Subnetting...Shortcuts 8

[bobbyforhire]

Ok so as i stated before i think i'm getting good on this..So good that i have decided to put all of my cisco equipment together and set it up right.

now, i will have a connection from my pix to my router. Just to make sure this is the correct standard for my subnetting.

Because I only need two hosts i have selected the following.

172.0.0.0 as my base. I will asign my pix 172.0.0.1 and my router 172.0.0.2 both using a subnet 255.255.255.252

Can anyone confirm that this is the correct subnet to use. Also to increase my time on the test, does anyone have any shortcuts that you use?


172.0.0.0
CLASS B
2 Host needed
128 64 32 16 8 4 2 1
i need 2 bits to get 2 hosts.

*Tip

HOSTS = 0'S (RIGHT TO LEFT)
NETWORK = 1'S (LEFT TO RIGHT)


255.255.0.0 = 11111111.11111111.00000000.00000000
turn that into 11111111.11111111.11111111.11111100

255 - 3 = 252
Lowest INC = 4

172.0.0.0 - 172.0.0.3 ( Giving me 1 and 2 to use)

 

[Cluebird]

Bobbyforhire,

You're getting confused. The original problem: 212.44.53.0. Subnet for 15 hosts with max networks. What is the broadcast of the 5th subnet?

This is typical of how Cisco may ask questions. The problem requires several pieces of knowledge to get the correct answer. It also only allows one answer that meets both the host requirement and the number of network requirement.

First, this is a Class C with starting mask 255.255.255.0. Second, the new mask extends the old mask but forces you to have an extra host bit so your new mask is 255.255.255.224. If you choose 255.255.255.240 you don't have enough hosts (16-2=14) so you have an "extra" host bit to give 32-2=30 hosts. You set three new bits in the mask which makes 8 total networks. Each bit set doubles the number of networks but cuts the size of the networks in halves. Your actual host address space is in blocks (increments) of 32. Therefore third, your networks look like this in the fourth octet:

0-31 1st subnet
32-63 2nd
64-95 3rd
96-127 4th
128-159 5th
160-191
192-223
224-255

The answer is: 212.44.53.159/27

 

[Cluebird]

bobbyforhire,

Just checking if you were paying attention...BTW, I didn't count #3 and #6 as problems so only 4 needed calculations? Tappity, tappity, tap, tap...Is that music I'm tap dancing to? ;-)

 

[bobbyforhire]

Sorry my mind is all over the place these days. Dealing with computers for the past 12 years just has me thinking of incroments of memory modules and hard drives. I need to start thinking smaller numbers associated with pc's. These days all i think about are gigs and tera's. It's very simple math. But i need to stop and watch my work.


212.44.53.0 - 212.44.53.31
212.44.53.32 - 212.44.53.63
212.44.53.64 - 212.44.53.95
212.44.53.96 - 212.44.53.127
212.44.53.128 - 212.44.53.159



As for the tapy tapy tapy...Dace all you want..Thanks for help!

 

[bobbyforhire]

3. 230.55.55.0 What mask gives 4 networks with max hosts?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.192.0
D. None of the above

The use of Max Networks and Max Hosts is just throwing me off.

230 = class C
1.1.1.0
need 4 net
1.1.1.11110000
240


I would say d because if you only need 4 networks and none of them give you just 4 networks.

4: IP Address?? please clarify?? an available public address?

5: Same thing as number 4.?@? it's got a 172.x.x.x but the RFC says private is only 172.16-172.31 so not 100%


6: Sure if your ISP gave you that IP and you have it routed to your server why not? wait a second.../28 starts gives you a 11111111.11111111.11111111.11110000 240 that IP has a 16 inc. you started with

197.23.15.0
16
32
48
64



Ok yeah your broadcast will never fall on that address i don't see why you couldn't. it's not a 10. or a 172.16-32. or a 192.168

my brain is fried..i'm going to get some sleep and take a second look at my answers in the morning

 

[N6750B]

I see some good stuff on this page. Good meat for training!

Consider this:

IP: 172.21.22.1 /21 so mask is 255.255.248.0

Stack the mask on top of the IP....

255.255.248.0
172. 21. 22.1

Find the octet in the mask that is not equal to 0 or 255.
Call that octet the "Interesting Octet".

So in the mask above, the value of the Interesting Octet is 248.

And in the IP address, the value in the Interesting Octet is 22.

Next do 256-248, which = 8 (Call this the Block Number.)

Next list the multiples of the Magic Number (8)until you list the multiple that is equal to or larger than 22 (The value of the Interesting Octet in the IP addy.)

So here's the multiples.....

0,8,16,24,32 and that's enough.

Remember we're looking for the multiple in the list that is greater than or equal to 22. That multiple is 16.

So the resident subnet of 172.21.22.1 /21 is:

172.21.16.0

The next subnet is 172.21.24.0 (24 is next multiple above 16 in the list of multiples.

The subnet broadcast address for subnet 172.21.16.0 is 1 less than the value of the next subnet (172.21.24.0)so it's 172.21.23.255.

The valid host addresses are all of the addresses between the subnet number 172.21.16.0 and the subnet broadcast address of 172.21.23.255 -

172.21.16.- 172.16.23.254

You'll notice that I found the Subnet Broadcast Address before I found the range of valid host addresses. Why? Because I sometimes make mistakes if I don't work backward from next Subnet Number to the Broadcast Address and then the last valid host IP.

I learned this in one of Wendell Odom's books and practiced it a lot. I missed none of the subnetting questions on the exam. So if it works for me I know it'll work for anyone else.

And hey - I never gave thought to the fact that broadcasts are always odd numbers. That helps!

 

[Cluebird]

bobbyforhire,

3. 230.55.55.0 What mask gives 4 networks with max hosts?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.192.0
D. None of the above

Answer: D The address is multicast Class D. Can't be subnetted. Gotcha!

4. What type of address is this: 194.255.172.64/27?

Answer: The address 194.255.172.64/27 represents a network address. This happens to be the same mask as #2. The way I phrase the questions is very common for Cisco exams.

5. What type of address is this? 172.56.33.255/23?

Answer: This is the directed broadcast address on the network 172.56.32.0/23. You're expected to understand and interpret that in the Cisco world when you're asked what type of address, you're looking for "host, network, or broadcast" for your answer.

6. Is 197.23.15.160/28 usable for my

http://WWW server?

Why or why not? 197.23.15.160 is a network address so, no I can't use it on my

http://www server.

Need some more?

 

[bobbyforhire]

Why am I not seeing any of this in any of my CCENT studies? I have the CBTNuggest videos and I have TheBryantAdvantage. I have alot of people giving me diffrent ways to get diffrent things?!?!


Are the questions your asking out of the CCENT leage? I feel like a dunce. But hey i have to be willing to fail sometimes to get past this. Yes MORE!!!

N675 -

How did you get the 16?

Next list the multiples of the Magic Number (8)until you list the multiple that is equal to or larger than 22 (The value of the Interesting Octet in the IP addy.)

So here's the multiples.....


0,8,16,24,32 and that's enough.
------------------------------

Is it because you went up 5 Digits?
1 2 4 8 16 32 64 128?

-------------------------------

Remember we're looking for the multiple in the list that is greater than or equal to 22. That multiple is 16.

-----------------------------------------------------
8*3 = 24 Why did we have to goto 32 if 24 gave us the more than 22?
-----------------------------------------------------

So the resident subnet of 172.21.22.1 /21 is:

172.21.16.0


So far i have got 5 people telling me to do things diffrent.

-------------------------------------------------------
What type of address is this: 194.255.172.64/27?

- a 27 is 128+_64_32 = 224
0,32,64

this is a network address..I don't know why but i can see these questions in the morning better than at night?!?

--------------------------------------------------------

What type of address is this? 172.56.33.255/23?

It's a broadcast pi address the last address useable for that subnet.

---------------------------------------------------------

Is 197.23.15.160/28 usable for my

http://WWW server?

Why or why not?

What told you right off the bat that this was a network address? is it because it ended in a 160?




Ok so far this is what you have to learn when learning subnets.


Find the solution for Hosts
Find the solution for Networks
Find the reverse solution for Hosts and Networks
Remember 1-126 = A , 128-191 = B , 192 - 223 = C

10000000 = 128 = /25
11000000 = 192 = /26
11100000 = 224 = /27
11110000 = 240 = /28
11111000 = 248 = /29
11111100 = 253 = /30
11111110 = 254 = /31
11111111 = 255 = /32

****READ YOUR MATH TWICE****
*****READ YOUR MATH TWICE****

 

[Cluebird]

I teach authorized curriculum as CCSI. These are topics I cover in my classes. I go into more depth than most books since they don't always cover the detail Cisco really tests.

You are so close on your understanding of subnetting. All the different techniques are merely ways to help you do the procedure of subnetting. They all work, some are just easier to wrap your brain around. I emphasize results. However, you get the result, IF you understand it and can repeat it quickly and accurately you are good to go. Some people do boolean addition. Others use exponents and lots of math. I like to teach patterns. I've taught my techniques to my 6-year old and she could subnet with them.

N6750B is showing a bit of what I teach but I split the subnetting instruction over 4 days, with 1 hour per day covering the foundational topics. I've had nearly 100% success teaching subnetting with learners who've struggled with it. I'm only covering a few points in this thread.

This thread is really getting long. How about starting another and I'll give some more practice problems there.