Subnetting Problem

[kiv]

Hello everybody?

Could somebody please help me with this problem:

All of your coporate servers have static ip addresses. Some users on your network are complaining that they cannot contact a server named MCSE2000, which was recently added to the network. You check the TCP/IP configuration on MCSE2000 and notice the following:

Connection-specific DNS Suffix.:
IP Address……………..192.168.35.4
Subnet Mask……………255.255.252.0
Default Gateway………..192.168.39.100

You notice that the IP address and subnet mask values are correct. Which of the following default gateway values would be correct?

a) 192.168.33.23
b) 192.169.35.1
c) 192.168.31.254
d) 192.168.36.1

Your help will be highly appreciated!

 

[ecafracs]

I beleive the address would be 192.168.33.23 as it falls under subnet 32. All of the other addresses woul either fall under network or broadcast address.

35.1 would be the broadcast address under subnet 32
31.254 - broadcast address for subnet 28
36.1 - network address for network 36

 

[iriscisco]

yes, 192.168.33.23 is correct. the subnet includes 192.168.32.0-191.168.35.255

 

[RMAKing]

a) 192.168.33.23
b) 192.169.35.1

Either one could be correct. iriscisco had the right range but excluded answer b. ecafracs explanation was wrong on all three accounts...sorry, but true.

 

[ecafracs]

I did not go about the details. I just gave the most accurate answer that is supposed to be chosen on the problem. Remember that when you are answering the certification exam, you would only choose the best answer.In cisco exams, only cisco answer is the correct answer.

 

[kiv]

Thank you for you quick response.

Sorry to be a burden, but could you please explain me how you arrived at the answer.

Thank you very much in advance

 

[amiak]

Hi,

I'll explain why answers C and D are incorrect. It helps to look at decimal IP address in binary form.

192 = 11000000
168 = 10101000
35 = 00100011
4 = 00000100

The subnet mask portion of this IP address tells us how many bits of this address are used to indentify the network and with the remaining bits used to identify hosts on the network. Counting from the left, the 1's indicate bits used to identify network and the 0's indicate bits used to identify hosts. So now let's convert the subnet mask to binary.

255 = 11111111
255 = 11111111
252 = 11111100
0 = 00000000

OK so 192.168.35.4 is our IP address
and 255.255.252.0 is our subnet mask.

In the third octet of our IP address you'll notice there is a 35 and in the third octet of the subnet mask you'll notice there is a 252.

35 = 00100011
252 = 11111100

Above the 1's there is only one bit turned on
so this gives us 00100000, or decimal 32. This
means the 32 is the beginning of our network,
I have to go but maybe I'll get to add more later.

G. French, CCNP

 

[kiv]

Thanx for your response.

I understand how you got to 32. How will I calculate the end of the range?

 

[amiak]

picking up where I left off...

35 = 00100011
252 = 11111100

the subnet mask is used to identify bits that represent the network portion of the IP address. So in the third octet pictured above, each bit above a one is used to represent the network ID.

Based on this, a decimal 36 in the third octet would represent a different network.

36 = 00100100
252 = 11111100

Notice the second 1 in the IP address above the 1 in the subnet mask represents a change in the network ID indicating
that hosts with an IP address of 192.168.36.* would be on a different network than hosts with an IP address of 192.168.35.*

for more info...

http://www.red.net/support/resourcecentre/leasedline/intro.php

take it easy

 

[kiv]

Thanks

Just one more question:

Suppose the IP address is 192.168.64.100
and the Default gateway is 192.168.79.1

The default subnet mask given is 255.255.255.0 which is incorrect,
how would I determine the correct subnet mask in this case?

Sorry to trouble you again, your help would be highly valued!

 

[amiak]

do you want to create a supernet that would include both addresses?

look for the common bits in the third octet of each adddress

64 = 01000000
79 = 01001111
mask = 11110000 = common bits

so the subnet mask would be 255.255.240.0

 

[ElijahBaley]

I am a bit confused with this whole thread, I thought subnetting was the process of 'borrowing' bits from the host portion of an address range and using them to create subnets, in the questioners example, there is a class C IP address with what looks to me like an illegal class B subnet mask

IP Address……………..192.168.35.4
Subnet Mask……………255.255.252.0
Default Gateway………..192.168.39.100

Please help!!

 

[kiv]

Hi Amiak

Please try this link, it refers to the above problem.
SUBNETTING EXAM 1

http://www.2000trainers.com/section.aspx?sectionID=5&tab=exams

The site says the correct subnet is 255.255.224.0

 

[ElijahBaley]

I am a bit confused with this whole thread, I thought subnetting was the process of 'borrowing' bits from the host portion of an address range and using them to create subnets, in the questioners example, there is a class C IP address with what looks to me like an illegal class B subnet mask

IP Address……………..192.168.35.4
Subnet Mask……………255.255.252.0
Default Gateway………..192.168.39.100

Am I missing something?

 

[amiak]

Hi,

Thanks for the site.

You notice that 255.255.255.240 is not an available option.

Answers A,B,C are wrong for the reason I attempted to explain above.

for more on supernetting

http://www.red.net/support/resourcecentre/leasedline/intro.php

 

[kiv]

Thanks Amiak for your great help!

For those that are still confused, try this link, it's clear and consise.

http://kazungu.com/navigate.shtml

I hope you'll find it useful!

 

[ElijahBaley]

Hi,

Reading through this thread has caused some confusion, I now understand that what we are dealing with here is 'Supernetting' (not subnetting) borrowing bits from the network portion of a address to create a inclusive range of addresses from the pool of the merged class C networks.

So my next question is, how can I identify the 4 subnets and the combined host range from the ip address and supernet mask below?

IP Address……………..192.168.35.4
Subnet Mask……………255.255.252.0

Thanks for any help,