1. DCN286 INTRODUCTION TO DATA COMMUNICATION TECHNOLOGY Network Basics Introduction Instructor: Ataur Rahman Ataur.Rahman@senecac.on.ca

2. What we know already…or should An IP address is 32 bits long – 4 separate bytes An IP Address is represented in dotted- decimal notation Each byte represents a decimal number separated by a period Example: 10.100.30.4 or (010.100.030.004) Each byte has a total of 256 values – 0-255

4. IP address 1. Traditionally, the IP address was classified in classes: Class A: network 1 – 126 (Network 127 is reserved for loopback and internal testing) Class B: network 128 – 191 Class C: network 192 – 223 Class D: network 224 - 239 (Reserved for multicast) Class E: network 240 – 255(Reserved for Experimental, used for research) 2. There are public IP address used in Internet and private IP address used in company/organization internal networks. Class A: 10.0.0.0 -10.255.255.255 Class B: 172.16.0.0 – 172.31.255.255 Class C: 192.168.0.0 – 192.168.255.255

5. IP Address Continued-  There are three (3) usable IP address classes - A, B and C The first byte identifies the class – “Classification” ClassExampleNetworksHosts A - 1 to 12624.0.0.012716,777,214 B- 128 to191150.18.0.016,38465,534 C- 192 to 223198.23.210. 02,097,152254 D – 224-239224.0.0.10Multicast E – 240-255DoD Reserved

6. Subnet Mask Subnet mask is to divide IP address into two sections: network ID and host ID. Say the IP address is 192.168.4.64 with subnet mask 255.255.255.0 Translate all decimal into binary: IP address would be 11000000.100101000.00000100.01000000 Subnet mask would be 11111111.11111111.11111111.00000000 Final result: Network id: 11000000.100101000.00000100 which equals to 192.168.4 Host id: 01000000 meaning 64

7. What we know already…or maybe not 10.1.0.255190.16.221.0 172.16.0.255128.255.1.255 10.0.255.0 Question: How many of these addresses are valid IP Number for a Host ? Correct Response? ------ How do you know?---

8. Practice: Classification – What Class? A 10.1.0.200 B 190.16.21.10 C 192.16.2.210 B 128.215.3.199 A 126.7.10.40 Nice Work!!!!

9. Practice: Class Boundary- Draw the line Network | Host Network | Host A  10.|1.0.200 B  190.16. | 21.10 Network | Host C  192.16.2. | 210

10. What is the Network address of 172.16.132.70/20?

11. Class B table # of bit borrowedSubnet Mask# of Usable Subnets # of available Hosts.11000000.00000000255.255.192.0416,382. 11100000.00000000 255.255.224.088,190. 11110000.00000000 255.255.240.0164,094