Caching 计算CPU缓存命中率

我正在努力理解计算缓存命中的过程/公式。因此，例如，如果我们有一个有16个条目的主存和一个有4个条目的高速缓存，CPU加载内存地址：0、1、2、8、9、2'，我如何计算命中数a）如果高速缓存是直接映射的，b）双向关联的？

假设没有预取器和LRU作为替换机制

a） 对于直接映射缓存，每个内存项只能位于一个缓存项中。 缓存将如下映射（默认情况下假定均匀分布）：

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.

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.

因此我们有1次命中和5次未命中，命中率为1/（5+1）=1/6=16%

b）对于两种关联方式，内存中的每个条目都有两个缓存条目可供访问。 因此，set0（缓存中的条目0,1）将保存所有偶数主内存条目，set1将保存所有奇数条目，因此如果我们对其进行跨距，我们将得到如下结果：

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 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.

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，命中率是相同的：1/（5+1）=1/6=16%