Consider we have n pointers that need to be swizzled, and swizzling one point will take time t on average. Suppose that if we swizzle all pointers automatically, we can perform the swizzling in half the time it would take to swizzle each separately. If the probability that a pointer in main memory will be followed at least once is p, for what values of p is it more efficient to swizzle automatically than on demand?

Considerwehavenpointers that need to be swizzled, and swizzling one point will take time t on average. Suppose that if we swizzle all pointers automatically, we can perform the swizzling in half the time it would take to swizzle each separately. If the probability that a pointer in main memory will be followed at least once is p, for what values of p is it more efficient to swizzle automatically than on demand?

Explanation:

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