Is supernetting allowed in Cisco Packet Tracer

Documentation date: IP addressing. Date: information. Version: 1.0. Minutes from page 1

IP addressing Author: Schmid Tobias Type: Information Version: 1.0 Protocol from 04/05/2016 page 1

Contents CONTENTS 1 First steps ... 3 2 Exercises for IP addressing ... 5 3 Subnetting theory ... 7 3.1 About these instructions ... 7 3.2 Subnetting A task with the solution .... 7 3.3 Subnetting calculation .. . 8 4 Subnetting calculation ... 12 5 Solutions ... 16 Protocol from 04/05/2016 page 2

1 First steps IP address: 193.193.144.12 with subnet mask 255.255.255.0 Alternative spelling of the subnet mask / ... Class ... Number of possible hosts in the subnet ... Number of possible subnets ... Network ID ... Broadcast .. . Host ID ... IP address: 88.153.166.182 with subnet mask 255.224.0.0 Alternative spelling of the subnet mask / ... Class ... Number of possible hosts in the subnet ... Number of possible subnets ... Network ID .. . Broadcast ... host ID ... IP address: 212.245.88.186 with subnet mask 255.255.255.248 Alternative spelling of the subnet mask / ... class ... number of possible hosts in the subnet ... number of possible subnets ... network -ID ... Broadcast ... Host-ID ... Protocol from 04/05/2016 page 3

IP address: 154.71.234.82 with subnet mask 255.255.252.0 Alternative spelling of the subnet mask / ... class ... number of possible hosts in the subnet ... number of possible subnets ... network ID ... broadcast ... host ID ... IP address: 77.88.99.234 with subnet mask 255.255.240.0 Alternative spelling of the subnet mask / ... class ... number of possible hosts in the subnet ... number of possible subnets ... network ID ... broadcast. .. Host ID ... IP address: 195.149.87.178 with subnet mask 255.255.255.252 Alternative spelling of the subnet mask / ... Class ... Number of possible hosts in the subnet ... Number of possible subnets ... Network ID. .. Broadcast ... Host-ID ... Log from 04/05/2016 page 4

2 exercises for IP addressing 1. Calculate the maximum number of network devices in the respective subnet if they have the following network mask or prefix: a) 255.255.255.0 g) 255.255.255.192 m) / 23 b) 255.255.0.0 h) 255.255. 224.0 n) / 20 c) 255.255.255.128 i) 255.252.0.0 o) / 12 d) 255.255.255.240 j) 255.255.255.252 p) / 30 e) 255.255.254.0 k) / 16 q) / 25 f) 255.255. 128.0 l) / 29 r) / 8 2. Calculate the minimum network mask (in decimal and prefix notation) for the networks with the specified number of network devices: a) 12 e) 4 i) 2 b) 36 f) 4'080 j ) 2'047 c) 170 g) 14 k) 224 d) 110 h) 32'430 l) 1 3. Calculate the associated network address, the broadcast address and the maximum number of network devices in the same network for the following network address specifications: a) 192.168 .1.30 255.255.255.0 h) 192.168.13.130/26 b) 160.85.160.255 255.255.224.0 i) 62.2.244.109/30 c) 176.255.13.22 255.255.255.224 j) 10.0.33.33/27 d) 179.248.222.18 255.255.255.252 k) 181.15.33.233/20 e) 10.10.0.255 255.248.0 .0 l) 10.55.0.4/12 f) 209.1.3.254 255.255.254.0 m) 33.33.32.0/19 g) 223.19.127.255 255.255.128.0 n) 62.3.256.13/23 Protocol from 04/05/2016 page 5

4. Which values ​​(decimal, binary) are permitted as the third byte for the network mask if the first two bytes are 255 and the last is 0? How many addresses can be freely assigned in the respective networks? 5. Which of the following network masks are permitted and how many addresses can be freely assigned in the networks? a) 255.255.0.0 b) 255.255.254.0 c) 255.255.240.0 d) 255.255.255.224 e) 254.255.255.0 f) 255.255.255.128 g) 255.255.255.250 h) 255.255.255.192 6. Which network mask do you need for the following IP Enter addresses? a) 160.215.39.14/20 b) 10.25.139.45/8 c) 200.5.90.124/24 d) 160.215.39.14/22 e) 60.15.9.4/29 7. You want to connect exactly two computers to a network in such a way that no third computer can be added. Which netmask do you choose? What is the IP address of one computer if the other has the IP 192.168.3.17? 8. A computer has the IP 192.168.242.255 and the network mask 255.255.252.0. What IP does the broadcast have and the network address? Minutes from April 5th, 2016, page 6

3 Subnetting Theory Subnetting is the division of a network into several smaller subnets. When calculating subnetting, the network share is expanded. This means that part of the host part is allocated to the network part, so to speak. 3.1 About this manual. There are several methods of calculating subnets with subnetting. These instructions are probably about the simplest form of calculation. So the target group are beginners. The aim of this guide is to help readers understand how a network is broken down into smaller networks. You cannot solve complicated subnetting tasks with these instructions. 3.2 Subnetting One task with the solution. Scenario: The employees of a department get the network 192.168.168.0/24 for their PCs, i.e. 256 IP addresses are available for the department: 1 IP address for the network address: 192.168.168.0 (first IP) 1 IP address for broadcast: 192.168.168.255 (last IP) 254 IP addresses for PCs, network printers, etc .: 192.168.168.1-192.168.168.254 Let's assume that the department is divided into 4 smaller departments. Each department should have its own network. Given 1 network: IP: 192.168.168.0 Net mask: 255.255.255.0 (Or written differently: 192.168.168.0/24) Task Divide the existing network into 4 subnets. I.e. each department should have its own subnet. Minutes from April 5th, 2016, page 7

3.3 Subnetting calculation protocol from 04/05/2016 page 8

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4 Subnetting calculation Step 1: So that a network can be divided into smaller subnets, the network portion should be expanded by a certain number of bits in the network mask. In the first step, this number of bits is calculated. The number of bits required depends on the number of subnets required. Number of bits: 1 2 3 4 5 6 7 8 Number of subnets: 2 4 8 16 32 64 128 256 With 1 bit, 2 ^ 1 = 2 subnets can be set up. But 4 subnets are necessary. With 2 bits we can build 2 ^ 2 = 4 subnets. The power supply unit must therefore be expanded by 2 bits. Intermediate step: Convert the host part into a binary number. Here, only those octets are converted into the binary number that are not equal to 255 in the network mask. (In this example: 255.255.255.0) IP address: 192.168.168.0 Net mask: 255.255.255.0 => 192.168.168.00000000 255.255.255.00000000 Step 2: Expand the network part by 2 bits. IP address: 192.168.168.00000000 Netmask: 255.255.255.00000000 => 192.168.168.00000000 255.255.255.11000000 The network portion was expanded by 2 bits in the netmask. (From left to right) Minutes from April 5th, 2016, page 12

This shifts the boundary between the host part and the network part to the right. (Ie the network share has now increased.) Step 3 If the network share is now expanded by 2 bits, we automatically have the subnet address of the 1st subnet: IP address: 192.168.168.00000000 Net mask: 255.255.255.11000000 The host share consists of 6 bits ( highlighted in red). This means that 2 ^ 6 = 64 IP addresses are available for each subnet: 1 for subnet 62 for hosts (host IP range) 1 for broadcast 1 IP 62 IPs 1 IP subnet address host IP range broadcast Step 4: The last octet of convert the subnet address back to a decimal number. IP address: 192.168.168.00000000 Netmask: 255.255.255.11000000 => IP address: 192.168.168.0 Netmask: 255.255.255.192 Aid decimal number 128 192 224 240 248 252 254 255 Binary number 10000000 11000000 11100000 11110000 11111000 11111100 from the 11111110 111111 Broadcast 1. Calculate the subnet. All host bits are set to 1 in the broadcast. 192.168.168.00000000 => 192.168.168.00111111 Step 6: Convert the last octet of broadcast to a decimal number: 192.168.168.00111111 => 192.168.168.63 Protocol from 04/05/2016 page 13

Step 7: Enter the subnet address and broadcast from the 1st subnet in the table: 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.63 The host IP range is the IP address range between subnet addresses and broadcast: 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.1-192.168.168.62 192.168.168.63 Step 8: If you increase the broadcast IP by 1, you get the subnet address from the next subnet (i.e. from Subnet 2): 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.1-192.168.168.62 192.168.168.63 192.168.168.64 Step 9: The IP address of the subnet address by 63 (62 host range IPs +1 Broadcast IP) increase = Broadcast: 192.168.168.64 + 63 = 127 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.1-192.168.168.62 192.168.168.63 192.168.168.64 192.168.168.127 Protocol from 04/05/2016 page 14

Step 10: Enter the host IP range in the table as in step 7: 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.1-192.168.168.62 192.168.168.63 192.168.168.64 192.168.168.65 -192.168.168.126 192.168.168.127 Step 11: Repeat steps 8 to 10 for subnets 3 and 4 = enter the subnet address, host IP range and broadcast in the table. 1 IP 62 IPs 1 IP subnet address host IP range broadcast 192.168.168.0 192.168.168.1-192.168.168.62 192.168.168.63 192.168.168.64 192.168.168.65-192.168.168.126 192.168.168.127 192.168.168.128 192.168.168.129-192.168.168.190 192.168.168 .168.191 192.168.168.192 192.168.168.193-192.168.168.254 192.168.168.255 Answer: The first department gets the subnet 1: IP: 192.168.168.0 Net mask: 255.255.255.192 The second department - the subnet 2: IP: 192.168.168.64 Net mask: 255.255.255.192 The third division - the subnet 3: IP: 192.168.168.128 Netmask: 255.255.255.192 The fourth division - the subnet 4: IP: 192.168.168.192 Netmask: 255.255.255.192 Log from 04/05/2016 page 15

5 solutions exercises on IP addressing (pages 5 and 6) 1. a) 254 g) 62 m) 510 b) 65,534 h) 8,190 n) 4,094 c) 126 i) 262,142 o) 1 ' 048'574 d) 14 j) 2 p) 2 e) 510 k) 65'534 q) 126 f) 32'766 l) 6 r) 16'777'214 2. a) 255.255.255.240 / 28 e) 255.255 .255.248 / 29 i) 255.255.255.252 / 30 b) 255.255.255.192 / 26 f) 255.255.240.0 / 20 k) 255.255.240.0 / 20 c) 255.255.255.0 / 24 g) 255.255.255.224 / 28 l) 255.255. 255.0 / 24 d) 255.255.255.128 / 25 h) 255.255.128.0 / 17 m) 255.255.255.252 / 30 3. a) N: 192.168.1.0 B: 192.168.1.255 S: 254 b) N: 160.85.160.0 B: 160.85.191.255 S: 8'190 c) N: 176.255.13.0 B: 176.255.13.31 S: 30 d) N: 179.248.222.16 B: 179.248.222.19 S: 2 e) N: 10.8.0.0 B: 10.15.255.255 S: 524'286 f) N: 209.1.2.0 B: 209.1.3.255 S: 510 g) N: 223.19.0.0 B: 223.19.127.255 S: 32'766 h) N: 192.168.13.128 B: 192.168.13.191 S : 62 i) N: 62.2.244.108 B: 62.2.244.111 S: 2 j) N: 10.0.33.32 B: 10.0.33.63 S: 30 k) N: 181.15.32.0 B: 181.15.47.255 S: 4,094 l ) N: 10.48.0.0 B: 10.63.255.255 S: 1'048'574 m) N: 33.33 .32.0 B: 33.33.63.255 S: 8'190 n) no valid IP address Protocol from 04/05/2016 page 16

4. Which values ​​(decimal, binary) are permitted as the third byte for the network mask if the first two bytes are 255 and the last is 0? How many addresses can be freely assigned in the respective networks? 0 00000000 2 high 16-2 128 10000000 2 high 15-2 192 11000000 2 high 14-2 224 11100000 2 high 13-2 240 11110000 2 high 12-2 248 11111000 2 high 11-2 252 11111100 2 high 10-2 254 11111110 2 to the power of 9-2 255 11111111 2 to the power of 8-2 5. Which of the following network masks are permitted and how many addresses can be freely assigned in the networks? a) 255.255.0.0 2 high 16-2 b) 255.255.254.0 2 high 9-2 c) 255.255.240.0 2 high 12-2 d) 255.255.255.224 2 high 5-2 e) 254.255.255.0 f) 255.255.255.128 2 high 7-2 g) 255.255.255.250 h) 255.255.255.192 2 high 6-2 6. Which network mask do you have to enter for the following IP addresses? a) 160.215.39.14/20 255.255.240.0 b) 10.25.139.45/8 255.0.0.0 c) 200.5.90.124/24 255.255.255.0 d) 160.215.39.14/22 255.255.252.0 e) 60.15.9.4/29 255.255.255.248 7. You want to connect exactly two computers to a network in such a way that no third computer can be added. Which netmask do you choose? What is the IP address of one computer if the other has the IP 192.168.3.17? Net mask 255.255.255.252 or / 30 IP 2nd PC 192.168.3.18 Net-ID 192.168.3.16 Network: 192.168.3.16/30 11000000.10101000.00000011.000100 00 HostMin: 192.168.3.17 11000000.10101000.00000011.000100 01 HostMax: 192.168.3.18 11000000.10101000.0000 Broadcast 10101000.0000 : 192.168.3.19 11000000.10101000.00000011.000100 11 hosts / network: 2 8. A computer has the IP 192.168.242.255 and the network mask 255.255.252.0. What IP does the broadcast have and the network address? Network: 192.168.240.0/22 ​​11000000.10101000.111100 00.00000000 HostMin: 192.168.240.1 11000000.10101000.111100 00.00000001 HostMax: 192.168.243.254 11000000.10101000.111100 11.11111110 Broadcast: 192.11168.243.255 11000000.10101000.111100 Protocol page: 192.168.243.255 11000000.10101000