假设没有预取器和LRU作为替换机制。

a） 对于直接映射的缓存，每个内存条目只能位于一个缓存条目中。缓存将如下所示映射（默认情况下假设均匀分布）：

Cache 0 --> can hold 0,4,8,12 of the main memory entries.
Cache 1 --> can hold 1,5,9,13 of the main memory entries.
Cache 2 --> can hold 2,6,10,14 of the main memory entries.
Cache 3 --> can hold 3,7,11,15 of the main memory entries.

重置后，缓存为空。

Load from 0 will be missed and will be cached in cache entry 0.
Load from 1 will be missed and will be cached in cache entry 1.
Load from 2 will be missed and will be cached in cache entry 2.
Load from 8 will be missed and will be cached in cache entry 0 (replaced load 0).
Load from 9 will be missed and will be cached in cache entry 1 (replaced load 1).
load from 2 will be hit and will be taken from cache entry 2.

所以我们有1次命中和5次未中，命中率为1 /（5 + 1）= 1/6 = 16％

b）对于2种关联方式，你需要为内存中的每个条目分配2个条目到缓存中。因此，set0（在缓存中的条目为0,1）将保存所有偶数主内存条目，而set1将保存所有奇数条目，因此如果我们将其扩展，则将具有以下内容：

cache 0 (set 0) --> can hold 0,2,4,6,8,10,12,14 of the main memory entries.
cache 1 (set 0) --> can hold 0,2,4,6,8,10,12,14 of the main memory entries.
cache 2 (set 1) --> can hold 1,3,5,7,9,11,13,15 of the main memory entries.
cache 2 (set 0) --> can hold 1,3,5,7,9,11,13,15 of the main memory entries.

重置后，缓存为空。

Load from 0 will be missed and will be cached in cache entry 0.
Load from 1 will be missed and will be cached in cache entry 2.
Load from 2 will be missed and will be cached in cache entry 1.
Load from 8 will be missed and will be cached in cache entry 0 (replaced load 0 because we replace the Least Recently Used).
Load from 9 will be missed and will be cached in cache entry 3.
load from 2 will be hit and will be taken from cache entry 1.

在这种情况下，点击率是相同的：1 /（5 + 1）= 1/6 = 16％