Module 4, Assignment 3-2 Lab 3.4 Basic Subnetting Question 3.4.1: 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. --------------------------------------------------------------------------- WORKSPACE: (Class C) NNNNNNNN NNNNNNNN NNNNNNNN SSSHHHHH 255 255 255 224 197 15 22 0 197 15 22 0 N = 24 S = 3 (2^3 2 = 6 usable subnets, 4 needed) H = 5 (2^5 2 = 30 usable hosts, 25 needed) --------------------------------------------------------------------------- 256 224 = 32 [magic number] 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? 0 Subnet 1 st Subnet 2 nd Subnet 3 rd Subnet 4 th Subnet 5 th Subnet 6 th Subnet 7 th Subnet 000 0 (197.15.22.0) 00000-11111 0-31 No (zero subnet) 001 32 (197.15.22.32) 00000-11111 32-63 YES 010 64 00000-11111 64-95 YES (197.15.22.64) 011 96 00000-11111 96-127 YES (197.15.22.96) 100 128 00000-11111 128-159 YES (197.15.22.128) 101 160 00000-11111 160-191 YES (197.15.22.160) 110 192 00000-11111 192-223 YES (197.15.22.192) 111 224 00000-11111 224-255 No (broadcast) (197.15.22.224)
Use the table just developed to help answer the following questions: 1. Which octet(s) represent the network portion of a Class C IP address? The first three 2. Which octet(s) represent the host portion of a Class C IP address? The last one 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: 11000001. 00001111. 00010110. 00000000 4. How many high-order bits were borrowed from the host bits in the fourth octet? 3 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? 8 7. What is the maximum number of useable subnets that can be created with this mask? 6 8. How many bits were left in the fourth octet for host IDs? 5 9. How many hosts per subnet can be defined with this subnet mask? 2^5 2 = 30 usable subnets 10. What is the maximum number of hosts that can be defined for all subnets with this scenario? Assume the lowest and highest subnet numbers and the lowest and highest host ID on each subnet cannot be used. 6 (# usable subnets) x 30 (# usable hosts/subnet) = 180 11. Is 197.15.22.63 a valid host IP address with this scenario? Why or why not? No. It s the broadcast address for the 1 st subnet (197.15.22.32) 12. Is 197.15.22.160 a valid host IP address with this scenario? Why or why not? No. It is the address for the 5th subnet. 13. 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? Why? No. Host A (126) is in the 3 rd subnet (96-127) and Host B (129) is in the 4 th subnet (128-159).
Lab 3.5 Subnetting a Class A Network Question 3.5.1: ----------------------------------------------- WORKSPACE: Address Network bits Subnet bits Host bits # of subnets in network 10.0.0.0 / 24 8 (Class A) 16 8 2^16 = 65,536 # if hosts per subnet 2^8-2 = 254 Subnet 16 Address: 10.0.16.0 Subnet 16 host range: 10.0.16.1 10.0.16.254 Subnet 16 broadcast: 10.0.16.255 Last subnet: 10.255.254.0 ----------------------------------------------- Given a Class A network address of 10.0.0.0 / 24 answer the following questions: 1. How many bits were borrowed from the host portion of this address? 16 2. What is the subnet mask for this network? Dotted decimal? 255.255.255.000 Binary? 11111111.11111111.11111111.00000000 3. How many usable subnetworks are there? 2^16 2 = 65,534 (can t use first or last) 4. How many usable hosts are there per subnet? 2^8 2 = 254 5. What is the host range for usable subnet sixteen? 10.0.16.1 10.0.16.254 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? 10.255.254.255 9. What is the broadcast address for the major network? 10.255.255.255 Lab 3.6 Subnetting a Class B Network Background / Preparation This is a written lab and is to be performed without the aid of an electronic calculator. 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.
Question 3.6.1: Given this Class B network address and these requirements answer the following questions 1. How many subnets are needed for this network? 70 2. What is the minimum number of bits that can be borrowed? 7 N = 16 (Class B) H = 9 (2^8-2 = 254 too few hosts) 3. What is the subnet mask for this network? Dotted decimal? 255.255.254.0 Binary? 11111111.11111111.11111110.00000000 4. Slash format /23 5. How many usable subnetworks are there? 2^7 2 = 126 (formula: 2^s 2) 6. How many usable hosts are there per subnet? 2^9 2 = 510 (formula: 2^h 2) NNNNNNNN NNNNNNNN SSSSSSSH HHHHHHHH 255 255 254 0 172 16 0 0 172 16 0 0 256-254 = 2 [magic number] Question 3.6.2: Complete the following chart listing the first three subnets and the last 4 subnets. Subnetwork # Subnetwork ID Host Range Broadcast ID 0 172.16.0.0 ZERO SUBNET 1 172.16.2.0 172.16.2.1 172.16.3.255 172.16.3.254 2 172.16.4.0 172.16.4.1 172.16.5.254 172.16.5.255 3 172.16.6.0 172.16.6.1 172.16.7.254 123 172.16.246.0 172.16.246.1 172.16.247.254 124 172.16.248.0 172.16.248.1 172.16.249.254 125 172.16.250.0 172.16.250.1 172.16.251.254 126 17.16.252.0 172.16.252.1 172.16.253.254 172.16.7.255 172.16.247.255 172.16.249.255 172.16.251.255 172.16.253.255 127 17.16.254.0 172.16.254.1 172.16.255.254 128 172.16.256.0 INVALID 172.16.255.255 (network broadcast)
1. What is the host range for subnet two? 172.16.4.1 172.16.5.254 2. What is the broadcast address for the 126th subnet? 172.16.253.255 3. What is the broadcast address for the major network? 17.16.255.255 Lab 3.7 Subnetting a Class C Network Background / Preparation This is a written exercise and is to be performed without the aid of an electronic calculator. 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: Question 3.7.1: Classroom #1 28 nodes Classroom #2 22 nodes Computer lab 30 nodes Instructors 12 nodes Administration 8 nodes Given this Class C network address and these requirements answer the following questions 1. How many subnets are needed for this network? 5 2. What is the subnet mask for this network? Dotted decimal? 255.255.255.224 Binary? 11111111.11111111.11111111.11100000 3. Slash format /27 4. How many usable hosts are there per subnet? 30 (2^5 2) NNNNNNNN NNNNNNNN NNNNNNNN SSSHHHHH 11111111 11111111 11111111 11100000 255 255 255 224 192 168 1 0 N = 24 S = 3 (2^3 = 8 subnets, 5 needed) H = 5 (2^5 2 = 30, 30 hosts per subnet needed) 256 224 = 32 [magic number]
Question 3.7.2: Complete the following chart Subnetwork # Subnetwork IP Host Range Broadcast ID 0 192.168.1.0 192.168.1.1-192.168.1.30 1 192.168.1.32 192.168.1.33-192.168.1.62 2 192.168.1.64 192.168.1.65-192.168.1.94 3 192.168.1.96 192.168.1.97-192.168.1.126 4 192.168.1.128 192.168.1.129-192.168.1.158 5 192.168.1.160 192.168.1.161-192.168.1.190 6 192.168.1.192 192.168.1.193-192.168.1.222 7 192.168.1.224 192.168.1.225-192.168.1.254 192.168.1.31 192.168.1.63 192.168.1.95 192.168.1.127 192.168.1.159 192.168.1.191 192.168.1.223 192.168.1.255 (network broadcast) 192.168.1.256 INVALID 1. What is the host range for subnet six? 192.168.1.193-192.168.1.222 2. What is the broadcast address for the 3rd subnet? 192.168.1.127 3. What is the broadcast address for the major network? 192.168.1.255