Overlapping VLSM Subnets – Speed Test 2 Answers

 In 200-301 V1 Appendices, CCENT-OLD, IPv4 VLSM

The answer for Overlapping VLSM Subnets Speed Test 2 are below the fold! Here’s a complete list of related posts:


No overlaps

While following the process helps, this one was pretty easy if you stepped back from the math and noticed that most of the IP addresses differed in the 2nd octet. The masks were all at least 16 bits long, so none of the subnet ID and subnet broadcast calculations would change the first two octets. Just to repeat the process, in case you’re learning how, here’s the process I suggest when learning:

1) Find the subnet IDs (if starting with IP address/prefix combinations)

2) List from lowest subnet ID to highest

3) List the broadcast address for each subnet next to the subnet ID

4) Compare adjacent entries to look for overlaps

5) If you find an overlap, when comparing the next item, compare to all subnets already known to overlap


Once sorted, you can find the first pair of overlaps working through the list sequentially. Here’s the sorted list.


Original Order Original Values Subnet ID Subnet Broadcast

Overlapping VLSM Subnets - Speed Test 2
Overlapping VLSM Subnets – Speed Test 3
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Hi Wendell
In the original question you listed (5) as:
which would overlap with (1), but in the answer for (5) you listed the 3rd octet as (209) which would not overlap with one, i solved it if the the 3rd octet is (205), so is it a typo or i did something wrong?


Well, I was confused by the table myself. So I expanded it to give more context. Note that the table lists the subnets in a different order – order based on subnet ID values, rather than the order in the original problem. I added the original problem order to the table, as well as those values.

So, you claimed that #5 was different here in the answer; I think you were looking at the wrong row. #5 (from the problem) has its subnet ID etc listed in the 2nd data row of the table here in the answers. My fault – sorry for the confusion.

That said, notice that the 2nd octet is different. So if their 2nd octets had been the same, I agree, there would have been an overlap. But the 2nd octets are not the same value, so no overlap.

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