VLSM check in.
Admiral Akmir
Member Posts: 40 ■■□□□□□□□□
in CCNA & CCENT
Hey guys, I was bored at work today and realized it had been a long time since I practiced VLSM, so I drew up a scenario and came up with the solution. If someone could check the work to make sure it's sound that would be great.
Requirements for each network
1) 650 hosts
2) 200 hosts
3) 78 hosts
4) 12 hosts
Block sizes for above networks
1) 1024
2) 256
3) 128
4) 16
The networks
1) 192.168.0.0 /22
2) 192.168.4.0 /23
3) 192.168.6.0 /25
4) 192.168.6.128 /28
I wanted to practice summarization too, and so I came up with 192.168.0.0 /21 as the summary address for all of those, which I think is right.
I'm really sorry if I forgot to include some information, but I have to be out the door right now.
Requirements for each network
1) 650 hosts
2) 200 hosts
3) 78 hosts
4) 12 hosts
Block sizes for above networks
1) 1024
2) 256
3) 128
4) 16
The networks
1) 192.168.0.0 /22
2) 192.168.4.0 /23
3) 192.168.6.0 /25
4) 192.168.6.128 /28
I wanted to practice summarization too, and so I came up with 192.168.0.0 /21 as the summary address for all of those, which I think is right.
I'm really sorry if I forgot to include some information, but I have to be out the door right now.
Comments
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verbhertz
Member Posts: 54 ■■□□□□□□□□
The mask for 2) should be /24.
Putting 3) at 192.168.5.0/25.
And 4) at 192.168.5.128/28.
I believe your summarization is correct. -
davenull
Member Posts: 173 ■■■□□□□□□□
/24 would give him 254 hosts, gotta remember to subtract 2 for network and broadcast addresses -
Admiral Akmir
Member Posts: 40 ■■□□□□□□□□
1) 192.168.0.0 - 192.168.3.255
2) 192.168.4.0 - 192.168.4.255
3) 192.168.5.0 - 192.168.5.127
4) 192.168.5.128 - 192.168.5.143
Remember that I didn't need 256 hosts, only 200, so /24 is more than enough. I either made the mistake of thinking I needed 256, or I got mixed up with place values ( 128, 64, 32, 16, 8, 4, 2, 1 ) Thinking that a /24 would somehow only give me 128, I've made that mistake in the past.