1. 5 - 1 5.5.2 Known IP Address and Ethernet Address Transmit from B => F (B knows F’s IP Address) PATHIP Source IP Destination Ethernet Source Ethernet Destination B => C 55.1263.2111-0111-44 C => G 55.1263.2111-AAA1-A1 G => D 55.1263.2111-0111-43 D => F 55.1263.2111-1233-33

2. 5 - 2 5.5.3 Unknown IP Address Transmit from B => F (B doesn’t know F’s IP address) PATHIP Source IP Destination Ethernet Source Ethernet Destination B => C 55.12254.611-0111-44 C => G 55.12254.611-AAA1-A1 G => DNS55.12254.62A-FF11-11 DNS => G254.655.1211-112A-FF G => C254.655.12A1-A111-AA C => B254.655.1211-4411-01 B => C 55.1263.2111-0111-44 C => G 55.1263.2111-AAA1-A1 G => D 55.1263.2111-0111-43 D => F 55.1263.2111-1233-33 DNS Request DNS Response

3. 5 - 3 5.5.4 Unknown Ethernet Address Transmit from B => F (doesn’t know F’s Ethernet address) PATHIP Source IP Destination Ethernet Source Ethernet Destination B => C 55.1263.2111-0111-44 C => G 55.1263.2111-AAA1-A1 G => D 55.1263.2111-0111-43 D => ARP Req63.163.2111-12Broadcast F => D63.2163.133-3311-12 D => F 55.1263.2111-1233-33

4. 5 - 4 IP Subnetting Practice 192.168.1.0/22 Based on needing 5 subnets, answer the following: Bits used for subnetting (borrowed bits) = 3 Total # of Subnets* = 2^ n = 2^ 3 = 8 subnets Bits available for Hosts = 10 – 3 = 7 Total # of Hosts per Subnet** = 2^ n – 2 = 2^ 7 – 2 = 126 Notes Since this IP has a CIDR of /22, this means that we can use the remaining 10 bits (32-22=10) to create our subnets. * n is equal to the number of bits used for subnetting ** n is equal to the remaining bits & minus 2 accounts for broadcast and network addresses

5. 5 - 5 IP Subnetting Practice 192.168.1.0/27 Based on needing 3 subnets, answer the following: Bits used for subnetting (borrowed bits) = 2 bits for subnets Total # of Subnets* = 2^ n = 2^ 2 = 4 subnets Bits available for Hosts = 5 – 2 = 3 bit left Total # of Hosts per Subnet** = 2^ n – 2 = 2^ 3 – 2 = 6 hosts Notes Since this IP has a CIDR of /27, this means that we can use the remaining 5 bits (32-27=5) to create our subnets. * n is equal to the number of bits used for subnetting ** n is equal to the remaining bits & minus 2 accounts for broadcast and network addresses

6. 5 - 6 IP Subnetting Practice 192.168.1.15/24 What is the network address? 192.168.1.0 (since this is /24 the first 24 bits will be the network address) What is the subnet mask for this address? To do this, put a 1 in each of the first 24 bits so: 11111111. 11111111. 11111111. 00000000 255.255.255.0