01-21-2006 10:35 AM - edited 03-13-2019 11:54 AM
I am in the process of manually summarizing some routes within my network.
Which of the addresses below would be included in the manual route
summarization of 153.25.200.0/21?
A. 153.25.198.0
B. 153.25.206.0
C. 153.25.208.0
D. 153.25.224.0
Should not the answer be A and B ?
01-21-2006 04:59 PM
Hi,
153.25.200.0/21 will include the following /24s:
153.25.200.0/24
153.25.201.0/24
153.25.202.0/24
153.25.203.0/24
153.25.204.0/24
153.25.205.0/24
153.25.206.0/24
153.25.207.0/24
This is how you work it out:
- the mask corresponding to /21 is 255.255.248.0
- since the third octet of themask is non-255, this is the one of interest
- the third octet of the network 153.25.200.0 is 200.
Converting 200 to binary gives: 11001000
Converting 248 to binary gives: 11111000
Since the last three bits of the mask are zero, it means that the last three bits of the third octet of the network address can take on any values, which therefore gives the following:
11001000 = 200
11001001 = 201
11001010 = 202
11001011 = 203
11001100 = 204
11001101 = 205
11001110 = 206
11001111 = 207
Therefore, the correct answers are B and C.
Hope that helps - pls rate the post if it does.
Paresh.
01-21-2006 08:07 PM
I did mistake when I said the answer should be A, B,,,,,it should be only B.
The 21 leftmost bits must match. If we study the 3rd oct
153.25.200.0 10011001 00011001 11001000 00000000
153.25.198.0 10011001 00011001 11000110 00000000
153.25.206.0 10011001 00011001 11001110 00000000
153.25.208.0 10011001 00011001 11010000 00000000
153.25.224.0 10011001 00011001 11100000 00000000
01-22-2006 12:49 AM
Whoops..you are right...I had it all worked out correctly.. just did not check the options too closely.
Yes, the answer is B.
Regards,
Paresh
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