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I'm studying for the CCNA exam and I want to create a VLSM scheme using RIPv2 for the following (CCNA practice example):

  • Network connecting 4 different locations
  • Each switch is fa 0/0 on the router
  • Every point-to-point serial connection is in the network is class C (192.168.1.0 )
  • Network 172.20.0.0 for each location
  • Use the lowest subnet number (zero is allowed)

Requirements for each location:

  • Dallas - 500 hosts
  • New York - 2000 hosts
  • Chicago - 1000 hosts
  • LA - 2000 hosts

Here's my design so far:
enter image description here

Do you see any flaws in this? How would you fix them? Or how would you design this?

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7  
Smells like homework. –  Tom O'Connor Apr 2 '11 at 18:45
    
Yeah, it was a practice question for the CCNA and I wanted to make sure I was starting off right. –  에이바 Apr 4 '11 at 4:25
    
gns3.net/phpBB/… –  에이바 Apr 8 '11 at 1:29
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3 Answers 3

It looks like you're asking us to do your homework... anyway, I do see some flaws in the subnetting.

Do you really understand how subnetting works ? It doesn't seem so.

This post may help you to get the picture, but let's improve your design & skills:

  • Dallas : 500 hosts => 512 => /23
  • New York : 2000 hosts => 2048 => /21
  • Chicago : 1000 hosts => 1024 => /22
  • LA : 2000 hosts => 2048 => /21

On this part you're right. But here is the fault :

 172.20.0.1/23 = 172.20.0.0 -> 172.20.1.255
 172.20.0.1/22 = 172.20.0.0 -> 172.20.3.255
 172.20.4.1/21 = 172.20.0.0 -> 172.20.7.255
 172.20.6.1/21 = 172.20.0.0 -> 172.20.7.255

So all your networks are overlapping, which isn't good. Each address must be unique. So a correct design would be like this :

 172.20.0.0/21  = 172.20.0.0  -> 172.20.7.255
 172.20.8.0/21  = 172.20.8.0  -> 172.20.15.255
 172.20.16.0/22 = 172.20.16.0  -> 172.20.19.255
 172.20.20.0/23 = 172.20.20.0 -> 172.20.21.255

You can start from the largest networks to the smallest, with the starting address being the last+1 from the previous network.

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The 172.20.16.0/22 should = 172.20.16.0 ...right? –  에이바 Apr 4 '11 at 16:49
1  
@에이바: indeed, copy/paste error :) –  petrus Apr 4 '11 at 17:34
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Your Serial Connections subnet is off: You are specifying a 192.168.1.0/30 for 4 point to point connections @ 2 addresses each = 8 needed IPs. A /30 CIDR will only yield 2 available host addresses (enough for only one connection). You should have a single /30 subnet for each point to point:

192.168.1.0/30  = 192.168.1.0  -> 192.168.1.3  (Use .1 & .2)
192.168.1.4/30  = 192.168.1.4  -> 192.168.1.7  (Use .5 & .6)
192.168.1.8/30  = 192.168.1.8  -> 192.168.1.11 (Use .9 & .10)
192.168.1.12/30 = 192.168.1.12 -> 192.168.1.15 (Use .13 & .14)
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Thanks, I actually just figured it out. –  에이바 Apr 11 '11 at 16:14
    
Your diagram is still technically wrong though. –  Cheekaleak Apr 11 '11 at 16:19
    
How so? I posted a new answer the the diagram. –  에이바 Apr 11 '11 at 16:22
    
Yes, the new diagram is still technically incorrect. You are only specifying one of the four serial connection subnets. –  Cheekaleak Apr 11 '11 at 16:29
    
i.imgur.com/VuhGX.png –  에이바 Apr 11 '11 at 16:40
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up vote 0 down vote accepted

Switches:

New York --> LGA 172.20.0.0/21
Los Angeles --> 172.20.8.0/21
Dallas --> DFW 170.20.20.0/23
Chicago --> ORD 172.20.16.0/22

Routers:

192.168.1.13 ORD s0/1
192.168.1.14 LAX s0/0

192.168.1.1 DFW s0/0
192.168.1.2 LAX s0/1

192.168.1.5 DFW 0/1
192.168.1.6 LGA s0/0

192.168.1.9 LGA s0/1
192.168.1.10 ORD s0/0

enter image description here

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