1. 17 November 2015 RE Meyers, Ms.Ed., CCAI CCENT Test Strategies Accurate and Fast IP Address Problem Solving Part 2: Critical Reading Skills

2. 17 November 2015 RE Meyers, Ms.Ed., CCAI Subnets and VLSM Problem Solving  This is not a presentation on how to subnet or how to develop VLSM IP addresses.  Both skills are presumed to be taught, practiced, and accomplished by now.  If you still can’t subnet or develop VLSM addresses, then you need to go back and practice those skills.  Practice is the only way to become proficient at subnetting and VLSM addressing. Very few people actually like to practice. Nobody likes to fail. Practice reduces failure to a small percentage. Your choice.

3. 17 November 2015 RE Meyers, Ms.Ed., CCAI Subnets and VLSM Problem Solving  Given enough practice and time, networking instructors and students will be able to solve all subnet and VLSM problems on an exam.  On a certification exam, the biggest challenge is time. Certification exam time management is a critical skill for success. Spending the least amount of time on each question to arrive at fast and accurate answers is vital. Under pressure of a ticking clock, we all can choke and freeze. This lesson proposes a way to solve time-consuming subnet and VLSM problems fast and accurately. It still requires patience and practice. Did I mention practice? You must practice. A lot.

4. 17 November 2015 RE Meyers, Ms.Ed., CCAI Critical Reading  Many complain that Cisco exams are “reading comprehension exams” and not technical exams.  Anyone who works at a help desk will confirm that clear communication is the first hurdle to solving any problem.  If you don’t understand the question, then any technical solution is lost.

5. 17 November 2015 RE Meyers, Ms.Ed., CCAI Critical Reading: Eliminate Distracters, Find the Keys  Word problems and logical topologies always give:  Exactly what you need, Keys, to solve the problem.  More than you need, called Distracters.  The first step to quick problem solving is eliminate the Distracters and pick out the Keys to the solution.

6. 17 November 2015 RE Meyers, Ms.Ed., CCAI IP Problem Solving: Eliminate Distracters, Find the Keys  Example:  A network engineer assigned the following addresses to serial connections on his enterprise network: 192.168.1.137/30, 192.168.1.209/30. 192.168.202/30, and 192.168.1.242/30. What class of addressing is being used? (select two) A B C RFC 1918 RFC 1919 PAT

7. 17 November 2015 RE Meyers, Ms.Ed., CCAI IP Problem Solving: Eliminate Distracters, Find the Keys  Red distracters. Blue keys:  A network engineer assigned the following addresses to serial connections on his enterprise network: 192.168.1.137/30, 192.168.1.209/30. 192.168.1.202/30, and 192.168.1.242/30. What class of addressing is being used? (select 2) Serial connections: Who cares? The question is about address class not connections. 1.137/30, 1.209/30, 1.202/30, 1.242/30: The third and fourth octet and CIDR do NOT determine the class of the address. 192.: The first octet number determines the IP Class. 192.168.1.x: The first octets of this IP will also tell you if it is a special reserved IP class.

8. 17 November 2015 RE Meyers, Ms.Ed., CCAI IP Problem Solving: Use the Keys  Since you can’t highlight the PC screen, you are allowed to use pen and paper on exams.  USE THEM and write down the keys from the question. 192. The first octet number determines the IP Class. –Class C 192.168. A special reserved IP class –Class C private address defined by RFC 1918

9. 17 November 2015 RE Meyers, Ms.Ed., CCAI IP Problem Solving: Pick the Correct Answers  A network engineer assigned the following addresses to serial connections on his enterprise network: 192.168.1.137/30, 192.168.1.209/30. 192.168.1.202/30, and 192.168.1.242/30. What class of addressing is being used? (select 2) : AA BB CC  RFC 1918  RFC 1919  PAT Any address beginning with 192 is Class C. Addresses beginning with 192.168. are private RFC 1918.

10. 17 November 2015 RE Meyers, Ms.Ed., CCAI 1) Finding Network Addresses: Subnetting a Subnet

11. 17 November 2015 RE Meyers, Ms.Ed., CCAI 1) Finding Network Addresses: Subnetting a Subnet  Eliminate Distracters, Find the Keys  Router names, port id’s (S0, E0)  192.168.1.64/26  192.168.1.128/26  “…valid VLSM network addresses for the serial link…”

12. 17 November 2015 RE Meyers, Ms.Ed., CCAI 1) Finding Network Addresses: Use the Keys  192.168.1.64/26 and 192.168.1.128/26 use up a lot of IP addresses.  What ranges are used and what ranges are available?  /26 means 2 bits are borrowed, so network addresses start at 0 and increment by 64: Subnet 0 = 192.168.1.0 – 63 /26 Unused; Range Available Subnet 1 = 192.168.1.64 – 127 /26 Used; unavailable Subnet 2 = 192.168.1.128 – 191 /26 Used; unavailable Subnet 3 = 192.168.1.192 – 255 /26 Unused; Range Available

13. 17 November 2015 RE Meyers, Ms.Ed., CCAI 1) Finding Network Addresses: Use the Keys  Solve the problem  Subnet 0 = 192.168.1.0 – 63 /26 Unused; Range Available Option 1: 192.168.1.4/30 increments the networks by 4 and falls inside the subnet 0 range which is available. Option 2: 192.168.1.8/30 is the next network after 4/30 and also falls inside the subnet 0 range which is available.  Subnet 1 = 192.168.1.64 – 127 /26 Used; Range Unavailable XOptions 3 & 4 are inside this range and NOT available.  Subnet 2 = 192.168.1.128 – 191 /26 Used; Range Unavailable XOption 5 is inside this range and NOT available

14. 17 November 2015 RE Meyers, Ms.Ed., CCAI 1) Finding Network Addresses: Pick the Correct Answers

15. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. A

16. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. A  Eliminate Distracters, Find the Keys:  10.118.197.55/20 assigned to Host A  “How many additional networked devices will this subnetwork support?”

17. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. A: Use the Keys  10.118.197.55/20  The IP address is not important.  /20 tells you that 12 bits are used for the host range.  2 12 – 2 = 4096 – 2 = 4094  4094 – 1 host already assigned = 4093

18. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. A: Pick the Answer

19. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. B

20. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. B  Eliminate Distracters, Find the Keys  Router Serial connection and cloud  192.168.65.32/27 is a subnetwork because of /27

21. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. B: Use the Keys  192.168.65.32/27  /27 indicates 3 bits were borrowed.  Network addresses start at 0 and increment by 32. Subnet 0 = 192.168.65.0 – 31 /27 Subnet 1 = 192.168.65.32 – 63 /27 –All host addresses from 192.168.65.33 through 192.168.65.62/27are available for use.

22. 17 November 2015 RE Meyers, Ms.Ed., CCAI 2) Finding a Range of Hosts, Ex. B: Pick the Answers All host addresses from 192.168.65.33 through 192.168.65.62/27are available for use.

23. 17 November 2015 RE Meyers, Ms.Ed., CCAI 3) Find Network Information from a Host IP

24. 17 November 2015 RE Meyers, Ms.Ed., CCAI 3) Find Network Information from a Host IP  Eliminate Distracters, Find Keys:  Distracters are also found in answer options!  In this case, all of the answers are very distracting, and three are very wrong!  Concentrate on what you can learn from the given IP “… information…from 192.168.2.93/29”

25. 17 November 2015 RE Meyers, Ms.Ed., CCAI 3) Find Network Information from a Host IP: Use the Keys  192.168.2.93/29  /29 indicates 5 bits were borrowed  /29 is CIDR for subnet mask 255.255.255.248  Networks increment by 8.  0 – 7, 8 – 15, 16 – 31, … 72 – 79, 80 – 87, 88 – 95, etc.  Usable hosts per network are 23 -2 = 6 .93 is the 5th host on the.88 –.95 network

26. 17 November 2015 RE Meyers, Ms.Ed., CCAI 3) Find Network Information from a Host IP: NOTE!  Use your correct work to eliminate wrong options Broadcast addresses are NEVER even numbers! /29 indicates 5 bits were borrowed /29 is CIDR for subnet mask 255.255.255.248 Networks increment by 8. 0 – 7, 8 – 15, 16 – 31…72 – 79, 80 – 87, 88 – 95, etc. Usable hosts per network are 23 -2 = 6.93 is the 5th host on the.88 –.95 network

27. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. A

28. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. A  Eliminate Distracters, Find Keys:  Few real distracters – ignore the router/cloud connection.  Keys Router Ethernet address: 192.133.219.30/27 Host configuration: 192.133.133.219.33 255.255.255.224 192.133.219.30

29. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. A: Use Keys  Compare router port and PC configuration:  IP of gateway matches router port.  Router CIDR /27 means 3 bits are borrowed 255.255.255.224 PC and router SM match.  CIDR /27 means networks increment by 32 0 – 31, 32 – 63, 64 – 95, etc. IP of router port is on subnet 0. IP of PC is on subnet 1.

30. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. A: Pick Answer

31. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. B

32. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. B  Eliminate Distracters, Find Keys:  H2 and H1 can’t communicate H1=192.168.22.30/28 H2=192.168.22.60/28  Switch is a distracter.

33. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. B: Use Keys  All you have are host IP’s and CIDR. So one is wrong!  CIDR /28 matches on both. /28 borrows 4 bits, so networks increment by 16 0 – 15, 16 – 32, 32 – 47, 48 – 63, etc  H1 192.168.22.30 is on subnet 1  H2 192.168.22.33 is on subnet 2  Host IP’s are not on the same subnet.

34. 17 November 2015 RE Meyers, Ms.Ed., CCAI 4) Solving a Configuration Problem, Ex. B: Pick Answer

35. 17 November 2015 RE Meyers, Ms.Ed., CCAI 5) Subnetting a Subnet: Applied VLSM  A network engineer is implementing a network design using VLSM for network 192.168.1.0/24. The engineer has decided to take one of the subnets, 192.168.1.16/28, and subnet it further for point-to-point serial link addresses.  What is the maximum number of subnets that can be created from the 192.168.1.16/28 subnet for serial connections?

36. 17 November 2015 RE Meyers, Ms.Ed., CCAI 5) Subnetting a Subnet: Applied VLSM  Distracters and Keys  A network engineer is implementing a network design using VLSM for network 192.168.1.0/24. The engineer has decided to take one of the subnets, 192.168.1.16/28, and subnet it further for point-to-point serial link addresses. What is the maximum number of subnets that can be created from the 192.168.1.16/28 subnet for serial connections?

37. 17 November 2015 RE Meyers, Ms.Ed., CCAI 5) Subnetting a Subnet: Applied VLSM:Use Keys  192.168.1.0/24  Complete Class C private address is used  192.168.1.16/28 /28 = 4 bits borrowed, subnets increment by 16 192.168.1.16/28 is the subnetwork address of the range 192.168.1.16 – 31/28.  Point to point serial connections require only 2 hosts /30 CIDR will provide 2 useable hosts per network. Start at 192.168.1.16 and end at 192.168.1.31 with /30.16 –.19,.20 –.23,.24 –.27,.28 –.31 Four subnets are available for serial connections in the 192.168.1.16/28 network.

38. 17 November 2015 RE Meyers, Ms.Ed., CCAI 5) Subnetting a Subnet: Applied VLSM: Pick Answer  A network engineer is implementing a network design using VLSM for network 192.168.1.0/24. The engineer has decided to take one of the subnets, 192.168.1.16/28, and subnet it further for point-to-point serial link addresses. What is the maximum number of subnets that can be created from the 192.168.1.16/28 subnet for serial connections?  Four

39. 17 November 2015 RE Meyers, Ms.Ed., CCAI Practice!

40. 17 November 2015 RE Meyers, Ms.Ed., CCAI End