Hi Alishah,
There is no single method to compute summarized ranges. It is interesting how often these types of questions pop up in exams but in the reality, the things are reversed: You as a network administrator are given a prefix which you can subnet anyway you like, but the original prefix is the summary network you can use to summarize all subnets you have created under it.
The ranges you have indicated can not be validly summarized because they do not represent address ranges that would go nicely one after another. For example, the first network 10.10.1.0/30 goes from 10.10.1.0 to 10.10.1.3. The second network goes from 10.10.4.0/30 to 10.10.4.3. The gap between 10.10.1.4 to 10.10.3.255 is not your space. We can create a network prefix that will cover all the subnets you have indicated but it will not be a valid summary address because it will also cover some addresses that do not belong to you.
Okay, with the addresses you have indicated, we need to compute the lowest and the highest address in the overall range. Obviously, the lowest address is 10.10.1.0 , the highest address is 10.10.16.3.
Then, find the leftmost octet that is different between the lowest and the highest address. In our case, it is the third octet that starts with 1 and ends with 16.
Now we need to find a suitable covering network size that will cover all values of the third octet between the lowest and highest address. Keep in mind that the network size within a single octet can be only a power of 2, that is, 1, 2, 4, 8, 16, 32, 64, or 128, and that the resulting network address must be a multiple of such power of 2.
With this in mind, the range of values in the third octet is 1-16, that would be 16 different values. So the covering network will be of size of at least 16 in the third octet. However, a network of size 16 must start at a multiple of 16, and 1 is not a multiple of 16. We must therefore find the nearest lower multiple of 16 to 1. Obviously, 0 is such number. So if we start with 0 in the third octet with the size of 16, we will cover the range 0 - 15 in the third octet. As you can see, this range is not large enough because it does not cover all values we need to cover (16 is uncovered). So let's double the range, from 16 to 32, and try it again: The nearest smaller multiple of 32 to 1 is 0 again. In this case, the range in the third octet would go from 0 to 31, safely covering all numbers inbetween. Therefore, that is our network size in the third octet, and the network address itself will contain 0 in the third octet as that is our multiple of 32 we've started with.
Finally, write down the mask. For the octets that have not changed, keep the value 255 in the netmask. For the leftmost octet where we were looking for the appropriate network size, compute the netmask as (256-size) = (256-32) = 224. For the remaining octets, set the netmask to 0. Thus, the resulting netmask is 255.255.224.0, and the network is 10.10.0.0.
Feel welcome to ask further.
Best regards,
Peter
