高速缓存块地址的访问序列的长度为N，并且包含n个唯一的块地址。两次连续访问同一块地址之间的唯一块地址数由k限制。什么是未命中率，是访问序列通过关联性A> = k的高速缓存传递，从而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： (A)
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