# Chapter 9a Intro to Routing & Switching.  Upon completion of this chapter, you should be able to:  Explain why routing is necessary for hosts on different.

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Chapter 9a Intro to Routing & Switching

 Upon completion of this chapter, you should be able to:  Explain why routing is necessary for hosts on different networks to communicate.  Describe IP as a communication protocol used to identify a single device on a network.  Given a network and a subnet mask, calculate the number of host addresses available.  Calculate the necessary subnet mask in order to accommodate the requirements of a network.  Describe the benefits of variable length subnet masking (VLSM)

9.1

 How many bits in an IPv4 address?  32  How many octets in an IPv4 address? 44  What’s the range of numbers in each octet?  0-255  What are the bit values?  128, 64, 32, 16, 8, 4, 2, 1  Convert 192.168.1.106  11000000.10101000.00000001.01101010

 11100101 to decimal  10001110 to decimal  11111000 to decimal  11111111 to decimal

 192 to binary  224 to binary  47 to binary  115 to binary

 Range:  Default Subnet Mask:  Which octets are Network & Host?  How many hosts available?  Give an example IP & SM:

 5 to binary  77 to binary  100 to binary  127 to binary  What’s in common with all of them?

 Range:  Default Subnet Mask:  Which octets are Network & Host?  How many hosts available?  Give an example IP & SM:

 128 to binary  142 to binary  191 to binary  What’s in common here?

 Range:  Default Subnet Mask:  Which octets are Network & Host?  How many hosts available, total & useable?  Give an example IP & SM:

 192 to binary  200 to binary  223 to binary  What’s common here?

 Class D  Multicasting  Class E  Experimental Use  Private Addresses  A-  B-  C-

9.2

 255.255.255.0  How many total bits are on? (1’s) 11111111.11111111.11111111.00000000 /24 notation  255.255.0.0  How many total bits are on? (1’s) 11111111.11111111.00000000.00000000 /16 notation  255.255.255.248  How many total bits are on? (1’s) 11111111.11111111.11111111.11111000 /29 notation

 Router ONLY knows which NETWORKS it is connected to!!!  Doesn’t care about individual hosts  It ANDs the IP & Subnet Mask  Result= DESTINATION NETWORK  Looks in routing table for destination network & sends it out the outgoing port

9.3

Network Subnetwork Hosts

 Borrow bits from host portion to make new networks  Ask yourself…  How many networks do you need?  How many hosts per network are there?

 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  2 2 = 4  2 3 = 8  2 4 = 16  2 5 = 32  2 6 = 64 Remember 2 you can’t use: Network & Broadcast

1. 199.72.101.0-31 2..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

 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 #

 221.17.125.46 /28  What class address? C: Only deal with the last octet!  255.255.255.240  11111111.11111111.11111111.11110000  How many bits borrowed? 4 2 4 = 16 networks  How many bits left over? 4 2 4 = 16 hosts per sub-network

 Based on the IP address & SM, identify…  The network address  The broadcast address  How many bits were borrowed  How many bits were left over  Is the address valid

 Are the hosts on the same network or separate?

 What’s the broadcast address for 201.78.90.0 /24?  201.78.90.255  Default 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

 Which class gives you the most hosts/network? AA  Which class give you the most networks? CC  Public IP addresses must be __________.

 CIDR  VLSM  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

 Instead of having multiple subnet entries for each router port, CIDR uses the common bits to make ONE routing table address per port.

 Complete the study guide handout  Take the quiz on netacad.com  Jeopardy review

In this chapter, you learned:

Chapter 9a Intro to Routing & Switching

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