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CCNA 640-801 suppernetting question
19 years 1 month ago #10503
by seanjuan
CCNA 640-801 suppernetting question was created by seanjuan
Q. Which of the following /24 subnets can be supernetted intoa larger network.
A. 10.10.20.0/24 and 10.10.21.0/24 into 10.10.20.0/23
B. 10.10.21.0/24 and 10.10.22.0/24 into 10.10.21.0/23
C. 10.10.130.0/24,10.10.131.0/24, 10.10.132.0/24, and 10.10.133.0/24 into 10.10.130.0/22
D. 10.10.164.0/24, 10.10.165.0/24, 10.10.166.0/24, and 10.10.167.0/24 into 10.10.165.0/22
A. 10.10.20.0/24 and 10.10.21.0/24 into 10.10.20.0/23
B. 10.10.21.0/24 and 10.10.22.0/24 into 10.10.21.0/23
C. 10.10.130.0/24,10.10.131.0/24, 10.10.132.0/24, and 10.10.133.0/24 into 10.10.130.0/22
D. 10.10.164.0/24, 10.10.165.0/24, 10.10.166.0/24, and 10.10.167.0/24 into 10.10.165.0/22
19 years 1 month ago #10504
by seanjuan
Replied by seanjuan on topic Re: CCNA 640-801 suppernetting question
I feel the answer is A
19 years 1 month ago #10535
by jwj
-Jeremy-
Replied by jwj on topic Re: CCNA 640-801 suppernetting question
It's all the answers but "C". I'll quiz you now, why is it not answer "D"?
-Jeremy-
19 years 1 month ago #10578
by seanjuan
Replied by seanjuan on topic Re: CCNA 640-801 suppernetting question
the answer is not "D" because the supernetted address cannot be a odd address and since four networks are being supernetted the summarized address has to must be a multiple of four.
19 years 1 month ago #10579
by jwj
-Jeremy-
Replied by jwj on topic Re: CCNA 640-801 suppernetting question
Sorry for my very confusing post, I had a typo, I meant "C". I am not the best at the explanation of this, but here it goes...
I do this in binary because you'll end of messing things up otherwise.
Supernetted network is 10.10.130.0/22
Network bits 00001010.00001010.10000.....Host bits 10.00000000
The size of the host bits I get from /22.
So if you zero out the host portion to get your network
00001010.00001010.10000 00.00000000
You'll get 10.10.128.0. So then you can find first and last hosts.
00001010.00001010.10000 00.00000001 = 10.10.128.1
00001010.00001010.10000 11.11111110 = 10.10.131.254
Back into relation with the question. The networks
10.10.130.0/24 (10.10.130.1 - .254)
10.10.131.0/24 (10.10.131.1 - .254)
10.10.132.0/24 (10.10.132.1 - .254)
10.10.133.0/24 (10.10.133.1 - .254)
Would exceed 10.10.130.0/22 because as you can see, 10.10.131.254 is last useable address, far below 10.10.133.254, the last useable address for 10.10.133.0/24.
Doing all of this in binary is the only way to go.
I do this in binary because you'll end of messing things up otherwise.
Supernetted network is 10.10.130.0/22
Network bits 00001010.00001010.10000.....Host bits 10.00000000
The size of the host bits I get from /22.
So if you zero out the host portion to get your network
00001010.00001010.10000 00.00000000
You'll get 10.10.128.0. So then you can find first and last hosts.
00001010.00001010.10000 00.00000001 = 10.10.128.1
00001010.00001010.10000 11.11111110 = 10.10.131.254
Back into relation with the question. The networks
10.10.130.0/24 (10.10.130.1 - .254)
10.10.131.0/24 (10.10.131.1 - .254)
10.10.132.0/24 (10.10.132.1 - .254)
10.10.133.0/24 (10.10.133.1 - .254)
Would exceed 10.10.130.0/22 because as you can see, 10.10.131.254 is last useable address, far below 10.10.133.254, the last useable address for 10.10.133.0/24.
Doing all of this in binary is the only way to go.
-Jeremy-
19 years 1 month ago #10610
by seanjuan
Replied by seanjuan on topic Re: CCNA 640-801 suppernetting question
I have never seen it broken down that way. Every Cybex, Cisco press, and Prep Logic matarial I have read does not go into detail about how to supernet, Thanks. I'll give it a try!!!
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