Refer to the exhibit.
What is the most efficient summarization that R1 can use to advertise its networks to R2?
Click on the arrows to vote for the correct answer
A. B. C. D. E.E
The top two (4.128/25 and 4.0/25) aggregate into 4.0/25, so now we have four /24 networks. if they are aligned right, we can merge these into a /22. If not we'd need to merge them into a /21. The other three subnets are 5.0/24, 6.0/24, 7.0/24.
That means we start at 4.0 and go up to 5.255, i.e. just below 8.0. So this fits neatly into a /22 as these go up in steps of 4 in the 3rd octet. Yes, we can summarise with a /22 that starts at 172.1.4.0, answer D.