Subnetting Made EZ

The following presentation will show you: 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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.

Brian Binary Sub Mask Says CIDR Class Host Has Been B Networks Canceled C Networks 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”.

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.

Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 128 192 192 224 224 240 240 248 248 252 252 254 254 255 255 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.

Binary 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 Sub Mask 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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.

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 - 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 32766 16382 8190 4094 2046 1022 510 254 126 62 30 14 6 2 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.

Check yourself 254 in Host is above 254 in B Networks 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - 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.

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - “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.

IP Subnet Zero is ENABLED 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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

Created by counting by 16 starting with 0, from left to right. Net ID 16 32 48 64 80 96 112 128 144 160 176 192 208 224 240 256 Broadcast 15 31 47 63 79 95 111 127 143 159 175 191 207 223 239 255 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.

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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)

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 CIDR /17 /18 /19 /20 /21 /22 /23 /24 /25 /26 /27 /28 /29 /30 /31 /32 + 2 Host 32766 16382 8190 4094 2046 1022 510 254 126 62 30 14 6 2 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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.) Answer: /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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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!!!!! We will need to count by 4s. Shortcut use a multiple of 4 to get you close to target number! 255.255.255.255 is TCP/IP Broadcast Address. 224.0.0.0 through 239.255.255.255 are Class D Address, designated for multicast. This is a MAC Address that is a Broadcast as well. 4 X 10 = Net ID 40 44 48 52 56 60 Net ID 0.0 64.0 128.0 192.0 256.0 Broadcast 63.255 127.255 Broadcast 191.255 43 47 255.255 51 55 59

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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 Next we will have to determine the network ID and the broadcast address for 192.168.15.19/28. We will need to count by 16 to determine Network IDs & Broadcast IDs, to help us determine valid hosts Net ID 16 32 48 64 80 96 112 128 144 160 176 192 208 224 240 256 Broadcast 15 31 47 63 79 95 111 127 143 159 175 191 207 223 239 255

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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - What is the subnet work number of the IP address 159.56.219.56/21? 159.56.219.56/21 A. 159.56.0.0 B. 159.56.219.0 C. 159.56.8.0 D. 159.56.52.0 E. 159.56.216.0 F. 159.56.199.0 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 = Net ID 200.0 208.0 216.0 224.0 Broadcast 207.255 215.255 223.255

IP subnet zero is not active. 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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) IP subnet zero is not active. A. 135.205.20.100 B. 135.205.40.0 C. 135.205.95.255 D. 135.205.96.255 E. 135.205.80.100 F. 135.205.225.150 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!!! Net ID 0.0 32.0 64.0 96.0 128.0 160.0 192.0 224.0 256.0 Broadcast 31.255 63.255 95.255 127.255 159.255 191.255 223.255 255.255

IP subnet has been applied. 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 + 2 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - Given the following list of available IP addresses you need to find which are valid on the 4 network given the network 159.56.0.0. 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. IP subnet has been applied. A. 159.56.50.255 B. 159.56.219.45 C. 159.56.8.0 D. 159.56.52.0 E. 159.56.216.50 F. 159.56.63.1 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!!! Net ID 0.0 16.0 32.0 48.0 64.0 Broadcast 15.255 31.255 47.255 63.255

IP subnet zero is not active. 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 128 192 224 240 248 252 254 255 128 192 224 240 248 252 254 255 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 B Networks 2 6 14 30 62 126 254 510 1022 2046 4094 8190 16382 - - C Networks - - - - - - - - 2 6 14 30 62 - - 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. IP subnet zero is not active. A. 192.168.5.184 B. 192.168.5.232 C. 192.168.5.201 D. 192.168.5.200 E. 192.168.5.8 F. 192.168.5.162 We will need to count by 16s in fourth octet. Using the value in the binary row the increment. 192 + 8 = 200 Net ID 16 32 48 64 80 96 112 128 144 160 176 192 224 Broadcast 15 31 47 63 79 95 111 127 143 159 175 191 223

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)

IP subnet zero is not active. 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!!! I IP subnet zero is not active. Net ID 0.0 16.0 32.0 48.0 64.0 160.0 192.0 224.0 256.0 Broadcast 15.255 31.255 47.255 63.255 159.255 191.255 223.255 255.255 We will need to count by 16s in fourth octet. Using the value in the binary row the increment. Net ID 40 44 48 52 56 60 Broadcast 43 47 51 55 59