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Home > Archive > CCNP > February 2002 > 503 Question
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| Hello everyone~ Here comes a 503 question.
Q. Given the route summarization entry 192.168.16.0/20, how many class C addresses can be summarized?
a. 4
b. 8
c. 16
d. 20
e. 32
f. 64
The ans is b. (or the ans is wrong?)
Why c is wrong? Is it related to the reserved network ID and broadcast addresses? This kind of questions always disturb me. >_< | |
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|  West
Here's my attempt to explain why I think the correct answer is 'C'.
The prefix /20 equates to a subnet mask of 255.255.240.0. The subnetting operates on the third octet of the IP address 192.168.16.0. Using the simple maths, 256-240=16, the class C address block is in groups of 16. Therefore, the class C address ranges defined by /20 is,
192.168.16.0
192.168.17.0
192.168.18.0
''''''''''''
192.168.31.0
In other words, 16 class C addresses can be summarized with a /20 prefix.
The 503 study guide I am using has a worked example of 204.1.64.0/20, and the result is 16 class C.
This will not be the first time that the GIVEN answer to a question is wrong. Sometimes, I think that is DELIBERATE, by the author to make you think harder and KNOW your stuff. Try sending the author an e-mail with your correction.
Cheers
Hippo
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| cross36 2002-02-20, 8:48 am |
| Granted the answer does look like "C". Unless you are specifying the reserves. | |
| RTRGOD 2002-02-20, 11:58 am |
| C is correct for a 20 bit subnet mask.
B would be correct for a 21 bit subnet mask.
16 is the starting point in the third octect:
00010000 = 16
00010001 = 17
00010010 = 18
00010011 = 19
00010100 = 20
00010101 = 21
00010110 = 22
00010111 = 23
00011000 = 24
00011001 = 25
00011010 = 26
00011011 = 27
00011100 = 28
00011101 = 29
00011110 = 30
00011111 = 31
00100000 = 32...............
The subnet mask of 255.255.240.0 is used to
summarize 192.168.16.0. Another to say this is: Starting with a value of 16 in the third octect, find all values in which the first 20 bits are identical. Clearly 16-31, or 16 addresses share the same first 20 bits. | |
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| Thank you all very much~
I just doubt with my own answer. You give me confidence. I will trust myself next time. Of course, if i can't, i will trust you. ^_^ | |
| bhatok 2002-02-21, 1:34 am |
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I'm new to this stuff but from what i learned about subnetting, what Hippo said is not correct. With a mask of 255.255.240.0, the simple math 256-240 =16. I'm with you up to there. But from what I understand the subnets shouldn't be
16.0
17.0
18.0
...
31.0
I thought they should be:
16.0
32.0
48.0
64.0
80.0
96.0
....
224.0
Can somebody please explain Hippo's answer to me and explain why my answer is incorrect. Thanks guys. | |
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| hoho~ You both are correct. bhatok, you are talking about the subnetting. If the subnet mask using 20 bits, the subnets should be 16.0, 32.0, 48.0 ...
Hippo is talking about the address summarization. You can think like that:
Subnets,
192.168.16.0/24
192.168.17.0/24
192.168.18.0/24
...
192.168.31.0/24
can be summarized as 192.168.16.0/20. That's it. | |
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|  Thanks West.
This just goes to prove that at this stage of Cisco Certification, candidates NEED to have IP addressing and subnetting down to a art form. When topics like CIDR, VLSM and summarization are introduced, subnetting takes on a different dimension; and all this before you even get to OSPF and BGP.
Hippo
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