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VLSM Questions!

Posted on 2010-09-14
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Last Modified: 2012-06-27
Hi EEpeeps,

I want to learn tough VLSM but iam confuse on these a friend on cisco forum asked me some question so that i can practice for it but iam only able to answer easy ones.

Pls tell me what can be answer of "Not so easy ones" and "MUST" describe/explain how you solved it yourself this is what real question is :)



You are given an address of 76.222.112.64/22

 First easy ones.

 

1. What are the network  and broadcast addresses of this network?

2. How many hosts can be in the network?

 

Now not so easy.

 

3. Suppose you need to subnet that network. You need 6 subnets with at least 30 addresses each, 4 subnets with at least 100 addresses each, and 1 subnet with no more than 6 and no less than 4 host addresses. Is that possible? If so, how would you do it?

4. You are asked to summarize your network with a 76.216.232.111/17 network. What will the summary address be like provided you need the smallest summary possible?

 
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Question by:acl-puzz
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14 Comments
 
LVL 6

Expert Comment

by:collins23
ID: 33670510
Answer to easy ones first.


You are given an address of 76.222.112.64/22

 First easy ones.

 

1. What are the network  and broadcast addresses of this network?

next boundary is 24 .

24-22 = 2

2^2 = 4

Block Size is 4  ( counting in block size using 3rd octect )

76.222.0.0
76.222.4.0
.......etc
76.222.100.0
76.222.104.0
76.222.108.0
76.222.112.0 - subnet (network address )

76.222.115.255 ( broadcast address )

76.222.116.0





2. How many hosts can be in the network?

32-22 = 10 ( 22 bits used for network so the rest are host addresses )

2^10 number of hosts :)


Let me scratch my head for the hard ones..
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LVL 6

Expert Comment

by:collins23
ID: 33670514
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LVL 34

Expert Comment

by:Istvan Kalmar
ID: 33670572
HI,

the easyest way if you don't want to compute:

http://www.subnet-calculator.com/
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LVL 12

Author Comment

by:acl-puzz
ID: 33670617
@ikalmar

I desperately want to compute/learn the hard ones and thats the real question is.
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LVL 17

Expert Comment

by:Kvistofta
ID: 33670637
You are given an address of 76.222.112.64/22

3. Suppose you need to subnet that network. You need 6 subnets with at least 30 addresses each, 4 subnets with at least 100 addresses each, and 1 subnet with no more than 6 and no less than 4 host addresses. Is that possible? If so, how would you do it?

You need these ranges:
1) 30 addresses each, times 6 -> 192 addresses total
2) 100 (128!) addresses each, times 4 = 512
3) 8 addresses (a range of 8 addreses gives you 6 host addresses available since the first address is the network address and the last is a broadcast address).

Sounds easy. First place the biggest chunks, 4x128 addresses:
76.222.112.0/25 (half of a c-class network)
76.222.112.128/25 (half of a c-class network)
76.222.113.0/25 (half of a c-class network)
76.222.113.128/25 (half of a c-class network)

Next chunks of 32 addresses:
76.222.114.0/26 (one quarter of a c-class network, from 0 to 64)
76.222.114.64/26
76.222.114.128/26
76.222.114.192/26
76.222.115.0/26
76.222.115.64/26

Now what we have left is the 115-network from .128 up to .255. That would easily fit the 8-addresses scope we need.
76.222.116.128/28

The addresses from 136 (128+8) are left for future uses.


4. You are asked to summarize your network with a 76.216.232.111/17 network. What will the summary address be like provided you need the smallest summary possible?

My network is 76.222.112.0/22, right. That is 76.222.[112-115].x. Now we want to summarize this with 76.216.232.111/17 which is 76.216.[232-233].x. So we need to find a common subnet for 76.216.[112-233].x which isnt that easy. If we start with a /16 we have 76.216.[whatever].x, which might be too easy. If we split that in halves (/17) we have 2 networks, 76.216.[0-127].x and 76.216.[128-255].x. We see that our span of third octet, 112-233 would be split by this netmask, so the smallest summary address would be /16, that is 76.216.0.0/16.



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LVL 17

Expert Comment

by:Kvistofta
ID: 33670644
I wouldnt recommend anyone to use tools for this. If you want to master subnetting you must be able to use nothing else than perhaps a pen to solve any questions like these...

/Kvistofta
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LVL 34

Expert Comment

by:Istvan Kalmar
ID: 33670754
Yes I agree with you, but if you want to check your computing you need this tool!
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LVL 12

Author Comment

by:acl-puzz
ID: 33671551
@Kvistofta

I did understand most of the part of yours answer except 4th answer ..actually i still dont understand this question yet how/why the hell we would summarize 76.222.112.64 with 76.216.232.111 i mean are we counting 76.0.0.0 as a common subnet and what is the need?

one more thing when we are dealing with this type of questions which given requirement we shold give preference if it says "500 hosts required per subnet" or  "10 subnets required" within same question.


Thanks
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LVL 17

Accepted Solution

by:
Kvistofta earned 2000 total points
ID: 33672242
@acl-puzz: why? Because we are told so. ;) How? We find a network/mask-combination that is as small as possible but that includes both networks in the question.

But NOW I see that I am wrong. Forget everything I wrote above regarding #4. I see that the difference is in the 2:nd octet, not the third!

We have these networks:
* 76.222.112.64/22, which fools us to believe that that is the network address, but it isn´t. With a /22, the network address is 76.222.112.0/22.
* 76.216.232.111/17, which also fools us. The network address with that mask is 76.216.128.0/17.

so:
76.222.112.64/22
76.216.232.111/17

The first difference fro left to right is in the 2:nd octet. So the network/netmask-boundary should be in that 2:nd octet, right?

222 and 216. How do we define them as narrow as possible? If you cant see it by eyes I suggest that you write the numbers down in binaries:

         128 64 32 16 8 4 2 1
222 = 1   1    0   1   1 1 1 0
216 = 1   1    0   1   1 0 0 0

So, how many bits are common for those numbers, from left to right? 11011, right which means 5 bits. So the network number is 11011 binary = 216.

Where is the boundary? First in 1:st octet we have 8 bits, then we have 5 bits in 2:nd octet, which give us 13 bits in the netmask. Conclusion: /13.

So the solution should be 76.216.0.0/13 which includes 76.[216-223].x.x

Sorry for bringing this confusion. :-)

/Kvistofta
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LVL 12

Author Comment

by:acl-puzz
ID: 33673058
its ok ;-)

So all except 4th were right?

And let me also tell you how i understands process of solving 4th question

1.Firstly we need to know what is classful network which is 76.0.0.0 then we need to consider right most octect comes after 76.X

2.Then we need to find common bits in binary in this first right-most octect and conclusion is 11011 then you added default classful network bits for the given network in question+common bits which is 8+5=13 is the answer

pls rectify me so that i wont have any confusion as iam studying for certification and is there any faster method for 2nd step u know binary method takes time and we have just 1:30 for everything

Thanks
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LVL 17

Expert Comment

by:Kvistofta
ID: 33673693
Exactly. See a subnet-mask as it is a boundary between all 1:s (to the left) and all 0:s (to the right). Where is that boundary? With /16 (255.255.0.0) it is between octet 2 and three for example. With /15, the rightmost bit in the 2:nd octet is turned into a 0 which means that that bit is part of the host-address, not the network-address. And so on.

I think you are pretty well on the spot. The exact way of thinking is different between all ov us. Each tech-guy has his own way of seeing/thinking/calculating stuff like this. It doesnt matter which method you use as long as the result is the correct. :)

Best way to do this is practice, practice and practice.

/Kvistofta
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LVL 12

Author Comment

by:acl-puzz
ID: 33673864
i wish there is other method then that u used in 4th question so that i can solve it faster during exam if any pls let me know :)
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LVL 17

Expert Comment

by:Kvistofta
ID: 33674009
You just need practice. That m.ethod would let you solve it within 30 seconds with some practice.

/Kvistofta
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LVL 12

Author Comment

by:acl-puzz
ID: 33674030
ok man thanks for help
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