VLSM summarization

daras

Hi,

Can you please help me with the summarization of these routes ?

1) 10.1.128.0/20
2) 10.1.32.64/26
3) 10.1.32.128/26
4) 10.1.64.0/20

I belive it should be 10.1.0.0 /16but i am not very sure

Another example

1) 172.1.4.0 /25
2) 172.1.4.128/25
3) 172.1.5.0/24
4) 172.1.6.0/24
5) 172.1.7.0/24

This is what i do

172.1.4.0/25 and 172.1.4.128/25

128=10000000
0= 00000000

Route summary is 172.1.4.0/24

Now summarize with the rest

4=00000100
5=00000101
6=00000110
7=00000111

so common bits 16+6=22 and route summary 172.1.0.0/22

but the answer should be 172.1.4.0/22 Whyyyyyy ?????



can you help ?

jsb515

1) 172.1.4.0 /25
2) 172.1.4.128/25
3) 172.1.5.0/24
4) 172.1.6.0/24
5) 172.1.7.0/24

you have 5 routes here. So do your powers of 2. 2,4,8. These 5 routes will fit within the power of 2^3 since it = 8. Now take the 3 and subtract it from /25. /25-3 = /22

172.1.4.0 /22

255.255.252.0
11111111.11111111.11111100.00000000

10 host bits = 2^10-2=1024 host
This is a class B address with a class C mask. To get our total of subnets do this
Subtract the C address 22 from the Class B address 16. 22-16=6. Put this to the power of 2. 2^6=64 subnets
our network bits stopped in the 3rd octet with only 2 host bits left. Put that to the power of 2. 2^2=4 this is our subnet count. which is 172.1.4.0

Hope this helps some.

jsb515

1) 10.1.128.0/20
2) 10.1.32.64/26
3) 10.1.32.128/26
4) 10.1.64.0/20

4 routes. Powers of 2. 2,4 =2^2=4
subtract the 2 from /26 = /24

10.1.0.0 /24 would be your summarized address.

d6bmg

^^ Last one's wrong, mate.

d6bmg

IP - subnet - broadcast
1) 10.1.128.0/20 - 10.1.128.0 - 10.1.143.255
2) 10.1.32.64/26 - 10.1.32.64 - 10.1.32.127
3) 10.1.32.128/26 - 10.1.32.128 - 10.1.32.191
4) 10.1.64.0/20 - 10.1.64.0 - 10.1.79.255

Range: 10.1.32.64 - 10.1.143.255
Summary: 10.1.0.0/16

jsb515

thanks d6bmg, thought my theory was good until that 10.1.0.0 series of subnets came up... that was a hard one to answer and counting the bits does seem to be the best method for this. Thanks for the correct answer.

jsb515

this correct way of doing it?

10.1.32.64
10.1.32.128
10.1.64.0
10.1.128.0

00001010.00000001|.00100000.01000000
00001010.00000001|.00100000.10000000
00001010.00000001|.01000000.00000000
00001010.00000001|.10000000.00000000

So count the common bits in blue 8+8=16

So it is 10.1.0.0 /16

theodoxa

Correct me if I'm wrong, but if an octet contains numbers on both sides of 128 (e.g. 128 and 64), then that octet and all following octets must be all zeroes when summarized.

1) 10.1.128.0/20
2) 10.1.32.64/26
3) 10.1.32.128/26
4) 10.1.64.0/20

When I see #1 with the rest being below 128, I know it can't be any smaller than /16 and since the second octet is the same...there is no need to go any larger than a /16.

d6bmg

theodoxa wrote: »

When I see #1 with the rest being below 128, I know it can't be any smaller than /16 and since the second octet is the same...there is no need to go any larger than a /16.

Yeah, that's a shortcut.

d6bmg

jsb515 wrote: »

this correct way of doing it?

10.1.32.64
10.1.32.128
10.1.64.0
10.1.128.0

00001010.00000001|.00100000.01000000
00001010.00000001|.00100000.10000000
00001010.00000001|.01000000.00000000
00001010.00000001|.10000000.00000000

So count the common bits in blue 8+8=16

So it is 10.1.0.0 /16

Yes, that's the exact way to do it. But you need to consider the subnet masks to get the range of the network.