Does anyone know a good shortcut for problems like these

dazerski

I really can't express how far I have come with subnetting in the last week. I went from not knowing how to really even do it, to doing most of the problems in my head within 25 seconds (still not fast enough).

But then I get to problems like this from subnettingquestions.com and it takes me way too long...

Question: How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

Answer: 4096 subnets and 14 hosts


I guess one solution would to be bone up on addition (that's embarrassing to even say) but you people have given me some very handy shortcuts in my last couple threads. I was wondering if anyone had one off the top of your head for these kinds.

Plus I just wanted to post that I officially scheduled my exam date for Sept.30th... I'm stoked... I feel confident in 85% of the stuff in Odom's and CBT as well as doing hands-on stuff on my cisco lab I built... If I keep my original 9/30 test-date, I'll have plenty of time to get everything down. I knew I'd never take the test unless I slapped down $125 and set a date. I'm totally excited right now

Thanks again for all your help!

Dale

iworms

This question shouldn't take long. /28 means 32-28=4 bits for hosts, so (2^4) - 2 = 16 - 2 = 14.

172.20 is the class B network, so you have 8+4=12 bits for subnets. 2^12 = 4096.

It shouldn't take you long to arrive at 2^4 and 2^12 (worry the algebra later). If it does then you need to do more problems to get used to it.

How do I get 4096 quickly? Well I'm used to seeing powers of 2, up to 4096 coincidentally: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096. The hard part is connecting which power of 2 is which number. To be honest I count with my fingers: say I want to know (2^6=?) or (64=2^?). So I hold down two fingers and start from 4. As I recite each number I hold down one more finger, and when I get to 64 I have six fingers down. So 2^6=64.

jovan88

when i was learning subnetting i thought i couldnt do it because im very very bad at maths, so i set out to find a way i could do it as easy as possible. I came up with a crazy little chart and technique that makes sense to nobody else but me lol

anyways the point im getting at is everyone has their own way of doing these questions you just have to find one that you personally can do fast and accurately, and keep practicing

dazerski

iworms wrote:

This question shouldn't take long. /28 means 32-28=4 bits for hosts, so (2^4) - 2 = 16 - 2 = 14.

172.20 is the class B network, so you have 8+4=12 bits for subnets. 2^12 = 4096.

It shouldn't take you long to arrive at 2^4 and 2^12 (worry the algebra later). If it does then you need to do more problems to get used to it.

How do I get 4096 quickly? Well I'm used to seeing powers of 2, up to 4096 coincidentally: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096. The hard part is connecting which power of 2 is which number. To be honest I count with my fingers: say I want to know (2^6=?) or (64=2^?). So I hold down two fingers and start from 4. As I recite each number I hold down one more finger, and when I get to 64 I have six fingers down. So 2^6=64.

I see! Very cool... I kept seeing these familiar answers over and over but I didn't understand how they were reaching them. THank you for your help!!! Much much appreciated.

Dale

nicklauscombs

for really quick answers to problems like that on the exam I wrote out the subnetting chart
however its crucial to understand why its the answer

tech-airman

ddzierzek wrote:

I really can't express how far I have come with subnetting in the last week. I went from not knowing how to really even do it, to doing most of the problems in my head within 25 seconds (still not fast enough).

But then I get to problems like this from subnettingquestions.com and it takes me way too long...

Question: How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

Answer: 4096 subnets and 14 hosts


I guess one solution would to be bone up on addition (that's embarrassing to even say) but you people have given me some very handy shortcuts in my last couple threads. I was wondering if anyone had one off the top of your head for these kinds.

Plus I just wanted to post that I officially scheduled my exam date for Sept.30th... I'm stoked... I feel confident in 85% of the stuff in Odom's and CBT as well as doing hands-on stuff on my cisco lab I built... If I keep my original 9/30 test-date, I'll have plenty of time to get everything down. I knew I'd never take the test unless I slapped down $125 and set a date. I'm totally excited right now

Thanks again for all your help!

Dale

ddzierzek,

The problem with any shortcut without the foundational understanding leads you to the apparent confusion that you're experiencing. Trust me when I say that Cisco is very good at confusing you if you don't firmly understand the foundation knowledge. With that said, you have to understand the "long way" before you can appreciate the "shortcut" as a shorter way of doing the long way instead of just trying to short circuit the long way of learning.

With that said, here's the long way that I used to successfully pass the 640-801 CCNA exam. I call it my "five questions" method. The five questions are:

1. What is the class of the address?
2. What is the default mask?
3. What is the relevant octet?
4. What is the subnet address?
5. What is the broadcast address?

Let's answer the five questions for 172.20.0.0/28.

Q: What is the class of the address?
A: To find the class of the address, convert the first octet from decimal to binary. I will use the "divide by two" method.

172/2 = 86 0
86/2 = 43 0
43/2 = 21 1
21/2 = 10 1
10/2 = 5 0
5/2 = 2 1
2/2 = 1 0
1/2 = 0 1

10101100.

Since the first bit is not a '0', this is NOT a Class A address.
Since the first two bits are "10", this is a Class B address.
Since the first three bits are not "110", this is NOT a Class C address.
Since the first four bits are not "1111", this is NOT a Class D address.

Q: What is the default mask?
A: Since this is a Class B address, the default mask is 255.255.0.0.

The way to remember default masks, think of the following:
A.0.0.0 = 255.0.0.0
A.B.0.0 = 255.255.0.0
A.B.C.0 = 255.255.255.0

Q: What is the relevant octet?
A: Since the default mask is 255.255.0.0 or /16 and the subnet mask in the problem is /28 or 255.255.255.240, the relevant octet is the fourth octet. The fourth octet converted to binary with the subnet boundary is .0000|0000 where the '|' is the subnet boundary.

Q: What is the subnet address?
A: Since the given address is 172.20.0.0 and the relevant octet is the fourth octet, the subnet address is 172.20.0.0000|0000 or 172.20.0.0 in full decimal.

Q: What is the broadcast address?
A: To find the broadcast address, fill up the host bits with all binary '1's. So in the above mentioned subnet address, it's 172.20.0.0000|1111 or 172.20.0.15.

Now you might be wondering "why did I just do all this work when the problem doesn't even ask these five questions?" Because you can use the answers to these five questions to figure out the answers to the problem. By taking the broadcast address and subtracting from it the subnet address, you can find out how many possible valid host addresses. So the range of valid host addresses is from 172.20.0.1 through 172.20.0.14. That means a total number of 14 valid hosts.

Now, as far as the number of subnets, this is easier somewhat. You take the default mask of 255.255.0.0 then you take the given subnet mask of 255.255.255.240. If you convert them to binary and compare them you get...

11111111.11111111.00000000.00000000 = 255.255.0.0
11111111.11111111.11111111.11110000 = 255.255.255.240
==========================
00000000.00000000.11111111.11110000 = difference

So you take the difference between /28 and /16 to get 12 subnet bits. So you start with 2 for the first bit then just count up by doubling like...

1 2 3 4  5  6  7   8   9   10   11   12 bits
2 4 8 16 32 64 128 256 512 1024 2048 4096 subnets

So by doing the "hard work" of the "long way", you can then have information to calculate the rest. But if you're only relying on shortcuts without understanding the fundamentals like thinking in binary, it's going to result in the confusion that you're feeling.

I hope this helps.

shednik

nicklauscombs wrote:

for really quick answers to problems like that on the exam I wrote out the subnetting chart
however its crucial to understand why its the answer

That made the exams so much easier having that next to me!

dazerski

tech-airman wrote:

ddzierzek wrote:

I really can't express how far I have come with subnetting in the last week. I went from not knowing how to really even do it, to doing most of the problems in my head within 25 seconds (still not fast enough).

But then I get to problems like this from subnettingquestions.com and it takes me way too long...

Question: How many subnets and hosts per subnet can you get from the network 172.20.0.0/28?

Answer: 4096 subnets and 14 hosts


I guess one solution would to be bone up on addition (that's embarrassing to even say) but you people have given me some very handy shortcuts in my last couple threads. I was wondering if anyone had one off the top of your head for these kinds.

Plus I just wanted to post that I officially scheduled my exam date for Sept.30th... I'm stoked... I feel confident in 85% of the stuff in Odom's and CBT as well as doing hands-on stuff on my cisco lab I built... If I keep my original 9/30 test-date, I'll have plenty of time to get everything down. I knew I'd never take the test unless I slapped down $125 and set a date. I'm totally excited right now

Thanks again for all your help!

Dale

ddzierzek,

The problem with any shortcut without the foundational understanding leads you to the apparent confusion that you're experiencing. Trust me when I say that Cisco is very good at confusing you if you don't firmly understand the foundation knowledge. With that said, you have to understand the "long way" before you can appreciate the "shortcut" as a shorter way of doing the long way instead of just trying to short circuit the long way of learning.

With that said, here's the long way that I used to successfully pass the 640-801 CCNA exam. I call it my "five questions" method. The five questions are:

1. What is the class of the address?
2. What is the default mask?
3. What is the relevant octet?
4. What is the subnet address?
5. What is the broadcast address?

Let's answer the five questions for 172.20.0.0/28.

Q: What is the class of the address?
A: To find the class of the address, convert the first octet from decimal to binary. I will use the "divide by two" method.

172/2 = 86 0
86/2 = 43 0
43/2 = 21 1
21/2 = 10 1
10/2 = 5 0
5/2 = 2 1
2/2 = 1 0
1/2 = 0 1

10101100.

Since the first bit is not a '0', this is NOT a Class A address.
Since the first two bits are "10", this is a Class B address.
Since the first three bits are not "110", this is NOT a Class C address.
Since the first four bits are not "1111", this is NOT a Class D address.

Q: What is the default mask?
A: Since this is a Class B address, the default mask is 255.255.0.0.

The way to remember default masks, think of the following:
A.0.0.0 = 255.0.0.0
A.B.0.0 = 255.255.0.0
A.B.C.0 = 255.255.255.0

Q: What is the relevant octet?
A: Since the default mask is 255.255.0.0 or /16 and the subnet mask in the problem is /28 or 255.255.255.240, the relevant octet is the fourth octet. The fourth octet converted to binary with the subnet boundary is .0000|0000 where the '|' is the subnet boundary.

Q: What is the subnet address?
A: Since the given address is 172.20.0.0 and the relevant octet is the fourth octet, the subnet address is 172.20.0.0000|0000 or 172.20.0.0 in full decimal.

Q: What is the broadcast address?
A: To find the broadcast address, fill up the host bits with all binary '1's. So in the above mentioned subnet address, it's 172.20.0.0000|1111 or 172.20.0.15.

Now you might be wondering "why did I just do all this work when the problem doesn't even ask these five questions?" Because you can use the answers to these five questions to figure out the answers to the problem. By taking the broadcast address and subtracting from it the subnet address, you can find out how many possible valid host addresses. So the range of valid host addresses is from 172.20.0.1 through 172.20.0.14. That means a total number of 14 valid hosts.

Now, as far as the number of subnets, this is easier somewhat. You take the default mask of 255.255.0.0 then you take the given subnet mask of 255.255.255.240. If you convert them to binary and compare them you get...

11111111.11111111.00000000.00000000 = 255.255.0.0
11111111.11111111.11111111.11110000 = 255.255.255.240
==========================
00000000.00000000.11111111.11110000 = difference

So you take the difference between /28 and /16 to get 12 subnet bits. So you start with 2 for the first bit then just count up by doubling like...

1 2 3 4  5  6  7   8   9   10   11   12 bits
2 4 8 16 32 64 128 256 512 1024 2048 4096 subnets

So by doing the "hard work" of the "long way", you can then have information to calculate the rest. But if you're only relying on shortcuts without understanding the fundamentals like thinking in binary, it's going to result in the confusion that you're feeling.

I hope this helps.

I just got your reply, I am at work right now but when I get home tonight I am going to start heading back and doing what you suggest; understand the long fundamental logic of what is happening and then look for a faster way. Thank you for your help, much appreciated

wbosher

To be honest I count with my fingers: say I want to know (2^6=?) or (64=2^?). So I hold down two fingers and start from 4. As I recite each number I hold down one more finger, and when I get to 64 I have six fingers down. So 2^6=64.

I agree with the finger method, a little embarrassing to admit but it works.