got a question on my homework that i'm having trouble with. i have to write a valid subnet mask for 10.x.y.z with 4500 subnets.. now the way we use to find it out is by using binary but i can't fathom using binary to find 4500 subnets.. we just started so i might also be stupid (the probable one) any help would be appreciated!
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subnet mask question
- Thread starter shyrjiin
- Start date
Some hints:
Start at the far right side; one full eight-bit octet will get you 256, right? Two octets (16 bits) will get you 65536. For every bit you move to the left, the number will double. Every bit you move to the right will half the number.
Good Luck
Scott
Start at the far right side; one full eight-bit octet will get you 256, right? Two octets (16 bits) will get you 65536. For every bit you move to the left, the number will double. Every bit you move to the right will half the number.
Good Luck
Scott
here's another one :/.. i believe this one requires a short explanation and please keep in mind i've only had 2 hours of lecture time on this subject...
How many people are on a subnet with the following subnet mask?
255.255.248.1 ________
255.255.249.0 ________
255.255.250.0 ________
255.255.251.1 ________
255.255.251.0 ________
255.255.252.0 ________
How many people are on a subnet with the following subnet mask?
255.255.248.1 ________
255.255.249.0 ________
255.255.250.0 ________
255.255.251.1 ________
255.255.251.0 ________
255.255.252.0 ________
Buddha Bart
Diamond Member
http://www.learntosubnet.com/
Its a series of four or five videos that present to you exactly how subnetting works. Its a lot of video... 90 minutes in all iirc, but it does a very good job.
bart
Its a series of four or five videos that present to you exactly how subnetting works. Its a lot of video... 90 minutes in all iirc, but it does a very good job.
bart
10.0.0.0/21 or 255.255.248.0
It's easy.
10.0.0.0 is a class A so the default mask is 255.0.0.0
So if you want to subnet you have to "borrow bits" from left to right without skipping.
You also have to start in the 2nd octet because the first one is already being used.
So...
12 bits = 4095 subnets
13 bits = 8191 subnets
12 isn't enough and 13 is way to damn many but there's nothing you can do about the wasted subnets. At least not on the CCNA level, but VLSM is a whole 'nother topic.
OK, so you have to borrow 13 bits starting at the first available bit after the subnet mask, which is the 2nd octet, and once you are done borrowing (start using zeros again) you use them till the end.
So it looks like this:
11111111 . 11111111 . 11111000 . 00000000
If you add ALL the ones, that's 21, which is where the /21 comes from.
Translate it back into decimal:
1st octet = 128+64+32+16+8+4+2+1 = 255
2nd octet = 128+64+32+16+8+4+2+1 = 255
3rd octet = 128+64+32+16+8+0+0+0 = 248
4th octet = 0+0+0+0+0+0+0+0 = 0
or... 255.255.248.0
It's easy.
10.0.0.0 is a class A so the default mask is 255.0.0.0
So if you want to subnet you have to "borrow bits" from left to right without skipping.
You also have to start in the 2nd octet because the first one is already being used.
So...
12 bits = 4095 subnets
13 bits = 8191 subnets
12 isn't enough and 13 is way to damn many but there's nothing you can do about the wasted subnets. At least not on the CCNA level, but VLSM is a whole 'nother topic.
OK, so you have to borrow 13 bits starting at the first available bit after the subnet mask, which is the 2nd octet, and once you are done borrowing (start using zeros again) you use them till the end.
So it looks like this:
11111111 . 11111111 . 11111000 . 00000000
If you add ALL the ones, that's 21, which is where the /21 comes from.
Translate it back into decimal:
1st octet = 128+64+32+16+8+4+2+1 = 255
2nd octet = 128+64+32+16+8+4+2+1 = 255
3rd octet = 128+64+32+16+8+0+0+0 = 248
4th octet = 0+0+0+0+0+0+0+0 = 0
or... 255.255.248.0