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Author Subnet Masking
crvillani

2001-04-29, 4:55 pm

Taking the exam on Tuesday and I am wondering if questions regarding subnet masking such as the following have appeared on anyone's test. Through the course of my studies I have been told I would need only a basic understanding of subnet masks. (Ie. Network and Host ID's) But as of late I have come across a few questions similar to the one below which requires you to actually figure out a range of IP addresses. Any help would be a appreciated.


You are assigning a range of IP addresses to hosts on your Windows 2000 network. You would like to use the network ID of 117.72.32.0/20. What is the available range of IP addresses that you can use given the network ID specified above?
dentonb2000

2001-04-29, 5:26 pm

belay my last....
Yeti-GBR1

2001-04-29, 5:36 pm

Dam missed that one...this time you got there 1st
Yeti-GBR1

2001-04-29, 6:02 pm

Ok if the bits used are 20 then the subnet mask is 255.255.240.0

this will give you 4096 subnets with 4094 hosts on each.

dentonb2000 go check your maths
Yeti-GBR1

2001-04-29, 6:08 pm

Ok if the bits used are 20 then the subnet mask is 255.255.240.0

this will give you 4096 subnets with 4094 hosts on each.

Here is the ans

Thus the range is 117.72.32.0 - 117.72.47.254 witha subnet mask of 255.255.240.0

dentonb2000 go check your maths
dentonb2000

2001-04-29, 6:11 pm

/20 =
11111111 11111111 1111|0000 00000000
(2*12)-2= 4094
(2*12)-2= 4094

4096/256 = 16

117.72.32.0 - 117.72.47.254
little math error on my part
rctechy

2001-04-29, 10:45 pm

Hate to ruin your party but---

117.72.32.0 is the subnet

valid hosts are 117.72.32.1 through

117.72.47.254

If you want a great tool for calculating IP addresses, go to Boson.com

Under free stuff/utilities they have an IP subnetter.

Works for me
regards
rc
dentonb2000

2001-04-30, 8:20 am

true, was a typo on my part...Staying up too late has caught up to me and sometimes my long division gets the best of me. I have never used an IP calc before...windows calculator works fine for me (when there is no user error involved).

Thanks for clarifying.
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