计算CPU缓存命中次数

Question

我很难理解计算缓存命中的过程/公式是什么。所以，如果我们有一个主内存，有16个条目，缓存内存有4个条目，CPU加载内存地址: 0，1，2，8，9，2‘，如果缓存是直接映射的，b)双向关联的，我如何计算命中次数a？

Answer 1

假设没有预取器和LRU作为替换机制。

a)用于直接映射缓存，每个内存项只能在一个缓存项中.缓存将像这样映射(默认假定分布一致)：

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.

：1次命中，5次未命中，命中率为1/(5+1) = 1/6 = 16%

b)用于2种方式的关联，您将对缓存中的内存中的每个条目有两个要访问的条目。因此，set0 (缓存中的条目0,1 )将保存所有偶数主内存条目，而set1将保存所有奇怪的条目，因此，如果我们跨越它，就会有如下所示：

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，命中率是相同的：1/(5+1) = 1/6 = 16%