Module 10 Subnetting Class A, B and C addresses Solutions to the Lab Exercises 10.3.5a, 10.3.5b, 10.3.5c and 10.3.5d
10.3.5a Basic Subnetting Use the following information and answer the following subnet related questions A company has applied for and received a Class C network address of 197.15.22.0. The physical network is to be divided into 4 subnets, which will be interconnected by routers. At least 25 hosts will be needed per subnet. A Class C custom subnet mask needs to be used and a router is needed between the subnets to route packets from one subnet to another. Determine the number of bits that need to be borrowed from the host portion of the network address and the number of bits that will be left for host addresses. Note: There will be 8 possible subnets, of which 6 can be used. (By borrowing two bits, only 4 subnets can be created (2^2 = 4) of which only 2 can be used. So three bits have to be borrowed.) Default subnet mask: 255.255.255.0 Subnet Mask in binary form: 11111111 11111111 11111111 00000000 Bits borrowed: 3 New Subnet Mask in binary form: 11111111 11111111 11111111 11100000 Decimal format: 255.255.255.224 Number of usable subnets: (2^3) 2 = 6 Number of usable hosts: (2^5) 2 = 30 Fill in the following table and answer the following questions: Subnet No. Subnet Bits Borrowed Binary Value Subnet Bits Decimal and Subnet No. Host Bits Possible Binary Values (Range) (5 Bits) Subnet/Host Decimal Range Use? Subnet 0 000 0 00000-11111 0-31 No Subnet 1 001 32 00000-11111 32-63 Yes Subnet 2 010 64 00000-11111 64-95 Yes Subnet 3 011 96 00000-11111 96-127 Yes Subnet 4 100 128 00000-11111 128 159 Yes Subnet 5 101 160 00000-11111 160 191 Yes
Subnet 6 110 192 00000-11111 192-223 Yes Subnet 7 111 224 00000-11111 224 255 No There are 32 hosts in each of the 8 subnets. Subnet Ranges: Subnet Number Subnet Range 1 st subnet 197.15.22.0 ~ 197.15.22.31 2 nd subnet 197.15.22.32 ~ 197.15.22.63 3 rd subnet 197.15.22.64 ~ 197.15.22.95 4 th subnet 197.15.22.96 ~ 197.15.22.127 5 th subnet 197.15.22.128 ~ 197.15.22.159 6 th subnet 197.15.22.160 ~ 197.15.22.191 7 th subnet 197.15.22.192 ~ 197.15.22.223 8 th subnet 197.15.22.224 ~ 197.15.22.255 1. Which octet(s) represent the network portion of a Class C IP address? First three octets 2. Which octet(s) represent the host portion of a Class C IP address? Last Octet
3. What is the binary equivalent of the Class C network address in the scenario? 197.15.22.0 Decimal network address: 197.15.22.0 Binary network address: 11000101 00001111 00010110 00000000 4. How many high-order bits were borrowed from the host bits in the fourth octet? Three 5. What subnet mask must be used? Show the subnet mask in decimal and binary. Decimal subnet mask: 255.255.255.224 Binary subnet mask: 11111111 11111111 11111111 11100000 6. What is the maximum number of subnets that can be created with this subnet mask? Eight 7. What is the maximum number of usable subnets that can be created with this mask? Six 8. How many bits were left in the fourth octet for host IDs? Five 9. How many hosts per subnet can be defined with this subnet mask? 30 hosts (usable) 10. What is the maximum number of hosts that can be defined for all subnets with this scenario? 30 hosts per subnet * 6 usable subnets = 180hosts 11. Is 197.15.22.63 a valid host IP address with this scenario? No 12. Why or why not? Broadcast address of 2 nd subnet 13. Is 197.15.22.160 a valid host IP address with this scenario? No
14. Why or why not? Network address of 6 th subnet 15. Host A has an IP address of 197.15.22.126. Host B has an IP address of 197.15.22.129. Are these hosts on the same subnet? No. Host A (197.15.22.126) belongs to 4 th subnet and Host B (197.15.22.129) belongs to 5 th subnet.
10.3.5b Subnetting a Class A Network Given a Class A network address of 10.0.0.0 / 24 answer the following questions The default subnet mask for Class A network is 255.0.0.0 Binary Format: 11111111 00000000 00000000 00000000 The new subnet mask has 24 1 s in it. New subnet mask in binary format: 11111111 11111111 11111111 00000000 It can be observed that 16 bits have been borrowed into the network portion. Number of subnets created: 2^16 = 65536 Number of usable subnets: 65536 2 = 65534 Number of host bits: 8 Number of hosts in each subnet: 2^8 = 256 Number of usable hosts: 256 2 = 254 Subnet Range for the first 6 subnets: Subnet Number Subnet Range 1 st subnet 10.0.0.0 ~ 10.0.0.255 2 nd subnet 10.0.1.0 ~ 10.0.1.255 3 rd subnet 10.0.2.0 ~ 10.0.2.255 4 th subnet 10.0.3.0 ~ 10.0.3.255 5 th subnet 10.0.4.0 ~ 10.0.4.255 6 th subnet 10.0.5.0 ~ 10.0.5.255
1. How many bits were borrowed from the host portion of this address? Sixteen 2. What is the subnet mask for this network? 255.255.255.0 3. How many usable subnets are there? 65534 usable subnets 4. How many usable hosts are there? 254 usable hosts 5. What is the host range for usable subnet sixteen? From the subnet range table above, the host range for 4 th usable subnet (5 th subnet) is 10.0.4.0 ~ 10.0.4.255 Likewise, the host range for 16 th usable subnet will be 10.0.16.0 ~ 10.0.16.255 6. What is the network address for usable subnet sixteen? 10.0.16.0 7. What is the broadcast address for usable subnet sixteen? 10.0.16.255 8. What is the broadcast address for the last usable subnet? The host range for the last subnet is 10.255.255.0 ~ 10.255.255.255. The network address for this subnet is 10.255.255.0. So the broadcast address of the previous subnet (last usable subnet) is 10.255.254.255.
10.3.5c Subnetting a Class B Network ABC Manufacturing has acquired a Class B address, 172.16.0.0. The company needs to create a subnetting scheme to provide the following: 36 subnets with at least 100 hosts 24 subnets with at least 255 hosts 10 subnets with at least 50 hosts It is not necessary to supply an address for the WAN connection since it is supplied by the Internet service provider. Regardless of the number of hosts in each subnet, the total number of subnets that need to be created is 70 (36+24+10) Number of bits to be borrowed to create atleast 70 subnets is 7. Default subnet mask: 255.255.0.0 Subnet mask for this network: 255.255.254.0 1. How many subnets are needed for this network? 70 2. What is the minimum number of bits that can be borrowed? Seven 3. What is the subnet mask for this network? Dotted decimal: 255.255.254.0 Binary format: 11111111 11111111 11111110 00000000 Slash format: 172.16.0.0/23 4. How many usable subnets are there? (2^7)-2 = 126 5. How many usable hosts are there per subnet? (2^9) 2 = 510 The first subnet range is 172.16.0.0 ~ 172.16.1.255. But the first usable subnet starts from 172.16.2.0 ~ 172.16.3.255
Subnet ranges for the 1 st three and the last four usable subnets: Subnetwork ID (Network Address) Subnet# Host Range Broadcast ID (Broadcast Address) 172.16.2.0 1 172.16.2.1 ~ 172.16.3.254 172.16.3.255 172.16.4.0 2 172.16.4.1 ~ 172.16.5.254 172.16.5.255 172.16.6.0 3 172.16.6.1 ~ 172.16.7.254 172.16.7.255 172.16.246.0 123 172.16.246.1 ~ 172.16.247.254 172.16.247.255 172.16.248.0 124 172.16.248.1 ~ 172.16.249.254 172.16.249.255 172.16.250.0 125 172.16.250.1 ~ 172.16.251.254 172.16.251.255 172.16.252.0 126 172.16.252.1 ~ 172.16.253.254 172.16.253.255 What is the host range for subnet 2? 172.16.4.1 ~ 172.16.5.254 What is the broadcast address for 126 th subnet? 172.16.253.255 What is the broadcast address for the major network? 172.16.255.255
10.3.5d Subnetting a Class C Network The Classical Academy has acquired a Class C address, 192.168.1.0. The academy needs to create subnets to provide low level security and broadcast control on the LAN. It is not necessary to supply an address for the WAN connection. It is supplied by the Internet service provider. The LAN consists of the following, each of which will require its own subnet: Classroom #1 28 nodes Classroom #2 22 nodes Computer lab 30 nodes Instructors 12 nodes Administration 8 nodes From above, it is understood that 5 subnets need to be created. 1. What is the subnet mask for this network? The default subnet mask is 255.255.255.0. By borrowing 3 bits into the network portion, the subnet mask for this network becomes 255.255.255.224 Binary: 11111111 11111111 111111111 11100000 Slash format: /27 2. How many usable hosts are there per subnet? (2^5) 2 = 30 hosts Subnet# Subnetwork ID Host Range Broadcast ID 0 192.168.1.0 192.168.1.1 ~ 192.168.1.30 192.168.1.31 1 192.168.1.32 192.168.1.33 ~ 192.168.1.62 192.168.1.63 2 192.168.1.64 192.168.1.65 ~ 192.168.1.94 192.168.1.95 3 192.168.1.96 192.168.1.97 ~ 192.168.1.126 192.168.1.127 4 192.168.1.128 192.168.1.129 ~ 192.168.1.158 192.168.1.159 5 192.168.1.160 192.168.1.161 ~ 192.168.1.190 192.168.1.191
6 192.168.1.192 192.168.1.193 ~ 192.168.1.222 192.168.1.223 7 192.168.1.224 192.168.1.225 ~ 192.168.1.254 192.168.1.255 3. What is the host range for subnet six? 192.168.1.193 ~ 192.168.1.222 4. What is the broadcast address for 3 rd subnet? 192.168.1.127 5. What is the broadcast address for the major network? 192.168.1.255