Can you solve this subnetting question?

DLSK12 · March 2015

I have a question, my teacher gave me a complete different answer and says i'm wrong.. Who is correct and why?

I got this answer:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.2.128
F: 128.1.2.129
L: 128.1.2.254
B: 128.1.2.255

The teacher got:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.1.0
F: 128.1.1.1
L: 128.1.1.126
B: 128.1.1.127

kurosaki00 · March 2015

Should be 128.1.2.128 the 2nd subnet? Maybe I did not get it?

For me his answer looks wrong, but lets see what the community says

Priston · March 2015

Your both wrong!

The 128.1.2.3 network is a class B network which is really 128.1.0.0/16

128.1.0.0/16 will be subnetted into multiple /25

128.1.0.0/25 is the first subnet

128.1.0.128/25 is the Second subnet
F 128.1.0.129
L 128.1.0.254
B 128.1.0.255

128.1.1.0/25 is the third subnet

128.1.2.128/25 is the sixth subnet

mistabrumley89 · March 2015

Your teacher is confused.

Deathmage · March 2015

DLSK12 wrote: »

I have a question, my teacher gave me a complete different answer and says i'm wrong.. Who is correct and why?

I got this answer:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.2.128
F: 128.1.2.129
L: 128.1.2.254
B: 128.1.2.255

The teacher got:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.1.0
F: 128.1.1.1
L: 128.1.1.126
B: 128.1.1.127

remember, you if you have a mask of 255.255.255.128 you can only fit 128 two times into a 8 bit octet of 256 (128+64+32+16+8+4+2+1=256) so you can only have two subnets, but you need to always minus 2 bits, one for the Network ID and one for the Broadcast ID.

Below in the graph for subnet 2: 255-129 = 126. You have 128 binary bits, and you minus 2 which equals 126.

it should be this:

Network ID
Range (1st)
Range (last)
Broadcast

128.1.2.0
128.1.2.1
128.1.2.126
128.1.2.127

128.1.2.128
128.1.2.129
128.1.2.254
128.1.2.255

Hope this helps

Priston wrote: »

Your both wrong!

The 128.1.2.3 network is a class B network which is really 128.1.0.0/16

128.1.0.0/16 will be subnetted into multiple /25

128.1.0.0/25 is the first subnet

128.1.0.128/25 is the Second subnet
F 128.1.0.129
L 128.1.0.254
B 128.1.0.255

128.1.1.0/25 is the third subnet

128.1.2.128/25 is the sixth subnet

To me this would only be valid if the mask was 255.255.0.0, but the 3rd octet is fully used up. if it was a 255.255.254.0 then yes it would be what you say above.

Since this one is a 255.255.255.128, the 3rd octet of 2 doesn't change, only the 4th octet does.

Lastly if you look above it says in bold 68 hosts, so we both know that going to a /26 would be 64 hosts and he needs 68, so the next one up is 128 or a /25 and that's perfect. If it said 68 subnets well now that's a different story....

Curious if this is a classless CIDR from the instructor.

tpasmall · March 2015

Deathmage wrote: »

it should be this:

Network ID
Range (1st)
Range (last)
Broadcast

128.1.2.0
128.1.2.1
128.1.2.126
128.1.2.127

128.1.2.128
128.1.2.129
128.1.2.255
128.1.2.256

That second broadcast IP is impossible. The OP has it right, his teacher is wrong.

quickman007 · March 2015

Your teacher is wrong. He didn't read the question very well, apparently.

jvrlopez · March 2015

Isn't this the second time he's been off on a subnetting question?

Deathmage · March 2015

tpasmall wrote: »

That second broadcast IP is impossible. The OP has it right, his teacher is wrong.

durr, you would be correct, should be 255.

ibn_shaddad · March 2015

DLSK12 wrote: »

I have a question, my teacher gave me a complete different answer and says i'm wrong.. Who is correct and why?

I got this answer:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.2.128
F: 128.1.2.129
L: 128.1.2.254
B: 128.1.2.255

The teacher got:

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)
Subnet Mask: 255.255.255.128
Subnet Address: 128.1.1.0
F: 128.1.1.1
L: 128.1.1.126
B: 128.1.1.127

you are right
the teacher is wrong

ibn_shaddad · March 2015

Priston wrote: »

Your both wrong!

The 128.1.2.3 network is a class B network which is really 128.1.0.0/16

128.1.0.0/16 will be subnetted into multiple /25

128.1.0.0/25 is the first subnet

128.1.0.128/25 is the Second subnet
F 128.1.0.129
L 128.1.0.254
B 128.1.0.255

128.1.1.0/25 is the third subnet

128.1.2.128/25 is the sixth subnet

it starts with 128, initially it would be a class B network, but it's a class less network because it is subnetted.

the subnet mask 255.255.255.128 tells us we are using 25 bits for the net id part and the rest 7 bits for the host part.
So in the last octet we have 1 0 0 0 0 0 0 0
where the last "1" bit starting from left will be the increment number that we will use to add to come with the subnet ids is equal to 128 according to it's position 2^7.

Then we will start incrementing with that number starting with zero so we can come up with subnets. remember the adding will be in the same octet the increment bit is:

128.1.2.0
128.1.2.128 <<<< so here it goes your second subnet!
128.1.3.0
128.1.3.128
...

so:
S: 128.1.2.128
F: 128.1.2.129
L: 128.1.2.254
B: 128.1.2.255

Priston · March 2015

ibn_shaddad wrote: »

128.1.2.0
128.1.2.128 <<<< so here it goes your second subnet!
128.1.3.0
128.1.3.128
...

Why did you skip 128.1.0.0, 128.1.0.128, 128.1.1.0, 128.1.1.128?

Deathmage wrote: »

To me this would only be valid if the mask was 255.255.0.0

The mask is 255.255.0.0

IPv4 subnetting reference - Wikipedia, the free encyclopedia

B
10
128.0.0.0
191.255.255.255
255.255.0.0
/16

With the Cisco curriculum you have to assume it's a classful network unless they tell you otherwise. If no subnet mask is given in the question that means it's a classful network and your converting it to a classless network.

I think you guys are just letting this trick question trick you.

Network: 128.1.0.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)

has the same answers as

Network: 128.1.2.3 (you need 68 hosts with as many subnets as posssible) (calculate for the 2nd available subnet)

They both are IPs that start out in the same network before you subnet the network.

Deathmage · March 2015

Priston wrote: »

Why did you skip 128.1.0.0, 128.1.0.128, 128.1.1.0, 128.1.1.128?

The mask is 255.255.0.0

IPv4 subnetting reference - Wikipedia, the free encyclopedia

B
10
128.0.0.0
191.255.255.255
255.255.0.0
/16

With the Cisco curriculum you have to assume it's a classful network unless they tell you otherwise. If no subnet mask is given in the question that means it's a classful network and your converting it to a classless network.

I think you guys are just letting this trick question trick you.
has the same answers as
They both are IPs that start out in the same network before you subnet the network.

But it's clearly being stated as a 255.255.255.128 in the OP original post. Unless it's stipulates that's it's a 255.255.0.0 then why else would you put it as a 255.255.255.128. Masks don't lie.

If I got into a Cisco exam with that presumption and Cisco says I'm wrong I'm finding you and beating you with a bat.

See maybe I'm just follow my training with exams, but if anything is in bold, that trumps everything, my logic is this, it says above he needs 68 hosts and to solve for the 2nd subnet so in my mind, if he only needs 68 hosts but doesn't stipulate subnets than subetting for hosts has the greater importance since it's a classless address. (as you know hosts right to left and subnets left to right, hosts of 68 would be in the 4th octet even for a classful Class

If it said in bold it needed 500 subnets I'm pretty safe to presume it would be 255.255.254.0. I'm not disagreeing with the outcome just in the directions being stated. I don't think the OP unintentionally bolded those values accidentally.

So my logic is it's following the 255.255.255.128 classless mask and not the 255.255.0.0 default classful Class B mask.

NetworkNewb · March 2015

The more people talk about this subnetting question the more I hate this teacher.

A) for not being clear on the question and

for not knowing the answer to his own question

TechGuru80 · March 2015

Priston wrote: »

Your both wrong!

The 128.1.2.3 network is a class B network which is really 128.1.0.0/16

128.1.0.0/16 will be subnetted into multiple /25

128.1.0.0/25 is the first subnet

128.1.0.128/25 is the Second subnet
F 128.1.0.129
L 128.1.0.254
B 128.1.0.255

128.1.1.0/25 is the third subnet

128.1.2.128/25 is the sixth subnet

Correct. The question is poorly worded if it does not state the context using a classful address scheme.

TechGuru80 · March 2015

I just reread this and OP you are correct if the subnet given is /24 to start...it seems like your teacher is thinking classful although he got it wrong.

nster · March 2015

The question doesn't specify a subnet mask does it? OP's answer has a subnet mask not the question. so in the case you have to take into account classes

I can't tell where the question ends and the answer starts. we need a but of a clarification

mistabrumley89 · March 2015

Even if it were classful, the teacher is still wrong with a 1 in the third octet

Deathmage · March 2015

it's so funny.... everyone learns subnetting differently and to see everyone responses I see subnetting in different way, lol

Priston · March 2015

Deathmage wrote: »

But it's clearly being stated as a 255.255.255.128 in the OP original post. Unless it's stipulates that's it's a 255.255.0.0 then why else would you put it as a 255.255.255.128. Masks don't lie.

The 255.255.255.128 in the OP is part of the answer not part of the question, if the original mask was 255.255.255.128 there wouldn't be a 2nd network because you can't subnet it any further than it already is and still have 68 hosts.

Here is a better illustration of the question in this thread http://www.techexams.net/forums/ccna-ccent/109009-subnetting-problem.html

And here in this duplicate thread http://www.techexams.net/forums/ccna-ccent/109026-who-correct-help-me-solve-subnetting-question.html
His teacher is accepting two answers because the Cisco Networking Academy teaches you 2^N-2 then they teach you 2^N-2 is really wrong and it's 2^N

Priston · March 2015

The funny thing is, that's not the answer the teacher gave. That's the answer Cisco Networking Academy gave the teacher and everyone is trying to prove Cisco wrong.

To find the answer to this poorly written question
Given the following Network address and conditions, calculate the information listed below for the 2nd available subet.
Network: 128.1.2.3 You need 68 hosts with as many subnets as possible

You need to know what the network your given is, 128.1.2.3 is not a network even though the questions says it's a network. 128.1.2.3 is an IP address. That IP address has 128 in the first octet which means it's a class B address which has a subnet mask of 255.255.0.0.

Therefore the IP address 128.1.2.3 is in the 128.1.0.0/16 (128.1.0.0 255.255.0.0) Network

Then you can subnet that 128.1.0.0/16 network into multiple /25 networks

The 2nd subnet being 128.1.0.128/25

and also the other 2nd subnet (which is really the third subnet) being 128.1.1.0/25 with this stupid 2^N-2 rule they still teach.

The reason they teach 2^N-2: https://learningnetwork.cisco.com/thread/32136

ibn_shaddad · March 2015

Priston wrote: »

Why did you skip 128.1.0.0, 128.1.0.128, 128.1.1.0, 128.1.1.128?

The mask is 255.255.0.0

IPv4 subnetting reference - Wikipedia, the free encyclopedia

B
10
128.0.0.0
191.255.255.255
255.255.0.0
/16

With the Cisco curriculum you have to assume it's a classful network unless they tell you otherwise. If no subnet mask is given in the question that means it's a classful network and your converting it to a classless network.

I think you guys are just letting this trick question trick you.
has the same answers as
They both are IPs that start out in the same network before you subnet the network.

Guys guys, clearly the question is not poorly written, but it is not written at all!

all what the OP gave us was the answers, not the question.
But it is fair enough for us to know the IP address and the subnet mask.
The subnet mask was mentioned two time, (Subnet Mask: 255.255.255.12

one by the teacher and another by the OP.

Since the S/M is the same, then we can take it as a given info and move on.

Deathmage · March 2015

ibn_shaddad wrote: »

Guys guys, clearly the question is not poorly written, but it is not written at all!

all what the OP gave us was the answers, not the question.
But it is fair enough for us to know the IP address and the subnet mask.
The subnet mask was mentioned two time, (Subnet Mask: 255.255.255.12 one by the teacher and another by the OP.

Since the S/M is the same, then we can take it as a given info and move on.

NetworkNewb · March 2015

The same problem was posted in another thread on the same day, but stated the question better than the OP did here. Same issue though, the teacher does it wrong and question is worded poorly. The question is to calculate the 2nd subnet...The other thread shows the beginning info given. This OP did not. Don't care enough to find other thread right now though since the end result is the same. The teacher is wrong

Jon_Cisco · March 2015

NetworkNewb wrote: »

The same problem was posted in another thread on the same day, but stated the question better than the OP did here. Same issue though, the teacher does it wrong and question is worded poorly. The question is to calculate the 2nd subnet...The other thread shows the beginning info given. This OP did not. Don't care enough to find other thread right now though since the end result is the same. The teacher is wrong

I believe if you read the entire question you might reconsider your answer. I can't believe this thread is still open but the 3rd subnet is the second usable subnet when given the requirements. They are instructing you not to use subnet zero. This is not how you would do it now but that was the question.

NetworkNewb · March 2015

Jon_Cisco wrote: »

I believe if you read the entire question you might reconsider your answer. I can't believe this thread is still open but the 3rd subnet is the second usable subnet when given the requirements. They are instructing you not to use subnet zero. This is not how you would do it now but that was the question.

I must of missed the part where it said not to count the first subnet when choosing the second available subnet. The teacher would be right then

Can you solve this subnetting question?

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