Jump to content


 


Register a free account to unlock additional features at BleepingComputer.com
Welcome to BleepingComputer, a free community where people like yourself come together to discuss and learn how to use their computers. Using the site is easy and fun. As a guest, you can browse and view the various discussions in the forums, but can not create a new topic or reply to an existing one unless you are logged in. Other benefits of registering an account are subscribing to topics and forums, creating a blog, and having no ads shown anywhere on the site.


Click here to Register a free account now! or read our Welcome Guide to learn how to use this site.

Photo

Help with Networking Class, please


  • Please log in to reply
7 replies to this topic

#1 SaintGrimm

SaintGrimm

  • Members
  • 2 posts
  • OFFLINE
  •  
  • Local time:09:02 PM

Posted 04 May 2016 - 02:41 PM

Hello! I'm new to the forum, joined to get a better understanding of my Networking class.

 

I currently have homework that I'm unable to understand and I'm hoping someone here can show me the way to find the answer.

 

Given information:

 

IP/MASk - 172.16.32.0/24

Subnet A - 75 hosts

Subnet B - 10 hosts

 

Questions:
 

Subnet A:
 

Number of bits in the subnet - would this be 8 bit? 255.255.255.0 = 8 host bits. Is that the same thing?
 

IP mask (binary) - Is IP mask and Subnet mask the same thing? If they are same thing, answer is 11111111.11111111.11111111.00000000 (/24=255.255.255.0 convert to decimal?)
 

New IP mask (decimal) - ??? (Why does converting mask to decimal give a "new" mask?)
 

Maximum number of usable subnets (including the 0th subnet) - ??? (no clue on this)
 

IP subnet - ??? (no clue)


Edited by SaintGrimm, 04 May 2016 - 03:12 PM.


BC AdBot (Login to Remove)

 


#2 Sneakycyber

Sneakycyber

    Network Engineer


  • BC Advisor
  • 6,123 posts
  • OFFLINE
  •  
  • Gender:Male
  • Location:Ohio
  • Local time:04:02 PM

Posted 04 May 2016 - 09:59 PM

Subnet A would be 172.16.32.0/25 This gives you 126 hosts and 128 subnets. In binary it's 25 mask bits and 9 subnet bits. Subnet B would be 172.16.32.0/28 This gives you 14 hosts and 16 subnets. Mask bits are 28 subnet bits are 12. The maximum number of Subnets for 172.16.32.0/24 is 256.

Edited by Sneakycyber, 04 May 2016 - 10:04 PM.

Chad Mockensturm 
Network Engineer
Certified CompTia Network +, A +

#3 SaintGrimm

SaintGrimm
  • Topic Starter

  • Members
  • 2 posts
  • OFFLINE
  •  
  • Local time:09:02 PM

Posted 05 May 2016 - 10:10 AM

I thank you for your reply, but I have gotten some of it figured out by researching and I'm looking for how to do it myself instead of just the answers.

 

 

This is what I've gotten so far:

 

IP - 172.16.32.0/24

Hosts in Subnet A - 75

 

To find the network address, I did IP binary + /24 binary. 1+0=0 0+1=0 1+1=1. This gave me network address of 172.16.32.0 (same thing as IP)

The closest to 75 hosts is 126 hosts, which is /25. Broadcast is 172.16.32.128 if I figured out the broadcast correctly.

 

(all questions pertain only to subnet A as I haven't started Subnet B yet)

 

Question - IP Subnet: 172.16.32.0/25

Question - First IP Host Address - 172.16.32.1/25
Question - Last IP Host Address - 172.16.32.126/25
Question - Number of Usable hosts per subnet - 128 - broadcast - network = 126 usable hosts

Question - Maximum number of usable subnets including the 0th subnet - 128?

 

Question - Number of bits in the Subnet - still haven't learned how to figure this out.

Question - IP Mask (Binary) - Still can't figure out what an IP mask is, is the same at subnet mask? /25?

Question - New IP Mask (Decimal) - I can't figure out why it gets a "new" mask... And it's very frusterating.



#4 Wand3r3r

Wand3r3r

  • Members
  • 2,027 posts
  • OFFLINE
  •  
  • Local time:01:02 PM

Posted 05 May 2016 - 05:12 PM

"Subnet B would be 172.16.32.0/28"

sorry but not quite right.  you can't have subnet A and B in the same space

 

subnet A at 172.16.32.0/25 gives you a range of 0-127

subnet B would start at 172.16.32.128

with a /28 would give you a range of 128-143

 

Network id is at the bottom and broadcast is at the top of the range.  You can not have 172.16.32.128 as broadcast for the /28 subnet since it is not included in that subnet.  Correct answer is 0 is network id and 127 is broadcast for the A subnet

 

but then I haven't done this by hand for many years.  Online subnet calculators are the way to go

For figuring the binary

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

 

For CIDR

http://www.subnet-calculator.com/cidr.php

 

If you google subnet cheat sheets you can see there are patterns to the subnets. They fall into precise ranges.


Edited by Wand3r3r, 05 May 2016 - 05:14 PM.


#5 Sneakycyber

Sneakycyber

    Network Engineer


  • BC Advisor
  • 6,123 posts
  • OFFLINE
  •  
  • Gender:Male
  • Location:Ohio
  • Local time:04:02 PM

Posted 05 May 2016 - 05:25 PM

Thanks for the correction Wand3r3r. I never was able to do submitting very well. The worst part is I used the subnet calc to get the answers but read them wrong! :(
Chad Mockensturm 
Network Engineer
Certified CompTia Network +, A +

#6 Wand3r3r

Wand3r3r

  • Members
  • 2,027 posts
  • OFFLINE
  •  
  • Local time:01:02 PM

Posted 05 May 2016 - 06:33 PM

We all have off days Sneakycyber :-)



#7 technonymous

technonymous

  • Members
  • 2,502 posts
  • OFFLINE
  •  
  • Gender:Male
  • Local time:01:02 PM

Posted 05 May 2016 - 07:42 PM

Using the magic line method is probably best if you're doing all this in your head.



#8 Wand3r3r

Wand3r3r

  • Members
  • 2,027 posts
  • OFFLINE
  •  
  • Local time:01:02 PM

Posted 05 May 2016 - 07:59 PM

magic box method :-)

 

http://kevinhodge.com/WP/?p=103






0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users