Exercise 3.10 Prove Theorem 3.8 R(ALG) >= k(s+1)/(k+s) Solution: ALG is any paging algorithm. k is the cache size. s >= 1 is the cost of bringing a page from slow to fast memory & cost of accessing pages in fast memory = 1 in the full access cost model. Theorem 3.6 states that R(ALG) >= k. LFD(σ) = cost of the paging algorithm LFD for a request sequence σ We will compare ALG(σ) and LFD(σ) to find the lower bound of R(ALG).

According to Lemma 3.2, LFD faults at most once every k requests. The cost of a fault is s+1 since the cost of fast memory access is 1. Hence, LFD(σ) <= |σ|(s+1)/k + (|σ| - |σ|/k) = |σ|(k+s)/k For ALG(σ), let’s assume the cruelest request sequence which faults every request. ALG(σ) = |σ|(s+1) Using the costs of LFD(σ) and ALG(σ) above, ALG(σ)/LFD(σ) >= |σ|k(s+1)/|σ|(k+s) = k(s+1)/(k+s)