You have been given the network ID of 172.24.8.0/22 from your ISP.
All of the routers in your network use either RIP V2, or OSPF. Each of the
two subnets you will be creating will contain only 75 computers. You want
to use the most specific number of bits and the first two available
network ID numbers in your subnet mask. Drag and Drop question with the
following Answer.
(choose 2)
A. 172.24.12.0/22
B. 172.24.16.0/22
C. 172.24.24.0/22
D. 172.24.8.128/25
E. 172.24.9.0/25
F. 172.24.16.0/25


Any takers on the answer to this and the reasoning behind it??

iceman

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This is one I definitely missed on the exam!
From the basic rules of subnetting, you would go with 'A' and 'B'. But with OSPF and RIP v2, I can see where 'D' and 'F' could work in theory. 'C' is not a good choice, because it is not the least significant compared to 'A' or 'B'. 'E' would not be right, because it would fall in the middle of an address range. I'd really just from a curiousity standpoint, like to know the answer.

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Did ponder over this quiz, strangely, it says you are given an network ID of 172.24.8.0/22 from your ISP. Then you see answers that are with /25 behind. I was thinking whether this quiz has some error in it. But I finally came think the answer is the last 2-simply because with /25, you have 128 hosts, though it is more than 75, but that's that. So I need 2 subnet, so I just choose the next /25. One of the answer has 128, I don't think it is a valid network ID.

Read jeff comments, realised that maybe RIP and ospf has something on it. Qutie a strange quiz. Suprised it came out in the exam, which you've said.

__________________
" When asked if he felt discouraged by the 1,073 failures he had before inventing the electric light bulb, "I did not fail 1,073 times; I found 1,073 ways not to do it."

Thomas Edison.

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Outside of OSPF and RIP v2 which do CIDR, it is almost a waste to have a mask of 255.255.255.128 as the all zeros netowrk would be used for network ID and the all ones would be the broadcast range. This would give you 126 hosts on no nets. But with CIDR the mask is sent along with the route so this mask is theoretically usefull. All in all it is a tough one for the exam.

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Ok. I`m new in subneting so i see if someone correct my iddea, i see that the correct answers are between C,D and E.

You say that two subnets must contain 75 computers each.

In your answers you have two available subnet masks: /22 & /25

If you use 22 you have 1022 hosts... this is too much....

If you use 25, you have a host id of 7 ceros = (2^7)-2= 126 possible hosts.

so, enough for your subnets with 75 hosts...

So the correct answers for me are between the D,E and F.

I`m a little confusing at this point. i don`t know what options to select!!!

I appreciate a very good answer and correct me if i`m wrong!!

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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quote:

---

Originally posted by cm2gj
Ok. I`m new in subneting so i see if someone correct my iddea, i see that the correct answers are between C,D and E.

You say that two subnets must contain 75 computers each.

In your answers you have two available subnet masks: /22 & /25

If you use 22 you have 1022 hosts... this is too much....

If you use 25, you have a host id of 7 ceros = (2^7)-2= 126 possible hosts.

so, enough for your subnets with 75 hosts...

So the correct answers for me are between the D,E and F.

I`m a little confusing at this point. i don`t know what options to select!!!

I appreciate a very good answer and correct me if i`m wrong!!

---

You're nearly there - if you think about it just a little bit more you should get the answer!!

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One Exam leads to another! Where will it ever end?

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Yo, frank, is my logic correct in my previous message?

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D y F?

__________________
Best Regards
Alex
alexisgarcia72@hotmail.com
Cuban in Mexico
www.cm2gj.com

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Im just thinking out loud, but I feel that you still have to work within the confines of 172.24.8.0/22 network ID you recieve from your ISP. Not only do the first 3 provide too many host, but they are outside of your range. Leaveing just D. 172.24.8.128/25 and
E. 172.24.9.0/25. But is you use a subnet mask of /25 doesnt the first and only network start at 128? If so what is the deal with answer E?

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E & D is correct.

The first half being:
172.24.9.0/25
10101100 00011000 00001001 00000000
subnet mask- 255.255.255.128

And the second half :

172.24.8.128/25
10101100 00011000 00001000 10000000
Subnet mask : 255.255.255.128

good luck !

__________________
" When asked if he felt discouraged by the 1,073 failures he had before inventing the electric light bulb, "I did not fail 1,073 times; I found 1,073 ways not to do it."

Thomas Edison.

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