Unfamiliar questions regarding subnetting.

Good evening guys.
I was recently given a quiz by my ccna instructor regarding subnetting. I am new to subnetting, as I am with networking, however I haven't seen these types of questions before.

Using the address 130.100.0.0

a) create a subnet mask for 1000 subnets, with maximum amount of hosts.
b) Work out the 600th subnet.
c) VLSM network address in part b, for as many subnets as possible with at least 12 hosts.

Been trying to read online and watch various subnetting videos, without much luck. A lot of the sources iv'e been looking at dont give examples of these types of questions.

Any help showing me the ropes will be much appreciated.

Find more posts tagged with

Comments

goldenlight

Google Dan's courses and the magic number trick. On YouTube this will help your understanding tremendously

also practicing at this site http://www.subnettingquestions.com/ help me as well

Also try to find a Subnetting calculator that way you can build confidence.. Good LUCk.

Will give more info once I get home.. hard to type on phone.

volfkhat

a)
First,
Identify the Subnet Mask your need. It must fit 1,000 subnets
https://danielmiessler.com/images/subnetting-table.png

thus:
255.255.255.192

with a block size = 64

b)
Next,
Start listing the first 600 subnets:

1st subnet- 130.100.0.0
2nd subnet- 130.100.0.64
3rd subnet- 130.100.0.128
4th subnet- 130.100.0.192

5th - 8th subnets:
130.100.1.0
130.100.1.64
130.100.1.128
130.100.1.192

....so forth ...

597th subnet- 130.100.149.0
598th subnet- 130.100.149.64
599th subnet- 130.100.149.128
600th subnet- 130.100.149.192

c)
Last,
VLSM the subnet with at least 12 hosts.

Which means... you need a block size of 16 (in order to fit the 12).
So, your subnets will be:
130.100.149.192
130.100.149.208
130.100.149.224
130.100.149.240

QED.

This is actually a fairly easy question.

But only IF your instructor is actually Competent in the art of Teaching.

Priston

I'll give you an example with different numbers

Using the address 161.25.0.0

a) create a subnet mask for 500 subnets, with maximum amount of hosts.
b) Work out the 5th subnet.
c) VLSM network address in part b, for as many subnets as possible with at least 20 hosts.

Since no default subnet mask is given, the default subnet mask will be the classful subnet mask. (class B - 255.255.0.0)

You've got a subnet of 161.25.0.0 (/16) 255.255.0.0 and you want to create 500 more networks out of it. The way to do this is my increasing the number of network bits.
nnnnnnnn . nnnnnnnn . hhhhhhhh . hhhhhhhh
2^<number of converted network bits> will give you the number of subnets you've created

you need atleast 500 subnets and you've got 16 host bits to play with.

We know 1 octet 2 ^ 8 will give us 256 subnets which is not enough
nnnnnnnn . nnnnnnnn . NNNNNNNN . hhhhhhhh

So lets try 2 ^ 9, 512 subnets (now we've got our 500 subnets!)
So we have to use 9 host bits to give us 500 subnets, which leaves us with 7 host bits for host
nnnnnnnn . nnnnnnnn . NNNNNNNN . Nhhhhhhh

1st subnet 161.25.0.0 - 161.25.0.127
2nd subnet 161.25.0.128 - 161.25.0.255
3rd subnet 161.25.1.0 - 161.25.1.127
4th subnet 161.25.1.128 - 161.25.1.255
5th subnet 161.25.2.0 - 161.25.2.127

Now i'm going to VLSM the 5th subnet 161.25.2.0 255.255.255.128
nnnnnnnn . nnnnnnnn . nnnnnnnn . nhhhhhhh

I've got 7 host bits to make as many subnets as I can while still leaving 20 hosts
2^<number of host bits>-2 will tell me how many hosts I have

2^1-2 = 0 - nnnnnnnn . nnnnnnnn . nnnnnnnn . nNNNNNNH
2^2-2 = 2 - nnnnnnnn . nnnnnnnn . nnnnnnnn . nNNNNNHH
2^3-2 = 6 - nnnnnnnn . nnnnnnnn . nnnnnnnn . nNNNNHHH
2^4-2 = 14 - nnnnnnnn . nnnnnnnn . nnnnnnnn . nNNNHHHH
2^5-2 = 30 - nnnnnnnn . nnnnnnnn . nnnnnnnn . nNNHHHHH
5 host bits and 2 network bits
30 hosts and 4 subnets

161.25.2.0 - 161.25.2.127 is now

161.25.2.0 - 161.25.2.31
161.25.2.32 - 161.25.2.63
161.25.2.64 - 161.25.2.95
161.25.2.96 - 161.25.2.127

pinkiaiii

volfkhat shouldnt mask be /18 255.255.192.0 at least according to this table ?

as always jumping too fast to conclusion,seen mistake i made 130>class b

58e958d3750848979e85916bb4fb99b9.jpg

volfkhat

pinkiaiii wrote: »

volfkhat shouldnt mask be /18 255.255.192.0 at least according to this table ?

Hmmm... pinkiaii,
actually, i LIKE your thinking :]

b)
So the subnets would start:
130.100.0.0
130.100.64.0
130.100.128.0
130.100.192.0

130.101.0.0
130.101.64.0
130.101.128.0
130.101.192.0

... so forth ...

130.249.0.0
130.249.64.0
130.249.128.0
130.249.192.0
(600th subnet)

c)
Then start breaking this into smaller subnets of blocksize 16:
130.249.192.0
130.249.192.16
130.249.192.32
130.249.192.48
130.249.192.64
130.249.192.80
... and so forth ...
130.249.193.0
130.249.193.16
130.249.193.32
130.249.193.48
130.249.193.64
130.249.193.80

... and so forth ...
130.249.255.192
130.249.255.208
130.249.255.224
130.249.255.240

finished :]

So, technically..... i would accept this as an alternative correct answer!

In fact,
the Original question actually says:

"a) create a subnet mask for 1000 subnets, with maximum amount of hosts."

AND

"c) VLSM network address in part b, for as many subnets as possible with at least 12 hosts."

Pinkiaii's solution actualy has the MAXIMUM number of Hosts & Subnets.

I'm going to have to disagree with Dollar. What is "implied" is irrelevant.
Great minds think Outside the Box.

pinkiaiii

What got me thinking once you mentioned block size and gave network mask-thus i have the same chart as you posted,but also have other class tables,thus looking at it seen that it would be /26 for +1000 subnets,according to it.Then when started listing networks i thought why you would start at last octet when 130.100.0.0 third was empty.
Either way still grass hopper

at large numbers,or figuring questions like that thus theres always to learn smth new or refresh memory which is good non the less.

Only question how do you calculate 600th subnet ? since imagine your not writing them out one by one

class b subnet table.jpg

theodoxa

pinkiaiii wrote: »

volfkhat shouldnt mask be /18 255.255.192.0 at least according to this table ?

as always jumping too fast to conclusion,seen mistake i made 130>class b

Just in case anyone else had the same question (as I see you realized the mistake), 128-191 is Class B. The default subnet mask is --

255.255.0.0

1000 Subnets requires 10 bits (2^10 = 1024, 2^9 = 512) to be borrowed. Since, 10 > 8, you can add another 255 for --

255.255.255.0

Now, you still have 2 more bits (10-

, which would be 128 + 64 = 192 --

255.255.255.192

pinkiaiii

for sake of arguments can we agree that using /18 or /26 would yeald same results unless question was more specific as to use only class c mask or smth like that that would require to choose between.

and how to get the 600th subnet or any other past 20 without writing it out ?

Lycans

Thanks for all the input guys. This may seem like an easy task for some if you, but for me my instructor just tells me that netacad will explain everything.

I also missed something out in my original post

Cidr 16 was given at the end of the address,

130.100.0.0/16

volfkhat

Lycans wrote: »

This may seem like an easy task for some if you, but for me my instructor just tells me that netacad will explain everything.

I was afraid of that. Some people should not be allowed to teach.

pinkiaiii wrote: »

and how to get the 600th subnet or any other past 20 without writing it out ?

So, it's actually a matter of basic algebra.

Take a look at the first subnets i listed:
130.100.0.0
130.100.0.64
130.100.0.128
130.100.0.192

If you look carefully, you should be able to identify a pattern:
After every 4 Subnets, the fourth octet repeats.

Don't see it? then take a look at the next 4 subnets:
130.100.1.0
130.100.1.64
130.100.1.128
130.100.1.192

Hopefully you see this pattern now.

So, here's the Logic:
After every 4 subnets... the third Octet increments by 1.

This can be written in another way:

 4    (subnets)
---    
 1    (increments)

okay,
So now we will solve for the 600th subnet. We do this by inserting an "x" variable.
here's the formula:

 4          600
---   =   -----
 1            [I][B]x[/B][/I]

In English, this formula is saying:
"If 4 subnets give me 1 increment. Then 600 subnets gives me how many increments?"

Solving for 'x' gives us:

[I][B]
 x[/B][/I]          600
       =   -----
             4

x = 150.

The Third octet value will be "150" increments.
But you Start counting at ZERO (not 1), so actually the 3rd octet is 149.

Thus,
130.100.149.192

This is Basic Algebra.
Let me know if you don't get it; i can give you some easier examples :]

pinkiaiii

I knew there was some sort of math equation,and yes i do see the pattern,

say if address was 130.100.0.0 255.255.224.0

then it would go like
130.100.0.0
130.100.32.0
130.100.64.0
130.100.96.0
130.100.128.0
130.100.160.0
130.100.192.0
130.100.224.0 thus 8 >1 thus second octet gets incremented by 1 and say we needed 400 subnets = 400 divided by eight =50 -1 = 49
____________
130.101.0.0 <increment part
____________
130.149.0.0 <400th subnet

only tricky part to remember to take away one.

volfkhat

Almost.

Your Math is correct.
"400 divided by eight =50 -1 = 49"

But the 400th subnet is actually
130.149.224.0

(every 8th subnet ends with '.224')

:]

volfkhat

here's my favorite/real-life scenario:

Suppose that i can eat 27 Twinkies (http://en.wikipedia.org/wiki/Twinkie) in 8.5 minutes.

Assuming that i can maintain the pace....
how many Twinkies can i eat in 120 minutes?

lol

pinkiaiii

volfkhat wrote: »

Almost.

Your Math is correct.
"400 divided by eight =50 -1 = 49"

But the 400th subnet is actually
130.149.224.0

(every 8th subnet ends with '.224')

:]

like your attention to detail,seemed all figured out and yet still a quirk that would cost points as imagine would be easy to sneak such question into any exam,ans say give
130.149.50.0
130.149.49.0
130.149.224.0

as options and easily get screwed over.thus seems you start knowing stuff yet get stuck on minor points afterwards.

as for twinkies it would be smth like 387/390 -120 min divided by 8.5 result=14.112323smth x multiplied by 27
als its trickier once you mention 8.5min since in fact its 8min 30sec ^

volfkhat

yaw,
it just takes practice and detail... but it Can be done :]

As for the twinkies... don't overthink it.
just Stick to the formula :]

the Ratio is:

 27          [B][I]twinkies[/I][/B]
-----   =   -------------
 8.5           [I][B]minutes[/B][/I]

So, just plug in your values:

 27             [B][I]x[/I][/B]
------   =   -----
 8.5           120

Solving for x:

  120 * 27             [B][I]x[/I][/B]
------------   =  
    8.5

This gives you the Exact answer. (no estimates required)

EDIT:
oh, i just re-read your post. Nevermind!!
you got it :]