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Home > Archive > CCNA > August 2000 > certmadman where have you been?
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certmadman where have you been?
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| Studying? I presume. Have you gotten subnetting down pat yet??? If you could, would you go over that with me soon I plan to take the test on the 6th of september. Haven't see you for the past few days Are you taking the test soon, or didn't you quit?? | |
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| If you need help with subnetting, why not post your questions? If certmadman is not available, I'm sure someone else would be glad to help. | |
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| It's just that we work at the same place, we'll give it a shot, how would you find the broadcast adress of the subnet address 10.254.255.19 255.255.255.248? that's right from todd's study guide (507) I've read the chapter and still don't understand it all the way, I've even had a ICRC class and still haven't gotten it | |
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| Evil question!
Okay, first thing to do is to take a deep breath and remind yourself how much you love computers. (Don't forget to exhale!)
I'm not sure where you are getting lost, so I'm going to bring in several areas of TCP/IP subnetting to work through the problem, hopefully in a logical order. (The ^ sign means "to the power of" since I don't know how to superscript numbers here.)
Now, in any given network, the very first host ID and the very last host ID are taken. (Subnets are chopped up networks, same principal.) The first host ID is actually defining the start of the network. The last host ID is the broadcast address for that network.
Pretend this was a normal, little Class C network: 192.68.1.19. The basic subnet would be 255.255.255.0.
What is the Network ID? 192,168.1
What is the Host ID? 19
Which two network addresses can not be assigned to a computer?
192.168.1.0 - effectively the subnet mask.
192.168.1.255 - the broadcast address.
Let's hop back to the address in your question: 10.254.255.19. What if the subnet mask was 255.255.255.0?
What is the effective Network ID? 10.254.255
What is the host ID? 19
What two network addresses can not be used?
10.254.255.0 - effectively the subnet mask.
10.254.255.255 - the broadcast address.
Hey, this is looking a lot like the Class C address, isn't it? So far, so good.
How do I know how many Host IDs are available? Well, I have a total of 32 bits available to me for an ID: 4 sets of 8 bits. If my subnet mask is 255.255.255.0, then I am using 24 of those 32 bits for the Network, so 8 bits are left for the Host IDs. 11111111.11111111.11111111.00000000 (The ones mean the bit is being used.)
The number of bits available for the Host IDs is called the Delta. In this example, 2^8 = 256. After the 256th address, a new subnetted network starts. 10.254.254.19 is in a different subnetted network than 10.254.255.19 while 10.254.255.16 is in the same network. (If you need, to, look at it in binary.)
Remember, the first and last Host ID are used for the subnet and the broadcast address, so I have only 254 Host IDs possible in this subnet. (2^8 - 2.)
Now, let's go and chop up that address even further with the subnet mask of 255.255.255.248. With this subnet mask, how many bits are being taken up with for the Network ID?
11111111.11111111.11111111.11111000 - 29 bits. How many bits are left for the Delta? 3. How many possible Host IDs can be assigned in each subnet? 6. (2^Delta - 2, in this case 2^3 - 2.)
[This message has been edited by Crescent (edited 08-22-2000).] | |
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| Take the value of the last octet in the mask, 248, and subtract this from 256. This gives you the "magic number," which is 8. Another way to do this would be to look at the value of the last bit borrowed, again it would be 8. This number tells you the increments that your subnet run. With this in mind you can see that the subnets would be:
0 - 7
8 - 15
16 - 23
24 - 31
32 - 39....
I think you get the idea. Anyway the number on the left is the subnets, while the number on the right is the broadcast for the subnet. The ip you gave has a host id of 19, thus it resides on the 16 subnet. The broadcast would be 23. 17 through 22 would be the valid host ids for the 16 subnet.
One other example. Let's say the host id was a larger number such as 170. Take your "magic number" 8 and divide it into the host id, 170. 170 divided by 8 is 21. 8 times 21 is 168. So 170 resides on the 168 subnet. The next subnet would be 176. 1 less than this gives you the broadcast for the 168 is 175. Valid ip would be 169 through 174.
Clear as mud!
Reamer | |
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| Now, let's put this puppy to bed. (Or take it out back and shoot it. It's up to you.)
Look at the Subnet Address: 255.255.255.248.
255.255.255.xxx: we can treat this like a Class C address and ignore the first 3 octets of the Address.
xxx.xxx.xxx.248: 5 extra bits are stolen away from the Host ID. 3 bits are left. This is our magical Delta number.
How many IDs are available per subnet? 8. (2^Delta)
How many can be assigned to computers? 6. (2^Delta - 2)
Let's count off some subnetted address blocks:
Since we are pretending this is a Class C Address, we'll start with 10.254.255.0.
10.254.255.0 - 10.254.255.7 (The first 8 IDs.)
10.254.255.8 - 10.254.255.15 (The next 8.)
10.254.255.16 - 10.254.255.23
Hmmm ... we're looking for information on 10.254.255.19
Well, that address falls within the subnet block of 10.254.255.16 through 10.254.255.23. The first address of the block indicates the start of the subnet. That would be 10.254.255.16. The last address of the block is the broadcast address. 10.254.255.23.
Ta Da! You now have your broadcast address for just that subnet.
My fingers hurt. Let me know if this helps, or if you need more clarification.
Crescent
MCP+I, MCSE, A+, N+, MOUS, MCT, CTT
Paper cert that! ;-) | |
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| Uhh, cough cough, I'm sick. :-)
But you presumed correct, I am on a MONSTER 5 day cramsession. 18 hours a day for five days. OOKS, that's 90 hours. :-0
Plan on taking the test this friday
Have read through Lammle and Exam Cram 5 times each and underlined. Playing with virtual routers and protocol analyzers. Starting on memorizing the glossaries. Then it's practice test time. That is 100% on all 12 Boson and all Techmindworks. Yikes!!
2 and 1/2 days left to go, see you friday.
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| certmadman,
Thats some cram schedule your doing. Good lord, that is some serious studying. I'm not awake for 18 hours a day, let alone studying for 18hrs. You are a "madman"!!
All I can say (with my jaw on the ground) is best of luck this Friday.
Sean34 | |
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| see you on friday if this is like the n+ and i+ you'll have nothing to worry about, and thanks guy that makes things a little clearer, thank alot you guys rock | |
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| Good luck!
18 hours a day for 5 days?!? You've got more dedication and stamina than I do. Wow!
I'll go and hang my head in shame now.
90 hours? Wow! |
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