Question

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ip/subnets

By infinity1495 ·
Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
A. 192.168.1.8/29
B. 192.168.1.32/27
C. 192.168.1.40/29
D. 192.168.1.16/28
E. 192.168.1.48/29

and if anyone can explain how they got the answer i will appreciate that..thanks

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is that

by PurpleSkys In reply to ip/subnets

school work?

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Reponse To Answer

by infinity1495 In reply to is that

no its not, what is the difference anyway?

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Reponse To Answer

by PurpleSkys In reply to is that

considering that your first question was and you outright admitted it was homework, one can only assume that it is. It matters because if we do it FOR you, you will not learn anything from it. Once again, Google is your friend as are your textbooks.

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Reponse To Answer

by infinity1495 In reply to is that

again i know the answer but the book seems to disagree. so i was just checking to see maybe i was wrong

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Answer

by jedi_k9master In reply to ip/subnets

Answer is C: 192.168.1.40/29

.248 mask uses 5 bits (1111 1000).
.42 IP in binary is (0010 1010)

The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP ...
1111 1000 AND
0010 1010 equals
0010 1000 - which is .40

/24 is standard class C mask.
adding the 5 bits from the .248 mask gives /29
thus answer C

As an aside: the broadcast address would be the HIGHEST binary value without changing the output of an AND operation of the subnet mask and IP ...

0010 1000 - Base subnet
0010 1111 - (.47) Last three 1's is the highest value without changing the AND operation .. as in .. making it .48 would result in 0011 0000 which changes the outcome of the AND operation against the subnet mask)
Thus broadcast address is: .47
usable addresses: .47 - .40 = 7 .. less 2 = 5 usable addresses in that range
(take away 2 because the base subnet and broadcast address can't be used as end device IPs)

Hope that makes sense.
I'm admittedly quite rusty on my IP theory so if I've got anything wrong, please excuse me .. 9and others please feel free to correct me.

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Reponse To Answer

by infinity1495 In reply to Answer

Thanks for the answer, i understand how binary numbers work and all that but i still dont get why you are or where you are getting the 42 from? thanks

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Reponse To Answer

by Charles Bundy In reply to Answer

i still dont get why you are or where you are getting the 42 from This sounds like a Douglas Adams quip. :)

The .42 was just a valid IP in the range determined by your subnet mask. As stated above by [jedi] 29 bits are network and 3 are host. Your host range is .41 through .46 and .42 is a valid host IP. EG

11000000.10101000.00000001.00101010 192.168.1.42
11111111.11111111.11111111.11111000 255.255.255.248 or 29+3
11000000.10101000.00000001.00101000 192.168.1.40 (after AND op)

Make more sense?

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Reponse To Answer

by Charles Bundy In reply to Answer

ADDENDUM: You get six host IP's in this example. 3 bits = 8 possible combinations or the range 0-7. 8-2=6 = 41,42,43,44,45 & 46 that can be assigned to host devices.

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Reponse To Answer

by infinity1495 In reply to Answer

So even though the block size is 8 since 248 you dont use that information? i am sorry but still dont see why the host range is 41 thru 46, where are you getting that from? Thanks

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Reponse To Answer

by infinity1495 In reply to Answer

I see i think i got it! Thanks alot!

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