Question

CIDR notation - First 2 answers given:

If the dotted decimal subnet mask is given then give me the CIDR notation

If the CIDR notation subnet mask is given give me the dotted decimal notation

1. 255.255.240.0            ______________________________________
2. /30                      ______________________________________

What network does this address belong to?

3- 200.198.44.220/26       _____________________________

4. 112.55.78.120/16        _____________________________

What is the broadcast address for this network? (The easiest way to get the broadcast address is to figure out the next network and subtract 1 or put all 111? in the host portion of the address

   5- 200.150.100.64/27   ________________________________

6. 209.100.43.180/16   _____________________________

If you have these many network bits, what is the number of useable host addresses?

Ex: I have a /27 network

N + H = 32 , so 27 + X =32 , X = 5 (number of host bits), 2 to the power of X = the total hosts, minus 2 = number of useable hosts

So if X = 5, then 2^5=32 total hosts, and if I subtract 2 then I have 30 useable host addresses on a /27 network.

7 - /30                   _______________________________

8. 255.255.255.240       _______________________________

If you have this many usable hosts, what is your subnet mask?

9. 14    ____________________
10. 126   ____________________

Answer 1

1. 255.255.240.0

Subnet mask in CIDR notation for the above decimal dotted notation:

11111111.11111111.11110000.00000000

The number of 1's is 20. Therefore the subnet mask in CIDR notation is: /20

2. /30

The subnet mask in decimal dotted notation for the above CIDR notation is:

For /30, there will be 30 1's ,

11111111.11111111.11111111.11111100

In decimal dotted notation: 255.255.255.252

3. The address 200.198.44.220/26 belongs to network:

With /26, the two msb of last octet belongs to network portion and remaining 6 bits are for host portion. Thus by putting the remaining 6 bits as zero, will give the network it belongs to:

200.198.44.192/26

4. 112.55.78.120/16

With /16, the remaining 16 lsb bits are host portion. Thus by making the 16 lsb bits to zero, will give the network it belongs to:

112.55.0.0/16

5. For the address 200.150.100.64/27, the broadcast address by putting the host portion all 1's, where the host portion is 32-27=5 lsb bits

200.150.100.95/27 is the broadcast address