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.

 

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