CSE 5314 On-line Computation Homework 1 Wook Choi Feb/26/2004

1. CSE 5314 On-line Computation Homework 1 Wook ChoiFeb/26/2004

2. if x is in front of rj otherwise (including x = rj) Lemma 1.1 (Pairwise Property lemma) An algorithm satisfies the pairwise property if and only if for every request σ, when ALG servers σ, the relative order of every two elements x and y in L is the same as their relative order in Lxywhen ALG servers σxy Proof using the partial cost model which does consider only the cost from negative comparisons until a requested item is found in the list: Consider any request sequence σ = r1 , r2, ,… ,rnfor the list L. σxy is defined to be σ after deletion of all the requests for items except for x and y where x, y  L (x ≠ y); Lxyis the projection of the list L over x and y, thus having the same initial order as in L.For each x, and integer 1 ≤ j ≤ n, the penalty cost attributed to x is denoted by ALG (x,j) in accessing jth request, rj where: Let ALG*(σxy)and ALGxy*(σ)be the total cost for processing σxy over Lxy andσover {x,y} L , respectively. Since rj can be either x or y, ALG*(σxy) = ALGxy*(σ) = ALG (x,j) + ALG (y,j). Then, ALG satisfies pairwise property if ALG*(σxy)= ALGxy*(σ). Since the relative order of any two elements x and yL is the same as their relative order in Lxy, ALG(σxy ) and ALGxy*(σ)are the same for all x and y pairs. For example, if rj = x, ALG*(σxy) = ALGxy*(σ) = 0 + 0 andif rj = y, ALG*(σxy) = ALGxy*(σ) = 1 + 0, assuming that x is ahead of y. Therefore, Lemma 1.1 holds. Exercise 1.9 Prove Lemma 1.1

3. TIMESTAMP Algorithm (each element x maintains 2 timestamps, Recentx and Previousy) When an element x is accessed ifRecentx = 0 then Recentx current time /* first time request */elsePreviousx Recentx , Recentx current time examine each item y preceding x from the front of the list if Previousy < Previousxthen move x before such a first item y. end-ifend-if Exercise 1.12 Prove that TIMESTAMP satisfies the pairwise property Proof using Lemma 1.1 Consider two elements x and y in the list L. Letσxy be a subsequence of σincluding all occurrencesofonly x and y. Let Lxybe a list containing only x and y in the same initial order as in L. If the timestamp algorithm is run on Lxy with σxy and {x, y}  L with σ, the relative order of x and y in L and Lxy depends only on which of Previousx and Previousy is greater. Thus, if and only if x is requested more recently than y, x is moved before y in both lists, L and Lxy which means the relative order of the two elements in Lxyis the same astheir relative order of x and y in the original list L. Therefore, according to Lemma 1.1, the timestamp algorithm satisfies the pairwise property.

4. Marking algorithm Each page in memory is associated with a bit called its mark. A page is marked when it is first requested and a marking algorithm never evicts a marked page from its cache during k pages accesses. Proof by contradiction: Assume that FIFO is a marking algorithm for k-phase partition of any fixed sequence σ. Then FIFO should not evict any marked page during some k phase. Without a loss of generality, let’s consider a cache with k slots which are indexed by i, 1  i  k, in the order of placement into the cache. Now, suppose the first request of page x during a k-phase is the ithpage in the cache. So, this first request is hit and page x is marked after the request is processed. Then, for any remaining k-1 page requests, page x should not be evicted. Let i = 1 and there are i page faults within remaining k-1 page requests. This makes FIFO evict page x before the end of the k-phase. This contradicts our hypothesis. Therefore, FIFO is not a marking algorithm. Exercise 3.7 Show that algorithm FIFO is not a marking algorithm