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subnetting problem
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| miami_dude 2003-10-21, 11:38 am |
| Can anyone help me out with this problem and show me how to figure that out???
With a subnet mask of 255.255.255.224 which of the following subnet address or host address falls within a valid range. (Choose three)
1. 10.5.6.236
2. 192.54.36.10
3. 223.16.0.86
4. 86.201.79.30
5. 40.255.255.35
6. 191.26.50.129
7. 145.25.68.2 | |
| Demijohn 2003-10-21, 12:12 pm |
| For a given major network, the first and last subnets possible with a given mask are invalid (in Cisco thinking)
For the 10.0.0.0 network and a /27 mask, the two invalid ranges are 10.0.0.0-31 and 10.255.255.224-255. 10.5.6.235 isn’t in either range so it is valid.
Figure the others out the same way. (Haven't you seen this before?) | |
| miami_dude 2003-10-21, 1:28 pm |
| I still cudn't figure it out ...cud u give me the binary way of doing that !!! | |
| Demijohn 2003-10-21, 4:12 pm |
| Okay, in a binary way.
Consider the first choice, 10.5.6.236 255.255.255.224.
This is a Class A network, 10.0.0.0. So you have 8 bits for network, 19 bits borrowed for subnets, and 5 bits left for hosts. Sort of like this.
nnnnnnnn.ssssssss.ssssssss.ssshhhhh
Now we know that the invalid subnets are when the s bits are all 1 or all 0. Substituting for n and s we get
00001010.00000000.00000000.000hhhhh
or
00001010.11111111.11111111.111hhhhh
Ignoring the host bits for now, convert these back to decimal, and you get
10.0.0.0 and 10.255.255.224 These are the base, or subnet addresses for the two invalid ranges.
At this point we could evaluate the 5 host bits, but we can see by inspection that 10.5.6.236 isn’t a match for either of these ranges, so it’s valid.
Or you could compare the high 27 bits of our candidate:
00001010.00000101.00000110.111xxxxx
and see they don’t match the invalid ranges this way either.
Is that binary enough?
You would evaluate each of the other options for validity in the same way, except that the number of network and subnet bits will vary depending on the network class, but the two always adding up to 27, and always 5 host bits.
Of course when you do this you may find that more than three of the choices are valid. | |
| joel316 2003-10-21, 7:24 pm |
| I'm pretty sure this is right:
You take the subnet mask 255.255.255.224.
Take 256 - 224= 32
So you will have 32-2=30 host per subnet.
There is only real octet we have to look at is the last Octet.
If you put your range
33-62 (32 subnet 63 broadcast)
65-94 (64 subnet 95 broadcast)
97-126 (96 subnet 127 broadcast)
129-158 (128 subnet 159 broadcast)
161-190 (160 subnet 191 broadcast)
193-222 (192 subnet 223 broadcast)
You look at the ranges and verify if they fall into them and just look the last number.
So you look at the
1. 10.5.6.236 = NO
2. 192.54.36.10 = NO
3. 223.16.0.86 = YES
4. 86.201.79.30 = NO
5. 40.255.255.35 = YES
6. 191.26.50.129 = YES
7. 145.25.68.2 = NO
So the answer is 3, 5 and 6.
Please correct anyone if you think I am wrong.
Thanks. | |
| joel316 2003-10-21, 10:16 pm |
| quote:
Can anyone help me out with this problem and show me how to figure that out???
With a subnet mask of 255.255.255.224 which of the following subnet address or host address falls within a valid range. (Choose three)
1. 10.5.6.236
2. 192.54.36.10
3. 223.16.0.86
4. 86.201.79.30
5. 40.255.255.35
6. 191.26.50.129
7. 145.25.68.2
I did some calculations and all of them are valid answers. Not too sure as to why they are only asking three because it seems as though all of them are right. I thought I was right with my previous answer but that doesn't make sense. I was basically making the calcualations for a Class C network. If you wouldn't be using IP class subzero, #2 would be wrong since the IP addresses start at 192.54.36.33 Since the subnet mask is 192.54.36.32. Other then that, I can figure out what the answer would be. | |
| OHCCNP2003 2003-10-22, 2:51 am |
| If you do the anding operation your subnets are:
10.5.6.224
192.54.36.0
223.16.0.64
86.201.79.0
40.255.255.32
191.26.50.128
145.25.68.0
Assuming you are excluding Subnet Zero and the All-Ones Subnet, the correct answers are 3, 5, and 6. If you use ip subnet-zero, then they are all valid. | |
| joel316 2003-10-22, 8:31 am |
| Alright so I was right the first time then. I was looking at the ending but since we had multiple classes of IP address I just thought that a Class A or B address would accept an address x.x.x.0 to x.x.x.32 or x.x.x.224 to x.x.x.255.
Thanks. I hope this helped Miami_Dude. If you need any explaination please let me know. | |
| Demijohn 2003-10-22, 8:57 am |
| 192.54.36.10 is the only address in the bunch that involves either the zero subnet or the all ones subnet. As such, it would be valid only if the subnet zero command is used. The others are all always valid.
Warning: For exam purposes, is Cisco asks if an address is valid, never assume that the subnet zero has been issued. | |
| Demijohn 2003-10-22, 1:07 pm |
| Joel goes wrong in his first post when he says: “There is only real octet we have to look at is the last Octet.” This is only true for the class C addresses (i.e. #2 & #3) (Of course you have to look at the first octet to determine the class) He got it right with his second post, but then backed out again the third time.
OHCCNP also got it wrong. He figured out the subnet addresses correctly, but then failed to identify which are zero and/or all 1’s subnets. Apparently he reverted to thinking in class C terms as well.
For the given addresses, the zero and all 1’s subnet ranges associated with a 27 bit mask are:
Address Class subnet zero all 1’s subnet:
10.5.6.236 A 10.0.0.0-31 10.255.255.224-255
192.54.36.10* C 192.54.36.0-31* 192.54.36.224-255
223.16.0.86 C 223.16.0.0-31 223.16.0.224-255
86.201.79.30 A 86.0.0.0-31 86.255.255.224-255
40.255.255.35 A 40.0.0.0-31 40.255.255.224-255
191.26.50.129 B 191.26.0.0-31 191.25.255.224-255
145.25.68.2 B 145.25.0.0-31 145.25.255.224-255
* a match, therefore invalid
So they’re all valid addresses, except for #2. #2 would be valid with the 'subnet zero' command, but again, for exam purposes, it would be a mistake to assume this command is used.
Conclusion: a bad question, but a good exercise.
Miami_Dude: Did any of this help? | |
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| miami_dude 2003-10-22, 3:56 pm |
| Yes it helped me a lot!!!Thank you everyone for helping me out with that example.I've my exam on 27th Oct. I wish I dont see that kind of question in the exam.
Gourav | |
| joel316 2003-10-22, 9:42 pm |
| I am writing the exam on the 25th so I will let you know how it goes. I doubt that we will get a question like that. I heard most questions like this involves more Class C IP but I could be wrong. I don't know where you took that question from but it seem that there is only one logical wrong and not three.
I think this is more of a question you will get:
What is the network address for a host with the IP address 123.200.8.68/28?
A. 123.200.8.0
B. 1231.200.8.32
C. 123.200.8.64
D. 123.200.8.65
E. 123.200.8.31
F. 123.200.8.1
If you can answer this, you should be good to go for subnetting. Good luck to you and I will let you know how I did on the 25th. |
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