IP addresses associated operations (such as VLSM, super network summary)

+ 1. The IP address of the subnet mask segment is calculated from the IP address (network number)

example:

IP address of 192.168.10.33, subnet mask: 255.255.255.240 (/ 28), where the number of write network

1. The results subnet block size : 256-240 = 16

2. Because the first three 8 bytes are 255, so (with 1 or with the operation itself), so this problem can focus on the fourth byte, that 192.168.10.x

3. The fourth 16 * 2 = 32, the next segment is 3 * 16 = 48, so the segment is located 192.168.10.32

4. The number of legitimate network host is 192.168.10.33 - 192.168.10.46 ( since the network number 32, 47 for the network broadcast address )

5. VLSM because it is divided, the IP address is performed according to the class C subnet, thus 28-24 = 4 bits Network subnets

Therefore, 2 ^ 4 = 16 subnets, each 16-2 = 14, there are number of hosts.

PS:

If Subnet in WAN links (point to point link), then, as a network of only two routers at both ends, so use a mask of 255.255.255.252 (ie, / 30), why can be used, because the point the point is not broadcasting.

 

2.VLSM design and distribution

Note: VLSM design should pay attention to each separate subnet, because it is not a conflict between subnets.

Background Condition: generally under the premise of a segment, with a different mask to a reasonable division of this segment, the number of hosts to adapt to different subnets have.

Step: general start with a large number of hosts to start dividing.

For example: 192.168.10.0/26 (64 units) 192.168.10.64/27 (32 units) 192.168.10.96/28 (16 units)

 

Private and public address range

private:

1. Class A: 10.0.0.0 - 10.255.255.255.255

2. Class B: 172.16.0.0 - 172.31.255.255

3. Class C: 192.168.0.0 - 192.168.255.255.

Public:

1. Class A: 1.0.0.0 - 127.255.255.255 (01 ....)

2. Class B: 128.0.0.0 - 191.255.255.255 (10 ....)

3. Class C: 192.0.0.0 - 223.255.255.255 (110 ..)

4. Class D: 224.0.0.0 - 239.255.255.255 (1110 ..)