Who is correct? Help me solve subnetting question

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

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Comments

Adam B

Not sure what your teacher is talking about. 128.1.2.0 /25. Second subnet, First 128.1.2.129 Last .254, Bcast.255. I almost wonder if your teacher was looking at a totally different question.

DLSK12

I know, he said something like after 128.1.0.255 its 128.1.1.0 i'm like huh, there can only be 2 subnets on this mask and hes making up like 50

DLSK12

And he attached this. also, Should i start counting my first subnet on 0-? or lets say if its increments of 4 should my first subnet be 0-3 or 4-7?

image00168 (2).jpg

Syntax_ninja*

I'm not exactly sure what network number he started with but if the network was 128.1.1.0/25, the first usable address is .1 and the last usable address is .126. Your next network will be 128.1.1.128/25.

I think your teacher was explaining that after the .255 in the fourth octet the third octet increments by 1 changing the address to 128.1.1.0, think of it like a car's odometer.

samwinchester

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

I didn't quite get that. What exactly is the question again?

Are you dividing 128.1.2.3 network into 68 hosts??

Ltat42a

Agree with Adam B. 123.1.2.3 is in the 123.1.2.0/25 network. The next subnet is 123.1.2.128. With this class A address and /25 mask, there will be 131072 subnets, 126 available hosts in each subnet.
Using bits, looks like this: 0nnnnnnn.ssssssss.ssssssss.shhhhhhh

Priston

First of all, separate the question from the answer, your confusing EVERYONE!!!!
255.255.255.128 is part of the answer, not part of the question.

According to the 2^N-2 rule the teacher is right.

The 128.1.2.3 IP address is a class B address which is in the 128.1.0.0/16 network so your subnetting 128.1.0.0/16 into multiple /25

According to the 2^N-2 rule
128.1.0.0/25 is not useable
128.1.0.128/25 is the 1st usable subnet
128.1.1.0/25 is the 2nd usable subnet

https://learningnetwork.cisco.com/thread/32136

With the ip subnet-zero
128.1.0.0/25 is the 1st usable subnet
128.1.0.128/25 is the 2nd usable subnet