如何从2路set associate Cache计算cache命中率

Question

在我的任务中，我们有两个问题:我们有一个2路组关联缓存。缓存总共有四个集合。主存储器由4K块组成，每个块8个字，并使用字寻址。第a部分)要求演示地址格式，我已将其解析为word =3位set =2位和field =7位。问题在b)部分:计算从位置8到位置51循环3次的程序的命中率。换句话说，这是一个汇编语言程序，它从位置8的操作码运行到位置51m的操作码，然后循环回到位置8。它总共执行三次这样的迭代。

现在，据我所知，在我做了研究之后，有一个标准，通常是某种速度或命中率？我在想，如果我不知道未命中率、未命中惩罚、缓存速度或其他信息，如何计算命中率？

Answer 1

我想我们在同一个班上笑我今晚要交作业时也有同样的问题..无论如何，我做了一些研究，在chegg上找到了类似问题的答案：

a. Given that memory contains 2K blocks of eight words.
2K can be distributed as 2K * 23 = 211* 23 = 214 so we have 14-bit addresses with 9 bits
in the tag field, 2 bits in the set field and 3 in the word field

b. First iteration of the loop:
    → Address 8 is a miss, then entire block brought into Set 1.9-15 are then hits.
    → 16 is a miss, entire block brought into Set 2, 17-23 are hits.
    → 24 is a miss, entire block brought into Set 3, 25-31 are hits.
    → 32 is a miss, entire block brought into Set 0, 33-39 are then hits.
    → 40 is a miss, entire block brought into Set 1 41-47 are hits.
    → 48 is a miss, entire block brought into Set 2, 49-51 are hits.

    For the first iteration of the loop, we have 6 misses, and 5*7 + 3 hits, or 38 hits.
    On the remaining iterations, we have 5*8+4 hits, or 44 hits each,for 88 more hits.
    Therefore, we have 6 misses and 126 hits, for a hit ratio of 126/132, or 95.45%.

希望这能帮上忙，祝你好运！