一个缓存块地址的访问序列长度为N，包含n个唯一的块地址。对同一块地址的两次连续访问之间的唯一块地址的数量以 k 为界。什么是未命中率是访问序列通过关联性 A >= k 的缓存传递，并执行最近最少使用的替换策略。
(A) 不适用
(B) 1/N
(C) 1/A
(D) k/n答案：(一)
解释：
Their are N access request for the cache blocks out this n
blocks are unique .
In between two access of the same block their are request of
(k-1) other block block.
And if their associativity >=k and use LRU, then
there will be only one cache miss for every unique block i.e.,
n and it will be the time when the enter the cahe for the first
time. Therefore Miss ratio =(Cache miss)/(No. of request) = n/N 这个问题的测验