Question about subnetting and router configuration

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  M_Boyd Member Posts: 31 ■■□□□□□□□□

  April 2004

  dan87951 wrote:

  Here’s my question hopefully you can help me out. I’m currently a second semester student as well. I was given a WAN IP address of 192.168.3.0 with a mask of 255.255.255.0 and a LAN IP address of 172.31.0.0.

  
  ok fourth semester student so ill see what I can do...
  

  I was told to use the second lowest usable WAN address which is 192.168.3.2 correct?

  
  Sounds good to me your host portion is .1 to .254 - add 1 to 1 and you get 2 [;)]
  

  On the local LAN it needed to handle 4000 host address per subnet so I borrowed 4 bits on the LAN address which gave me a subnet mast of 255.255.240.0

  
  This sounds good - this mask will allow 4096 using formula 2 to the power x minus 2......
  

  On the LAN address I was told to use the 3rd usable subnet which would be 172.31.48.0 correct?

  
  Looking at the third octet and the four bits stops just after 16 so therefore it goes in incrediments of 16 - 16 x 3 = 48 very good
  

  I was also told my IGRP AS number is 3? Does this mean when I’m configuring IGRP in global config mode that I just type in "router IGRP 3"?

  
  yup
  

  Here are some of the guidelines for the IP assigning. The production workstations will be assigned the lower-half of the IP address. All network devices and management stations will be assigned the upper half of the IP address numbers and the Ethernet router interface is to be assigned the highest usable address.
  
  Here is what I have.
  
  Network address
  172.31.48.0
  

  Network address is correct
  

  Range of address for that subnet
  172.31.48.1-172.31.63.254

  
  Yup assuming you really mean range of valid IP addresses for that subnet
  

  Broadcast
  172.31.63.255

  
  sounds good to me
  

  Ethernet interface IP
  172.31.63.254??

  
  This forfills the 'Ethernet router interface is to be assigned the highest usable address'
  

  Production host range
  172.31.48.1 - 172.31.55.???
  How do I figure out what number will be in the last octet? Could it be any number?

  
  Well you have to dig a little deeper here into your understanding of binary. To have the lower range the bit after the 4 bits you have assigned in your mask for the subnet will either be 1 or 0 giving you your upper range or lower range.
  
  Your criterior
  The production workstations will be assigned the lower-half of the IP address
  
  So.... basically you need to think of defining your range with keeping that bit after which will seperate the upper from the lower the same - in this case we will use the 0 value afterwards because we want the lower range
  
  So our range will be 172.31.48.1 and 172.31.55.255 remember this is not the broadcast so its still valid....
  
  Heres the binary conversion of the third octet 1111|0|111. with the line going through the host portion....and the bit that defines lower range
  

  Management Host range
  172.31.56.???- 172.31.63.254

  with doing the same idea as above but setting that bit to 1 to define the higher addresses and getting the lowest number out the host portion to define the lowest valid IP
  
  172.31.56.0 not a network - 172.31.63.254
  
  third octet - 1111|1|000
  

  So one of the production hosts could be:
  IP address 172.31.48.1
  Subnet mask 255.255.240.0
  Default gateway 172.31.63.254

  
  yup
  

  and on of the management hosts could be:
  IP address 172.31.56.???
  Subnet mast 255.255.240.0
  Default gateway 172.31.63.254

  
  Ip addresses anywhere between 172.31.56.0 to 172.31.63.253 are all valid.....with 254 as your gateway
  
  

  Thanks for the help I know this is long.

  
  Your welcome - subnetting is one of my stronger areas....

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