# Mistake #1

This is the problem that was randomly generated with my answers:

## IP=152 . 207 . 137 . 102 / 27

Mask = 255 . 255 . 1 1 1 1 1 1 1 1 . 1 1 1 0 0 0 0 0

Convert 137 to bin:   1 0 0 0 1 0 0 1
----------------
Do bit-wise AND: 1 0 0 0 1 0 0 1 (= 137) Magic = 32

## Find the Answers for each

1. What is the Dotted Decimal Subnet Mask? 255 . 255 . 255 . 224
2. Find this SubNetwork 152 . 207 . 137 . 0
3. Find the First useable IP address 152 . 207 . 137 . 1
4. Find the Broadcast address 152 . 207 . 168 . 255
5. Find the Last useable IP address 152 . 207 . 168 . 254
6. What is the Total Number of IPs per Subnet? 32 - 27 = 5 ; 25 = 32
7. How many useable Hosts Addresses per subnet? 32 - 2 = 30
8. What is the Next Subnetwork Number? 152 . 207 . 169 . 0
9. What IP Class is this? Class B
10. How many possible Subnetworks with this IP Class and mask? 27 - 16 = 11 ; 211 = 2048

As you can see, I got numbers 2, 3, 4, 5 and 8 wrong. The correct answers should have been:

2. 152 . 207 . 137 . 96

3. 152 . 207 . 137 . 97

4. 152 . 207 . 137 . 127

5. 152 . 207 . 137 . 126

8. 152 . 207 . 137 . 128

### What did I do wrong?

When I calculated this subnetwork number, I was working in the wrong octet of the given dotted quad!

Always, always work in the octet that contains the subnet / host crossover! So, instead of using the third octet to find the subnetwork address in this case, I should have been using the fourth octet to find the subnetwork address.

<-- Back to the main Subnetting page