| Author |
subneting explanation
|
|
| cm2gj 2002-10-13, 12:10 am |
| You are creating a DHCP scope for your 192.168.1.32/28 subnet. The subnet consists of Windows 2000, Windows 98, and Windows 95 computers.
You have two UNIX computers on this subnet that will be assigned the two highest available static IP addresses. The subnet's default gateway will be assigned the lowest available IP address on the subnet.
Which scope should you create on your DHCP server?
a- 192.168.1.33 - 192.168.1.45
b- 192.168.1.34 - 192.168.1.61
c- 192.168.1.34 - 192.168.1.46
d- 192.168.1.34 - 192.168.1.44
i know the answer but i need to make SEVERAL calculus to resolve simple issues. what is the most easy way or give me some explanation to understand more easy this easy subneting questions...
 | |
| Slinky 2002-10-13, 12:35 am |
| Answer should be D. The host range for the 192.168.1.32 subnet is 192.168.1.33-192.168.1.46. So the router will be assigned the lowest IP which is 192.168.1.33, and the UNIX computers will be assigned the 2 highest which will be 192.168.1.45 and 192.168.1.46. So this leaves 192.168.1.34-192.168.1.44. Make sense? | |
| cm2gj 2002-10-13, 12:37 am |
| quote:
Originally posted by Slinky
Answer should be D. The host range for the 192.168.1.32 subnet is 192.168.1.33-192.168.1.46. So the router will be assigned the lowest IP which is 192.168.1.33, and the UNIX computers will be assigned the 2 highest which will be 192.168.1.45 and 192.168.1.46. So this leaves 192.168.1.34-192.168.1.44. Make sense?
yes.
have sense.
and is the correct answer.
my doubt is regarding how to calculate in a fast way all the ranges when i have the subnets.......... i don`t know why i become confused with some subneting questions... some are extremely easy for me and another ones that appear super easy are complex for my brain!!!  | |
| Slinky 2002-10-13, 12:52 am |
| That's actually a very basic subnetting question. The only thing thats different is that you have to get rid of the ones that will be used and find out whats left over. | |
|
| quote:
Originally posted by Slinky
That's actually a very basic subnetting question. The only thing thats different is that you have to get rid of the ones that will be used and find out whats left over.
i donŽt have problems getting raid of the ip for the gateway and the ips for the server..... but iŽbeen lost with this kind of questions. is very basic but i lost with this kind..... | |
| jeff_j_black 2002-10-13, 8:13 am |
| /28 means that the significant value you are working with is 16.
code:
128 64 32 16 8 4 2 1
/25 /26 /27 /28 /29 /30 /31 /32
Your address is x.x.x.32
sixteen addresses in the range inclusive of x.x.x.32 = x.x.x.32 - x.x.x.32+15
or x.x.x.32 to x.x.x.47
you cant use the .32 or .47 addresses for hosts so your router or 'default gateway' would get x.x.x.33 and your UNIX boxes would get x.x.x.45-46, leaving you 192.168.1.34 to 192.168.1.44 . | |
| Zaraspook 2002-10-13, 10:37 am |
| quote:
Originally posted by jeff_j_black
/28 means that the significant value you are working with is 16.
code:
128 64 32 16 8 4 2 1
/25 /26 /27 /28 /29 /30 /31 /32
Your address is x.x.x.32
sixteen addresses in the range inclusive of x.x.x.32 = x.x.x.32 - x.x.x.32+15
or x.x.x.32 to x.x.x.47
you cant use the .32 or .47 addresses for hosts so your router or 'default gateway' would get x.x.x.33 and your UNIX boxes would get x.x.x.45-46, leaving you 192.168.1.34 to 192.168.1.44 .
Fantastic explanation and formatting Jeff! | |
| jeff_j_black 2002-10-13, 11:30 am |
| Thanks, CCNA hangover. I started this method with 216, but can work it in my head since CCNA, just hope no one sees me counting my fingers!!! | |
| Slinky 2002-10-13, 11:34 am |
| quote:
Originally posted by Slinky
That's actually a very basic subnetting question.
I wasn't insulting your intelligence as it seems. I don't think you thought I was, but I just wanted to make sure. You said that you already had subnetting down, so I just thought that maybe you were missing something and making it more difficult than it really was. So if you thought that or anyone else did, I apologize.  | |
| Slinky 2002-10-13, 11:38 am |
| quote:
Originally posted by jeff_j_black
/28 means that the significant value you are working with is 16.
What do you mean by this? | |
| jeff_j_black 2002-10-13, 12:15 pm |
| /28 corresponds with the bit in the mask that has a value of 16, so all of your nets are going to be at intervals of 16.
x.x.x32-47 for example.
Thats what the chart is for. If you just eliminate the full classes from the cidr then you know the address range of each subnet is the range. cidr = /28 - 24 = /4 or the fourth bit from the left or the bit value 16.
In this case we're dividing a Class C network beginning at 32 counting 16 places including 32 = 32 + 15 = 47. All zeroes address is network id, all ones address is broadcast address.
Also by identifying the significant bit you can know immediately how many host addresses are available: 16 - 2 = 14 | |
| Slinky 2002-10-13, 12:29 pm |
| Ok, I never made that correlation. I always subtracted 256 from the subnet mask to find the increment. In that case it would be 256 - 240 = 16. Result is the same, but your way is a little easier. | |
| jeff_j_black 2002-10-13, 12:39 pm |
| Hey, you say tomato, I say tomatoe. What works for you is best, I like this way for making a quick chart in the test booth. | |
|
| quote:
Originally posted by jeff_j_black
Hey, you say tomato, I say tomatoe. What works for you is best, I like this way for making a quick chart in the test booth.
this correlation is the thing i had missing. thanks again to everyone!!!!! |
|
|
|