Thanks to Randy of our CCNA program for this suggestion. Randy wanted some guidance on how to solve the subnetting questions in ICND1 and ICND2 very quickly. The ability to do this is often the difference between a passing score and a failed attempt.
WARNING: You must master subnetting using our course or some other trusted materials before you start using these shortcut approaches. It is a common issue for Cisco candidates to move directly to subnetting shortcuts for the exams without fully understanding exactly how subnetting functions.
For this series of posts, we will use simulated exam questions from ICND1 and ICND2. Well, with all that out of the way - let's have some fun. You will find that once you "turn the corner" on subnetting, you will pray for many of these questions in the exam. It is an opportunity to solve questions quickly and be 100% convinced that your response is "spot on".
Question 1: What is the last usable address in the subnet of a host with the address 192.168.1.134 and the subnet mask of 255.255.255.240?
Step 1: Since the is the first subnetting question I have encountered in my exam, I am going to use this as my opportunity to build my Powers of Two reference chart on my scratch paper.
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: 192 in the first octet tells me I have a Class C address. I memorized these number ranges; some students like to list those on the scratch paper as well. The default subnet mask for the Class C space is 255.255.255.0. You also need to memorize these, and again, many students like to list these on scratch paper as well.
Step 3: How many bits of subnetting are used in the fourth octet here? My Power of Two chart tells me. 1 bit = 128; 2 bits = 192; 3 bits = 224; 4 bits = 240.
Step 4: I go four bits deep (from left to right) in the Power of Two chart. This tells me that the subnets increment on 16:
Step 5: We can see that this host lives 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 is 142. Our answer - 192.168.1.142. Wooohooo! Bring on more of these!
Did you have an even easier way to arrive at the answer? Let us know in the comments.