Is this Subnetting Question/Answer correct?...

[jsingh7]

3. Which one of the following does not belong to the same subnet as the other three when using 255.255.224.0 as the subnet mask?

a . 172.16.66.24
b . 172.16.65.33
c . 172.16.64.42
d . 172.16.63.51

Answer(s): 172.16.63.51

Explanation:
224 = 3 bits, (2*2*2)-2=6 subnets, the remaining 5 bits from the third octet = 2*2*2*2*2=32, hosts in the first subnet use IP addresses from the range 172.16.32.1 to 172.16.63.254, second subnet 172.16.64.1 to 172.16.95.254 (.0 is (sub-)network ID, .255 = broadcast address)

**I think there is something wrong with either the answer or the explanation...Can someone explain a little better possibly...

 

[svermill]

You arrived at the correct answer but the explanation is a little off. The "-2" part only comes into play with hosts - not subnets. Yes, there are still routers out there that can't handle subnet zero so it might occasionally be "-1." As far as I know, there isn't any limitation against subnet "255" so to speak. There is a LOT of misleading material out there in this regard.

An individual going by the handle "CCIEWANNABE" has several times posted a very nice subnetting explanation. Look around for some of his (her?) prior posts. What I have burned into my memory is the binary value of each bit position. That is to say, eight bit positions are as follows:

1 6 3 1 8 4 2 1
2 4 2 6
8

So with three bits, you fall under the 32 position. Your subnets will increment on this value. Stands to reason then that you will have 0-31, 32-63, 64-95, and so on. If it had been four bits (.240), it would have been 0-15, 16-31, and so on. Eventually, it will become second nature that 1 bit = 128, 2 bits = 192, 3 bits = 224, etc (they are simply additive from left to right using the values of the eight bit postions mentioned above).

Good luck. This is one of those humps to be gotten over that leave you wondering what all the confusion was about. But it's a pain getting to where it's instinct.

 

[jsingh7]

Okay... I'm sorry if I'm just not SEEING how you arrived at the answer... I took this question out of an online test... So I was just asking how THEY came up with the answer...

I understand the bit positions, etc... But I don't understand how the Answer: 172.16.63.51 doesn't fall betweeen the range 172.16.32.1 to 172.16.63.254, or the range 172.16.64.1 to 172.16.95.254?

I know it doesn't fall between 172.16.64.1 to 172.16.95.254, but doesn't it fall between 172.16.32.1 to 172.16.63.254? Maybe I'm just missing one of the crucial concepts in Subnetting or something...please help.

Also, about the 2^n - 2 ... You are saying that you only use that formula for Hosts and not Subnets??? I'm studying from the Sybex book and it states that you use that formula to calculate the # of Subnets where n=# of 1's, AND to calculate the # of Hosts where n=# of 0's. So my second question is...Which method should I follow to pass the CCNA Exam?

 

[rburke]

"same subnet as the other three"

It looks to me that you understood how to solve the question, but you didn't pick up on the details of the question. I may be wrong, but it looks like the question wanted the one IP address that was NOT in the same subnet as the other 3. The answer is the 172.16.63.51. From the subnets that you have figured out 32-63, 64-95... the other 3 IP's fall in the subnet of 64-95, while only 172.16.63.51 falls in the subnet of 32-63.

burke

 

[Taraporter]

ITA. Make sure that you pay attention to the small little details of the question. That will be the difference between pass or fail on the exam. You will see several correct answers on the exam but make sure that you read the question carefully and pick out the small little details to arrive at the absulute correct answer for that particular question. Your question asked which of the following IP addresses does NOT fall in the same range as the other 3. All answers are valid IP addresses, but only one of them belongs to a separate subnet.

As for your question about the test. SVERMILL is correct when stating that you don't have to subtract 2 to figure out the available subnets (thanks to the command 'ip subnet zero'). However, on the test, I believe they are still looking for the formula where you subtract 2 to come up with the answer for available subnets. Again, read the question correctly. If it states anything about 'ip subnet zero', take that into consideration. Also, if in doubt, look at each answer to the question. Figure out why "A" is not right, then look to see if "B" could be right. Look at each individual choice to determine if that could or could not be the answer.

The CCNA test can be tricky with the wording. Again, pay attention to details of the question and each answer. You will be suprised how many answers will look correct at first glance.
HTH, Tara

 

[wybnormal]

Tara has nailed this one very well. This is something that should be hit on every day for the CCNA candiate. The devil is in the details and the CCNA test works that angle hard.

A trick I read about that seems to help is to read the answers FIRST and then read the question. It helps to keep mind from focusing on the 1st answer that *looks* right..

MikeS
Find me at

http://www.packetattack.com

"The trouble with giving up civil rights is that you never get them back"

 

[jsingh7]

Man, you guys (& gals) are good... I did NOT read the question thoroughly and carefully. Now I understand why the answer is what it is. Thanks! I will probably ask a few more questions in the few days to come before I take the exam. Somebody just drop me a note if you think I'm asking too many questions...

J. Singh
j.singh7@verizon.net

 

[rburke]

Too many questions???? What do you think this forum is for.... Ask away(I'm just not promising to answer, hahaha)...

burke

 

[CCIEWANNABE]

Follow the color Patterns carfully.
Address Class : B Class A, B and C addresses have a self encoded or default subnet mask built in
Network Address : 172.16.0.0
Subnet Mask : 255.255.224. 0
Network ID : 10nnnnnn.nnnnnnnn
Subnet bit mask : 10nnnnnn.nnnnnnnn.sss. hhhhh.hhhhhhhh
Subnet Bits : 10nnnnnn.nnnnnnnn.sss = 19 A self encoded or default subnet mask built for B = 16 + 3 Sunbent Bits = 19
Host Bits : 10nnnnnn.nnnnnnnn. sss. hhhhh.hhhhhhhh = 13
Possible Number of Subnets : 8 = sss .or 2*2*2
Hosts per Subnet : 8190



10101100.00010000.01000010.00011000 = 172.16.66.24
10101100.00010000.01000001.00100001 = 172 16.65.33
10101100.00010000.01000000.00101010 = 172.16.64.42
10101100.00010000.00111111.00110011 = 172.16.63.51
Notice that the last value becomes all ones and would become a broadcast value address for the suubnet mask value below
11111111.11111111.111XXXXX.XXXXXXX = 255.255.224.XXX
A broadcast is sent to all nodes.


....................................n
..............Max nodes = 2 - 2
Were n represents the available host bits in host portion of an Octet.
An IP address consists of four octets (1 octet = 8 bits), or 32 bits or 4 octets
IPv4 has four Octets
Route once; switch many

 

[svermill]

..."10101100.00010000.00111111.00110011 = 172.16.63.51
Notice that the last value becomes all ones and would become a broadcast value address for the suubnet mask value below
11111111.11111111.111XXXXX.XXXXXXX = 255.255.224.XXX"...

The last value is not all ones. The host bits are one entity regardles of octet boundary. The broadcast address for that range would have been:

10101100.00010000.00111111.11111111 = 172.16.63.255

The difference between the first three addresses and the last is that the subnet bits don't match. That is to say, they are in different ranges. 172.16.63.51 is just a regular old host address in the range 172.16.0.0 - 172.16.63.255.

Did I just misunderstand the point you were trying to make?

 

[CCIEWANNABE]

Thank you for correcting my statement. What I should have said was the mask value would change the multicast value for example: 172.16.63.51 and network value to all ones.

Netmask: 255.255.255.0 = 24 11111111.11111111.11111111 .00000000. Stating a broadcast is incorrect on my part. Only the subnet mask would convert to all ones and create a discontigous network value and new
Multicast value.

Broadcasting refers to sending a message to everyone.
Multicasting refers to sending a message to a select group. There are various ways to interpret multicast meanings.

Link:

http://www.cisco.com/warp/public/614/17.html

Route once; switch many

 

[CCIEWANNABE]

Address: 172.16.63.51 .............10101100.00010000.00111111.00110011
Netmask: 255.255.224.0 = 19 11111111.11111111.11100000.00000000
Now you will see why it changes the mask value Route once; switch many

 

[svermill]

In classes A, B, & C, there are no "multicast" addresses. There are network addresses (e.g. .0), host addresses, and broadcast addresses (e.g. .255). Multicast addresses are in the class D range.

172.16.63.51/19 is a perfectly legitimate host address. It just isn't in the same subnet as the other three. I think you might be confusing things a bit. The host bits have nothing at all to do with the subnet bits. The only sigfificant host bit values are all zeros and all ones - and host bits do not recognize octet boundaries.

10101100.00010000.00111111.00110011

This is a completely insignificant bit arrangement. Only

10101100.00010000.00111111.11111111

has any significance.

 

[CCIEWANNABE]

Your right off topic but multicast has many meanings hence the link to cisco. Route once; switch many