Quick Question about VLSM

redgunner

Hello all,

I would just appreciate some clarification on something when working with Variable Length Subnet Masks

So this is the example I have created myself

Sales team need 37 IP Addresses and HR needs 7 IP Addresses

Sales Team
Subnet 192.168.0.0/26
Broadcast 192.168.0.64
Mask 255.255.255.192
Range 192.168.0.1 to 192.168.0.63

So I am using my block table so if HR team needs 7 IP Addresses must I plus 2 (reserved for subnet and broadcast)

If I look at my block table 192.168.0.65/29 would be wrong because I can only get 8 IP addresses max out of it so if I wanted 7 I have to bear in mind that I must reserve subnet and broadcast? so the answer would be to move up a block and say 192.168.0.65/28 would be the answer because I can get 16 IP addresses out of it maximum so HR needs 7 IP Addresses + 2 for subnet and broadcast?

Thanks for your help in advance!!

GDaines

redgunner wrote: »

Hello all,

I would just appreciate some clarification on something when working with Variable Length Subnet Masks

So this is the example I have created myself

Sales team need 37 IP Addresses and HR needs 7 IP Addresses

Sales Team
Subnet 192.168.0.0/26
Broadcast 192.168.0.64
Mask 255.255.255.192
Range 192.168.0.1 to 192.168.0.63

So I am using my block table so if HR team needs 7 IP Addresses must I plus 2 (reserved for subnet and broadcast)

If I look at my block table 192.168.0.65/29 would be wrong because I can only get 8 IP addresses max out of it so if I wanted 7 I have to bear in mind that I must reserve subnet and broadcast? so the answer would be to move up a block and say 192.168.0.65/28 would be the answer because I can get 16 IP addresses out of it maximum so HR needs 7 IP Addresses + 2 for subnet and broadcast?

Thanks for your help in advance!!

Network address is 192.168.0.0 /26
Broadcast address is 192.168.0.63 (0-63, 64-127, 128-191 and 192-255)
Mask 255.255.255.192
Valid Host Range 192.168.0.1 to 192.168.0.62

64 is the network address of the next block so be careful with that.

Your other assumptions are correct, the number of valid hosts is block size minus 2 (reserved for network and broadcast addresses).

Block size of 8 = 6 valid host addresses, so you correctly assumed a block size of 16 giving 14 valid host addresses is needed for 7 clients. And the same with the other block - a block size of 32 gives 30 valid host addresses which is short so you have to use a block size of 64 giving you 62 valid host addresses when dealing with class C addresses.

DPG

Keep in mind that in real life you will want to reserve at least one IP for a gateway if you plan on routing the network. So block size minus 3.

Example: /29 = 5 usable IP's for hosts, /28 = 13 usable, /28 = 29 usable, etc.