summarization of 153.25.200.0/21 and correct answer

[zillah2004]

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 ?

[pkhatri]

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.

[zillah2004]

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

[pkhatri]

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