17 november 2015 re meyers, ms.ed., ccai ccent test strategies accurate and fast ip address problem...

April 20, 2023 RE Meyers, Ms.Ed., CCAI

CCENT Test StrategiesCCENT Test Strategies

Accurate and Fast IP Address Problem Solving

Part 2: Critical Reading Skills


Subnets and VLSM Problem SolvingSubnets 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.


Subnets and VLSM Problem SolvingSubnets 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.


Critical ReadingCritical 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.


Critical Reading: Eliminate Critical Reading: Eliminate DistractersDistracters, Find the , Find the KeysKeys

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.


IP Problem Solving: Eliminate IP Problem Solving: Eliminate DistractersDistracters, Find the , Find the KeysKeys

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


IP Problem Solving: Eliminate IP Problem Solving: Eliminate DistractersDistracters, Find the , Find the KeysKeys

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.


IP Problem Solving: IP Problem Solving: Use the KeysUse 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


IP Problem Solving: Pick the Correct AnswersIP 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) :ABCRFC 1918RFC 1919PAT

Any address beginning with 192 is Class C.

Addresses beginning with 192.168. are private RFC 1918.


1) Finding Network Addresses: Subnetting a Subnet1) Finding Network Addresses: Subnetting a Subnet


1) Finding Network Addresses: Subnetting a Subnet1) Finding Network Addresses: Subnetting a Subnet

Eliminate Distracters, Find the Keys Router names, port id’s (S0, E0)192.168.1.64/26192.168.1.128/26“…valid VLSM network addresses for the serial link…”


1) Finding Network Addresses: Use the 1) Finding Network Addresses: Use the KeysKeys

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


1) Finding Network Addresses: Use the 1) Finding Network Addresses: Use the KeysKeys

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 UnavailableX Options 3 & 4 are inside this range and NOT available.

Subnet 2 = 192.168.1.128 – 191 /26 Used; Range UnavailableX Option 5 is inside this range and NOT available


1) 1) Finding Network Addresses: Pick the Correct AnswersFinding Network Addresses: Pick the Correct Answers


2) Finding a Range of Hosts, Ex. A2) Finding a Range of Hosts, Ex. A


2) Finding a Range of Hosts, Ex. A2) 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?”


2) Finding a Range of Hosts, Ex. A: Use the 2) Finding a Range of Hosts, Ex. A: Use the KeysKeys

10.118.197.55/20The IP address is not important. /20 tells you that 12 bits are used for

the host range.212 – 2 = 4096 – 2 = 40944094 – 1 host already assigned =

4093


2) Finding a Range of Hosts, Ex. A: Pick the Answer2) Finding a Range of Hosts, Ex. A: Pick the Answer


2) Finding a Range of Hosts, Ex. B2) Finding a Range of Hosts, Ex. B


2) Finding a Range of Hosts, Ex. B2) Finding a Range of Hosts, Ex. B

Eliminate Distracters, Find the KeysRouter Serial connection and cloud192.168.65.32/27 is a subnetwork because of /27


2) Finding a Range of Hosts, Ex. B: Use the 2) Finding a Range of Hosts, Ex. B: Use the KeysKeys

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.


2) Finding a Range of Hosts, Ex. B: Pick the Answers2) 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.


3) Find Network Information from a Host IP3) Find Network Information from a Host IP


3) Find Network Information from a Host IP3) 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”


3) Find Network Information from a Host IP: Use the 3) Find Network Information from a Host IP: Use the KeysKeys

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


3) Find Network Information from a Host IP: 3) Find Network Information from a Host IP: NOTE!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


4) Solving a Configuration Problem, Ex. A4) Solving a Configuration Problem, Ex. A


4) Solving a Configuration Problem, Ex. A4) 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


4) Solving a Configuration Problem, Ex. A: Use 4) Solving a Configuration Problem, Ex. A: Use KeysKeys

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.


4) Solving a Configuration Problem, Ex. A: Pick Answer4) Solving a Configuration Problem, Ex. A: Pick Answer


4) Solving a Configuration Problem, Ex. B4) Solving a Configuration Problem, Ex. B


4) Solving a Configuration Problem, Ex. B4) 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.


4) Solving a Configuration Problem, Ex. B: Use 4) Solving a Configuration Problem, Ex. B: Use KeysKeys

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 1H2 192.168.22.33 is on subnet 2Host IP’s are not on the same

subnet.


4) Solving a Configuration Problem, Ex. B: Pick Answer4) Solving a Configuration Problem, Ex. B: Pick Answer


5) Subnetting a Subnet: Applied VLSM5) 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?


5) Subnetting a Subnet: Applied VLSM5) Subnetting a Subnet: Applied VLSM

Distracters and KeysA 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?


5) Subnetting a Subnet: Applied VLSM:Use 5) Subnetting a Subnet: Applied VLSM:Use KeysKeys

192.168.1.0/24Complete Class C private address is used192.168.1.16/28

• /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.


5) Subnetting a Subnet: Applied VLSM: Pick Answer5) 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


Practice!


EndEnd

17 november 2015 re meyers, ms.ed., ccai ccent test strategies accurate and fast ip address problem...

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