CISCO NETWORKING ACADEMY Chabot College ELEC 99.08 Subnet Review.

CISCO NETWORKING ACADEMYCISCO NETWORKING ACADEMY

Chabot CollegeChabot College

ELEC 99.08ELEC 99.08Subnet Review


Address ClassesAddress Classes

Class A (0 - 127)

Class B (128-191)

Class C (192-223)

Network Host Host Host

Network Network Host Host

Network Network Network Host

1st octet 2nd octet 3rd octet 4th octet


IP Address CompositionIP Address Composition

IP address may have an additional field, the subnet field, that is used to identify smaller networks within a larger network.

NETWORK NUMBER

NETWORK NUMBER

SUBNET NUMBERSUBNET NUMBER

HOST NUMBER

HOST NUMBER


The Subnet MaskThe Subnet Mask

11001111 00010101 00110110 00001000

207 21 54 8Address

Mask


11111111 11111111 11111111 00000000

255 255 255 0

Mask has:

1s in the network/subnet portion

0s in the host portion


The Default Subnet MasksThe Default Subnet Masks

Class A or /8

Class B or /16

Class C or /24

255 0 0 0

255 255 0 0

255 255 255 0

If any additional bits in the subnet mask are set to “1,” then subnets have been created.


Octet - binary/decimal equivalents Octet - binary/decimal equivalents (memorize)(memorize)

Bit

27 26 25 24 23 22 21 20 =

128 64 32 16 8 4 2 1 =DecimalValue

1 0 0 0 0 0 0 0 = 128

1 1 0 0 0 0 0 0 = 192

1 1 1 0 0 0 0 0 = 224

1 1 1 1 0 0 0 0 = 240

1 1 1 1 1 0 0 0 = 248

1 1 1 1 1 1 0 0 = 252

1 1 1 1 1 1 1 0 = 254

1 1 1 1 1 1 1 1 = 255


Class C : 1-bit Mask (/25)Class C : 1-bit Mask (/25)

11001111 00010101 00110110 00001000

207 21 54 8Address

Mask


11111111 11111111 11111111 10000000

255 255 255 128

Masks can be written different ways. When the phrase “1-bit” mask is used, that means 1 more bit than the default. This example can also be called a 25-bit mask, or /25.


All Zeros / All OnesAll Zeros / All Ones

• The address of the subnet itself (the wire) hasa host portion that is all zeros.

• The broadcast address for the subnet hasa host portion that is all ones.

000000000 =

128 =

255 =

01111111127 =

10000000

11111111

Subnet #0

Subnet #1

Decimal Binary

Subnet addr.

Broadcast addr.

Subnet addr.

Broadcast addr.


Class C : 1-bit Mask (/25)Class C : 1-bit Mask (/25)

11001111 00010101 00110110 00000000

207 21 54 0Address

Mask


11111111 11111111 11111111 10000000

255 255 255 128

Because this is a class C, all nodes on this network will share the first three octets. Those numbers are not an issue.


Last Octet 1-bit Mask (/25)Last Octet 1-bit Mask (/25)

00000000

0Address

Mask

Host

10000000

128

How many subnets can we have with 1 bit?

21 = 2 different possibilities

1 bit


Last Octet 1-bit Mask (/25)Last Octet 1-bit Mask (/25)

000000000 =

128 =

255 =

How many hosts can be on each subnet?

01111111127 =

10000000

11111111

Subnet #0

Subnet #1

Decimal Binary

27 = 128 different possibilities MINUS TWO.

7 bits


Test Question Type 1Test Question Type 1

IP Address 169.199.69.137

Subnet Mask 255.255.255.192

• How many subnets are created by this mask?

• How many hosts can be on each subnet?

Steps:

• Identify address class & network bits

• Count subnet bits; express as power of 2 (-2)

• Count host bits; express as power of 2 (-2)


Step 1: Identify Class & Network BitsStep 1: Identify Class & Network Bits

169 199 69 137Address

Mask

Network Network Subnet Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10001001

CLASS B - ignore network octets

16 bits


Step 2: Count Subnet BitsStep 2: Count Subnet Bits

169 199 69 137Address

Mask


255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10001001

10 bits

210 = 1,024 possible subnets MINUS TWO


Step 3: Count Host BitsStep 3: Count Host Bits

169 199 69 137Address

Mask


255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10001001

6 bits

26 = 64 possible hosts / subnet MINUS TWO.


Test Question Type 2Test Question Type 2Long MethodLong Method

IP Address 169.199.69.137

Subnet Mask 255.255.255.192

• What is this host’s subnetwork number? • What is the broadcast address for the subnet that contains this host?

Method:• Count host bits; determine how many host addresses/subnet• List the host ranges in each subnet (Multiples of 2N)• Find the subnet containing the target address• The first address will have all 0s in the host portion• The last address will have all 1s in the host portion


Step 1: Count Host BitsStep 1: Count Host Bits

169 199 69 137Address

Mask


255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10001001

6 bits

26 = 64 host address / subnetIncrement host ranges by 64.


Remaning Steps: Remaning Steps: List host ranges in each subnetList host ranges in each subnetFind subnet containing target addressFind subnet containing target addressFirst address is subnet, last is broadcastFirst address is subnet, last is broadcast

Subnet Subnet mask Subnetwork Address

Range of possibleHost IP Addresses Broadcast Address

0 255.255.255.192 169.199.69.0 169.199.69.1 - 62 169.199.69.63

1 255.255.255.192 169.199.69.64 169.199.69.65 - 126 169.199.69.127

2 255.255.255.192 169.199.69.128 169.199.69.129 - 190 169.199.69.191

3 255.255.255.192 169.199.69.192 169.199.69.193 - 254 169.199.69.255

Subnet address is: 169.199.69.128Subnet broadcast address is: 169.199.69.191


IP Address 169.199.69.137

Subnet Mask 255.255.255.192

• What is this host’s subnetwork number?

Steps:

1. Convert address & mask to binary

2. Replace host portion of address with all 0s

3. Convert each octet back to dotted decimal• Tip: Ignore any octets with a mask that is all 1s

Test Question Type 2Test Question Type 2Quick MethodQuick Method


Tip: Ignore octets masked with all 1s:Tip: Ignore octets masked with all 1s:

169 199 69 137Address

Mask255 255 255 192

11111111 11111111 11111111 11000000

The values of these octets are the same for the subnet network & broadcast address.


Step 1: Convert to binaryStep 1: Convert to binary

169 199 69 137Address

Mask255 255 255 192

11111111 11111111 11111111 11000000

10001001

Convert the remaining octet(s).


Step 2: Set host portion of address to all 0sStep 2: Set host portion of address to all 0s

169 199 69Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10000000

(Value of this octet will change)


Step 3: Step 3: Convert octet back to dotted decimalConvert octet back to dotted decimal

169 199 69 128Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10000000

Subnet address is: 169.199.69.128


Quick Method - Broadcast addr.Quick Method - Broadcast addr.

IP Address 169.199.69.137

Subnet Mask 255.255.255.192

• What is the broadcast address for the subnet that contains this host?

Same method, but use all 1s in step 2:

1. Convert address & mask to binary

2. Set host portion of address to all 1s

3. Convert each octet back to dotted decimal


Tip: Ignore octets masked with all 1s:Tip: Ignore octets masked with all 1s:

169 199 69 137Address

Mask255 255 255 192

11111111 11111111 11111111 11000000



169 199 69 137Address

Mask255 255 255 192

11111111 11111111 11111111 11000000

10001001


Step 2: Set host portion of address to all 1sStep 2: Set host portion of address to all 1s

169 199 69Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10111111

(Value of this octet will change)


Step 3: Step 3: Convert each octet back to dotted decimalConvert each octet back to dotted decimal

169 199 69 191Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10111111

Subnet broadcast address is: 169.199.69.191


Second Example, Quick MethodSecond Example, Quick Method

IP Address 169.199.69.137

Subnet Mask 255.255.248.0

• What is this host’s subnet number?

• What is the broadcast address for that subnet?

Steps:

• Convert address & mask to binary

• Replace host portion of address with all 0s for subnet

• Replace host portion of address with all 1s for broadcast

• Convert each octet back to dotted decimal


Ignore octets masked with all 1s:Ignore octets masked with all 1s:

169 199 69 137Address

Mask255 255 248 0

11111111 11111111 11111000 00000000



169 199 69 137Address

Mask255 255 248 0

11111111 11111111 11111000 00000000

01000101 10001001


Subnet AddressSubnet AddressStep 2: Set host portion of address to all 0sStep 2: Set host portion of address to all 0s

169 199Address

Mask

Host

255 255 248 0

11111111 11111111 11111000 00000000

01000000 00000000

(Value of these octets will change)

Subnet


169 199Address

Mask

Host

255 255 248 0

11111111 11111111 11111000 00000000

01000000 00000000


64 0

Subnet address is: 169.199.64.0

Subnet


Broadcast AddressBroadcast AddressStep 2: Set host portion of address to all 1sStep 2: Set host portion of address to all 1s

169 199Address

Mask

Host

255 255 248 0

11111111 11111111 11111000 00000000

01000111 11111111

(Value of these octets will change)

Subnet


169 199Address

Mask

Host

255 255 248 0

11111111 11111111 11111000 00000000

01000111 11111111


71 255

Subnet broadcast address is: 169.199.71.255

Subnet


Network 200.199.69.0

Subnet Mask 255.255.255.192

• What is the fourth usable host address on the second usable subnet?

Steps:

1. Ignore octets masked with all ones.

2. Replace host portion of address with binary four

3. Replace subnet portion with binary two

3. Convert each octet back to dotted decimal

Test Question Type 3Test Question Type 3Quick MethodQuick Method


Step 1: Ignore octets masked with all 1s:Step 1: Ignore octets masked with all 1s:

200 199 69 137Address

Mask255 255 255 192

11111111 11111111 11111111 11000000

These octets won’t change.


Step 2: Set host portion to binary fourStep 2: Set host portion to binary four

200 199 69Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 000100

000000000001000010000011000100

Reserved

Fourth usable host

First usable host


Step 3: Set subnet portion to binary twoStep 3: Set subnet portion to binary two

200 199 69Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10000100

00011011

Reserved

Second usable subnetFirst usable subnet


Step 3: Step 3: Convert octet back to dotted decimalConvert octet back to dotted decimal

200 199 69 132Address

Mask

Host

255 255 255 192

11111111 11111111 11111111 11000000

10101001 11000111 01000101 10000100

Fourth usable address in second usable subnet is: 169.199.69.132


1. Ignore octets masked 255.

2. Convert remaining octets to binary.

2. Segment binary address into portions:network | subnet | host

4. Use your understanding of binary addressing to answer the problem.

5. Convert each octet back to dotted decimal.

Quick Method - General StrategyQuick Method - General Strategy

CISCO NETWORKING ACADEMY Chabot College ELEC 99.08 Subnet Review.

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