 4 0 The mask is now 255.255.252.0. Since we used 22 bits the mask can be written as “/22” Supernet boundary Any address that matches “11001000.01111010.000001” will be advertised

29. 200.122.4.0/22 ADVERTISE 200.122.4.0/22 =200.122.4.0/24 200.122.5.0/24 200.122.6.0/24 and 200.122.7.0/24 11001000. 01111010.00000100.00000000 = 200.122.4.0 11001000. 01111010.00000101.00000000 = 200.122.5.0 11001000. 01111010.00000110.00000000 = 200.122.6.0 11001000. 01111010.00000111.00000000 = 200.122.7.0

30. 11001000. 01111010.00000010.00000000 = 200.122.2.0 11001000. 01111010.00000011.00000000 = 200.122.3.0 11001000. 01111010.00000100.00000000 = 200.122.4.0 11001000. 01111010.00000101.00000000 = 200.122.5.0 11001000. 01111010.00000110.00000000 = 200.122.6.0 11001000. 01111010.00000111.00000000 = 200.122.7.0 11001000. 01111010.00001000.00000000 = 200.122.8.0 11001000. 01111010.00001001.00000000 = 200.122.9.0 Note that subnetting and supernetting are collectively referred to as CIDR Why won’t other addresses work??? No match! Match!

31. Supernet the following addresses: 220.5.0.0/23 220.5.1.0/23 11011100.00000101.00000000.00000000 11011100.00000101.00000001.00000000 The Supernet Address is 220.5.0.0/23 Step one: Convert to binary and find the common bits 220.5.0.0/24 220.5.1.0/24 Step two: Apply ANDing 11011100.00000101.00000000.00000000 – 220.5.0.0 11111111.11111111.11111110.00000000 – 255.255.254.0 11011100.00000101.00000000.00000000 220 5 0 0

32. Supernet the following addresses: 130.0.0.0/16 130.1.0.0/16 130.2.0.0/16 ……. 130.7.0.0/16 10000001.00000000.00000000.00000000 10000001.00000001.00000000.00000000 10000001.00000010.00000000.00000000 10000001.00000111.00000000.00000000 The Supernet Address is 220.5.0.0/13 Step one: Convert to binary and find the common bits 130.0.0.0/16 thru 130.7.0.0/16 Step two: Apply ANDing 10000001.00000000.00000000.00000000 – 130.0.0.0 11111111.11111000. 00000000.00000000 – 255.248.0.0 11011100.00000101.00000000.00000000 220 0 0 0

33. Which addresses are in the 199.175.8.0/21 supernet? Step one: Convert 199.175.8.0 to binary 199.175.8.0 = 11000111.10101111.00001000.00000000 Step two: Isolate the number of network bits Step three: Calculate the number of possible combinations of hosts 11000111. 10101111.00001000.00000000 = 199.175.8.0 11000111. 10101111.00001001.00000000 = 199.175.9.0 11000111. 10101111.00001010.00000000 = 199.175.10.0 11000111. 10101111.00001011.00000000 = 199.175.11.0... etc 11000111. 10101111.00001111.11111111 = 199.175.15.255 Therefore; 199.175.8.0/21 covers 199.175.8.0 thru 199.175.15.255

34.  This is the heart and soul of Subnetting Made Easy ◦ Why not teach it first?

35. Now that you know the fundamentals, here’s a shortcut... Subnet BitsPower of 2 N/W – Host 1–128–72 0 1 2–192–62 1 2 3–224–52 2 4 4–240–42 3 8 5–248–32 4 16 6–252–22 5 32 7–254–12 6 64 8–255–02 7 128 2 8 256 2 9 512 2 10 1024 This cheat sheet can be easily re-created from memory and will solve most subnetting problems

36. Given a class “C” address, what is the subnet mask needed to provide at least 12 hosts per subnet? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.2 4 will yield 16 hosts 2.Find the corresponding number 4 on the “hosts” side of the center table on the cheat sheet 3.A 240 mask is required 4.Answer: 255.255.255.240 Time to break out the cheat sheet..

37. Given a class “C” address, what is the subnet mask needed to provide at least 50 hosts per subnet? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.2 6 will yield 64 hosts 2.Find the corresponding number 6 on the “hosts” side of the center table on the cheat sheet 3.A 192 mask is required 4.Answer: 255.255.255.192 Time to break out the cheat sheet..

38. Given a class “B” address, what is the subnet mask needed to provide at least 60 hosts per subnet? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.2 6 will yield 64 hosts 2.Find the number 6 on the “hosts” side of the center table on the cheat sheet 3.A 192 mask is required 4.Answer: 255.255.255.192 Time to break out the cheat sheet..

39. Given a class “C” address, what is the subnet mask needed to provide at least 20 subnets? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.2 5 will yield 32 subnets 2.Find the number 5 on the “N/W” (network) side of the center table on the cheat sheet 3.A 248 mask is required 4.Answer: 255.255.255.248 Time to break out the cheat sheet..

40. Will a class “C” address with a /27 bit mask provide at least 6 subnets with a minimum of 20 hosts per subnet? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.Lets look up the subnets first. 2.2 3 will yield 8 subnets.. 3.Find the number 3 on the “N/W” (network) side of the center table on the cheat sheet 4.A 224 mask will provide 3 subnet bits and 5 host bits 5.Find the value for 2 5 6.Is 2 3 >= 6 and is 2 5 >= 20? 7.Yes! Time to break out the cheat sheet..

41. Will a class “C” address with a /26 bit mask provide at least 3 subnets with a minimum of 70 hosts per subnet? Subnet BitsPower of 2 N/W - Hosts 1-128-72 0 1 2-192-62 1 2 3-224-52 2 4 4-240-42 3 8 5-248-32 4 16 6-252-22 5 32 7-254-12 6 64 8-255-02 7 128 2 8 256 2 9 512 2 10 1024 1.Lets look up the subnets first 2.2 2 will yield 4 subnets.. 3.Find the number 2 on the “N/W” (network) side of the center table on the cheat sheet 4.A 192 mask will provide 2 subnet bits and 6 host bits 5.Find the value for 2 6 6.Is 2 2 >= 3 and is 2 6 >= 70? 7.No! Time to break out the cheat sheet..

42.  OK, we can subnet and supernet….what now????  Two more IP address rules: ◦ The first address of every IP address range is called the network number ◦ The last address of every IP address range is called the broadcast address  Neither of these are usable for network hosts  To find the number of hosts in a network we look at the subnet mask ◦ Host bits = total bits – network bits  i.e. a “/28” mask will leave 4 host bits (32 – 28 = 4)  4 bits = 2 4 = 16 so each network will have 16 hosts (14 usable addresses)  Now that we know that….  Now we tie up the loose ends…

43.  This is why 200.32.15.0 /28 subdivides into the following 16 networks Network NumberUsable RangeBroadcast 200.32.15.0200.32.15.1 - 200.32.15.14200.32.15.15 200.32.15.16200.32.15.17 - 200.32.15.30200.32.15.31 200.32.15.32200.32.15.33 - 200.32.15.46200.32.15.47 200.32.15.48200.32.15.49 - 200.32.15.62200.32.15.63 200.32.15.64200.32.15.65 - 200.32.15.78200.32.15.79 200.32.15.80200.32.15.81 - 200.32.15.94200.32.15.95 200.32.15.96200.32.15.97 - 200.32.15.110200.32.15.111 200.32.15.112200.32.15.113 - 200.32.15.126200.32.15.127 200.32.15.128200.32.15.129 - 200.32.15.142200.32.15.143 200.32.15.144200.32.15.145 - 200.32.15.158200.32.15.159 200.32.15.160200.32.15.161 - 200.32.15.174200.32.15.175 200.32.15.176200.32.15.177 - 200.32.15.190200.32.15.191 200.32.15.192200.32.15.193 - 200.32.15.206200.32.15.207 200.32.15.208200.32.15.209 - 200.32.15.222200.32.15.223 200.32.15.224200.32.15.225 - 200.32.15.238200.32.15.239 200.32.15.240200.32.15.241 - 200.32.15.254200.32.15.255

44.  Now you try it: Subnet 199.45.20.0 /27 ◦ Step one 32 – 27 =5 ◦ Step two 2 5 = 32 ◦ Step three fill in the chart BroadcastUsable RangeNetwork Number 199.45.20.255199.45.20.225 - 199.45.20.254199.45.20.224 199.45.20.223199.45.20.193 - 199.45.20.222199.45.20.192 199.45.20.191199.45.20.161 - 199.45.20.190199.45.20.160 199.45.20.159199.45.20.129 - 199.45.20.158199.45.20.128 199.45.20.127199.45.20.97 - 199.45.20.126199.45.20.96 199.45.20.95199.45.20.65 - 199.45.20.94199.45.20.64 199.45.20.63199.45.20.33 - 199.45.20.62199.45.20.32 199.45.20.31199.45.20.1 - 199.45.20.30199.45.20.0

45.  Now you try it: Subnet 134.55.77.0 /25 ◦ Step one 32 – 25 = 7 ◦ Step two 2 7 = 128 ◦ Step three fill in the chart BroadcastUsable RangeNetwork Number 134.55.77.255134.55.77.129 - 134.55.77.254134.55.77.128 134.55.77.127134.55.77.1 - 134.55.77.126134.55.77.0

46.  Subnetting is hard because it is abstract and we do not use it every day  Start from the scratch – assume nothing  Go slow and take “mental stops”  Watch for the light bulbs – you will know that you are on the right track

47.  Ordered this book as a compliment to my other recent purchases in prep for the CCNA test. Just got it and in skimming, I am glad I did. It looks very concise, easy to follow and very understandable; much like when I took the course at a CC many years ago. (and something many 'bigger' books lack)  I am into the second chapter and already I am VERY please with it. I know that by the time I finish reading and applying what I learned sub-netting is going to be a breeze.  GREAT LITTLE BOOK AND IT HELPED ME PASS THE COURSE.  Excellent product, by an excellent author. Mr Kowalski takes time to make sure the foundations are laid correctly in this book.  I finally get it after years of not understanding subnetting. This book is a must if you are having problems understanding this concept.

48.  If you know nothing about subnetting, this is the book for you. If you know how to subnet, get this book and you will know it even better.  Great book:-) If you want to understand Subnetting - this is the book. Easy to understand, well paced and focused. Highly Recommended :-))  Book is an excellent source for getting a handle on subnetting. An easy read. The only book you will need to learn subnetting.  The book has very clear information. No fancy no elegant just what you need if you need to learn IP subnetting...  Really Outstanding! I have purchased several books on subnetting and just could not "get it" Now I get it. Thanks for your plain English translation.

49.  Questions?