# Subnetting Practice Exercises ## Level 1 - Easy Exercise 1: Network: 192.168.1.0/24 Task: Find network address, broadcast address, first host, last host, and number of usable hosts. Answer: Network address: 192.168.1.0 Broadcast address: 192.168.1.255 First host: 192.168.1.1 Last host: 192.168.1.254 Usable hosts: 254 Exercise 2: IP address: 192.168.5.100/26 Task: Identify which subnet this host belongs to. Answer: /26 block size = 64. Third multiple of 64 below 100 is 64. Network address: 192.168.5.64 Broadcast: 192.168.5.127 Host is in subnet 192.168.5.64/26. Exercise 3: Convert /28 to dotted decimal subnet mask. Answer: 28 ones followed by 4 zeros in binary. 11111111.11111111.11111111.11110000 = 255.255.255.240 ## Level 2 - Medium Exercise 4: Network: 10.0.0.0/8 Task: Create subnets of size /24. How many subnets? How many hosts per subnet? Answer: Bits borrowed = 24 - 8 = 16 bits. Number of /24 subnets = 2^16 = 65,536 subnets. Host bits per subnet = 32 - 24 = 8. Usable hosts per subnet = 2^8 - 2 = 254. Exercise 5: A company needs subnets supporting 200 hosts, 100 hosts, 50 hosts, and 25 hosts from 192.168.0.0/24. Answer using VLSM allocate largest first: 200 hosts needs /24 which gives 254 hosts but that uses entire network. Actually 200 hosts needs at least 8 host bits giving /24, not possible within /24 parent. Recalculate: 200 hosts needs 2^h >= 202, so h=8 giving /24. Cannot fit in /24 parent. Corrected using 10.0.0.0/22 instead: 200 hosts: /24 - 10.0.0.0/24, hosts .1 to .254 100 hosts: /25 - 10.0.1.0/25, hosts .1 to .126 50 hosts: /26 - 10.0.1.128/26, hosts .129 to .190 25 hosts: /27 - 10.0.1.192/27, hosts .193 to .222 Exercise 6: IP: 172.20.55.67/21 Find network address and broadcast address. Answer: /21 means interesting octet is third octet with 5 network bits and 3 host bits. Block size = 2^3 = 8. 55 / 8 = 6 remainder 7, subnet starts at 6 x 8 = 48. Network: 172.20.48.0 Broadcast: 172.20.55.255 ## Level 3 - Hard Exercise 7: A router has the following networks in its routing table: 192.168.16.0/24 192.168.17.0/24 192.168.18.0/24 192.168.19.0/24 Summarize these into a single route. Answer: Third octet values: 16=00010000, 17=00010001, 18=00010010, 19=00010011. Common bits: first 22 bits are 00010000 to 00010011, first 6 bits of third octet are common 000100. Summary route: 192.168.16.0/22 Exercise 8: How many subnets and hosts does 172.16.0.0/12 have when subnetted to /20? Answer: Bits borrowed = 20 - 12 = 8 bits. Number of subnets = 2^8 = 256. Host bits = 32 - 20 = 12. Usable hosts = 2^12 - 2 = 4094.