Chapter 5b. Upon completion of this chapter, you should be able to: Subnet an IPv4 address to make...

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5.1.3

Transcript of Chapter 5b. Upon completion of this chapter, you should be able to: Subnet an IPv4 address to make...

IP ADDRESSING & SUBNETTING

Chapter 5b

OBJECTIVES Upon completion of this chapter, you

should be able to:Subnet an IPv4 address to make separate

networks out of one address givenUse the new subnet mask to identify the

subnetAssign IPv4 addresses to hosts within your

network Identify each subnet’s new network ID and

broadcast addressFind the network, broadcast, or check if the

address is a valid host address by working backwards

CLASSFUL SUBNETTING

5.1.3

PARTS OF THE NETWORK

Network

Subnetwork

Hosts

SUBNETTING-SPLIT UP THE IP ADDRESS Borrow bits from host portion to make

new networks Ask yourself…

How many networks do you need?How many hosts per network are there?

SOME RULES FIRST… You MUST borrow at least 2 bits or leave at

least 2 bits Class C has 1 octet to borrow from Class B has 2 octets to borrow from Class A has 3 octets to borrow from

Remember the powers of 2 22= 4 23= 8 24= 16 25= 32 26= 64

Remember 2 addresses you

can’t use: Network & Broadcast

199.72.101.0

199.72.101.0

1. 199.72.101.0-312. .32-.63 (.33-.62)3. .64-.95 (.65-.94)4. .96-.127 (.97-.126)5. .128-.159 (.129-.158)6. .160-.191 (.161- .190)7. .192-.223 (.191-.222)8. .224-.255

Total Range #3 Useable Range #2

Network ID 199.72.101.64 /27

Broadcast Address 199.72.101.95 /27

ASSIGN ADDRESSES

ROUTER…ACTION! A packet with a destination IP of

199.72.101.85 255.255.255.224 goes to a router It ANDs to come up with the NETWORK #

HOW TO WORK BACKWARDS 221.17.125.46 /28

What class address? C: Only deal with the last octet!

255.255.255.24011111111.11111111.11111111.11110000How many bits borrowed?

4 24= 16 networks

How many bits left over? 4 24= 16 hosts per sub-network

WORK BACKWARDS Based on the IP address & SM, identify…

The network addressThe broadcast addressHow many bits were borrowedHow many bits were left over Is the address valid

WHAT’S WRONG? PC1 has an address of 199.15.1.40 /28.

PC2 has an address of 199.15.1.50 /28. Both are connected to a switch but they cannot communicate.The switch is working normally as are the PC’s NICs and Ethernet cables.Are the hosts on the same network or separate?

QUESTIONS What’s the broadcast address for

201.78.90.0 /24?201.78.90.255Default SM, no subnetting

Sam’s Beef Hut uses network 215.67.106.0 & 255.255.255.240 to create subnets. How many useable hosts can be created per network?14

QUESTIONS Which class gives you the most

hosts/network?A

Which class give you the most networks?C

Public IP addresses must be __________.

CLASSLESS SUBNETTING

5.1.5

CLASSLESS SUBNETTING Allows you to use part of the address for

the network portion Called partial subnetting or VLSM CIDR allows this

Helps reduce wasted addresses

You can subnet for each unequal network Your address is 210.1.17.64 /26

Net A needs 37 hosts Net B needs 15 hosts Net C needs 100 hosts

TEST ON ADDRESSING & SUBNETTING

IP ADDRESSING & SUBNETTING

Chapter 5b