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Subnetting Made EZ

Subnetting Made EZ. By Brian D'Andrea. 32768. 16384. 8192. 4096. 2048. 1024. 512. 256. Binary. 128. 64. 32. 16. 8. 4. 2. 1. 128. 64. 32. 16. 8. 4. 2. 1. Sub Mask. 192. 224. 240. 248. 252. 254. 255. 128. 192. 224. 240. 248. 252. 254. 255. 128. /25. /26.

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Subnetting Made EZ

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  1. Subnetting Made EZ By Brian D'Andrea

  2. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - Subnetting is a major part of your CCNA exam. The chart above will speed the time taken when answering subnetting questions you will encounter on your exam. The following presentation will show you: • How to create the above chart. • How to use the chart when answer subnetting questions.

  3. Binary Brian Sub Mask Says CIDR Class Host Has B Networks Been C Networks Canceled Your subnetting chart made EZ foundation is remembering the row titles. The acronym I use to remember the row titles is: “Brian Says Class Has Been Canceled”. Feel free however, to use whatever you like. The title rows from top to bottom are as follows: “Binary”, “Sub Mask”, “CIDR”, “Host”, “B Networks”, & “C Networks”.

  4. Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask CIDR Host B Networks C Networks Next complete the row titled “Binary”. The “Binary” row is going to use the same chart we used to convert decimal numbers to binary and vice versa. With our subnetting chart we focus on the 3rd and 4th octets of a TCP/IP address so we write this chart out twice.

  5. Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 240 248 252 Sub Mask 128 240 248 252 254 254 255 255 128 224 192 192 224 CIDR Host The row “Sub Mask” was abbreviated for subnet mask. To determine the subnet mask row, start with the number “128” it is easy to remember because it falls in line with “128” from the “Binary” row in the 3rd octet just above it. B Networks C Networks Then take the first column in “Sub Mask” row, which has a value of “128”, and add it to next column in the “Binary” row. In this case, the number is “64”. Adding these two together will determine the value of the “Sub Mask” row in the second column. 128 + 64 = 192. Repeat this until you reach the end of the 3rd octet numeric value 255. You should be adding values: 192 + 32 = 224, 224 + 16 = 240, 240 + 8 = 248, 248 + 4 = 252, 252 + 2 = 254, & 254 + 1 = 255. Then the line you just created gets copied from the 3rd octet into the 4th octet.

  6. Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 240 248 252 254 255 224 128 192 192 224 240 248 252 254 255 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 Host B Networks C Networks Moving on to the row titled “CIDR”. It is always important to remember our TCP/IP addresses are 32 bits long, and with our chart we concentrate on the last 16 bits. /1 /2 /3 /4 /5 /6 /7 /8 /9 /10 /11 /12 /13 /14 /15 /16 /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 “CIDR” row is simple count from 1 through 32. We drop 1 through 16 because these numbers are found in the first 2 octets. Then take 17 through 32 counting by 1 left to right.

  7. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 240 248 252 254 255 224 128 192 192 224 240 248 252 254 255 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks C Networks “Host” row is short for Host per Network. To start we need to add a row above the binary line in the 3rd octet. To create this, just double the previous number. Now, we have to take the numbers we just created in the 3rd octet, the numbers in the “Binary” row in the 4th octet, and subtracting 2 from each to calculate the number of host per network.

  8. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 240 248 252 254 255 224 128 192 192 224 240 248 252 254 255 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 Host 32766 16382 8190 4094 2046 1022 510 254 126 62 30 14 6 2 0 0 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 C Networks “B Networks” row is to answer questions when subnetting a Class B network address to determine how many networks are created. To create the “B Networks” row, take the numbers right to left from the “Host” row after dropping the first zero. Copy these numbers into the “B Networks” left to right. Check yourself 254 in Host is above 254 in B Networks You will not need to calculate the last 2 positions in this row because neither are capable of holding any host as you can see in the row above.

  9. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 240 248 252 254 255 224 128 192 192 224 240 248 252 254 255 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 Host 32766 16382 8190 4094 2046 1022 510 254 126 62 30 14 6 2 0 0 510 1022 2046 4094 8190 16382 - - B Networks 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - “C Networks” row is to answer questions when subnetting a Class C network address to determine how many networks are created. To create the “C Networks” row, take the “B Networks” section in the 3rd octet and copy it to the 4th octet in “C Networks”. You will not need to calculate the last 2 positions in this row because neither are capable of holding any host. In the “C Networks” row under the third octet will be blank because a Class C address will begin to be subnetted in the 4th octet.

  10. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 240 248 252 254 255 224 128 192 192 224 240 248 252 254 255 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 Host 32766 16382 8190 4094 2046 1022 510 254 126 62 30 14 6 2 0 0 510 1022 2046 4094 8190 16382 - - B Networks 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - One last topic to understand with this chart is what to do when IP Subnet Zero is ENABLED. When IP Subnet Zero is ENABLED be sure to add 2 to the number of networks for both rows “B Networks” & “C Networks” IP Subnet Zero is ENABLED B Networks + 2 C Networks + 2

  11. 256 0 16 32 48 64 80 96 112 128 144 160 176 192 208 224 240 Net ID 15 31 47 63 79 95 111 127 143 159 175 191 207 223 239 255 Broadcast Another chart I highly recommend creating is “Count by 16”. This will prove very helpful when determining Valid Hosts, Network IDs, and Broadcast Addresses. Created by counting by 16 starting with 0, from left to right. Then by taking the next Network ID subtract 1 and this will give you the Broadcast address for the previous network. Now any IP address between the Network ID and Broadcast ID is a Valid Host.

  12. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - 1. How many host will I have per network using a /30 address? IP subnet zero has been configured. Answer: 2 Remember a /30 mask allows for 2 hosts per network, commonly used for serial connections. (Point to Point)

  13. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 + 2 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - How many networks are created using a 192.168.0.0 with a subnet mask 255.255.255.224? IP Subnet Zero is applied. Answer: 8

  14. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 You need to subnet 160.200.3.0 to support 30 site locations and as many hosts supported per subnet as possible. What CIDR would you recommend to use? (IP subnet zero is not applied.) Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - Answer: /21

  15. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - You have been assigned the network of 191.5.5.0. How many hosts do you have using a mask of 255.255.240.0? IP subnet zero is applied. Answer: 4094

  16. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - Your company has a class C network address. The company requires 5 usable subnets. Each subnet mask must accommodate at least 18 hosts. Which subnet mask should you use? (IP subnet zero is not applied) Answer: 255.255.255.224

  17. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - Which of the following addresses is representative of a unicast address? A. 224.1.5.2 B. 255.255.255.255 C. FFFF.FFFF.FFFF D. 192.168.24.59/30 E. 172.31.128.255/18 We will need to count by 64s in the 3rd octet. Don’t forget the fourth octet, a 0 in Network ID, and a 255 as a Broadcast Address. Example 64.0 minus 1 from both the third and forth octets to calculate the Broadcast is 63.255!!!!! 224.0.0.0 through 239.255.255.255 are Class D Address, designated for multicast. We will need to count by 4s. Shortcut use a multiple of 4 to get you close to target number! This is a MAC Address that is a Broadcast as well. 255.255.255.255 is TCP/IP Broadcast Address. 4 X 10 = 40 44 48 52 56 60 Net ID 0.0 64.0 128.0 192.0 256.0 Net ID Broadcast 63.255 127.255 191.255 43 47 255.255 51 55 59 Broadcast

  18. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - Consider the address 192.168.15.19/28, which of the following addresses are valid host addresses on this subnet? (Select two options.) A. 192.168.15.17 B. 192.168.15.0 C. 192.168.15.29 F. 192.168.15.35 D. 192.168.15.16 E. 192.168.15.31 We will need to count by 16 to determine Network IDs & Broadcast IDs, to help us determine valid hosts Next we will have to determine the network ID and the broadcast address for 192.168.15.19/28. 256 0 16 32 48 64 80 96 112 128 144 160 176 192 208 224 240 Net ID 15 31 47 63 79 95 111 127 143 159 175 191 207 223 239 255 Broadcast

  19. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 What is the subnet work number of the IP address 159.56.219.56/21? Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - 159.56.219.56/21 A. 159.56.0.0 B. 159.56.219.0 C. 159.56.8.0 F. 159.56.199.0 D. 159.56.52.0 E. 159.56.216.0 We will need to count by 8s in the 3rd octet. Don’t forget the fourth octet, a 0 in Network ID, and a 255 as a Broadcast Address. Example 224.0 minus 1 from the 3rd & 4th octets to calculate the Broadcast is 223.255!!! 8 X 25 = 200.0 208.0 216.0 224.0 Net ID Broadcast 207.255 215.255 223.255

  20. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Which of the following are valid host addresses given the IP address 135.205.0.0/19? IP subnet zero is not active. (Choose 3) Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - IP subnet zero is not active. A. 135.205.20.100 B. 135.205.40.0 C. 135.205.95.255 F. 135.205.225.150 D. 135.205.96.255 E. 135.205.80.100 We will need to count by 32s in the 3rd octet. Don’t forget the fourth octet, a 0 in Network ID, and a 255 as a Broadcast Address. Example 224.0 minus 1 from the 3rd & 4th octets to calculate the Broadcast is 223.255!!! 160.0 192.0 224.0 256.0 0.0 32.0 64.0 96.0 128.0 Net ID 159.255 191.255 223.255 255.255 Broadcast 63.255 95.255 127.255 31.255

  21. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 + 2 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 Given the following list of available IP addresses you need to find which are valid on the 4 network given the network 172.30.0.0. This address space has already been divided by Brian into 16 equal subnets, because he said it would make it harder to figure out. Choose 3 IP addresses that meet the defined requirements. IP subnet has been applied. Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - IP subnet has been applied. A. 159.56.50.255 B. 159.56.219.45 C. 159.56.8.0 F. 159.56.63.1 D. 159.56.52.0 E. 159.56.216.50 We will need to count by 16s in the 3rd octet. Don’t forget the fourth octet, a 0 in Network ID, and a 255 as a Broadcast Address. Example 16.0 minus 1 from the 3rd & 4th octets to calculate the Broadcast is 15.255!!! 0.0 16.0 32.0 48.0 64.0 Net ID Broadcast 15.255 31.255 47.255 63.255

  22. 32768 16384 8192 4096 2048 1024 512 256 Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 128 /25 /26 /27 /28 /29 /30 /31 /32 CIDR /17 /18 /19 /20 /21 /22 /23 /24 You have been assigned the task of configuring a serial interface of a router. You were given the IP addresses listed below, use the eighth IP address on the 12 network. 192.168.5.0/24 is the network used by your organization. They have chosen to subnet that network allowing 14 hosts per subnet. IP subnet zero is not active. Host 126 62 30 14 6 2 0 0 32766 16382 8190 4094 2046 1022 510 254 B Networks 510 1022 2046 4094 8190 16382 - - 0 2 6 14 30 62 126 254 0 2 6 14 30 62 - - C Networks - - - - - - - - IP subnet zero is not active. A. 192.168.5.184 B. 192.168.5.232 C. 192.168.5.201 F. 192.168.5.162 D. 192.168.5.200 E. 192.168.5.8 We will need to count by 16s in fourth octet. Using the value in the binary row the increment. 192 + 8 = 200 0 16 32 48 64 80 96 112 128 144 160 176 192 224 Net ID 15 31 47 63 79 95 111 127 143 159 175 191 223 Broadcast

  23. Congratulations • Practice • Watch the octet subnetting occurs • Remember if trying to find Network ID, Broadcast Address, or Valid Host you need to find your increment (the number in binary row)

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