Summarization. 7th ed. ch. 5

[stud]

For the life of me I must not be understanding summarization. Could somebody please explain the answers to questions 2, 4, and 5 on the written lab on page 198 in the 7th edition for me? I can figure out the rest of the answers in that written lab, but those ones there have me quite perplexed.

I get the block sizes, but the mask is evading me (particularly 2 and 5), even though once you get the block size it's usually pretty easy to get the mask. I'm assuming 4 the answer is supposed to be 203.168, not 192.168. Is this right?

summarization
2. 172.148.0.0/13 - 172.156.0.0/13 ANSWER IN BOOK - 172.144.0.0/16
4. 203.168.6.0/24 - 203.168.60.0/24 ANSWER IN BOOK- 192.168.96.0 255.255.240.0
5. 66.66.0.0 - 66.66.15.0 ANSWER IN BOOK- 66.66.0.0/16

[Fuzz]

Are the questions asking you to summarise the range of subnets? If so, here's the block sizes you'll need for each one.

2. Block size 16
4. Block size 64
5. Block size 16

The subnet mask is 255 - block size in the interesting octet. Remember the summary address must be the block starting point, which is always a multiple of the block size. For example, a block of 16 can start at 0, 16, 32, 48 etc but not 8, 12, 24 etc.

Does this help at all?

[stud]

Yes, you are to determine the summary address and the mask used to summarize the subnets.

I understand that much, but for questions 2:

172.148.0.0/13 - 172.156.0.0/13
I can see the block size is 16, starting at 172.144.0.0 and that would be advertised as the summary address, but how is the mask /16? In my mind it's should be a /12. Right? 255.240.0.0, not 255.255.0.0

And question 5:
66.66.0.0 - 66.66.15.0
Block size of 16, 66.66.0.0 summary address. Again, shouldn't the mask be 255.255.240.0? Why is it 255.255.0.0? What am I missing here?

[Fuzz]

You're not missing anything, a block size of 16 uses a .240 mask.

[joewoody]

I too have been quite perplexed by this lab. After working through this lab on several occasions both checking and double checking, I believe that:

2. 172.144.0.0 /12 (NOT /16)
4. 203.168.0.0 /18 (255.255.192.0)
5. 66.66.0.0 /20 (NOT /16)

I am also of the conclusion that (and I may possibly be wrong):

6. 192.168.0.0 /17 NOT 192.168.0.0 /25
7. 172.16.0.0 255.255.248.0 NOT 172.16.1.0 255.255.248.0

I, too am seeking clarification on this subject matter, since I have worked through all the problems using the same methodology and I have no qualms with any of the other answers. Maybe, I too am missing something!

Regards,

JW

[Big Evil]

Fuzz is the subnet king here and if i explain this wrong i am sorry. I know/why but explain math has never been a strength (unless it is beer tokens :))

Take down the eight binary numbers of the octet that is different in the examples it is the third octet and this is the only part we need to work with, i have give the decimal at the end also.

128 64 32 16 8 4 2 1

1001|0100 - 148
|
1001|1100 - 156

Draw a line through where the bits are common, here this would be the forth bit Block 16, now look at the eigh binary numbers, you see that the number is 16 (Block 16)

00|000110 - 6
|
00|111100 - 60

Draw a line through where the bits are common, here this would be the second bit, now look at the eigh binary numbers, you see that the number is 64 (Block 64)


0000|0000 - 0
|
0000|1111 - 15

Same as the first example.

HTH.

[dieshonest]

Hi there.

Could someone please clarify these exercises, 2, 4 and 5 are driving me insane.(summarization, 7th edition)

"For each of the following sets of networks, determine the summary address and the mask to be used that will summarize the subnets."

2) 172.148.0.0/13 through 172.156.0.0/13
Answer given on the book: 172.144.0.0/16

4) 203.168.6.0/24 and 203.168.60.0/24
Answer given on the book: 192.168.96.0 255.255.240.0

5) 66.66.0.0 through 66.66.15.0
Answer given on the book: 66.66.0.0/16

I think the answers are wrong, and they are really confusing me.

Thanks in advanced

[Guinnesshead]

198Text correction: Errors In Written Lab 5
Question 2. should read 172.144.0.0 through 172.159.0.0. The answer should read 172.144.0.0 255.240.0.0.

Question 4. should read 192.168.96.0 through 192.168.111.0.

The answer to 5. should read 66.66.0.0 255.255.240.0.

The answer to 6. should read 192.168.0.0/17

[lammle]

The errata is also in the Announcement section of the Forum.
These errors have been identified and were fixed after the first printing.
Thank you,
Todd

[booya1]

Thanks for clarifying errata for questions 2,4,5 and 6; but no one has talked about question #7...

the book answer is 172.16.1.0

I think this is another typo for the 3rd octet.

Shouldn't the answer be 172.16.0.0 since the range is in the zero subnet?

Could some please clarify... Thanks

[lammle]

yes, that was discussed and fixed in the 2nd printing along with the others.
172.16.0.0 is the correct answer.

[pcameron1]

Fuzz is the subnet king here and if i explain this wrong i am sorry. I know/why but explain math has never been a strength (unless it is beer tokens :))

Take down the eight binary numbers of the octet that is different in the examples it is the third octet and this is the only part we need to work with, i have give the decimal at the end also.

128 64 32 16 8 4 2 1

1001|0100 - 148
|
1001|1100 - 156

Draw a line through where the bits are common, here this would be the forth bit Block 16, now look at the eigh binary numbers, you see that the number is 16 (Block 16)

00|000110 - 6
|
00|111100 - 60

Draw a line through where the bits are common, here this would be the second bit, now look at the eigh binary numbers, you see that the number is 64 (Block 64)


0000|0000 - 0
|
0000|1111 - 15

Same as the first example.

HTH.

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

This is a really good explanation - I'm learning the process exactly like this too.I still need to practice a lot, but this method makes it much easier..Thanks Big Evil!!