Given the following IP address/CIDR: 193.16.20.35/29
- what is the Network IP
- number of hosts
- range of IP addresses
- broadcast IP from this subnet
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To solve this you have to first figure out the subnet mask from the given CIDR (/29)
Note: The CIDR indicates the total number of active bits (1's) which also indicates the network portion of the IP address while the 0's indicate the host portion of the network.
Converting it to binary then gives:
Network Portion: 1's
Host Portion: 0's
Netmask Binary: 11111111.11111111.11111111.11111000
Then inorder to convert the subnet mask address from binary to decimal apply the following formular
Note: In the binary system there are only 1s and 0s. Depending on their position in the octet, they get different values. Each position is a power of 2. To get the decimal number you have to sum up those numbers.
| First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|
| 2^7 | 2^6 | 2^5 | 2^4 | 2^3 | 2^2 | 2^1 | 2^0 |
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
Total no. of octets in binary: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
= 255
| First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Total: 255
| First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Total: 255
| First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Total: 255
| First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Total: 248
Octet Sum = 255.255.255.248
Therefore: 11111111.11111111.11111111.11111000 in binary = 255.255.255.248
Wild card = subtract the subnet mask from 255.255.255.255
= 255.255.255.255 - 255.255.255.248
= 255 - 248
= 7
... Wild card = 0.0.0.7
Where
Subnet Mask = 11111111.11111111.11111111.11111000
Given IP = 11000001.00010000.00010100.00100011
To find the network ID, simply do binary and operation between the given IP address and the subnet mask:
Binary and operation between (255 & 193) or (11111111 & 11000001)
| N/A | First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|---|
| Subnet Mask | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| Given IP | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
| Result | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
Total Sum = 193
Binary and operation between (255 & 16) or (11111111 & 00010000)
| N/A | First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|---|
| Subnet Mask | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| Given IP | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| Result | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
Total Sum = 16
Binary and operation between (255 & 20) or (11111111 & 00010100)
| N/A | First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|---|
| Subnet Mask | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| Given IP | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
| Result | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
Total Sum = 20
Binary and operation between (248 & 35) or (11111000 & 00100011)
| N/A | First Octet | Second Octet | Third Octet | Fourth Octet | Fifth Octet | Sixth Octet | Seventh Octet | Eight Octet |
|---|---|---|---|---|---|---|---|---|
| Subnet Mask | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
| Given IP | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
| Result | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
Total Sum = 32
... network IP address = 193.16.20.32
Number of Hosts = 2^n - 2
Where n = number of host bits minus two
= number of host bits - 2
This is because the first and last IP addresses are always reserved for the network and broadcast ID's respectively
Note In this case we have to count the number of host's bits (0's) in the subnet mask binary starting from the right, which will give a total of 3
Number of hosts = 2^3 - 2
= 8 - 2
= 6
From the details above we can then calculate the range of IP addresses and the broadcast:
Note The first & last IP adress will be reserved for the network and broadcast so:
Given IP: 193.16.20.35/29
Network IP: 193.16.20.32
Number of Hosts: 6
Range of IP Addresses: 193.16.20.33 - 193.16.20.38
- min range of IP's = 193.16.20.33
- max range of IP's = 193.16.20.38
Broadcast IP: 193.16.20.39