be weary of certain sites

I have found a few sites that have exam questions for free to be kind of not so trusty.

for example one of them has this question and answer from subnettingquestions.com

Question: How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.255.128?
Answer: 512 subnets and 126 hosts

that is not true becausw 255.255.255.128 is a /25 which is 2 subnets with 128-2 for hosts. not 512 subnets. Because that would be 65536 hosts which is a /16. I know that 172 is a class b prefix BUT it has a class C mask. Welcome to CIDR and VLSM. Just do the problem and you can see that it does not span across 512 networks. 172.31.0.0/25 is 172.31.0.0-172.31.0.127 and 172.31.0.128-172.31.0.255. Two subnets with 126 usable hosts each

Also simulartionexams.com is not very good either. Notice on the picture that it asks you to determine the GW address for host a and c. host a belongs to vlan 1 and c to vlan 3. Notice in the routers config that they have it named vlan 1, 100 and 200. so you cant answer the question because they suck at congiuring stuff.

However subnetting questions.com did have some that were correct. just use good judgment.

vlan question copy.jpg

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blackninja

LT72884 wrote: »

I have found a few sites that have exam questions for free to be kind of not so trusty.

for example one of them has this question and answer from subnettingquestions.com

Question: How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.255.128?
Answer: 512 subnets and 126 hosts

that is not true becausw 255.255.255.128 is a /25 which is 2 subnets with 128-2 for hosts. not 512 subnets. Because that would be 65536 hosts which is a /16. I know that 172 is a class b prefix BUT it has a class C mask. Welcome to CIDR and VLSM. Just do the problem and you can see that it does not span across 512 networks. 172.31.0.0/25 is 172.31.0.0-172.31.0.127 and 172.31.0.128-172.31.0.255. Two subnets with 126 usable hosts each

Also simulartionexams.com is not very good either. Notice on the picture that it asks you to determine the GW address for host a and c. host a belongs to vlan 1 and c to vlan 3. Notice in the routers config that they have it named vlan 1, 100 and 200. so you cant answer the question because they suck at congiuring stuff.

However subnetting questions.com did have some that were correct. just use good judgment.

I've been using subnettingquestions.com for ages now and I've never seen an error.

I think the questions is asking in class B sense not in CIDR/VLSM

LT72884

blackninja wrote: »

I've been using subnettingquestions.com for ages now and I've never seen an error.

I think the questions is asking in class B sense not in CIDR/VLSM

I was thinking the very same thing but still it would not work because the mask is what determines the amount of subnets and users

Let me explain before the flame. I know that 10-127 is class A and 128-191 is class b and 192-whatever is class C. But since 1992 the masks is what determines the class now because it is classes ip addressing. a /25 is a class C subnet mask. No matter what IP you put that on. It will have the same range. 10.1.1.0/25 is :

10.1.1.0-127 and 10.1.1.128-255. Even though that is a class A address it has a class C mask. Before 1992 it would have a mask of /8 and thats it. That is what CIDR and VLSM are for. to put a different class mask on a differnt class ip address to save space.

If the site did infact mean class B then it needs to state that because the mask is what determines the amount of subnets and hosts per subnet.

2^n = subnets

where n = number of bits in subnet portion. Which happens to be one
2^1 = 2

LT72884

im not saying your wrong. Trust me. i would not do that. Im just stating my opinion is all. I would like understanding on this because i feel that it is a confusing topic.

thanx

Kaminsky

Look at the first octet of the ip address first before you look at the subnet mask when going through subnetting or you can really mess your head up.

The first octet of the ip address implies how many bits will be assessed for the subnet. The class B implies the last two octets so although it may be a /25, the class B address (which defaults at /16) will be using 9 bits for network and 7 bits for hosts in this instance .

If the first octet of the ip address were 192 - 223.255.255.255 (class C range) then you would be right.

Neeko

With these questions you've just got to assume you're working from the classful boundary according to as Kaminsky has said, the first octet. If VLSM is assumed the original netmask could be anything so there is no right answer.

skrpune

LT72884 wrote: »

I have found a few sites that have exam questions for free to be kind of not so trusty.

for example one of them has this question and answer from subnettingquestions.com

Question: How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.255.128?
Answer: 512 subnets and 126 hosts

that is not true becausw 255.255.255.128 is a /25 which is 2 subnets with 128-2 for hosts. not 512 subnets. Because that would be 65536 hosts which is a /16. I know that 172 is a class b prefix BUT it has a class C mask. Welcome to CIDR and VLSM. Just do the problem and you can see that it does not span across 512 networks. 172.31.0.0/25 is 172.31.0.0-172.31.0.127 and 172.31.0.128-172.31.0.255. Two subnets with 126 usable hosts each

I used a subnet calculator and the answer given on subnetting questions.com is what comes up as the answer.

There are many ways to calculate number of subnets/hosts, and I'll just run through some not all of them.
Hosts:
32 bits - 25 bits for subnet address = 7 bits used for hosts
hosts = 2^n - 2 = 2^7 - 2 = 128 - 2 = 126 hosts per subnet

Subnets:
subnets = 2^n - 2
(where n is the number of bits in the subnet being switched from 0 to 1)
In this case, we've got a class B address. The normal subnet would be 255.255.0.0 or 11111111.11111111.00000000.00000000
BUT we've got a subnet of
255.255.255.128 or 11111111.11111111.11111111.10000000
...sooooo, nine 0's were switched to 1's. Back to our formula:
subnets = 2^n - 2 = 2^9 - 2 = 512-2 = 510
(NOTE: the 512 figure as given on subnettingquestions.com and on subnet calculators includes the network address and the broadcast address....but technically, the answer is 510 subnets.)

LT72884

Kaminsky wrote: »

Look at the first octet of the ip address first before you look at the subnet mask when going through subnetting or you can really mess your head up.

The first octet of the ip address implies how many bits will be assessed for the subnet. The class B implies the last two octets so although it may be a /25, the class B address (which defaults at /16) will be using 9 bits for network and 7 bits for hosts in this instance .

If the first octet of the ip address were 192 - 223.255.255.255 (class C range) then you would be right.

Yeah no kidding. It will mess with your head..

Classfull addressing is a confusing topic. LOL

This is how i would work this problem:

144.66.35.0/26

144.66.35.0-63
144.66.35.64-127
144.66.35.128-191
144.66.35.192-255

4 class c subnets and 62 usable ip's each subnet.

if i did:

144.66.35.0/15

144.66.35.0 - 144.67.255.255

or how would a router summarize the following range:

192.168.0.0 - 195.255.255.255

192.168.0.0/13
192.176.0.0/12
192.192.0.0/10
193.0.0.0/8
194.0.0.0/7

I think what is going on is i am confusing terms with other terms.

If you will. Please expound on your above statement. I think it will help me.

Once again thank you for your patience.. I do appreciate the help

LT72884

skrpune wrote: »

I used a subnet calculator and the answer given on subnetting questions.com is what comes up as the answer.

There are many ways to calculate number of subnets/hosts, and I'll just run through some not all of them.
Hosts:
32 bits - 25 bits for subnet address = 7 bits used for hosts
hosts = 2^n - 2 = 2^7 - 2 = 128 - 2 = 126 hosts per subnet

Subnets:
subnets = 2^n - 2
(where n is the number of bits in the subnet being switched from 0 to 1)
In this case, we've got a class B address. The normal subnet would be 255.255.0.0 or 11111111.11111111.00000000.00000000
BUT we've got a subnet of
255.255.255.128 or 11111111.11111111.11111111.10000000
...sooooo, nine 0's were switched to 1's. Back to our formula:
subnets = 2^n - 2 = 2^9 - 2 = 512-2 = 510
(NOTE: the 512 figure as given on subnettingquestions.com and on subnet calculators includes the network address and the broadcast address....but technically, the answer is 510 subnets.)

With classless ip addressing. the mask is what determines the subnets and hosts because you have placed an entirely differnent mask on a class B address so it makes is classless. with a /25 mask, only one bit is in the subnet portion 2^1 = 2 class C subnets

112.128.0. 00000000
255.255.255.100000000
logically AND them together to get 112.128.0.0 as the network now replace the 0's in the host portion with 1's and then AND it with a/32 to find the bcast or range of subnet
112.128.0.01111111
255.255.255.11111111
is 112.128.0.127 = bcast

now take net address 112.128.0.128 logically AND with /25
112.128.0.10000000
255.255.255.10000000
logically AND and your second subnet is 112.128.0.128

112.128.0.11111111
255.255.255.11111111

and you get 112.128.0.255 as the bcast

so you have

112.128.0.0-127
112.128.0.128-255

skrpune

LT72884 wrote: »

With classless ip addressing. the mask is what determines the subnets and hosts because you have placed an entirely differnent mask on a class B address so it makes is classless. with a /25 mask, only one bit is in the subnet portion 2^1 = 2 class C subnets

112.128.0. 00000000
255.255.255.100000000
logically AND them together to get 112.128.0.0 as the network now replace the 0's in the host portion with 1's and then AND it with a/32 to find the bcast or range of subnet
112.128.0.01111111
255.255.255.11111111
is 112.128.0.127 = bcast

now take net address 112.128.0.128 logically AND with /25
112.128.0.10000000
255.255.255.10000000
logically AND and your second subnet is 112.128.0.128

112.128.0.11111111
255.255.255.11111111

and you get 112.128.0.255 as the bcast

so you have

112.128.0.0-127
112.128.0.128-255

I agree that the subnet mask is what determines how you address a network & what the number of subnets & hosts is, but I think you're getting mixed up with how to figure out your starting point to count your "borrowed" bits. I realize that this example is classless IP addressing, but to find out how many bits were "borrowed" you need to look at what the normal subnet address would be for the given network address, which is in this case a class B network...so the classFUL subnet would be 255.255.0.0. We're changing things up by subnetting, but we have to start somewhere to find out what bits were changed over, and that somewhere is with what the subnet would normally be under a classful situation.

One way to see it is to put the IP address & the subnet address in binary. Then look at the last 1 in the IP address vs the last 1 in the subnet address, and you'll see what bits are what...then I think it will make some more sense:
network IP address 172.31.0.0 =
10101100.00011111.00000000.00000000
subnet address 255.255.255.128 =
11111111.11111111.11111111.10000000

Normally, you would use those last two octets for host addresses. But you've now got 9 1's that have been "borrowed" from what would be used for the host address if this were a regular classful addressing scheme. So you'd use the 2^n - 2 formula using 9 as your n value to calculate the number of subnets.

Forgive me if I'm not explaining it clearly, it's been a while since I've done subnetting. But I'm pulling out the books to make sure that I'm going through the calculations correctly, and I checked myself with a subnet calculator so I'm certain that the answers as given on subnettingquestions.com were correct.

blackninja

skrpune wrote: »

(NOTE: the 512 figure as given on subnettingquestions.com and on subnet calculators includes the network address and the broadcast address....but technically, the answer is 510 subnets.)

When subnetting with cisco, the ip subnet zero command is assummed now so the broadcast and network addresss are now valid subnets, so it is 512 subnets.

LT72884 wrote: »

im not saying your wrong. Trust me. i would not do that. Im just stating my opinion is all. I would like understanding on this because i feel that it is a confusing topic.

thanx

I do know what you mean I have a CCNA so have studied and understand subnetting. You and subnettingquestions.com are correct.

If you are subnetting a network using VLSM then you are correct but the question is asking in a class sense.

Yes very confusing, but has to taken in context.

LT72884

hmmm interesting.

Neeko

Sometimes I just like to remember that we could in theory have an IP address belonging to any class, with the smallest mask possible. This helps detatch my subnetting logic from the classful nature installed into us.

Give me 192.0.0.0 /2 and talk to me about classful

Subnetting questions are typically put forwarded in a classful sense, where the student needs to assume a previous classful mask in order to do the math. It would help if the original mask was provided, that way nothing is left to assumptions and whether CIDR or VLSM is in use doesn't matter because the math still works the same.

networker050184

SUBnetting = breaking networks into smaller pieces. So start with the major network and break it up. You guys are making this WAY more complicated than it actually is.

nevolved

Do you want us to grow tired of certain sites? Or are you wanting us to be cautious of some sites?

aordal

I'd just like to throw in. subnettingquestions.com is never wrong. i've spent hours and hours and hours on it, and when i think they have a wrong question and i break it down, i'm the one who's wrong.

briangl

LT72884 wrote: »

Question: How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.255.128?
Answer: 512 subnets and 126 hosts

This is correct. I used subnettingquestions.com a lot when I was studying for my CCNA tests. Another one I used a lot was IP Subnet Practice

Slowhand

Let me start off by saying that a /25 address is not, I repeat not reflective of a class C address. Class C addresses have /24 masks. Now, this problem involves subnetting the address 172.31.0.0 with a /25 mask. Being that they're giving us an obvious clue that this is going to involve breaking down a private class B address into all usable subnets, (just like the CCNA exam will give you obvious clues, if you know to look for them,) it's safe to say that there are, in fact, going to be 512 subnets with 126 hosts per subnet. (See the first image at the bottom of this post.)

I will say, however, that if this had been a usual, real-world case, then two subnets would be correct. If it were a public address issued by an ISP, you'd have to break down the address to figure out what your block is, as shown in the second image at the bottom of the post. In most cases, though, subnetting within the scope of CIDR comes up on the exam mainly in the form of "given the address 172.31.0.45/25, give the network address, subnet mask, usable host range, and broadcast address." If the question had been asked in this manner, specifically with testing the student on CIDR in mind, then the original poster's answer would be correct.

classful.jpg

CIDR.jpg

jmc012

aordal wrote: »

I'd just like to throw in. subnettingquestions.com is never wrong. i've spent hours and hours and hours on it, and when i think they have a wrong question and i break it down, i'm the one who's wrong.

I have seen it wrong exactly one time and it was a very easy one. But all the other times it was me.

Vogon Poet

nevolved wrote: »

Do you want us to grow tired of certain sites? Or are you wanting us to be cautious of some sites?

I'm weary from certain sites, reading long threads, and staying up too late.

LT72884

nevolved wrote: »

Do you want us to grow tired of certain sites? Or are you wanting us to be cautious of some sites?

Cautious because if you looked at my jpg i posted. youll notice that it is completely wrong. Im just saying that certain sites seem to have alot of wrong answers.

LT72884

jmc012 wrote: »

I have seen it wrong exactly one time and it was a very easy one. But all the other times it was me.

I have seen it wrong a few times. Like the example above. I dont use a subnet calculator because when it comes down to it, you cant use them on the CCNA so why bother.

It just bothers me when it says 512 subnets.

Slowhand

LT72884 wrote: »

I have seen it wrong a few times. Like the example above. I dont use a subnet calculator because when it comes down to it, you cant use them on the CCNA so why bother.

It just bothers me when it says 512 subnets.

Do you understand why the answer is right, and why there are two pictures in my post with output from a subnet calculator? You can't use a calculator on the CCNA exam, but it's very handy for providing quick output to a screen or to double-check an answer that's in dispute.

aordal

LT72884 wrote: »

I have seen it wrong a few times. Like the example above. I dont use a subnet calculator because when it comes down to it, you cant use them on the CCNA so why bother.

It just bothers me when it says 512 subnets.

Well, it is right though.

blackninja

LT72884 wrote: »

It just bothers me when it says 512 subnets.

Please re-read all the above posts.

If you still think it is incorrect, I would advise you to re-learn how to subnet, it's not all that complicated.

Well for most anyway

Think we need to draw a line under this one?

mikej412

LT72884 wrote: »

I know that 172 is a class b prefix BUT it has a class C mask.

Huh?

A Class B address with a /25 netmask has 9 subnet bits.

2^9=512

The first few bits of an address determines the class of the address (and defines the address ranges), not the subnet mask.

LT72884 wrote: »

It just bothers me when it says 512 subnets.

In that case, I'm guessing the Cisco Exams will really piss you off then.

LT72884

blackninja wrote: »

Please re-read all the above posts.

If you still think it is incorrect, I would advise you to re-learn how to subnet, it's not all that complicated.

Well for most anyway

Think we need to draw a line under this one?

LOL not yet. Trust me i know subnetting very well.

Ill find what im trying to say and ill post.

LT72884

mikej412 wrote: »

Huh?

A Class B address with a /25 netmask has 9 subnet bits.

2^9=512

The first few bits of an address determines the class of the address (and defines the address ranges), not the subnet mask.

In that case, I'm guessing the Cisco Exams will really piss you off then.

Before 1992 yes that is true that the first, second, or thid octet determined the class but not for the last 15 years with CIDR.
172.128.0.0/25 is an example of CIDR. not classfull addressing because you have a class C mask on a Class B address. So with CIDR you use the mask to determine the range.

Cidr & Vlsm [Archive] - Petri.co.il forums by Daniel Petri

CIDR means that you are ignoring the original classes (A, B, C) and that you can use whatever subnet mask you what with whatever range of IP addresses.

With CIDR, you use a bit mask to determine the network and host portions of an address, which are no longer restricted to using an entire octet.

ok right from my book cisco net acad CCNA2 companion guide.

figure 7-16 shows 144.16.0.0 with 255.255.255.128. The above figure(7-16) shows an example of classless addressing using CIDR using a structure of a class B address with a class C mask. In a classless address the subnet part ( also called the prefix) contains WHAT WOULD have been the combined network and subnet parts. Or in other words the borrwed bits from a classful address

(the 9 bits every one talks about)

figure 7-7

172.16.32.0/26 has 4 subnets and 64 ip's each
172.16.32.64/26
172.16.32.128/26
172.16.32.192/26

(I did the above example from the cisco CCNA 2 and 3 books and the calculater gets 1024 subnets not 4)

Classfull addressing means that, when analizing a IP address, the address MUST contain a ONE, TWO, or THREE network part. No more no less. Table 7-15 summerizes the class rules of classful addressing

calss A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0

Therefore a class A can and only will have a /8. Classfull addressing could not be subnetted before 1992. The IETF new that they would soon run out of space because of the classfull rules. Thats why they created Classless so that the engineer could use any mask they wanted to on any ip address regardless of the first, second, or third octet number.

networker050184

Well, don't say we didn't try to help you when you fail the exam.

Good luck!