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i am a noob to cisco packet tracer and i am trying VLSM a subnet

evanmckay927
Level 1
Level 1

i am a noob to cisco packet tracer and i am trying VLSM a subnet with 41 host, 19 hosts, a 2 wan links. 

here are the ip's address 

192.168.0.0    255.255.255.0

192.168.0.64  255.255.255.128

192.168.0.96  255.255.255.192

192.168.0.100 255.255.255.224

and packet tracer is not taking the subnet mask and i have no idea why. can anyone help?

 

 

1 Accepted Solution

Accepted Solutions

You were missing default gateways on PCs so ping ISP failed; I added those. Also, flipped #41 hosts with 19 hosts as /26 and /27 were in wrong places; if you need to put them as they were; then u need to change IPs on router and PCs around; Switches were missing  ip default-gateway x.x.x.x command.

Regards, ML  

View solution in original post

17 Replies 17

@evanmckay927 h, what is the exact command you tried? and what is the exact IP and subnet you entered.

Please rate this and mark as solution/answer, if this resolved your issue
Good luck
KB

i am trying to get the ip address's correct at the moment  

Joseph W. Doherty
Hall of Fame
Hall of Fame

You have overlapping subnets.  Depending on how you're trying to configure them, that can be a problem.

For instance, 192.168.0.0 255.255.255.0 covers 192.168.0.0..255, i.e. it overlaps the next three subnets.

ah i see is my math off which throw my network off. Right??


@evanmckay927 wrote:

ah i see is my math off which throw my network off. Right??


I'm unable to follow your image's logic.

To review, assuming you're starting with 192.168.0.0/24 and allocating largest to smallest subnets, you first need one that will allow for 41 hosts, which will be a 64 address block or a /26

i.e. 192.168.0.0/26 or 192.168.0.0 255.255.255.192

your next largest subnet needs to support 19 hosts which will be a 32 IP address block, or a /27.

i.e. 192.168.0.64/27 or 192.168.0.0 255.255.255.224

finally you need two p2p subnets, which can be provided by /30s

i.e. 192.168.0.96/30 or 192.168.0.0 255.255.255.252

i.e. 192.168.0.100/30 or 192.168.0.0 255.255.255.252

So, in your initial post, you have all the right subnets IPs, but all have incorrect network masks.

Martin L
VIP
VIP

My guess you were given 192.168.0.0/24 (255.255.255.0) to be used for VLSM.  U can use idea of blocks for VLSM and u must start with the largest requirement for hosts; U take original network/subnet divide it by a block subnet mask, and next left over block can be used for another subnet or divided into smaller blocks; i.e. /24 can be divided into 2x /25,  one of both can be further divided into /26, then /27 and so on; 

So, 41 falls into block of 64, that is 62 hosts;  that is 192.168.0.0/26 aka 192.168.0.1 thru .62  with 255.255.255.192; if u used that 1st network (and usually u start with 0 network or given one), the next free network/mask is 192.168.0.64/26; Next requirement is 19 hosts; that is 32 block; which is /27; 64+32=96;  Net 192.168.0.64/27 with 1st IP 192.168.0.65 and last one 0.94.  Next free is 192.168.0.96 /27 which can be divided into many /30 for your WAN links (2 hosts per link);  192.168.0.96/ 30 and 192.168.0.100/30

So, you have error in 192.168.0.64 255.255.255.128; it should be .192 not 128;

I recommend using Magic number table ; link here 

Regards, ML
**Please Rate All Helpful Responses **

I suspect @Martin L has identified the goal of a possible student assignment, which may be, given the 192.168.0.0/24 address block, subnet it to provide a subnet for 41 hosts, 19 hosts and two p2p subnets.

If that's goal, as Martin explains, the forgoing might be provided by a /26, a /27 and two /30s (or possibly /31s), respectly.

As Martin also explains, you've done that incorrectly which is why PT (as would real Cisco devices) reject your configuration changes.

Martins further explains how to subnet correctly, but note the technique for allocations he is using, although very common, and fantastic for providing answers to this kind of assignment, is often not the best way to manage address space allocations in "real" networks.

Oh, thinking about my prior reply, it might seem that I'm implying there is something less that perfect about @Martin L 's reply.  Well, yes, I believe that's the case, but only for "real" world address space management.  For an assignment, or exam question, his approach is "best"!

The "real" world issue with address space management, there's often some "churn".  New subnets are needed, some may need an increase in size, some are no longer needed.  If you don't plan your address space management to allow for this, you can find you've "painted yourself into a corner".

Consider, when you allocate a /30 anywhere within a /24, not only is the /24 no longer available for allocation, but so are all the larger address blocks that contain that /30.

Say you need to allocate two /30s.  If, within a /24, one is allocated anywhere in the top/first /25 and the other allocated anywhere in the bottom/last /25, you cannot allocate either /25 (and 3 /26s and etc.)

The technique Martin describes is fine (perfect) for an initial address space allocation but not often ideal for on-going address space allocations.

i tried the address you suggested and i was invalid. Could it routing protocol in packet tracer?

 

dndnfdnfsfnsfnsdv.pngScreenshot 2024-07-28 230408.png

192.168.0.64/26 is Network ID; 1st IP from it is 192.168.0.65, same as network ID is 192.168.0.0/26 , 1st IP is 192.168.0.1;  

Copy and paste this into your PT 2911 router

interface GigabitEthernet0/0

ip address 192.168.0.1 255.255.255.192

!

interface GigabitEthernet0/1

ip address 192.168.0.65 255.255.255.224

!

interface GigabitEthernet0/2

no ip address

!

interface GigabitEthernet0/2.1

encapsulation dot1Q 1 native

ip address 192.168.0.97 255.255.255.252

!

interface GigabitEthernet0/2.2

encapsulation dot1Q 2

ip address 192.168.0.101 255.255.255.252

Result is 

Router#show ip route connected

C 192.168.0.0/26 is directly connected, GigabitEthernet0/0

C 192.168.0.64/27 is directly connected, GigabitEthernet0/1

C 192.168.0.96/30 is directly connected, GigabitEthernet0/2.1

C 192.168.0.100/30 is directly connected, GigabitEthernet0/2.2

Regards, ML
**Please Rate All Helpful Responses **

 

I do not have interface interface GigabitEthernet0/2.1, GigabitEthernet0/2.2. on the 2 switches that have the only interface i have running on each is vlan 

those are sub-interfaces of a router, 2911 in PT. you do not have it; not needed it; so take 192.168.0.97 255.255.255.252 and move to your serial interfaces going to ISP router on left side

you can attach your PT file here but it must be in  a zip format 

Regards, ML

could you please show me though a video of some sort on what you did?   

 

aedsd.png

You were missing default gateways on PCs so ping ISP failed; I added those. Also, flipped #41 hosts with 19 hosts as /26 and /27 were in wrong places; if you need to put them as they were; then u need to change IPs on router and PCs around; Switches were missing  ip default-gateway x.x.x.x command.

Regards, ML  

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