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Home > Archive > CCNA > April 2003 > IP subnetting.
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| b_baruah 2003-03-25, 2:52 am |
| Guys just take a look at this --
Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240
1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175
I think non of the results are correct as I feel the address range for the network(192.168.27.0) with the mask 255.255.255.240 is from
192.168.27.1 to 192.168.27.14
I would like to know the comments from every one from the group
Regards
Bhaskar | |
| uncle8 2003-03-25, 3:32 am |
| b_baruah
I thought :
1. 192.168.27.33 - valid host addresses in subnet 192.168.27.32
2. 192.168.27.112 - subnet address
3. 192.168.27.119 - valid host address in subnet 192.168.27.112
4. 192.168.27.126 - valid host address in subnet 192.168.27.112
5. 192.168.27.175 - subnet broadcast addres in subnet 192.168.27.160
Uncle8 | |
| davidbeecken 2003-03-25, 3:43 am |
| I think that the question is just VERY poorly worded.
After reading this twice, the second answer I listed here was my first choice, but I think what they are asking, is in the network 192.168.27., with subnetmask .240 what are valid host ips ON ANY NETWORK inside the subnetted networks of 192.168.27.
This would be 1,3,4
Explination below after what else the question could be asking:
OR I think they are asking this: imagine that you have the class c network address 192.168.27.
Now with the subnet 255.255.255.240, what three hosts are valid in a subnetted network. Now the only one close to being correct, and I saw close because it is NOT correct is 2,3,4. They are all in the network of 192.168.27.12-26 but they are NOT ALL VALID HOSTS
So in conclusion, they are asking in the network 192.168.27. with subnet of ./28, after removing all the network and broadcast IPS, what are valid hosts, and you would choose what IPS are valid hosts on any subnetted network in 192.168.27.
How to do this quickly, just find the interval number, and start going through the list(interval number is talked about here www.ciscotrack.com/subetting.html)
btw, sorry for any small mistakes, its 4am and I just work up from sleeping a very long time | |
| b_baruah 2003-03-25, 9:41 am |
| Thanks for the answers. | |
| tedd07 2003-03-27, 8:19 am |
| Reading it quickly, you assume the same network, which would be answers 2,3,4, except that 2 is a network address, not a host. That should prompt you to re-read it and realize they are asking for host addresses in ANY network, which is 1,3,4. 2 is a network address and 5 is a broadcast. | |
| djmaplethorpe 2003-03-27, 10:55 am |
| The first question, How many people were able to complete this in less than two minutes?
This brings up another very good question, are there going to be questions like this on the test? and if so if you only get a minute or so for each question how are you suppose to run through all the numbers to see which is which? Even if you can subnet in your head, which I am working on everyday, how can you go through all of these and and still complete the exam? | |
| davidbeecken 2003-03-27, 12:29 pm |
| The acuall solving only took a min or so, it was the figureing out what they wanted that took a little time, for the most part the questions, at least in relation to subnetting are fairly strightforward on what they want, and if you can subnet, they are the easiest questions on the test | |
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| Anyone who thinks they are ready for the CCNA should be able to do this Q in under 2 min. Actually you will get more time than 2 min. The OSI and protocal Q's will provide extra time for the math Q's because they take all of 20 seconds to read and answer.
To answer this question in a fast manner, i would simply find my divisor (which is the last bit borrowed). In this case, 240= the "16" position. So just divide 16 into each answer choices. If it divides without leaving a remander then you know it's not a subnetwork (the "line") which can't be a valid host and then take into consideration the last host bit in each range is the broadcast address and also can't be used.
The first and last subnetwork in the network are not valid along with the host in those subnets.
Give me 60 subnetworking questions and I would feel better about my chances come MONDAY @ 4:30. | |
| KeiWarrior 2003-03-28, 11:27 pm |
| Here's my explination!
Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240
1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175
Ask Questions!
How many Subnets? 2^4-2=14
How many Host? 2^4-2=14
What is the valid subnet? 256-240=16
Subnets:
16 32 48 64 80 96 112 128 144 160
176 192 208 224 240
Now you can see that #2 and #7 cannot be the answer. #2 is a subnet address and #5 is a broadcast.
Subnet 192.168.27.160
First valid host address 192.168.27.161
Last valid host address 192.168.27.174
Broadcast address 192.168.27.175
Hope this clears up the confusion.
LOL
FSW | |
| djmaplethorpe 2003-03-31, 11:42 am |
| quote:
Originally posted by KeiWarrior
Here's my explination!
Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240
1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175
Ask Questions!
How many Subnets? 2^4-2=14
How many Host? 2^4-2=14
What is the valid subnet? 256-240=16
Subnets:
16 32 48 64 80 96 112 128 144 160
176 192 208 224 240
Now you can see that #2 and #7 cannot be the answer. #2 is a subnet address and #5 is a broadcast.
Subnet 192.168.27.160
First valid host address 192.168.27.161
Last valid host address 192.168.27.174
Broadcast address 192.168.27.175
Hope this clears up the confusion.
LOL
FSW
I know some people can so this in their sleep, but everytime I try it I just see dancing girls, but that's a whole different story. Anyway, can you run through how you got the subnets and host for me. I have read every article on this and the minute I think I understand it someone else will throw something in that takes me back to square one.
TIA | |
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| 255.255.255.240 converted to bits =
1111 0000 (the last octet is all we are concerned with) the last bit borrowed is in the 16 position. You have to understand that part! So use 16 as your divisor. Every multiple of 16 is a subnet address or "wire address". For example: 192.168.27.16, 192.168.27.32, 192.168.27.48, etc. Those addresses can't be used as hosts. Also, the last address before the the next subnet address can't be a host. It is the broadcast address. So, using the examples above, 192.168.27.31, 192.168.27.47, and 192.168.27.63 are the broadcast addresses for the above, respectively. That leaves you with the hosts. The hosts range for the examples are as such:
192.168.27.16
hosts= 192.168.27.17 - 192.168.27.30
192.168.27.32
hosts= 192.168.27.31 - 192.168.27.46
192.168.27.48
hosts= 192.168.27.49 - 192.168.27.62
Although you may run into questions where you may need to know such questions as how many valid subnets and/or how many valid hosts per subnet, it is not needed here. | |
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| I see I errored on that second host range. | |
| djmaplethorpe 2003-04-01, 10:37 am |
| thanks for that!! So I take it this will work for all ranges whether they are different classes or not or how you subnet them. For example if I took a class B and subnetted it to class C i.e. 137.161.56.0 with a mask of 255.255.240.0 would not the . 56 be a host in the subnet of 137.161.32.0. Although this may not be used this way or even recommended, is it correct? | |
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| thats correct.
.56.0 is a host in the .32.0 subnet.
But keep in mind that .32.0 has many more hosts in it than a class C ending with .32
The broadcast address for that particular subnet address is x.x.63.255 and the last host in that range is x.x.63.254. It may help if you convert octets into bits at first to see how it works or maybe not.
You can see how that if you add one more bit to the broadcast address, that last octet resets back to zeros and the third octet is set to 64. | |
| KeiWarrior 2003-04-02, 2:49 pm |
| Sorry I've been away, and didn't answer your post.
Wiza answered it pretty good. Allways think what class, and what is the subnet mask?
 FSW | |
| KeiWarrior 2003-04-02, 3:05 pm |
| Here's my explination!
Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240
1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175
Ask Questions!
(all 1's) (all 0's)
240 = 11110000 4bit=subnets, 4bits=host
How many Subnets? 2^4-2=14
How many Host? 2^4-2=14
What is the valid subnet? 256-240=16
Subnets:
16 32 48 64 80 96 112 128 144 160
176 192 208 224 240
Now you can see that #2 and #5 cannot be the answer. #2 is a subnet address and #5 is a broadcast.
Subnet 192.168.27.160
First valid host address 192.168.27.161
Last valid host address 192.168.27.174
Broadcast address 192.168.27.175
Hope this clears up the confusion.
LOL P.S. I corrected a typo!
FSW |
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