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Author Topic: Whats the answer? Place your betts!  (Read 1526 times)
dday
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« on: December 06, 2001, 08:29:21 AM »

A collegue at work came to me with this question any ideas?

Your network has 1900 hosts and requires internet connectivity, your network is not routed, xcept to the internet.
You have been assigned the follwoing 8 ip address
192.243.32.0/24
192.243.33.0/24
192.243.34.0/24
192.243.35.0/24
192.243.36.0/24
192.243.37.0/24
192.243.38.0/24
192.243.39.0/24

You want to minimise the complexity of your routing table which sunbnet mask should you use?

A 255.255.252.0
B 255.255.248.0
C 255.255.255.248
D 255.255.240.0

The answer he's got is B but I think it might be D.  A subnet mask of 240 gives 14 (valid) subnets.  Something like:-

192.243.16.0 to 192.243.31.254
192.243.32.0 to 192.243.47.254
192.243.48.0 to 192.243.61.254
192.243.62.0 to 192.243.77.254
192.243.78.0 to 192.243.93.254
192.243.94.0 to 192.243.109.254
192.243.110.0 to 192.243.125.254
192.243.126.0 to 192.243.141.254
192.243.142.0 to 192.243.157.254
192.243.158.0 to 192.243.173.254
192.243.174.0 to 192.243.189.254
192.243.190.0 to 192.243.205.254
192.243.206.0 to 192.243.221.254
192.243.222.0 to 192.243.238.254


A subnet mask of 248 gives 30 which I guess would work ok too but with more subnets and less hosts which for some reason I think is a bad thing, Wouldnt it give more entries in the routing table?


Am I right or sheeeite?
My TCP/IP is a bit weak.

If your viewing this and understand IP addressing (I dont!) then dont be lazy just reply!!

Cheers
DD:p
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Paisleyskye
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« Reply #1 on: December 06, 2001, 12:26:22 PM »

I believe the answer is B.
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Peace Out!
Paisleyskye
dday
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« Reply #2 on: December 06, 2001, 01:22:28 PM »

Thats 2 for B, I still think it could be D.  Any explaination for why it could be B?  Why would you want more subnets then necessary?  D gives enough subnets and more hosts!  (dosen't it?)  Come on this is annoying me, dont peek in here and bug out if you know the answer tell me and put me out of my misery!

Laters
DDay:cool:
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StewartH
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« Reply #3 on: December 06, 2001, 01:30:10 PM »

If I'm reading it right it stands to reason that B would create less entries in the routing table as there would be less subnets to route to.
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sparky74
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« Reply #4 on: December 06, 2001, 01:38:34 PM »

The answer is B. You have to look at the distance between .32 - .39 the closest is 8 which would take your IP's to .39.254 with .255 being the broadcast for that network. Take 256-248 = 8 then you see that the range of the subnet becomes:

.32.0 - .39.255
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csiszerd
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« Reply #5 on: December 06, 2001, 01:51:23 PM »

Binary
Answers
A.11111111.11111111.11111100.00000000
B.11111111.11111111.11111000.00000000
C.11111111.11111111.11111111.11111000
D.11111111.11111111.11110000.00000000

the 8 IP's
11000000.11110011.00100000.00000000
11000000.11110011.00100001.00000000
11000000.11110011.00100010.00000000
11000000.11110011.00100011.00000000
11000000.11110011.00100100.00000000
11000000.11110011.00100101.00000000
11000000.11110011.00100110.00000000
11000000.11110011.00100111.00000000

"your network is not routed"

A would require 2 segments
B would require 0 routers
C would require 7+ segments
D would require 0 routers

You want to minimize the complexity of your routing table which sunbnet mask should you use?
B and C requires 0 routing

DC
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dday
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« Reply #6 on: December 06, 2001, 02:41:50 PM »

Sooo Everyone things B and I think (thought D)  Looks like Im wrong then.  Clearly Im hazy on this.  Ive printed out your responses and will give them some thought.  Thanks for your time guys its people like you who keep this site as great as it is!
Cheers,
I hope I get this ip stuff clear by the 20th (the big day!)
cheers
DD.:cool:
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dday
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« Reply #7 on: December 06, 2001, 03:03:43 PM »

Right, Ive studied the replies, I understand what sparky says the range for .248 will be 8 which does map to the 32 - 39. Cheers Sparky.  So that kinda makes sense.  Dont really understand how csiszerd worked out the amount of routing they would take though. Any ideas?

Thanks for you patience!

David
 :confused:
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rosco9473
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« Reply #8 on: December 10, 2001, 11:37:12 PM »

Look at it from another point of view. Everyone is looking at it from the perspective of the number of the networks needed. That really is irrelevant in this scenario since you are combining subnets instead of separating them. The important thing is the number of hosts that you need.

Remember the formula for the number of hosts.
hosts=2^X-2 (with the X representing the # of bytes in the host portion of the address)
You currently have 8 subnets of 256 addresses (I know its only 254 valid addresses) for 2048 addresses (2046 valid addresses once they are combined). When applying this to the formula:
2046=2^X-2
X=11
Meaning the network portion should consist of 21 bytes or a subnet mask of 255.255.248.0.
This subnet mask would produce a valid ip address range of 192.243.32.1 - 192.243.39.254.
You are right, a subnet mask of 255.255.240.0 would combine the subnets into one subnet. However this would result in a valid ip address range of 192.243.32.1 ΤΗτ 192.243.47.254. Which means that part of the internet would not be accessible from your network.

Hope this helps.
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zEpS
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« Reply #9 on: December 11, 2001, 09:44:33 AM »

its 100 percent B
its 100 percent B
its 100 percent B
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