Data structure:

HASHTABLE<LinkedList>:
-- For each set there is a corresponding index. The number of indexes equals to the number of sets. if it is a direct mapped Cache
-- then it creates number of total blocks indexes
-- the hashtable arrays is composed of linkedList where the number of blocks according to associativity is added to each index

LinkedList:
-- each linked list contains the following variables
1) data: used to store the actual memory in binary format
2) tag: used to store the actual tag bits in binary format
3) next: to use the multiple blocks if needed



The Ratio:

The ratio seems to be fluctuating, thus it will be inaccurate to assume one to give better ratio than the other. In appears for small data Cache A seems to give better ration, and for enormous amount of data Cache B appears to give better ratio over all. However, still there are instances where Cache A gives better ration than Cache B on large amount of data, and Cach B gives better ration than Cache A on small amount of data.

Reason:
It is all about the memory size and indexing. since block off set are taken out first by both Cache A and Cache B. The only difference remains between them is the computation of their index numbers. Where Cache A  computes its on the right most n bits, and Cach B computes its index of the left most n bits. Thus their tags are obviously different. So the important thing to notice is the most memory addresses given are atleast 8 less than 48 bits, and those 8 bits are filled with 0s as a result. So for any block that ranges from 0-8 and number of sets that ranges from 0-8, will have an index of 0's, thus will always choose index 0 for Cache B since the indeices are computed from the left most n bits. I believe this causes disadvantage to the Cache B since many other indexes are left useless. I believe that is the reason why Cache B gives better ratio on larger amount of data rather than smaller amount of data.