Help with finding subs and hosts

gonX

I thought I had a grasp on this asked my instructor to look at some practice questions I worked out and I'm told this is wrong can't figure out why nor understand my instructors reasoning for why....

Anyway here's the question.

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

10.0.0.0 - 255.0.0.0

10.0.0.0 - 255.255.240.0



Masks
11111111.00000000.00000000.00000000
11111111.11111111.11110000.00000000


Net bits on 12
Host bits on 12


Number OF Subnets possible 2x2x2x2x2x2x2x2x2x2x2x2=4096
Number OF hosts possible 2x2x2x2x2x2x2x2x2x2x2x2=4096-2= 4094

-DeXteR-

gonX wrote: »

I thought I had a grasp on this asked my instructor to look at some practice questions I worked out and I'm told this is wrong can't figure out why nor understand my instructors reasoning for why....

Anyway here's the question.

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

10.0.0.0 - 255.0.0.0

10.0.0.0 - 255.255.240.0



Masks
11111111.00000000.00000000.00000000
11111111.11111111.11110000.00000000


Net bits on 12
Host bits on 12


Number OF Subnets possible 2x2x2x2x2x2x2x2x2x2x2x2=4096
Number OF hosts possible 2x2x2x2x2x2x2x2x2x2x2x2=4096-2= 4094

There are 2^12 subnets and 2^12-2 hosts per subnet
P.s - Sorry about that i'm also learning subnetting!! xD

phobophile

gonX wrote: »

I thought I had a grasp on this asked my instructor to look at some practice questions I worked out and I'm told this is wrong can't figure out why nor understand my instructors reasoning for why....

Anyway here's the question.

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

10.0.0.0 - 255.0.0.0

10.0.0.0 - 255.255.240.0



Masks
11111111.00000000.00000000.00000000
11111111.11111111.11110000.00000000


Net bits on 12
Host bits on 12


Number OF Subnets possible 2x2x2x2x2x2x2x2x2x2x2x2=4096
Number OF hosts possible 2x2x2x2x2x2x2x2x2x2x2x2=4096-2= 4094

Sorry, so the instructor said that the answer of 4096 subnets and 4094 hosts per subnet is incorrect?

If that was your answer, then you are correct.

The response above me is incorrect, as you do not count the first octet.

You have 12 network bits and 12 host bits.

2^12 = 4096 subnets.
(2^12) - 2 = 4094 hosts per subnet (the network address and broadcast address can not be assigned to hosts)

-DeXteR-

Chris_ wrote: »

16 subnets each with 4094 useable hosts.

Hey how is that 16 subnets bro :

gonX

What I put is how I worked it out the method seems to work fine for other practice questions similar to this one in class b and c from subnettingquestions.com

phobophile

gonX wrote: »

What I put is how I worked it out the method seems to work fine for other practice questions similar to this one in class b and c from subnettingquestions.com

Yeah, you're fine. Your instructor is wrong.

Chris_

Shows how easy it is to be wrong!!! I don't even look at the 10.0. Been playing with class B networks all day - my head has stamped them into my brain!!

gonX

I get confused a lot as well....My instructor thought the portion about the subnets was wrong. Thanks for the help guys

Ryuksapple84

Your answer is correct.

255.255.240.0
11111111.11111111.11110000.00000000

Thus you have 12 "0" which means you have 12 bits left.

2^n=Network

---

2^12= 4096
2^n-2=Hosts ---- 2^12-2= 4094

Futura

I always found that 2^10=1024 was an easy one to remember. i always work it backwards or forwards from there. Works for me.:D.

cyberguypr

Futura wrote: »

I always found that 2^10=1024 was an easy one to remember. i always work it backwards or forwards from there. Works for me.:D.

I do the same. I use 2^6=64 and 2^10=1024. The fact that the 6 and the 10 are repeated in the result make it easy to remember.

hermeszdata

A mask of 255.255.240.0 may also be expressed as /20 CIDR. This is a block of 16 and therefore 16 subnets, unless you are working in a no subnet zero situation. in that case it would be 14 subnets.

12 host bits yeilds 4096-2= 4094.

Subnets = 16 (or 14 depending subnet zero status)
Hosts/subnet = 4094

When calculating subnets, we normally only consider the interesting octet. in this case the 3rd. 4 bits define the mask, the balance go to hosts.

It seems that your instructor is correct. you were only half right!

phobophile

hermeszdata wrote: »

A mask of 255.255.240.0 may also be expressed as /20 CIDR. This is a block of 16 and therefore 16 subnets, unless you are working in a no subnet zero situation. in that case it would be 14 subnets.

12 host bits yeilds 4096-2= 4094.

Subnets = 16 (or 14 depending subnet zero status)
Hosts/subnet = 4094

When calculating subnets, we normally only consider the interesting octet. in this case the 3rd. 4 bits define the mask, the balance go to hosts.

It seems that your instructor is correct. you were only half right!

??

The question specified a Class A address, not a Class B address. 4096 subnets was correct.

-DeXteR-

hermeszdata wrote: »

A mask of 255.255.240.0 may also be expressed as /20 CIDR. This is a block of 16 and therefore 16 subnets, unless you are working in a no subnet zero situation. in that case it would be 14 subnets.

12 host bits yeilds 4096-2= 4094.

Subnets = 16 (or 14 depending subnet zero status)
Hosts/subnet = 4094

When calculating subnets, we normally only consider the interesting octet. in this case the 3rd. 4 bits define the mask, the balance go to hosts.

It seems that your instructor is correct. you were only half right!

you toatally confused over their mate!!

Ryuksapple84

hermeszdata wrote: »

A mask of 255.255.240.0 may also be expressed as /20 CIDR. This is a block of 16 and therefore 16 subnets, unless you are working in a no subnet zero situation. in that case it would be 14 subnets.

12 host bits yeilds 4096-2= 4094.

Subnets = 16 (or 14 depending subnet zero status)
Hosts/subnet = 4094

When calculating subnets, we normally only consider the interesting octet. in this case the 3rd. 4 bits define the mask, the balance go to hosts.

It seems that your instructor is correct. you were only half right!

Whaaaa??????

gonX

Can someone check me on this. Its the way we where shown to do subnetting in cisco 1 but its been awhile.



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okplaya

A mask of 255.255.255.224 allows for 30 hosts.

A mask of 255.255.255.128 allows for 126 hosts

gonX

Ok thanks I can see how in binary with mask comparison.

11111111.11111111.00000000.00000000
11111111.11111111.11111111.10000000

hermeszdata

hermeszdata wrote: »

A mask of 255.255.240.0 may also be expressed as /20 CIDR. This is a block of 16 and therefore 16 subnets, unless you are working in a no subnet zero situation. in that case it would be 14 subnets.

12 host bits yeilds 4096-2= 4094.

Subnets = 16 (or 14 depending subnet zero status)
Hosts/subnet = 4094

When calculating subnets, we normally only consider the interesting octet. in this case the 3rd. 4 bits define the mask, the balance go to hosts.

It seems that your instructor is correct. you were only half right!

OK guys, I stand corrected ... in part. I had 5 thiings going on at once and failed to give the balance of the shortcut I use to find subnets.

As mentioned before, I use the /* notation (CIDR) which defines the number of mask bits.

In the case of the problem originally stated, a mask of 255.255.240.0 contains 20 mask bits (8 because of the Class A network, the remaining 12 defining the subnet portion) and 12 host bits.

We may all agree that the number of hosts are 4094 (4096 less the network/subnet and broadcast addresses)

When we look at the interesting octet, *.*.240.* the bit pattern is 11110000. The first 4 bits when totaled gives us the block size I referred to above which is 16. (Had this been the last octet we would use this to give us the range of total addresses within the subnet subtracting 2 to end with the usable address of 14)

Still looking at the interesting octet, the subnet numbers in that octet would be:
0, 16, 32, ... 224, 240 giving us a total of 16 subnets within that octet.

Now, moving to the 2nd octet, because the mask portion in this octet is 255 or 11111111 (binary) we know that the range of subnet numbers is between 0 and 255 or a total of 2^8 or 256 subnet numbers.

We also know that 10.*.*.* is the classful network number so that must stay the same so I basically ignore it.

I hated binary math since in college studying engineering in the mid 1970's. If we were allowed to use calculators with y^x capability during the exam it would be no problem but we can't.

So, I needed to find a way to do calculate total subnets without having to do the 2x2x2x2x2x2...2x routine. While sitting here staring at a problem it finally sunk into my thick skull how very simple the math was!

Multiply the decimal numbers contained within the subnet mask ignoring any octet(s) defining a classful network (or assigning them a value of 1) and the host portion of the mask (anything that is 0)!

For this problem the equation becomes:
1+(Octet2value)x Octet3BlockSize = total number of subnets

(1+255) x 16 = 4096!

Once you understand the /* notation, the resulting block size for any given interesting octet, and can remember the relational equivilents i.e. 240 = block of 16, calculating total subnets is easy. Also, you will notice that you may use the same process for calculating the range of hosts within a subnet!

Back to the original problem. There are 2 correct answers to the number of hosts. The correct answer will depend, as I mentioned in my first post, whether ip subnet-zero is in effect. If it is, all subnets are usable! If not (the command no ip subnet-zero was issued on the router) you treat the range of subnets just as you do for the range of hosts, subtracting 2 from the answer becuase the all 0's and all 1's or 0 and 255 subnets are considered reserved.

no ip subnet-zero answer = 4094
ip subnet-zero answer = 4096

This is one thing to watch and be careful of on the exam.

Hopes this helpes and sorry for the original ommissions.