| Author |
Spid's Thu (2/13) Win2k Pro QoD
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| A little simple subnetting never hurt anyone 
You are the network administrator of your companies Windows 2000 network. The network currently supports about 10,000 hosts comprised of Windows 2000 Servers, HP laserjet printers and Windows 2000 Professional clients.
Your company has been assigned an IP address of 136.18.32.0. You expect the network to increase in size to approximately 20,000 hosts within the next couple of years.
To accomodate for this expanison, you'd like to setup the network with 10 subnets containing 2000 hosts/subnet.
How should you configure the IP addressing structure to allow for this configuration?
A. 136.18.32.0/20
B. 136.18.32.0/22
C. 136.18.32.0/24
D. 136.18.32.0/26
Good luck and see you tomorrow for the answer!! | |
| hairy51 2003-02-13, 12:37 pm |
| Not very good at subnet q's, but i would hazard a guess at A.
Hopefully someone will come along and explain!!! | |
| deehitchcock 2003-02-13, 1:54 pm |
| Taking a stab at it....C? | |
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| Slinky 2003-02-13, 2:39 pm |
| It's A. That will give you a block of 16 IPs ranging from those listed below, and 4094 hosts/network.
136.18.32.0
136.18.33.0
136.18.34.0
...........
136.18.48.0 | |
| Slinky 2003-02-13, 2:43 pm |
| quote:
Originally posted by tharg
Isn't 10 a strange number of subnets to have?
Am going with B but I think it should be E. /21
http://www.telusplanet.net/public/sparkman/netcalc.htm
You're absolutely right Tharg, using /21 will give you 2,046 hosts/network, but you went the wrong way by picking /22. That will cut the number in half to 1022 hosts/network which doesn't meet the requirement of 2000. So you need to goto /20 which will give you 4096 hosts which is 2x what's needed. Make sense? | |
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| quote:
Originally posted by tharg
Isn't 10 a strange number of subnets to have?
Am going with B but I think it should be E. /21
http://www.telusplanet.net/public/sparkman/netcalc.htm
One of the answers listed satisfies the requirement. /21 would work as well, but it's not one of the answers | |
| foxmedia 2003-02-13, 2:58 pm |
| Two weeks until test time. If this question were to have come up on my exam I would have chosen............A | |
| WPFossil 2003-02-13, 3:43 pm |
| Have to admit that subnetting is a weak point for me. Any good links to a decent explanation of the whole mess?
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| hairy51 2003-02-13, 4:09 pm |
| i'm not sure that my way of working it out was very accurate, but here is how i worked it out, and if anyone can correct me/give me some advice it would be much appreciated:-
converted 136.18.32.0 to bin.
1001000.00010010.00100000.00000000
started with /20 so that =
11111111.11111111.11110000.00000000
255.255.255.240
which leaves 12 bits for host ID(?)
i then tried to work out how many host addresses this would give me, and i originally worked it out as 3825, which i don't think is right, but it left me with the best answer for the question.
What is the easiest way to work out how many hosts a given subnet mask could have? | |
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| gcw123 2003-02-13, 10:27 pm |
| Ace Ace Ace Ace Ace ....... | |
| cramersaunders 2003-02-14, 1:49 am |
| A | |
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| Excellent link Slinky!  Thanks!! | |
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| adam salam 2003-02-14, 8:52 am |
| This Q for me 
"A"
20 bits for sub netting
136.18.32.0 is "class B" means start play from third octet
8 + 8 + "4"
2^4= 16 subnet, if you like 16-2 (eliminate all zeros and all ones subnets)= 14 net subnets.
The rest of bits 2^12= 4096 hosts
Unfortunately, there will be no sub netting Q's in the exam, will be a piece of cake
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| MartyMcFly 2003-02-14, 9:28 am |
| quote:
Originally posted by gcw123
Ace Ace Ace Ace Ace .......
nice poker hand, but i smell a rat | |
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| quote:
Originally posted by adam salam
136.18.32.0 is "class B" means start play from third octet
I'd assume this question is is dealing with RFC1812 - CIDR, rather than the old rfc 950 (Class A,B,C etc.) | |
| adam salam 2003-02-14, 9:54 am |
| quote:
Originally posted by tharg
I'd assume this question is is dealing with RFC1812 - CIDR, rather than the old rfc 950 (Class A,B,C etc.)
Sorry, didn't get you | |
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| quote:
Originally posted by MartyMcFly
nice poker hand, but i smell a rat
I don't want this to be taken the wrong way but we like to try and keep the technical/exam question discussion threads on track in the 210 forum. You guys can keep your confusion tactics and changing subjects of threads to postings in the General forum.
Thanks | |
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| quote:
Originally posted by tharg
I'd assume this question is is dealing with RFC1812 - CIDR, rather than the old rfc 950 (Class A,B,C etc.)
Sorry, I should have specified that this is not a CIDR question. | |
| rama1179 2003-02-14, 10:17 am |
| all the answers given here i think are wrong
136.18.32.0 ip address belongs to class b
subnet mask 255.255.0.0
2^11=2048
eliminating the first and last addresses we get 2046 hosts per subnet
32-11=21
since 2000 hosts will be accomplished by this number
so the answer must be 136.18.32.0/21
which is not given as choices | |
| MartyMcFly 2003-02-14, 10:27 am |
| quote:
Originally posted by Spid
I don't want this to be taken the wrong way but we like to try and keep the technical/exam question discussion threads on track in the 210 forum. You guys can keep your confusion tactics and changing subjects of threads to postings in the General forum.
Thanks
 | |
| Slinky 2003-02-14, 10:27 am |
| quote:
Originally posted by rama1179
so the answer must be 136.18.32.0/21
which is not given as choices
Once again, /21 is definately the better option however its not an answer. Like I said before /20 will cover the needed 2000 hosts and then some.
Spid just give us the freaking answer.  | |
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| quote:
Originally posted by rama1179
all the answers given here i think are wrong
136.18.32.0 ip address belongs to class b
subnet mask 255.255.0.0
2^11=2048
eliminating the first and last addresses we get 2046 hosts per subnet
32-11=21
since 2000 hosts will be accomplished by this number
so the answer must be 136.18.32.0/21
which is not given as choices
No the answer does not have to be /21.
As stated before, /21 would work, but it is not one of the answers provided in this particular answer set.
Can you please explain why /20 is not a viable answer when it gives you (2^4)-2 subnets=(16-2=14) and (2^12)-2 hosts=(4096-2=4094).
Both conditions are satisfied.
10 subnets are satisfied by (2^4)-2 = 14
2000 hosts are satisfied by (2^12)-2 = 4094 | |
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| quote:
Originally posted by Spid
A little simple subnetting never hurt anyone
You are the network administrator of your companies Windows 2000 network. The network currently supports about 10,000 hosts comprised of Windows 2000 Servers, HP laserjet printers and Windows 2000 Professional clients.
Your company has been assigned an IP address of 136.18.32.0. You expect the network to increase in size to approximately 20,000 hosts within the next couple of years.
To accomodate for this expanison, you'd like to setup the network with 10 subnets containing 2000 hosts/subnet.
How should you configure the IP addressing structure to allow for this configuration?
A. 136.18.32.0/20
B. 136.18.32.0/22
C. 136.18.32.0/24
D. 136.18.32.0/26
Good luck and see you tomorrow for the answer!!
And the answer is.....A
136.18.32.0/20 will satisfy both subnet and host requirements in this question. | |
| tharg 2003-02-14, 10:46 am |
| Great Q Spid! Can we have another one of these next week? | |
| foxmedia 2003-02-14, 10:49 am |
| quote:
Originally posted by tharg
Great Q Spid! Can we have another one of these next week?
Agree tharg...there is a lot to learn on this subject. Maybe at least one a week? Good question spid. Thanks.
Fox | |
| adam salam 2003-02-14, 10:58 am |
| quote:
Originally posted by tharg
/8 is like the old Class A
/16 is like the old Class 24
etc.
I think we're being asked to figure
/whatever
...rather than saying, yeah, that's Class B.
RFC 950 (1985 - the old way)
http://www.faqs.org/rfcs/rfc950.html
RFC 1812 (1995 - the new way)
http://www.faqs.org/rfcs/rfc1812.html
Even the links give an interesting info about sunetting I couldn't go through them.
but I want to explain some thing here, when we say class a, b or c what we main is to know what you have to do with the given bits.
in this Q for example: if you don't know the class how you know the specific bits to use for calculation.
what I mean here we have 20 bits for calculating, but we calculate only the bits resides on the third octet only we eliminate the first and second octets bits 20-(8+8)=4 bits we can work with to find the possible number of subnetting 2^"4 bits"= 16 subnets
say if the range of IP addresses given here are 10.18.32.0/20
we here just change the class from b to A here the calculation will be deferent 2^12= nuber of subnets
instead of 2^2= number of subnets, hen we calculate subnets in class b
see the deference when qwe change the class.
hope you got me | |
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| quote:
Originally posted by tharg
Great Q Spid! Can we have another one of these next week?
If you guys want subnetting type questions, I'll put some together. I know that they are a little outside the scope of the material covered in the 210 exam, but if you want them, I'll try my best to whip some up. Maybe something like 1 a week.
I'll try to keep them straight forward and fair as well, focusing on the core concepts of subnetting instead of making them overly difficult and/or tricky.
Thanks | |
| Tsakali 2003-02-15, 3:41 am |
| quote:
Originally posted by WPFossil
Have to admit that subnetting is a weak point for me. Any good links to a decent explanation of the whole mess?
good luck
http://www.mcsefreak.com/subnetting.htm | |
| rama1179 2003-02-15, 10:48 pm |
| quote:
Originally posted by Spid
No the answer does not have to be /21.
As stated before, /21 would work, but it is not one of the answers provided in this particular answer set.
Can you please explain why /20 is not a viable answer when it gives you (2^4)-2 subnets=(16-2=14) and (2^12)-2 hosts=(4096-2=4094).
Both conditions are satisfied.
10 subnets are satisfied by (2^4)-2 = 14
2000 hosts are satisfied by (2^12)-2 = 4094
EXPLAINATION
10 subnets needed and 2000 hosts in each subnet
therefore
1)ip address 136.18.32.0 subnet mask 255.255.0.0
if we consider option a as the answer
136.18.32.0/20
20 bits for host-id and remaining 32-20=12 for network-id
2^12-2=4094 which means 4094 hosts are possible in each subnet
but we want 2000 hosts in each subnet remaining ip addresses
4094-2000=2094 are gone waste which we don't want in
pratically
2)explantion for the answer which i had given
136.18.32.0/21
21 bits are for host-id and remininig 32-21=11 for network-id
2^11-2=2046 which means 2046 hosts are possible in each subnet
we want 2000 hosts in each subnet 2046-2000=46 only 46 ip addresses
are gone waste which is better than option 136.18.32.0/20 | |
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| quote:
Originally posted by rama1179
EXPLAINATION
10 subnets needed and 2000 hosts in each subnet
therefore
1)ip address 136.18.32.0 subnet mask 255.255.0.0
if we consider option a as the answer
136.18.32.0/20
20 bits for host-id and remaining 32-20=12 for network-id
2^12-2=4094 which means 4094 hosts are possible in each subnet
but we want 2000 hosts in each subnet remaining ip addresses
4094-2000=2094 are gone waste which we don't want in
pratically
2)explantion for the answer which i had given
136.18.32.0/21
21 bits are for host-id and remininig 32-21=11 for network-id
2^11-2=2046 which means 2046 hosts are possible in each subnet
we want 2000 hosts in each subnet 2046-2000=46 only 46 ip addresses
are gone waste which is better than option 136.18.32.0/20
Grrrr....For the LAST time, /21 was NOT one if the answers given to choose from.
/21 is a better answer in your opinion. What do you mean we are wasting hosts with a /20? It's not like you are charged a cost for each host on the network.
Plain and simple:
Does /20 provide at least 10 subnets? - Yes.
Does /20 provide for at least 2000 host per subnet? - Yes.
Will /21 meet the same criteria? - Yes.
But, Oh wait, /21 is not given as an answer, I guess I'll have to throw my opinions grudgingly aside and go with /20 for this question.
What is so difficult about this line of reasoning to grasp??
Now if /20 and /21 were given as possible selections in the answer set provided, we could have a nice debate about efficiency and practicality. But we don't, so let's drop the /21 arguements. | |
| rama1179 2003-02-17, 9:24 am |
| quote:
Originally posted by Spid
If you guys want subnetting type questions, I'll put some together. I know that they are a little outside the scope of the material covered in the 210 exam, but if you want them, I'll try my best to whip some up. Maybe something like 1 a week.
I'll try to keep them straight forward and fair as well, focusing on the core concepts of subnetting instead of making them overly difficult and/or tricky.
Thanks
it will be very good .
try to find good tricky
question which will confuse us |
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