Project 10-3: Research Subnetting Calculation Shortcuts Instructors, network tec

Question

Project 10-3: Research Subnetting Calculation Shortcuts Instructors, network technicians, and IT students have developed a plethora of Enter HANDS-ON PROJECTS to find the method that clicks for you. In this project, you will research some of the options these various methods provide, and create your own "cheat sheet" to help pre pare you for subnetting on the CompTIA Network+ exam. 1. Start with a search engine and, if you know you prefer videos instead of written instructions, limit your search to videos or images, or conduct your search on youtube.com. Do a search for topics such as "subnetting," "subnetting made easy," "subnetting tutorial," or "subnetting for dummies."

Explanation / Answer

A Sample Subnetting Scenario

Let us stick with the example where we begin with the private IP address space of a Class B address of 172.16.0.0. We examine our networking needs and realize that we need to accommodate 100 different subnetworks in our organization. From the previous information in this article, we know that in order to create 100 subnets, we need to borrow 7 bits (2^7 = 128). This is perfect; we have the number of subnetworks that we need, plus a few extra that we can call upon when the network inevitably grows.

What will the subnet mask be in this scenario? This mask will be the one that is used by all of the hosts in all of the different subnetworks. It is critical that we calculate this number correctly, of course.

Notice that our Class B address originally had 16 bits that made up the network ID portion. In this sample scenario we are going to borrow 7 bits. Now we have a network ID that is made up of 23 (16+7). We can write out the 23 bits of the subnet mask now:

11111111.11111111.11111110.00000000

So converting to our convenient dotted decimal notation, we can see our mask:

255.255.254.0

A cooler way to write the mask is just /23 after the IP address. This is called prefix notation.

So what would the first subnetwork network ID be? Well, we know it will start 172.16, but what will the value be in that third octet where we have some bits (7) representing the subnet and one bit representing the host portion.

To answer this in the longhand method, we write out the mask and the address from that octet in binary and do some analysis.

Mask: 11111110

IP: 00000000

Notice the first subnetwork will be 172.16.0.0. We can use all zeros in the first 7 bit positions of the third octet, and we have a zero in the last bit position which is used to identify hosts.

What would the first host address be in this network? Let’s write those last two octets out longhand: 00000000.00000001

The first host address on the 172.16.0.0 network would be:

172.16.0.1

What would the broadcast address be for that network? To get this you fill all the host bits with a 1: 00000001.111111111

172.16.1.255

How about the last usable host address on this subnetwork? Easy. We will turn all the host bits to 1, except for the last one.

00000001.11111110

172.16.1.254

What is the next subnetwork in this scheme? Well, let us turn one of those subnetwork bits on. We will start with the least significant (rightmost):

Mask: 11111110

IP: 00000010

Ahh, so the next network is 172.16.2.0/23.

Shortcuts

Sample Question 1:

What is the last usable address in the subnet of a host with the address 192.168.1.136 and a subnet mask of 255.255.255.240?

Step 1 I build a Powers of Two reference chart on the scratch paper Cisco provides.

2^7=128 | 2^6=64 | 2^5=32 | 2^4=16 | 2^3=8 | 2^2-=4 | 2 ^1=2 | 2^0=1

Step 2

How many bits of subnetting are used in the fourth octet here? My Powers of Two chart tells me. 1 bit = 128; 2 bits = 192; 3 bits = 224; 4 bits = 240. Yes, the forth octet of the subnet mask looks like this in binary—11110000.

Step 3

Now the magic of the shortcut—we go four bits deep (from left to right) in the Powers of Two chart. This tells us the value that the subnets increment on. In our example it is 16:

So our subnets are:

Step 4

This host, with the address of 192.168.1.136 must live on the 192.168.1.128 subnet. The broadcast address for this subnet is one less than the next subnet of 144, so that is 143. The last usable address is 142. We have arrived at our answer—192.168.1.142.

Sample Question 2

Your IT Junior Administrator has provided you with the address and mask of 192.168.20.102 and 255.255.255.224. You Junior Admin has asked you to tell him how many hosts can be created on your subnet?

Step 1 Here I begin by referencing the Powers of Two chart I created on my scratch paper. Adding 128 + 64 + 32, I get the 224 value used in the fourth octet of the subnet mask. Therefore, I can see that there are 3 bits used for subnetting in that octet. This leaves 5 bits for host addressing.

2^7=128 | 2^6=64 | 2^5=32 | 2^4=16 | 2^3=8 | 2^2-=4 | 2 ^1=2 | 2^0=1

Step 2

As we discussed earlier, the equation for the number of hosts per subnet is 2^h – 2 where h is the number of host bits. From the chart I see that 2^5 = 32. 32-2 = 30 hosts per subnet.

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