| Author |
Another subnet question
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| iceman2001 2002-02-11, 5:25 am |
| You have been given the network ID of 172.24.8.0/22 from your ISP.
All of the routers in your network use either RIP V2, or OSPF. Each of the
two subnets you will be creating will contain only 75 computers. You want
to use the most specific number of bits and the first two available
network ID numbers in your subnet mask. Drag and Drop question with the
following Answer.
(choose 2)
A. 172.24.12.0/22
B. 172.24.16.0/22
C. 172.24.24.0/22
D. 172.24.8.128/25
E. 172.24.9.0/25
F. 172.24.16.0/25
Any takers on the answer to this and the reasoning behind it??
iceman | |
| jeff_j_black 2002-02-11, 2:19 pm |
| This is one I definitely missed on the exam!
From the basic rules of subnetting, you would go with 'A' and 'B'. But with OSPF and RIP v2, I can see where 'D' and 'F' could work in theory. 'C' is not a good choice, because it is not the least significant compared to 'A' or 'B'. 'E' would not be right, because it would fall in the middle of an address range. I'd really just from a curiousity standpoint, like to know the answer. | |
| unreal 2002-02-11, 8:30 pm |
| Did ponder over this quiz, strangely, it says you are given an network ID of 172.24.8.0/22 from your ISP. Then you see answers that are with /25 behind. I was thinking whether this quiz has some error in it. But I finally came think the answer is the last 2-simply because with /25, you have 128 hosts, though it is more than 75, but that's that. So I need 2 subnet, so I just choose the next /25. One of the answer has 128, I don't think it is a valid network ID.
Read jeff comments, realised that maybe RIP and ospf has something on it. Qutie a strange quiz. Suprised it came out in the exam, which you've said. | |
| jeff_j_black 2002-02-12, 7:02 am |
| Outside of OSPF and RIP v2 which do CIDR, it is almost a waste to have a mask of 255.255.255.128 as the all zeros netowrk would be used for network ID and the all ones would be the broadcast range. This would give you 126 hosts on no nets. But with CIDR the mask is sent along with the route so this mask is theoretically usefull. All in all it is a tough one for the exam. | |
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| Ok. I`m new in subneting so i see if someone correct my iddea, i see that the correct answers are between C,D and E.
You say that two subnets must contain 75 computers each.
In your answers you have two available subnet masks: /22 & /25
If you use 22 you have 1022 hosts... this is too much....
If you use 25, you have a host id of 7 ceros = (2^7)-2= 126 possible hosts.
so, enough for your subnets with 75 hosts...
So the correct answers for me are between the D,E and F.
I`m a little confusing at this point. i don`t know what options to select!!!
I appreciate a very good answer and correct me if i`m wrong!! | |
| wbafrank 2002-02-13, 3:45 am |
| quote:
Originally posted by cm2gj
Ok. I`m new in subneting so i see if someone correct my iddea, i see that the correct answers are between C,D and E.
You say that two subnets must contain 75 computers each.
In your answers you have two available subnet masks: /22 & /25
If you use 22 you have 1022 hosts... this is too much....
If you use 25, you have a host id of 7 ceros = (2^7)-2= 126 possible hosts.
so, enough for your subnets with 75 hosts...
So the correct answers for me are between the D,E and F.
I`m a little confusing at this point. i don`t know what options to select!!!
I appreciate a very good answer and correct me if i`m wrong!!
 You're nearly there - if you think about it just a little bit more you should get the answer!! | |
| jeff_j_black 2002-02-13, 9:25 am |
| Yo, frank, is my logic correct in my previous message? | |
| cm2gj 2002-02-13, 12:36 pm |
| D y F? | |
| bluhen99 2002-02-14, 12:38 am |
| Im just thinking out loud, but I feel that you still have to work within the confines of 172.24.8.0/22 network ID you recieve from your ISP. Not only do the first 3 provide too many host, but they are outside of your range. Leaveing just D. 172.24.8.128/25 and
E. 172.24.9.0/25. But is you use a subnet mask of /25 doesnt the first and only network start at 128? If so what is the deal with answer E? | |
| unreal 2002-02-14, 1:56 am |
| E & D is correct.
The first half being:
172.24.9.0/25
10101100 00011000 00001001 00000000
subnet mask- 255.255.255.128
And the second half :
172.24.8.128/25
10101100 00011000 00001000 10000000
Subnet mask : 255.255.255.128
good luck ! | |
| wbafrank 2002-02-14, 6:05 am |
| Unreal you're spot on - well done!! | |
| unreal 2002-02-14, 7:52 am |
| YES ! Thanks ! I needed that badly  Well it seems I've moved up a gear into understanding the magic of subnet. | |
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| unreal 2002-02-15, 10:34 pm |
| Great subneting Guardian site | |
| Zaraspook 2002-02-16, 3:10 am |
| The answer is E & D
The subnet mask is 255.255.255.128
It is often believed that the 172.24.8.128/25 and the 172.24.9.0/25 subnets would be unusable; however, this is a common misconception, which is perpetuated because some routers do not support subnet zero. Although two host are lost for each subnet through subnetting, no network blocks are lost in the process.
Zaraspook | |
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| I don`t understand yet this answer...
i Appreciate very well a good detailed explanation... | |
| Zaraspook 2002-02-16, 8:39 am |
| Alexis Garcia
/25 yields 2^7= (2x2x2x2x2x2x2) or 128 host, which is the most specific per our parameters, and because the number of host are determined by the number of host bits remaining, we can only have 128 host per subnet; also (256-128 (minus the first bit taken for our network IDs)=128 bits remaining for our host). The number of subnets is determined by the number of bits used for the network IDs, and in this case, a /25 network would use the first bit of the last octet for the network ID, yielding 2^ 1= (2x1) or 2, which gives us our two subnets. Therefore, 172.24.8.128/25 and 172.24.9.0/25 would be valid for each of our 2 subnets.
Zaraspook | |
| Zaraspook 2002-02-16, 8:54 am |
| Alexis Garcia
/25 yields 2^7= (2x2x2x2x2x2x2) or 128 host, which is the most specific per our parameters, and because the number of host are determined by the number of host bits remaining, we can only have 128 host per subnet; also (256-128 (minus the first bit taken for our network IDs)=128 bits remaining for our host). The number of subnets is determined by the number of bits used for the network IDs, and in this case, a /25 network would use the first bit of the last octet for the network ID, yielding 2^ 1= (2x1) or 2, which gives us our two subnets. Therefore, 172.24.8.128/25 and 172.24.9.0/25 would be valid for each of our 2 subnets.
Zaraspook | |
|
| I know the formula / the net id and hosts id stuff, etc.. i donīt understand why 172.24.16.0 /25 is not a correct answer... | |
| chodan 2002-02-16, 5:15 pm |
| quote:
Originally posted by unreal
E & D is correct.
The first half being:
172.24.9.0/25
10101100 00011000 00001001 00000000
subnet mask- 255.255.255.128
And the second half :
172.24.8.128/25
10101100 00011000 00001000 10000000
Subnet mask : 255.255.255.128
good luck !
I would agree.
With the routing protocols given there shouldn`t be a problem.
When I first did the MCSE track
microsoft didn`t recognize subnet zero
for their exams, now with cidr it is not a problem. In those days it would have been a different question. | |
| jeff_j_black 2002-02-16, 6:07 pm |
| cm2gj, sixteen does not have any bits in common with eight, nine does, that's why in this instance the network with sisteen in the third octet does not work. This is definitely a question in the exam, word for word. | |
|
| D. 172.24.8.128/25
10101100.00011000.00001000.10000000
E. 172.24.9.0/25
10101100.00011000.00001001.00000000
F. 172.24.16.0/25
10101100.00011000.00010000.00000000
the CIDR mask 25 is:
11111111.11111111.11111111.10000000
25 bits for network id
7 bits for hosts ids...
you can see that D & E are in differents network ID too!!!!
maybe iīm miss something.. i donīt understand the answer yet!!  | |
| Zaraspook 2002-02-16, 10:49 pm |
| Alexis Garcia
The CIDR mask is the key:
11111111.11111111.11111111.10000000 or 255.255.255.128
All the bits are being used in the first three octets plus the first bit in the fourth octet for the Network ID portion of the IP address. Therefore, only seven bits (represented by the 0s) remain for use with the host IDs. (Remember, that the Subnet Mask identifies which portion of an IP address is for the Network ID and which portion of an IP address is for the Host ID.)
In Binary Math, with each bit we get two addresses, (either 0 or 1), and since we have seven available bits, (indicated by the 0s above), we compute our number of available addresses as 2 ^ 7 = 128 or (2 x 2 x 2 x 2 x 2 x 2 x 2 = 128).
Another method is 32 total bits contained in an IP address minus the total number of bits used for the network ID, which in this case is 25, and yields 7 bits remaining for the Host IDs, (32 25 = 7). Therefore, again 2 ^ 7 = 128 or (2 x 2 x 2 x 2 x 2 x 2 x 2 = 128).
And another is the last octet consist of 8 bits of which the first bit is being used for the Network ID portion of the IP address, leaving 7 bits remaining for use with the Host IDs, (8 1 = 7) and again, 2 ^ 7 = 128 or (2 x 2 x 2 x 2 x 2 x 2 x 2 = 128).
Still another is the total number of Host IDs, which is 256, (0 255), minus 128 = 128 remaining for the Host IDs, (256 128 = 128). (The 128 that is subtracted represents the decimal conversion for the first bit of the last octet borrowed for the Network ID portion of the IP address.)
The number of subnets is determined by the number of bits used for the network IDs, and in this case, a /25 network would use the first bit of the last octet for the network ID portion of the IP address, yielding 2 ^ 1 = 2 or (2 x 1 = 2), which gives us our two subnets. (Remember from Binary Math each bit yields two addresses either 0 or 1.)
Therefore, 172.24.8.128/25 and 172.24.9.0/25 would be the only two addresses listed in the question that are valid for each of our 2 subnets and, again each of our 2 subnets can only have 128 Host IDs, (from above).
Hope this helps!
Zaraspook | |
| wirechild 2002-02-16, 11:29 pm |
| cm2gj,
I think you are overlooking something in the question. You only have 172.24.8.0/22 to work from this would include:
172.24.8.0 through 172.24.11.255 we can then
further subnet this down however you need to but the only two choices in the question that do fall in this range is D & E which is
D. 172.24.8.128/25
E. 172.24.9.0/25
I could also use
172.24.8.0 /25
172.24.9.196 /26
172.24.9.128 /26
or how about
172.24.10.0 /30
172.24.10.4 /29
Or what ever you need to dish out as long it is within you original /22 that was gave to you. | |
| cm2gj 2002-02-16, 11:41 pm |
| I know all this concepts!!!! but i donīt understand this paragraph!!!!!:
"Therefore, 172.24.8.128/25 and 172.24.9.0/25 would be the only two addresses listed in the question that are valid for each of our 2 subnets and, again each of our 2 subnets can only have 128 Host IDs, (from above)."
Sorry............ | |
| cm2gj 2002-02-16, 11:47 pm |
| wyrechild...
Why between 172.24.8.0 and 172.24.11.255 ?
for the subnet?
22 subnet 2^10 = 1024-2 1022 hosts??? | |
| wirechild 2002-02-17, 12:10 am |
| cm2gj asks:
"Why between 172.24.8.0 and 172.24.11.255 ?
for the subnet?
22 subnet 2^10 = 1024-2 1022 hosts???"
The original question stated you were given 172.24.8.0 /22. which is indeed
172.24.8.0 - 172.24.11.255 giving you
172.24.8.0 network or wire address
172.24.8.1 - 172.24.11.254 as usables
172.24.11.255 Broadcast address
1022 total hosts to work with | |
| Zaraspook 2002-02-17, 12:45 am |
| Alexis Garcia
A /22 network gives us:
11111111.11111111.11111100.00000000 or 255.255.252.0
It yields many more subnets than the two specified and as you have already determined yourself, way too many Host IDs
Answer (A) 172.24.12.0/22 yields more Host IDs than we need.
Answer (B) 172.24.16.0/22 yields more Host IDs than we need.
Answer (C) 172.24.24.0/22 yields more Host IDs than we need.
Answers (A), (B), & (C) are invalid for the network and therefore incorrect.
Answers (D) & (E) give us the two required subnets and the right number of Host IDs to accommodate our 75 computers per subnet. Answer (D) 172.24.8.128/25 falls within our valid range of Host IDs on the first subnet and answer (E) 172.24.9.0/25 falls within our valid range of Host IDs on the second subnet.
172.24.8.0/25 - 172.24.8.128/25
172.24.9.0/25 - 172.24.9.128/25
Answer (F) 172.24.16.0/25 is outside our valid range and is therefore incorrect.
Zaraspook | |
|
| Zaraspook...
Thanks for your time...
Thanks for your help...
I understand your words... my only doubt its the same:
"why Answer (F) 172.24.16.0/25 is outside our valid range and is therefore incorrect??????"
SORRY!!!  | |
|
| quote:
Originally posted by jeff_j_black
cm2gj, sixteen does not have any bits in common with eight, nine does, that's why in this instance the network with sisteen in the third octet does not work. This is definitely a question in the exam, word for word.
WHAT???
8 --- 00001000
9 --- 00001001
16 --- 00010000
this is what you say...
8 and 9 have a "1" in common...
but the mask put all this bits in the network ID section!!! are remote!!!! not local!!! | |
| wirechild 2002-02-17, 1:24 am |
| cm2gj,
The valid range for 172.24.8.0 /22 is:
172.24.8.0 through 172.24.11.255..........
Where in that range is 172.24.16.0/22?
It isn't that is why F is incorrect.... | |
|
| quote:
Originally posted by wirechild
cm2gj,
The valid range for 172.24.8.0 /22 is:
172.24.8.0 through 172.24.11.255..........
Where in that range is 172.24.16.0/22?
It isn't that is why F is incorrect....
UFFF  I UNDERSTAND NOW.........
THANKS YOU!!!! 
THANKS FOR ALL YOUR HELP
THANKS YOU ALL OF YOU GUYS....
I KEEP STUDYNG SUBNETING BECAUSE I SEE THAT I HAVE SEVERAL PROBLEMS YET... I TAKE SEVERAL TESTS / BOOKS / SUBNETING GUIDES BUT WHEN I SEE SCENARIOS I LOST MY MIND!! IN SOME QUESTIONS!!
I NEED TO RE-SCHEDULE THE 70216 FOR THAT!!! | |
| jeff_j_black 2002-02-17, 10:09 am |
| Nope, the point is the the 172.24.9.0/25 is contained within the original 172.24.8.0/22 network. A network of 172.24.8.0/22 would contain all the addresses within the range 172.24.8.0-172.24.11.254 so the 172.24.16.0/25 would not be in the original range.You just manipulate the subnet mask to divide your original pie slice into small slices of pie. Using RIP v2 and OSPF which send the mask along with the route, you can use 255.255.255.128 as a mask.
So, anyone correct me if I'm wrong, you would have the following networks available from you original pie slice of 172.24.8.0/22:
172.24.8.0-172.24.8.127
172.24.8.128-172.24.8.254
172.24.9.0-172.24.9.127
172.24.9.128-172.24.9.254
172.24.10.0-172.24.10.127
172.24.10.128-172.24.10.254
172.24.11.0-172.24.11.127
172.24.11.128-172.24.11.254
Using a mask of 255.255.255.128 or '/25.
So you can see there are two answers that would have been better in this question, except they weren't given. | |
| wirechild 2002-02-17, 10:13 am |
| quote:
Originally posted by jeff_j_black
Nope, the point is the the 172.24.9.0/25 is contained within the original 172.24.8.0/22 network. A network of 172.24.8.0/22 would contain all the addresses within the range 172.24.8.0-172.24.11.254 so the 172.24.16.0/25 would not be in the original range.You just manipulate the subnet mask to divide your original pie slice into small slices of pie. Using RIP v2 and OSPF which send the mask along with the route, you can use 255.255.255.128 as a mask.
So, anyone correct me if I'm wrong, you would have the following networks available from you original pie slice of 172.24.8.0/22:
172.24.8.0-172.24.8.127
172.24.8.128-172.24.8.254
172.24.9.0-172.24.9.127
172.24.9.128-172.24.9.254
172.24.10.0-172.24.10.127
172.24.10.128-172.24.10.254
172.24.11.0-172.24.11.127
172.24.11.128-172.24.11.254
Using a mask of 255.255.255.128 or '/25.
So you can see there are two answers that would have been better in this question, except they weren't given.
you are almost correct except your 254's should be 255's in the context that you wrote them, ie.
172.24.11.128-172.24.11.254 should be written as 172.24.11.128-172.24.11.255 with usable hosts ip's being 172.24.11.129-172.24.11.254 | |
| jeff_j_black 2002-02-17, 10:20 am |
| That's right, I did say networks, huh. Then I realize that if I meant host address ranges, I should have made all the zeroes into ones. Thanks. All in all, this question alone has made one heck of a topic. | |
| adesemmyk 2002-02-18, 6:56 am |
| Hey guys,
The line below simply struck out choice F.
You want
to use the most specific number of bits and the
first two available network ID numbers in your subnet mask.
D, E stand
-Hackade | |
|
| Hey
I don't use this stuff everyday, so it gets a little fuzzy. Check this out. If its a 22 bit class b subnet it super subnetted. that means you can't use a 22 subnet, so the last two are the only two.
dw
You have been given the network ID of 172.24.8.0/22 from your ISP.
All of the routers in your network use either RIP V2, or OSPF. Each of the
two subnets you will be creating will contain only 75 computers. You want
to use the most specific number of bits and the first two available
network ID numbers in your subnet mask. Drag and Drop question with the
following Answer.
(choose 2)
A. 172.24.12.0/22
B. 172.24.16.0/22
C. 172.24.24.0/22
D. 172.24.8.128/25
E. 172.24.9.0/25
F. 172.24.16.0/25 | |
| bfattima 2002-02-18, 7:58 am |
| So what's the answres.
I think will be A,b because it's part of the the subnet?????? | |
| mscott36 2002-02-18, 7:59 am |
| This can be a tricky question, but if you look at the address and mask given the answer is clear. The ISP tells us the block we are given is 172.24.8.0 our block size is 4 because the 22bit mask equates to 252 and if we subtract that from the total number of possible IP addresses 256 (0-255)we get 4. Now we know our range is 172.24.8.0 our network address to 172.24.11.255 which is our broadcast address. Any answer that does not fall in this range is incorrect. Answers D and E are the only ones that do. Therefore thoses are your answers. Question done. The rest is there to confuse you. Yes RIP2 does send subnetmask and the 128 mask does get routed. | |
|
| I'm having some difficulty getting my head round subnet masks, and this q. nearly blew my mind! Is my logic correct in the following:
ISP-given network address has as its 3rd octet 00001000, so anything 'higher' than that (e.g. the 00010000 of B and E) are out? | |
| jeff_j_black 2002-02-18, 10:59 am |
| A,B,C and F are out, for the reason you have given. E is in the original network range of 172.24.8.0/22. This corresponds to a range of addresses: 172.24.8.0 to 172.24.11.255. So you see 192.24.9.0 falls within the original range. | |
| trogers5 2002-02-18, 11:40 am |
| Answers are "D" and "E"
If any questions will be happy to post reasoning!! | |
|
| Back again - yes I think I see now:
The 3rd octet of the network address has to be 000010.., so 8, 9, 10 and 11 are OK because they just fill in the last 2 places whereas 12 is 000011.. Yes?
Thanks for the help. | |
| jeff_j_black 2002-02-18, 5:38 pm |
| You got it! | |
| Zaraspook 2002-02-18, 10:09 pm |
| 172.24.8.0/22
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1
128 192 224 240 248 252 254 255
0 2 6 14 30 62 126 254
/22 equals 255.255.252.0
Using the above chart, the range incrementing value for 252 is 4, (from top row above 252). Therefore, our range is 172.24.8.0 - 172.24.11.255.
Using a /25 or 255.255.255.128 subnet mask, translates: 11111111.11111111.11111111.10000000
In Binary Math, with each bit we get two addresses, (either 0 or 1), and since we have seven available bits, (indicated by the 0s above), we compute our number of available addresses as 2 ^ 7 = 128 or (2 x 2 x 2 x 2 x 2 x 2 x 2 = 128).
Therefore, the following are the ranges for /25.
172.24.8.0 - 172.24.8.127
172.24.8.128 - 172.24.8.255
172.24.9.0 - 172.24.9.127
172.24.9.128 - 172.24.9.255
172.24.10.0 - 172.24.10.127
172.24.10.128 - 172.24.10.255
172.24.11.0 - 172.24.11.127
172.24.11.128 - 172.24.11.255
The first address in each range is the address for the network itself and the last address is the broadcast address for that subnet.
(a) 172.24.12.0/22, (b) 172.24.16.0/22, and (c) 172.24.24.0/22 all fall outside our 172.24.8.0 - 172.24.11.255 range for 172.24.8.0/22; (f) 172.24.16.0/25 falls outside the range for /25.
Therefore, the network ID addresses (d) 172.24.8.128/25 and (e) 172.24.9.0/25 are the only two valid addresses.
Special thanks to wirechild! | |
| cm2gj 2002-02-18, 10:37 pm |
| very good explanations!!!
thanks you!! | |
| hblindell 2002-02-24, 9:49 am |
| As far as I can see the answer can only be D and E. The subnet 172.24.8.0/22 will only give you from 172.24.8.0 up to 172.24.12.255.255 (subtracting the network and broadcast addresses of course) | |
| jeff_j_black 2002-02-24, 5:39 pm |
| Or 172.24.8.0-172.24.11.255 actually. |
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