Need help on VLSM
Please help me solve this sample lab correct me if I am wrong with my solution.
Given network is 192.168.24.0/22 or mask of 252
Subtract 256 – 252 = gives me block size of 4
These arethe Subnets 0, 4, 8, 12, 16, 20 up to 252, 256 that ISP distribute.
192.168.4.0
192.168.8.0
192.168.20.0
Up to
192.168.252.0
ISP provide me an address of 192.168.24.0/22
I need to create an IP addressing scheme using this given address.
I have 4 routers which needs the following number of Hosts.
Router A needs 400 hosts
Router B needs 200 hosts
Router C needs 50 hosts
Router D needs 50 hosts
So my first step is to get the highest number which is router A 400 hosts
For Router A IP addressing scheme
I gave the IP addressing scheme of:
Subnet:192.168.24.0/23
Available IP’s: 192.168.24.1 to 192.168.25.254
Broadcast: 192.168.25.255
Next Router B needs 200 hosts
Subnet I used is 192.168.26.0/24
Subnet: 192.168.26.0
Available IP’s:192.168.26.1 to 192.168.26.254
Broadast: 192.168.26.255
Now I am having problem what IP will I assign to router C and D since they need both 50 hosts.
I am thinking if I can use the broadcast of 27 since I can enable the IP subnet zero on the router. 24, 25, 26 is already used in Router A and B.
Is it possible? If yes then
The subnet for router C is:
192.168.27.0/18 mask of 192
192.168.27.1 to 192.168.27.62
192.168.27.63
For Router
192.168.27.64 is my subnet
192.168.27.65 to 192.168.27.126
192.168.27.127 is my broadcast
Still I have plenty of available IP address 192.168.27.128 to 192.168.27.254 broadcast 192.168.27.255 and the next subnet that ISP will give is 28
32
And to summarize address this network to the ISP if I am using RIP version 2
192.168.4.0/22 or 255.255.252.0
Just want to ask you if my solution to this lab is correct.
Thanks
Given network is 192.168.24.0/22 or mask of 252
Subtract 256 – 252 = gives me block size of 4
These arethe Subnets 0, 4, 8, 12, 16, 20 up to 252, 256 that ISP distribute.
192.168.4.0
192.168.8.0
192.168.20.0
Up to
192.168.252.0
ISP provide me an address of 192.168.24.0/22
I need to create an IP addressing scheme using this given address.
I have 4 routers which needs the following number of Hosts.
Router A needs 400 hosts
Router B needs 200 hosts
Router C needs 50 hosts
Router D needs 50 hosts
So my first step is to get the highest number which is router A 400 hosts
For Router A IP addressing scheme
I gave the IP addressing scheme of:
Subnet:192.168.24.0/23
Available IP’s: 192.168.24.1 to 192.168.25.254
Broadcast: 192.168.25.255
Next Router B needs 200 hosts
Subnet I used is 192.168.26.0/24
Subnet: 192.168.26.0
Available IP’s:192.168.26.1 to 192.168.26.254
Broadast: 192.168.26.255
Now I am having problem what IP will I assign to router C and D since they need both 50 hosts.
I am thinking if I can use the broadcast of 27 since I can enable the IP subnet zero on the router. 24, 25, 26 is already used in Router A and B.
Is it possible? If yes then
The subnet for router C is:
192.168.27.0/18 mask of 192
192.168.27.1 to 192.168.27.62
192.168.27.63
For Router
192.168.27.64 is my subnet
192.168.27.65 to 192.168.27.126
192.168.27.127 is my broadcast
Still I have plenty of available IP address 192.168.27.128 to 192.168.27.254 broadcast 192.168.27.255 and the next subnet that ISP will give is 28
32
And to summarize address this network to the ISP if I am using RIP version 2
192.168.4.0/22 or 255.255.252.0
Just want to ask you if my solution to this lab is correct.
Thanks
Comments
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EdTheLad Member Posts: 2,111 ■■■■□□□□□□Ching01 wrote:
And to summarize address this network to the ISP if I am using RIP version 2
192.168.4.0/22 or 255.255.252.0
Just want to ask you if my solution to this lab is correct.
Thanks
Yes vlsm correct, but to summarize you will use 192.168.24.0/22Networking, sometimes i love it, mostly i hate it.Its all about the $$$$ -
EdTheLad Member Posts: 2,111 ■■■■□□□□□□Ching01 wrote:Thanks ah sorry it's 192.168.24.0/22 typo error
You stated at the beginning of your question the isp provided you with the block 192.168.24.0/22, you then used vlsm to breakup the block.You then
talk about summarizing the block using 192.168.4.0/22, this would summarize networks 192.168.4.0 -> 192.168.7.255, you are using .24 not .4
so you would summarize what you were given in the first place, not that there is any point to summarize back to the isp.Networking, sometimes i love it, mostly i hate it.Its all about the $$$$ -
BubbaJ Member Posts: 323Ching01 wrote:Given network is 192.168.24.0/22 or mask of 252
Subtract 256 – 252 = gives me block size of 4
These arethe Subnets 0, 4, 8, 12, 16, 20 up to 252, 256 that ISP distribute.
2 networks, 512 addresses, 510 hosts
4 networks, 256 addresses, 254 hosts
8 networks, 128 addresses, 126 hosts
16 networks, 64 addresses, 62 hosts
32 networks, 32 addresses, 30 hosts
64 networks, 16 addresses 14 hosts
128 networks, 8 addresses, 6 hosts
256 networks, 4 addresses, 2 hostsChing01 wrote:I am thinking if I can use the broadcast of 27 since I can enable the IP subnet zero on the router. 24, 25, 26 is already used in Router A and B.
CIDR was designed to conserve address space and allow an ISP to assign the proper size address space, ignoring class. In order to facilitate address conservation, there is no subnet zero/broadcast subnet reservation. You always use those subnets. -
Ching01 Member Posts: 83 ■■□□□□□□□□I remember I think from my Sybex CCNA that using CIDR ignoring the ip subnet zero and broadcast. I can now use this address to assign on my network.
Just want to clear VLSM is used for private network IP addressing scheme and CIDR is used by ISP to conserve and assign IP address to an organization or company.
Anyway, last night I tried to do some NAT thing but I got a message Expired in transit, and check it 0 % loss. I just follow the lab setup on the CCNP academy program.