Expected Performance and Worst Case Scenario Analysis of the Divide-and-Conquer Method for the 0-1 Knapsack Problem
aa r X i v : . [ c s . D S ] A ug Expected Performance and Worst Case Scenario Analysisof the Divide-and-Conquer Methodfor the 0-1 Knapsack Problem
Fernando A Morales a , Jairo A Mart´ınez b a Escuela de Matem´aticas Universidad Nacional de Colombia, Sede Medell´ınCarrera 65 b Departamento de Ciencias Matem´aticas, Universidad EAFIT.Carrera 49
Abstract
In this paper we furnish quality certificates for the Divide-and-Conquer method solving the 0-1 KnapsackProblem: the worst case scenario and an estimate for the expected performance. The probabilistic setting isgiven and the main random variables are defined for the analysis of the expected performance. The performanceis accurately approximated for one iteration of the method then, these values are used to derive analyticestimates for the performance of a general Divide-and-Conquer tree. Most of the theoretical results are verifiedvs numerical experiments for a wider illustration of the method.
Keywords:
Divide-and-Conquer Method, Quality Certificates, Probabilistic Analysis, Monte Carlosimulations, method’s efficiency.
1. Introduction
The 0-1 knapsack problem (0-1KP) is one of the most widely discussed problems in the combinatorialoptimization literature and is certainly the simplest prototype of a maximization problem [1]. It is defined asfollows: given a set of n items, each item j having a weight w ( j ) and a profit p ( j ), the problem is to choosea subset of items such that the sum of profits is maximized, while the sum of weights does not exceed theknapsack capacity δ . The simplicity of its formulation (see Problem 1) contrasts with its surprising theoreticaland practical relevance: its decision version is one of Karp’s 21 NP-complete problems [2], 0-1KP itself or someof its well-known variants is used in the modeling of important practical problems such as portfolio managementand container optimization [3, 4]. In addition, it appears as a subproblem when applying some decompositiontechnique to large problems, for example, solving material cutting models using a column generation method[5, 6]. It has also played an interesting role in the development of cryptographic systems [7].The Divide-and-Conquer method for solving the 0-1KP was recently introduced by Morales and Mart´ınez in[8]. The method seeks to reduce the computational complexity of a large instance of the problem, by executinga recursive subdivision into smaller instances, so that the process can be visualized as the construction of abinary tree whose nodes are knapsack subproblems. As it was emphasized in the original work, the method doesnot compete with the existing algorithms, it complements them (observe that in Example 1, it is not specifiedhow to solve the defined subproblems). The experimental results presented in [8], show that the method is a ✩ This material is based upon work supported by project HERMES 45713 from Universidad Nacional de Colombia, Sede Medell´ın. ∗ Corresponding Author
Email address: [email protected] (Fernando A Morales)
Preprint submitted to arXiv August 11, 2020 ood middle grounds alternative, halfway between computational complexity and quality of the solution. Sofar, the quality performance of Divide-and-Conquer has been measured only empirically. The aim of this paperis to analyze theoretically its quality performance from two points of view: the worst-case scenario and itsexpected/average performance.
In the last three decades of the 20th century, the algorithms implemented for the resolution of 0-1KPreached a great maturity, standing out the primal and dual variants of branch and bound, [9, 10, 11], dynamicprogramming [12, 13], the core -type algorithms [14, 15] and hybrid procedures like the
Combo algorithm[16]. Although in terms of worst case time complexity, the best bounds achieved are pseudpolynomial, thecombined application of different techniques made it possible to effectively solve a large number of benchmarkinstances, which led to the designation of the knapsack problem as one of the “easy to solve” NP-hard problems.Consequently, the research line directed at understanding the characteristics of the most computationallychallenging instances [17], was develoled.The discrepancy observed between the good performance of simple heuristics and exact methods whenapplied on pure random instances, and the high complexity pointed out by the worst case analysis, startedto be explained theoretically through probabilistic analysis. In this respect, Kellerer et al. [1] classified thecontributions depending on whether the results are: • structural , if they give a probabilistic statement e.g. onthe optimal solution value. • Expected performance of algorithms, which produce an optimal solution with acertain probability. • Expected running time of algorithms, which always produce solutions of a certain quality.A probabilistic model for the knapsack problem widely used in the literature is the one proposed by Lueker[18], in which it is assumed that weights and profits are uniformly selected from the interval [0 , n items can be understood as the random location of n points in the unit square.The knapsack capacity should be specified as δ = βn , where β is some constant in the interval (0 , β parameter, see for exaple [19, 20, 21].The structural result presented by Lueker [18] consisted in estimating the expected value of the linear relax-ation gap, to formally explain the empirically observed good performance of the branch and bound algorithms(B&B). The fundamental result was that integrality gap has order O (log ( n ) /n ) which means it decreaseswith problem size increase. Regarding the analysis of the exact solution, we highlight two works: Frieze andClarke [21] conducted a probabilistic analysis of 0-1KP and obtained an interesting bound for the behavior ofthe objective value, showing that it is asymptotically equal to p n/ n tends toinfinity. Mamer et al. [22] carried out a similar analysis for a very large class of joint distributions and deductedthe same upper bound as Frieze and Clarke.Since the 80’s, of the last century the probabilistic method was applied to the study of different versions ofthe greedy algorithm. Szkatula & Libura [23] obtained moments and distribution functions for some parametersof the greedy algorithm without ordering, obtaining recursive equations for the distribution function of theaccumulated weight in any iteration. Under slightly different hypotheses than those in the standard model,Calvin and Leung [19] proved, using convergence in distribution, that the sorted greedy algorithm producesresults that differ from the optimum value by order 1 / √ n . Diubin et al. [20] address the analysis of theminimization version of the 0-1KP and proved that the primal and dual greedy methods for the minimizationknapsack problem are also asymptotically good. They showed that despite the complementarity between theminimization and maximization problems, the result concerning the former cannot be obtained from the resultaddressing the latter problem. It is worth noting that most of the mathematical analyses involved in theseinvestigations exploit the geometric interpretation of the extended greedy algorithm, in particular the critical orsplitting ray.There are also very relevant works related to the expected running time of exact and approximation al-gorithms. Beier and V¨ocking [24] presented the first average-case analysis proving a polynomial upper boundon the expected running time of a sparse dynamic programming algorithm for the 0-1KP; originally proposedby Nemhauser and Ullman [25]. The algorithm iteratively extend non-dominated or Pareto-efficient subset,contained in the set of the first i items. The main conclusion is that the number of Pareto-efficient knapsack2llings is polynomially bounded in the number of available items. The random input model used in this studyis more general, the weights of the items are chosen by an adversary and their profits are chosen accordingto arbitrary continuous or discrete probability distributions with finite mean, allowing to address the effectsof correlation between parameters. It is interesting to point out that when using discrete distributions, theywere able to prove a trade-off, ranging from polynomial to pseudo-polynomial running time, depending on therandomness of the specified instances.In a later work Beier and V¨ocking [26] studied the average-case performance of core algorithms for the0-1KP. They proved an upper bound of O ( n polylog( n )) on the expected running time of a core algorithmon instances with n items, whose profits and weights are drawn random and independently from a uniformdistribution. Unlike previous works such as Goldberg and & Marchetti-Spaccamela [27], the degree of thepolynomial involved is relatively low, but the probabilistic analysis is complicated due to the dependence betweenrandom variables.More recent research has attempted to theoretically understand, the efficiency of simple and successfulheuristics such as rollout algorithms. These iterative methods use a base policy , whose performance is evaluatedto obtain an improved policy , by one-step look ahead. Rollout algorithms are easy to implement and guaranteea not worse , and usually much better results than corresponding base policies. Bertazzi [28] proved minimumand worst case performance ratio when the greedy, full greedy and the extended greedy algorithms are chosen asbase policies, respectively. In all cases the analysis was applied to only the first iteration, showing furthermore,that for the algorithms considered there exist an instance in which the worst-case performance ratio is obtainedat the first iteration, so that the expected value deducted cannot be subsequently improved. The worstcase performance ratios was improved from 0 to 1 / / / First, a worst case performance ratio of 1 / • Discreteuniform probability distributions are assumed for the parameters. • A very simple relation is defined betweenthe number of items n and the knapsack capacity δ . • The profits are defined by means of the weights andthe efficiencies, which in turn are given in terms of random variables called the increments. We point outthat according to the literature review, discrete distributions were considered only in Beier and V¨ocking’s work[24]. The adopted model allowed to obtain structural results for the (sorted) greedy and the eligible first itemalgorithm, which are difficult to approach from the usual model (see for example Bertazzi [28]). Similarly to theMastin & Jaillet proof strategy [29], the theoretical analysis of the Divide-and-Conquer method concentrateson its first iteration. Asymptotic relationships are presented, these permit to define and evaluate numerically,the performance ratios for the entire solution process (see Theorem 13, Lemmas 20 and 21 and Corollaries 23,26).
2. Preliminaries
In this section the general setting and preliminaries of the problem are presented. We start introducingthe mathematical notation. For any natural number µ ∈ N , the symbol [ µ ] def = { , , . . . , µ } indicates the3orted set of the first µ natural numbers. In the same fashion [0 , , ,
3] stands for the set containing thementioned elements in the order 0 , , ,
3. Greek lowercase letters ( δ, λ, µ, ν, . . . ), are used for important fixedconstants. For any set E we denote by E its cardinal and ℘ ( E ) its power set. Given an event E ⊆ Ω, wedenote its indicator function by E : Ω → { , } , with E ( ω ) = 1 if ω ∈ E and zero otherwise. Randomvariables will be represented with upright capital letters, e.g. X , Y , Z , ... and its respective expectations with E ( X ) , E ( Y ) , E ( Z ) , ... . Vectors are indicated with bold letters, namely p , g , ... etc. Particularly importantcollections of objects will be written with calligraphic characters, e.g. A , D , E to add emphasis. A particularlyimportant set is S N , where S N denotes the collection of all permutations in [ N ]. For any real number x ∈ R the floor and ceiling function are given (and denoted) by ⌊ x ⌋ def = max { k : ℓ ≤ x, k integer } , ⌈ x ⌉ def = max { k : k ≥ x, k integer } , respectively. In the current section we introduce the 0-1 Knapsack Problem and review a list of greedy algorithms, to beused in the analysis of the Divide-and-Conquer method for both ends: attain a quality certificate in the worstcase scenario and compute the expected performance of the method.
Problem 1 (0-1KP).
Consider the problem z ∗ def = max µ X i = 1 p ( i ) x ( i ) , (1a)subject to µ X i = 1 w ( i ) x ( i ) ≤ δ, (1b) x ( i ) ∈ { , } , for all i ∈ [ µ ] . (1c)Here, δ is the knapsack capacity and (cid:0) x ( i ) (cid:1) µi =1 is the list of binary valued decision variables . In addition,the weight coefficients (cid:0) w ( i ) (cid:1) µi =1 , as well as the knapsack capacity δ are all positive integers. In the sequel z ∗ denotes the objective function optimal solution value . We refer to the parameters (cid:0) p ( i ) (cid:1) µi =1 ⊆ (0 , ∞ ] µ asthe profits and introduce the efficiency rate g ( i ) def = p ( i ) w ( i ) . Finally, in the sequel the problem is indicated by theacronym and we denote by Π = (cid:10) δ, ( p ( i )) i ∈ [ µ ] , ( w ( i )) i ∈ [ µ ] (cid:11) one of its instances.Before we continue our analysis, the next hypothesis is adopted. Hypothesis 1.
In the sequel we assume that the instances Π of the 0-1KP satisfy the following(i) The items of Problem 1 are sorted according to their efficiencies in decreasing order i.e., g (1) ≥ g (2) ≥ . . . ≥ g ( µ ) . (2)(ii) The weights of the items satisfy w ( i ) ≤ δ, for all i ∈ [ µ ] , µ X i = 1 w ( i ) > δ. (3) Remark 1 (0-1KP Setting).
We make the following observations about the setting of the problem 1.(i) The condition (2) in Hypothesis 1 is assumed to ease the algorithm analysis later on.4ii) The condition (3) in Hypothesis 1 guarantees two things. First, every item is eligible to be chosen. Second,the complete set of items is not eligible. Both conditions are introduced to prevent trivial instances ofProblem 1.(iii) Due to the condition (3), the split item and the greedy algorithm solutions of Definition 1 are well-defined.Next, we recall a catalog of greedy algorithms for the solution of Problem 1, to be used in the probabilisticanalysis of the Divide-and-Conquer method.
Definition 1 (Greedy Solutions).
Let Π = (cid:10) δ, (cid:0) p ( i ) (cid:1) µi =1 , (cid:0) w ( i ) (cid:1) µi =1 (cid:11) be an instance of Problem 1. Let { J } be the indicator function of the singleton { J } , with J ∈ N . Define the following(i) The split item is the index s ∈ [ µ ] satisfying s − X i = 1 w ( i ) ≤ δ, s X i = 1 w ( i ) > δ. (4)(ii) The greedy algorithm solution to the problem 1 and its corresponding objective function values are givenby x G ( i ) def = ( , i = 1 , . . . , s − , , i = s, . . . , µ, z G def = s − X i = 1 p ( i ) . (5)(iii) The extended-greedy algorithm solution yields the following objective function value and correspondingsolution to the problem 1 z eG def = max (cid:8) z G , max i ∈ [ µ ] { p ( i ) : i ∈ [ µ ] } (cid:9) , x eG ( i ) def = ( x G ( i ) , z eG = z G , { J } ( i ) , z eG > z G . (6)Here, J def = min (cid:8) j ∈ [ µ ] : p ( j ) = max ℓ ∈ [ µ ] p ( ℓ ) (cid:9) .(iv) The eligible-First greedy algorithm solution defines the following set E def = n i > s : w ( i ) ≤ δ − s − X i = 1 w ( i ) o , (7a)to yield the following objective function value and corresponding solution to the problem 1 z eF def = ( z G , E = ∅ ,z G + z J , J = min E, x eF ( i ) def = ( x G ( i ) , E = ∅ ,x G ( i ) + { J } ( i ) , J = min E. (7b)(v) Finally we describe the full-greedy algorithm solution for solving problem 1 with the following pseudocode5 lgorithm 1 Greedy Algorithm, returns feasible solution (cid:0) x ( i ) (cid:1) µi =1 and the associated value z fG = P µi =1 p ( i ) x ( i )of the objective function for Problem 1. procedure Greedy-Algorithm pseudo-code(Input: Capacity: δ , Profits: ( p ( i )) µi =1 , Weights: ( w ( i )) µi =1 . Theitems’ efficiencies satisfy g (1) ≥ g (2) ≥ . . . ≥ g ( µ ).) w fG def = 0 ⊲ w fG is the total weight of the currently packed items z fG def = 0 ⊲ z fG is the profit of the current solution for j = 1 , . . . , µ do if w fG + w ( j ) ≤ δ then x ( j ) = 1 ⊲ put item j into the knapsack w fG = w fG + w ( j ) z fG = z fG + p ( j ) else x ( j ) = 0 end if end for end procedureRemark 2 (Greedy Algorithms). It is direct to see that z G ≤ min { z eF , z eG } ≤ z fG for any instance of 0-1KPand that all the algorithms are of the same order in terms of computational cost. Therefore, only the full-greedy algorithm should be implemented in practice however, it is very hard to analyze it from the probabilisticpoint of view. The extended-greedy algorithm furnishes a quality certificate for the worst case scenario, asit can be seen in Theorem 1 (ii), however its probabilistic performance analysis is as hard as in the previouscase. On the other hand, the probabilistic analysis of the greedy algorithm is tractable (see Theorem 10)and it characterizes the linear programming relaxation of 0-1KP (see Theorem 1 (i)), which contributes tothe probabilistic analysis of the latter problem (see Theorem 12). Finally, the eligible-first greedy algorithm isintroduced because its probabilistic analysis is tractable at the time of furnishing better approximation estimatesto the optimal solution, than the greedy algorithm, see Section 3.3. Definition 2.
The natural linear programming relaxation of Problem 1, is given by
Problem 2 (0-1LPK). max µ X i = 1 p ( i ) x ( i ) , (8a)subject to µ X i = 1 w ( i ) x ( i ) ≤ δ, (8b)0 ≤ x ( i ) ≤ , for all i ∈ [ µ ] , (8c)i.e., the decision variables (cid:0) x ( i ) (cid:1) µi =1 are are now real-valued.In the sequel this problem the acronym will stand for the associated linear relaxation problem.We close this section recalling a couple of classical results for the sake of completeness Theorem 1.
Let
Π = (cid:10) δ, (cid:0) p ( i ) (cid:1) µi =1 , (cid:0) w ( i ) (cid:1) µi =1 (cid:11) be an instance of Problem
1, then i) The optimal of the problem 2 (0-1 LPK) is given by x LP ( i ) = , i = 1 , . . . , s − , w ( s ) (cid:0) δ − s − P i =1 w ( i ) (cid:1) , i = s, , i = s + 1 , . . . , µ, (9a) with the corresponding objective function value z LP = s − X i = 1 p ( i ) + (cid:16) δ − s − X j =1 w ( j ) (cid:17) p ( s ) w ( s ) . (9b) (ii) Let z ∗ , z eG be respectively, the optimal and the extended greedy algorithm objective values for Problem
1. Then, z ∗ ≤ z eG , (10) i.e., the extended greedy algorithm has a relative performance quality certificate of .Proof. (i) See Theorem 2.2.1 in [1].(ii) See Theorem 2.5.4 in [1]. The Divide-and-Conquer method for solving the 0-1KP was introduced in [8]. Here was presented anextensive discussion (theoretical and empirical) on the possible strategies to implement it and conclude thatthe best strategy is given by the following algorithm
Definition 3 (Divide-and-Conquer pairs and trees).
Let Π = (cid:10) δ, (cid:0) p ( i ) (cid:1) µi =1 , (cid:0) w ( i ) (cid:1) µi =1 (cid:11) be an instance ofProblem 1(i) Let V be a subset of [ µ ] and δ V ≤ δ with δ V ∈ N . A subproblem of Problem 1 is an integer problemwith the following structure max X i ∈ V p ( i ) x ( i ) , subject to X i ∈ V w ( i ) x ( i ) ≤ δ V ,x ( i ) ∈ { , } , for all i ∈ V. In the the sequel, the subproblem will be denoted by Π V def = (cid:10) δ V , (cid:0) x ( i ) (cid:1) i ∈ V , (cid:0) w ( i ) (cid:1) i ∈ V (cid:11) (ii) Let ( V , V ) be a set partition of [ µ ] and let ( δ , δ ) be an integer partition of δ (i.e., δ = δ + δ ). Wesay a Divide-and-Conquer pair of Problem 1 is the couple of subproblems (cid:0) Π b : b ∈ { , } (cid:1) , each withinput data Π b = (cid:10) δ b , (cid:0) p ( i ) (cid:1) i ∈ V b , (cid:0) w ( i ) (cid:1) i ∈ V b (cid:11) . In the sequel, we refer to (cid:0) Π b , b = 0 , (cid:1) as a D&C pair and denote by z ∗ b the optimal solution value of the problem Π b .(iii) A D&C tree (see Example 1 and Figure 1 below) for Problem 1 is defined recursively by Algorithm 2. Itsinput is an instance Π = (cid:10) δ, ( p ( i )) i ∈ [ µ ] , ( w ( i )) i ∈ [ µ ] (cid:11) of Problem 1 and a minimum size of subproblems ξ .It satisfies the following properties 7. Every vertex of the tree is in bijective correspondence with a subproblem Π of Π .b. The root of the tree is associated with Problem 1 itself.c. Every internal vertex Π (which is not a leave) has a left and right child , Π left , Π right respectively. Itschildren make a D&C pair for the subproblem Π, whose generation is given by Algorithm 2.(iv) Let Π = (cid:10) δ, ( p ( i )) i ∈ [ µ ] , ( w ( i )) i ∈ [ µ ] (cid:11) be an instance of a 0-1KP and let T be a D&C tree. The methoduses the search space and objective values x T def = [ L is a leave of T x L , z T def = X L is a leave of T z L . (11)Here, we introduce some abuse of notation, denoting by x L a feasible solution of Π L and using the samesymbol as a set of chosen items (instead of a vector) in the union operator. In particular, the maximalpossible value occurs when all the summands are at its max i.e., the method approximates the optimalsolution by x ∗T def = S { x ∗ L : L is a leave of T } with objective value z ∗T def = P { z ∗ L : L is a leave of T } . Algorithm 2
Divide-and-Conquer tree generation branch function, returns a D&C tree T of Problem 1. function Branch( Subproblem: Π = h δ, ( p ( i )) i ∈ V , ( w ( i )) i ∈ V i , D&C Tree: T , Minimum problem size: ζ ) compute s (split item), z G = P s − i =1 p ( i ) (objective function value), k = δ − P s − i =1 w ( i ) (slack) forproblem Π ⊲ Greedy Algorithm, Definition 1 (ii) compute z eG for problem Π ⊲ Extended Greedy Algorithm, Definition 1 (iii) if z G ≥ z eG and | V | ≥ ζ then ⊲ Branching condition V left def = (cid:2) i : i ∈ V, i is in odd relative position (cid:3) ⊲ Computing the left child indexes V right def = (cid:2) i : i ∈ V, i is in even relative position (cid:3) ⊲ Computing the right child indexes δ left def = ⌈ × k ⌉ + P { w ( i ) : i ∈ V right } ⊲ Computing the right capacity δ right def = ⌊ × k ⌋ + P { w ( i ) : i ∈ V right } ⊲ Computing the left capacity Π left def = (cid:10) δ left , ( p ( i )) i ∈ V left , ( w ( i )) i ∈ V left (cid:11) ⊲ Defining the left child problem Π left Π right def = (cid:10) δ right , ( p ( i )) i ∈ V right , ( w ( i )) i ∈ V right (cid:11) ⊲ Defining the right child problem Π right Π left ֒ → V ( T ), (Π , Π left ) ֒ → E ( T ), Π right ֒ → V ( T ), (Π , Π right ) ֒ → E ( T ) ⊲ Pushing problems Π left , Π right as nodes and (Π , Π left ) , (Π , Π right ) as arcs of the D&C tree T Branch(Π left , T , ζ ) ⊲ Recursing for the left subtree
Branch(Π right , T , ζ ) ⊲ Recursing for the right subtree return T ⊲ output D&C tree else return T ⊲ output D&C tree end if end functionRemark 3 (Divide-and-Conquer pairs and trees). Observe the following about the algorithm 2 defined above(i) The instance of Problem 1, Π = (cid:10) δ, ( p ( i )) i ∈ [ µ ] , ( w ( i )) i ∈ [ µ ] (cid:11) , to be solved with the Divide-and-Conquermethod is assumed to satisfy Hypothesis 1.(ii) When the Branch function is called for the first time, the D&C tree T must be initizalized as V ( T ) def = { Π } , E ( T ) def = ∅ .(iii) When defining the ordered sets V left the sentence “is in odd relative position” is used; it means, thoseindexes which occupy odd positions in the sorted set V (the analogous holds for V right ). For instance,8bserve the subproblem Π in Example 1, Figure 1. Here the indexes 1 , , , , , V left = [1 ,
5] and V right = [3 ,
7] (subsets for Π and Π subproblems in Example 1,Figure 1).(iv) The definition of V left , V right subdividing the list of eligible items V for each node of the tree T , is adoptedbecause it has been observed empirically in [8] (balanced left-right subtrees, Section 4.2) that the Divide-and-Conquer method is expected to produce better results with this branching process.(v) The condition for branching: ( z G ≥ z eG and | V | ≥ ξ ) states that a subproblem will not be furthersubdivided if z G < z eG or if the number of items | V | < ξ . The first condition is discussed in Theorem3 and Remark 5 below, while the second aims to ensure that no problem will be smaller that ξ . This,because it has been observed empirically in [8] that the Divide-and-Conquer method no longer producesgood results beyond a problem size threshold, namely ξ . Example 1 (Divide-and-Conquer tree).
Consider the 0-1KP instance described by the table 1, with knapsackcapacity δ = 7 and number of items µ = 8. i w ( i ) 3 2 3 3 4 7 1 5 p ( i ) 11.7 7.0 9.3 8.4 8.4 9.1 0.7 1.0 g ( i ) 3.9 3.5 3.1 2.8 2.1 1.3 0.7 0.2 Table 1: 0-1KP problem of Example 1, knapsack capacity δ = 7, number of items µ = 8. In this particular case s = 3 , x eG = x G = [1 , , , , , , , , z G = 14 . z eG , k = 7 − X i = 1 w ( i ) x G ( i ) = 2 , x ∗ = [1 , , , , , , , , z ∗ = 21 . . Here, k denotes the slack in the knapsack. Hence, due to Algorithm 2 it follows thatΠ left : V left = [1 , , , , δ left = 3 + 1(3 from item 1 and 1 from the slack ⌈ k ⌉ ) , x ∗ left = [1 , , , , z ∗ left = 12 . , x Gleft = x eGleft = [1 , , , , z Gleft = z eGleft = 11 . . Π right : V right = [2 , , , , δ right = 2 + 1(2 from item 2 and 1 from the slack ⌊ k ⌋ ) , x eGright = x ∗ right = [0 , , , , z eGright = z ∗ right = 8 . . x Gright = [1 , , , , z Gright = 7 . . In this case z ∗ > z ∗ left + z ∗ right . Next, given that z Gleft = z eGleft we repeat the same procedure for Π left , however wedo not branch on Π right since z Gright < z eGright ; this is observed in Table 2 and Figure 1.
Theorem 2.
Let
Π = (cid:10) δ, (cid:0) p ( i ) (cid:1) µi =1 , (cid:0) w ( i ) (cid:1) µi =1 (cid:11) be an instance of the 0-1KP 1. Let ( V n ) Nn =1 be a partition of [ µ ] and let ˜ x def = (cid:0) ˜ x ( i ) (cid:1) µi =1 be a fixed feasible solution to the 0-1 KP problem. Hence, if δ = N X n = 1 δ n , X i ∈ V n w ( i ) ˜ x ( i ) ≤ δ n , for all n ∈ [ N ] , (12)9 tem Vertex V V V V V δ Table 2: Algorithm 2, D&C tree gener-ated for the 0-1KP instance describedin Table 1. V = [1 , , , , , , , δ = 7 ! ≡ Π V = [1 , , , δ = 3 ! ≡ Π V = [1 , δ = 3 ! ≡ Π V = [3 , δ = 0 ! ≡ Π V = [2 , , , δ = 4 ! ≡ Π Figure 1: Algorithm 2 D&C tree generated for Table 2. Every vertex Π ℓ is a sub-problem of the 0-1 KP instance Π = h δ , ( p ( i )) i ∈ V , ( w ( i )) i ∈ V i . then µ X i = 1 p ( i ) ˜ x ( i ) ≤ N X n = 1 z ∗ n . (13) Here z ∗ n is the optimal solution of the subproblem Π n = (cid:10) δ n , (cid:0) p ( i ) (cid:1) i ∈ V n , (cid:0) w ( i ) (cid:1) i ∈ V n (cid:11) for all n = 1 , . . . , N . Inthe following we refer to ˜ x as the control solution .Proof. It is direct to see that that (cid:0) ˜ x ( i ) (cid:1) i ∈ V n is a feasible solution of Π n for all n = 1 , . . . , N , due to thecapacities condition (12). Hence, P i ∈ V n p ( i ) ˜ x ( i ) ≤ z ∗ n for each n = 1 , . . . , N then, N X n = 1 X i ∈ V n p ( i ) ˜ x ( i ) ≤ N X n = 1 z ∗ n . Given that ( V n ) Nn =1 is a partition of [ µ ], the inequality (13) follows. Remark 4.
Notice that if an optimal solution (cid:0) x ∗ ( i ) (cid:1) µi =1 of Problem 1 satisfies the set of capacities constraint(12) then z ∗ ≤ P Nn = 1 z ∗ n i.e., the D&C collection of subproblmes (Π n ) Nn =1 reduces the computational complexityof Problem 1 at no expense of precision, which is the ideal scenario. Theorem 3.
Let Π be a 0-1KP instance,(i) Let Π left , Π right be a D&C pair for the 0-1KP instance Π . Let x G , x Gleft , x Gright and z G , z Gleft + z Gright , be theircorresponding solutions and objective function values, furnished by the greedy algorithm. Then x G and V left , V right satisfy the hypothesis of Theorem
2. Moreover, z G ≤ z Gleft + z Gright , (14) where z Gleft , z
Gright are the greedy algorithm solutions for Π left and Π right respectively.(ii) Let z ∗T be the optimal approximation value furnished by a D&C tree T of Π , generated by Algorithm ≤ z ∗T z ∗ , (15) where z ∗ is the optimal value for the problem Π .Proof. (i) It is direct to see that x G and V left , V right satisfy the hypothesis of Theorem 2, because of how δ left and δ right are defined in Algorithm 2. Moreover, such definition ensures that the inequality (14) holds.10ii) Let x eG def = (cid:0) x eG ( i ) (cid:1) µi =1 be the extended-algorithm solution for the problem Π; observe that if x G = x eG then T = { Π } , due to the method’s definition (see Algorithm 2) and the result is obvious. Hence, fromnow on we assume that x G = x eG .Consider (cid:8) Π L = (cid:10) δ L , (cid:0) p ( i ) (cid:1) i ∈ V L , (cid:0) w ( i ) (cid:1) i ∈ V L (cid:11) : L is a leave of T (cid:9) , due to the theorem 4 in [8], the collec-tion (cid:8) V L : L is a leave of T (cid:9) is a partition of [ µ ]. Then, in order to prove the result, it suffices to show that x G and (cid:8) V L : L is a leave of T (cid:9) satisfy the hypothesis of Theorem 2. We prove this by induction on thenumber of Divide-and-Conquer iterations used to generate the tree. Denote by { Π } def = T , T , . . . , T n = T be the colection of trees attained by subsequent iterations of the Divide-and-Conquer method, with T the original problem and T n the tree of interest. For T the result is obvious and for T this was proved inthe previous part. Denote by (Π j ) Jj =1 the leaves of T n − , due to the induction hypothesis, the solution x G and ( V j ) Jj =1 satisfy the hypothesis of Theorem 2. But then, due to the first part, for each problem Π j , itholds that δ j = δ j left + δ j right , X i ∈ V j left w ( i ) x G ( i ) ≤ δ j left , X i ∈ V j right w ( i ) x G ( i ) ≤ δ j right . Hence, δ = J X j = 1 δ j = J X j = 1 δ j left + δ j right = X L leave of T δ L and recalling that { L : L is a leave of T } is in bijective correspondence with (cid:8) Π j side : j = 1 , . . . , J, side ∈{ left , right } (cid:9) , we conclude that x G and (cid:8) V L : L is a leave of T (cid:9) satisfy the hypothesis of Theorem 2.Hence, z eG = z G = µ X i = 1 p ( i ) x ( i ) ≤ ν X n = 1 z ∗ n = z ∗T . But then, z ∗T z ∗ ≥ z G z ∗ ≥
12 , where the last bound holds due to the inequality (10) from Theorem 1 part (ii).
Remark 5.
We observe two facts in Theorem 3 above(i) It is possible to have a strict inequality in the expression (14). To see this, let s be the split items for Πthen, w ( s ) > k = δ − P s − i = 1 w ( i ) which stops the algorithm. However, it is possible that w ( s + 1) ≤ ⌈ k ⌉ for s even, or w ( s + 1) ≤ ⌈ k ⌉ for s odd. In these cases we would necessarily have z G < z Gleft + z Gright ,because one more item could be packed by the greedy algorithm in the problem Π side (side ∈ { left , right } ),for which the item s is not assigned.(ii) When x G = x eG , this is a control solution for any D&C tree built by Algorithm 2. In order to have thisglobal control solution, there is no need to require that z GΠ = z eGΠ for every node Π of T as the algorithmrequires for branching. However, it has been observed empirically, that removing this requirement, heavilydeteriorates the quality of the solution in a Divided-and-Conquer iteration.(iii) Finally, if z G < z eG a rule for assigning capacities δ left , δ right different from the one used by Algorithm 2could be defined. However, given that the extended-greedy algorithm is intractable from the probabilisticpoint of view (as mentioned in Remark 2), this would also make intractable the probabilistic analysis ofthe Divide-and-Conquer method.(iv) In the proof of Theorem 3 we introduced a slight inconsistency with the notation adopted so far, byswitching from subindex to superscript to denote a particular family of problems Π j and its associatedelements δ j , V j . This was done out of necessity for this one time throughout the paper.11 .3. Results from Combinatorics and Probabiilty We devote this subsection to recall some previous background necessary to analyze the 0-1KP from theprobabilistic point of view. We begin with a concept from combinatorics
Definition 4 (Compositions).
Let ( a , . . . , a m ) be a sequence of integers satisfying P mi = 1 a i = n . If a i ≥ i = 1 , . . . , m , the sequence is said to be a composition of n in m parts. (Naturally m should be less orequal that n .) Theorem 4.
Let n, m be two natural numbers with m ≤ n then(i) (cid:18) nm (cid:19) = (cid:18) m − m − (cid:19) + (cid:18) mm − (cid:19) + . . . + (cid:18) n − m − (cid:19) . (16) (ii) The number of compositions of n into m parts is (cid:18) n − m − (cid:19) .(iii) The following identity holds (cid:18) nm (cid:19) = nm (cid:18) n − m − (cid:19) . (17) Proof. (i) See Theorem 4.5 in [30].(ii) See Corollary 5.3 in [30].(iii) By direct calculation.
Proposition 5.
Let A be the set of compositions of n in m parts. Denote by α = ( a , . . . , a m ) , β =( b , . . . , b m ) , the elements of A and define the quantities Σ odd def = X α ∈ A X i odd a i , Σ even def = X α ∈ A X i even a i . (18) (i) If m even, then Σ odd ≡ Σ even .(ii) If m odd, then Σ odd ≡ Σ even + n +1 ℓ +1 (cid:0) n ℓ +1 (cid:1) = Σ even + n +1 ℓ +1 A , where m = 2 ℓ + 1 .Proof. (i) Since m = 2 ℓ , consider the permutation σ ∈ S ([ m ]) defined by σ : [ m ] → [ m ] , σ ( i ) def = ( i + 1 , i is odd ,i − , i is even . Define the map B : A → A α = ( a , a , . . . , a ℓ − , a ℓ ) ( a , a , . . . , a ℓ , a ℓ − ) = ( a σ (1) , a σ (2) , . . . , a σ (2 ℓ − , a σ (2 ℓ ) ) . It is direct to see that B is a bijection, then P α ∈ A m P i = 1 a i = P β = B ( α ) α ∈ A m P i = 1 b i . MoreoverΣ odd = X α ∈ A X i odd a i = X β = B ( α ) α ∈ A X i odd b i = X α ∈ A X i odd a σ ( i ) = X α ∈ A X i even a i = Σ even , which concludes the first part. 12ii) Since m = 2 ℓ + 1, consider the permutation σ ∈ S ([ m ]) defined by σ : [ m ] → [ m ] , σ ( i ) def = ℓ + 1 , i = 2 ℓ + 1 ,i + 1 , i is odd, i = 2 ℓ + 1 ,i − , i is even . As in the previous part, define the map B : A →A α = ( a , a , . . . , a ℓ − , a ℓ , a ℓ +1 ) ( a , a , . . . , a ℓ , a ℓ − , a ℓ +1 )= ( a σ (1) , a σ (2) , . . . , a σ (2 ℓ − , a σ (2 ℓ ) , a σ (2 ℓ +1) ) . As before, this is a bijection, however if we are to use it for computing the difference between Σ left andΣ right further specifications need to be done. Observe that the range of a ℓ +1 ∈ { , . . . , n − ℓ } and define A i = { α ∈ A : a ℓ +1 = i } . Observe that B : A i → A i is also a bijection and that there is a bijectionbetween A i and the set of compositions of n − i in 2 ℓ parts. In particular, it has (cid:0) n − i ℓ (cid:1) elements and dueto the previous part, we have Σ odd ( A i ) = Σ even ( A i ) + i (cid:18) n − i ℓ (cid:19) . Here, Σ odd ( A i ) = P α ∈ A i P i odd a i and Σ even ( A i ) = P α ∈ A i P i even a i . ThereforeΣ odd = n − ℓ X i = 1 Σ odd ( A i ) = n − ℓ X i = 1 Σ even ( A i ) + n − ℓ X i = 1 i (cid:18) n − i ℓ (cid:19) = Σ odd + n − ℓ X i = 1 i (cid:18) n − i ℓ (cid:19) . We focus on the last sum n − ℓ X i = 1 i (cid:18) n − i ℓ (cid:19) = n − X j = 2 ℓ ( n − j ) (cid:18) j ℓ (cid:19) =( n + 1) n − X j = 2 ℓ (cid:18) j ℓ (cid:19) − n − X j = 2 ℓ ( j + 1) (cid:18) j ℓ (cid:19) =( n + 1) n − X j = 2 ℓ (cid:18) j ℓ (cid:19) − (2 ℓ + 1) n X m = 2 ℓ +1 (cid:18) m ℓ + 1 (cid:19) =( n + 1) (cid:18) n ℓ + 1 (cid:19) − (2 ℓ + 1) (cid:18) n + 12 ℓ + 2 (cid:19) . In the expression above, the second equality is a convenient association of summands, the third equalityuses the identity (17) to adjust the binomial coefficient, while the fourth equality applies the expression(16). Further simplifying the latter and combining with the previous we haveΣ odd = Σ even + 12 n + 1 ℓ + 1 (cid:18) n ℓ + 1 (cid:19) , which is the desired result.Next we recall some results from basic discrete probability Theorem 6.
Let (Ω , P ) be a discrete probability space and let (Ω n ) Nn = 1 be a partition of Ω then i) Let A, B ⊆ Ω be two events then P ( A, B ) = P ( A ∩ B ) = P ( A (cid:12)(cid:12) B (cid:1) P ( B ) , (19a) P ( A ) = N X n = 1 P (cid:0) A (cid:12)(cid:12) Ω n (cid:1) . (19b) (ii) Let X : Ω → R be a discrete random variable, let A ⊆ Ω be an event then E (cid:0) X (cid:12)(cid:12) A (cid:1) = X x ∈ X(Ω) x P (cid:0) X = x (cid:12)(cid:12) A (cid:1) , (20a) E (cid:0) X A (cid:1) = E (cid:0) X (cid:12)(cid:12) A (cid:1) P ( A ) , (20b) E ( X ) = N X n = 1 E (cid:0) X (cid:12)(cid:12) Ω n (cid:1) P (Ω n ) . (20c) In the expression (20a) , X (Ω) stands for the range of the random variable X .Proof. (i) For (19a) see Definition 1.3.7 in [31]. For (19b) see Theorem 1.3.9 in [31].(ii) For (20a) see Section 2.3.9, page 49 in [31]. For (20b) see Theorem 2.3.1 in in [31]. Finally, noticingthat E ( X ) = P Nn = 1 E (cid:0) X Ω n (cid:1) and the identity (20b), the equation (20c) follows.
3. Probabilistic Analysis of 0-1KP
In this section, we present the probabilistic analysis of the Divide-and-Conquer method. We begin introducingthe probabilistic model.
Hypothesis 2 (The Random Model).
The random instances (cid:10) δ, (cid:0) W ( i ) (cid:1) µi =1 , (cid:0) P ( i ) (cid:1) µi =1 (cid:11) of the knapsack prob-lem to be analyzed satisfya. The capacity δ and the number of items µ with µ = δ + 1 are fixed.b. The weights (cid:0) W ( i ) (cid:1) µi =1 are i.i.d. random variables, uniformly distributed on [1 , δ ] for all i ∈ [ µ ].c. The profits (cid:0) P ( i ) (cid:1) µi =1 are define by means of the weights and the efficiencies (cid:0) G ( i ) (cid:1) µi =1 . To define theefficiencies we introduce a set of random variables named the increments (cid:0) T ( i ) (cid:1) µi =1 , which are i.i.d.,uniformly distributed on [0 ,
1] for all i ∈ [ µ ]. Hence, the efficiencies G ( i ) and profits P ( i ) are defined by G ( i ) = µ X t = i T ( t ) , P ( i ) = G ( i ) W ( i ) , for all i ∈ [ µ ] . (21) Definition 5 (The Random Model).
With the random model introduced in the hypothesis 2 above, we definethe following problems(i) The random version of the problem 1 is given bymax µ X i = 1 P ( i ) x ( i ) , (22a)14ubject to µ X i = 1 W ( i ) x ( i ) ≤ δ, (22b) x ( i ) ∈ { , } , for all i ∈ [ µ ] . (22c)From now on we refer to it as (ii) The random version of problem 2 is analogous to how 0-1RKP is generated. In the sequel, we refer to itas . Remark 6. (i) It is direct to see that the random instances of Problem (22) satisfy the conditions of Hy-pothesis 1. In particular the efficiencies (cid:0) G ( i ) (cid:1) µi =1 verify the monotonicity condition G (1) ≥ G (2) ≥ . . . ≥ G ( µ ) . (23)(ii) Since W ( i ) ≥ i = 1 , . . . , µ , it follows that the number of packed items is at most δ (i.e., P µi = 1 x ( i ) ≤ δ ), hence we adopt µ = δ + 1 for mathematical convenience.(iii) In the figure 2 we depict two random realizations for the weights, profits and efficiencies, according to theproposed probabilistic model, Table 3 summarizes the values of the random variables for both realizations.In order to compute expected values for the Greedy Algorithm, two important random variables have to beintroduced Definition 6.
Let (cid:10) δ, (cid:0) W ( i ) (cid:1) i =1 , (cid:0) P ( i ) (cid:1) i =1 (cid:11) be a random instance satisfying the hypothesis 2, define(i) The split item random variable S is the value of the index s (introduced in Definition 1 (i)) for the randominstance.(ii) The slack random variable is defined by K def = δ − S − X j = 1 W j , (24)where S is the split item random variable. In this section we compute the expectations of the most important random variables related to the proba-bilistic model introduced in Section 3.1. We begin presenting a result which turns out to be the cornerstone ofour whole construction despite its simplicity
Lemma 7 (Cornerstone Lemma) . Let S and K be the split item and the slack random variables defined above,then P (cid:0) K = k, S = s (cid:1) = δ − kδ s (cid:18) δ − k − s − (cid:19) , (25) for s = 2 , . . . , µ and k = 0 , . . . , δ − s + 1 . Item024681012 (a) Weights (cid:0) W( i ) (cid:1) i =1 for two random realizations. Item0102030405060 (b) Profits (cid:0) P( i ) (cid:1) i =1 for two random realizations. Item0123456 (c) Efficiencies (cid:0) G( i ) (cid:1) i =1 for two random realizations. Item W( i ) P( i ) G( i ) i Table 3: Numerical values for the two random realizations de-picted in the graphs above.Figure 2: Two random realizations according to the probabilistic model introduced in Definition 5, capacity δ = 12, number ofitems µ = 13. Figure (a) displays the weights (W( i )) i =1 , while figure (b) depicts the profits (P( i )) i =1 and (c) portrays the valuesof the efficiencies (G( i )) i =1 . The blue color indicates the first realization while the orange stands for the second realization. Allthe corresponding numerical values are summarized in the table 3. Proof.
Observe the following equivalence of events P (cid:0) K = k, S = s (cid:1) = P (cid:16) δ − s − X j = 1 W ( j ) = k, W ( s ) > k (cid:17) = P (cid:16) s − X j = 1 W ( j ) = δ − k (cid:17) P (cid:16) W ( s ) > k (cid:17) . The last equality uses the independence of the weight random variables. For the first factor we observe thatthe event occurs if and only if ( W ( i )) s − i =1 is a composition of δ − k . According to Theorem 4 (ii) there are (cid:0) δ − k − s − (cid:1) of these compositions and since W ( i ) are uniformly distributed over [ δ ], the probability for each ofthese compositions to occur is δ s − . Next, the event [ W ( s ) > k ] has probability δ − kδ again due to the uniformdistribution of the variable. Combining the previous observations, the equality (25) follows.Finally, for the range of the variables, it is direct to see that S , K are intertwined, then regarding S as theindependent and K the dependent, the first can range freely inside { , . . . , δ + 1 } while the second only takesvalues within { , . . . , δ − s + 1 } because P s − j = 1 W ( j ) ≥ s − S .16 heorem 8. Let S be the splitting item random variable defined above, then its distribution and expectationare given by P ( S = s ) = s − δ s (cid:18) δ + 1 s (cid:19) , for all s = 2 , . . . , µ, (26a) E ( S ) = (cid:0) δ (cid:1) δ , (26b) V ar( S ) = (cid:0) δ (cid:1)(cid:0) δ (cid:1) δ − − (cid:0) δ (cid:1) δ , (26c) where µ = δ + 1 .Proof. Due to the cornerstone lemma 7 if S = s , the slack k = δ − P s − j = 1 W ( j ) runs from 0 to δ − s + 1.Hence, we split the event { S = s } , according to the range of the slack, i.e. P ( S = s ) = δ − s +1 X k = 0 P ( K = k, S = s ) = δ − s +1 X k = 0 δ − kδ s (cid:18) δ − k − s − (cid:19) = δ X m = s − mδ s (cid:18) m − s − (cid:19) . The first equality holds due to the cornerstone identity (25) while the second is mere reindexing of the sum.Recalling that ( m + 1) (cid:0) ms − (cid:1) = ( s − (cid:0) m +1 s − (cid:1) due to the identity (17), we have P ( S = s ) = s − δ s δ X m = s − (cid:18) ms − (cid:19) = s − δ s (cid:18) δ + 1 s (cid:19) , where the last equality holds due to the combinatorial identity (16). This proves the identity (26a). Next, inorder to compute E ( S ), first recall that µ = δ + 1 and get µ X s = 2 s P ( S = s ) = δ +1 X s = 2 s ( s − δ s (cid:18) δ + 1 s (cid:19) = δ +1 X s = 0 s ( s − δ s (cid:18) δ + 1 s (cid:19) = ( δ + 1) δδ (cid:0) δ (cid:1) δ − . Applying some basic algebraic manipulations, the identity (26b) follows. Next, in order to find the variance firstwe compute E ( S ) = µ X s = 2 s P ( S = s )= δ +1 X s = 2 s s − δ s (cid:18) δ + 1 s (cid:19) = δ +1 X s = 2 s ( s − s − δ s (cid:18) δ + 1 s (cid:19) + 2 δ +1 X s = 2 s ( s − δ s (cid:18) δ + 1 s (cid:19) = ( δ + 1) δ ( δ − δ (cid:0) δ (cid:1) δ − + 2 (cid:0) δ (cid:1) δ . Here, the third equality is a convenient association of summands and the fourth equality simply uses thederivatives of the Newton’s binomial identity. Therefore, V ar( S ) = E ( S ) − E ( S ) = (cid:0) − δ (cid:1)(cid:0) δ (cid:1) δ − + 2 (cid:0) δ (cid:1) δ − (cid:0) δ (cid:1) δ . From here, the identity (26c) follows directly.Before computing the expectation of Z G a technical lemma is needed.17 emma 9. With the definitions above, the following identities hold P (cid:0) W ( j ) = w , S = s (cid:1) = 1 δ s ( s − δ + 1) + ws − (cid:18) δ − ws − (cid:19) , j = 1 , . . . , s − , (27a) E (cid:0) W ( j ) (cid:12)(cid:12) S = s (cid:1) = δs + s − s − , j = 1 , . . . , s − . (27b) Proof.
For the first equality, observe that if W ( j ) = w then P s − i = 1 W ( i ) ≥ ( s −
2) + w , consequently the slack K , can take values only in the set { , . . . , δ − ( s − − w } . Hence, P (cid:0) W ( j ) = w , S = s (cid:1) = δ − ( s − − w X k = 0 P (cid:0) W ( j ) = w , K = k, S = s (cid:1) = δ − s +2 − w X k = 0 P (cid:16) W ( j ) = w , X m ∈ [ s − − j W ( m ) = δ − k − w , W ( s ) > k (cid:17) = δ − s +2 − w X k = 0 δ δ s − (cid:18) δ − k − w − s − (cid:19) δ − kδ = 1 δ s δ − w − X ℓ = s − ( ℓ + 1 + w ) (cid:18) ℓs − (cid:19) . Here, the second equality is a direct interpretation of the event P (cid:0) W ( j ) = w , K = k, S = s (cid:1) . The thirdequality is the application of the basic identity (25), while the fourth equality is a reindexing of the sum. Wecompute the latter sum as follows δ − w − X ℓ = s − ( ℓ + 1 + w ) (cid:18) ℓs − (cid:19) = δ − w − X ℓ = s − ( ℓ + 1) (cid:18) ℓs − (cid:19) + w δ − w − X ℓ = s − (cid:18) ℓs − (cid:19) =( s − δ − w − X ℓ = s − (cid:18) ℓ + 1 s − (cid:19) + w δ − w − X ℓ = s − (cid:18) ℓs − (cid:19) =( s − δ − w X m = s − (cid:18) ms − (cid:19) + w δ − w − X ℓ = s − (cid:18) ℓs − (cid:19) =( s − (cid:18) δ − w + 1 s − (cid:19) + w (cid:18) δ − ws − (cid:19) . In the expression above, the second equality uses the identity (17) for shifting indexes, the third equality is amere reindexing of the first sum and the fourth equality applies the identity (16). From here, using again theidentity (cid:0) δ − w +1 s − (cid:1) = δ − w +1 s − (cid:0) δ − ws − (cid:1) and performing further algebraic simplifications, the equation (27a) follows.Next, we prove the identity (27b). Recalling the identity (20a) for conditional expectation, we get E (cid:0) W ( j ) (cid:12)(cid:12) S = s (cid:1) = δ − ( s − X w = 1 w P (cid:0) W ( j ) = w (cid:12)(cid:12) S = s (cid:1) = 1 s − δ s (cid:0) δ +1 s (cid:1) δ − ( s − X w = 1 w P (cid:0) W ( j ) = w , S = s (cid:1) = 1 s − δ s (cid:0) δ +1 s (cid:1) δ s δ − ( s − X w = 1 ( s − δ + 1) w + w s − (cid:18) δ − ws − (cid:19) . u def = δ − w we get δ − ( s − X w = 1 ( s − δ + 1) w + w s − (cid:18) δ − ws − (cid:19) = 1 s − δ − X u = s − (cid:8) ( s − δ + 1)( δ − u ) + ( δ − u ) (cid:9)(cid:18) us − (cid:19) . Appealing to the polynomial identity( s − δ + 1)( δ − u ) + ( δ − u ) =( δ + 1) ( s − − ( sδ + s + 1)( u + 1) + ( u + 1)( u + 2) , we have, δ − X u = s − (cid:8) ( δ + 1) ( s − − ( sδ + s + 1)( u + 1) + ( u + 1)( u + 2) (cid:9)(cid:18) us − (cid:19) =( δ + 1) ( s − δ − X u = s − (cid:18) us − (cid:19) − ( sδ + s + 1) δ − X u = s − ( u + 1) (cid:18) us − (cid:19) + δ − X u = s − ( u + 1)( u + 2) (cid:18) us − (cid:19) . Now, listing the three sums of the left hand side we have δ − X u = s − (cid:18) us − (cid:19) = (cid:18) δs − (cid:19) , δ − X u = s − ( u + 1) (cid:18) us − (cid:19) = ( s − δ − X u = s − (cid:18) u + 1 s − (cid:19) = ( s − δ X r = s − (cid:18) rs − (cid:19) = ( s − (cid:18) δ + 1 s (cid:19) , δ − X u = s − ( u + 2)( u + 1) (cid:18) us − (cid:19) = s ( s − δ − X u = s − (cid:18) u + 2 s (cid:19) = s ( s − δ +1 X r = s (cid:18) rs (cid:19) = s ( s − (cid:18) δ + 2 s + 1 (cid:19) . Combining the above with the previous gives s − δ s (cid:18) δ + 1 s (cid:19) δ s E (cid:0) W ( j ) (cid:12)(cid:12) S = s (cid:1) =( δ + 1) (cid:18) δs − (cid:19) − ( sδ + s + 1) (cid:18) δ + 1 s (cid:19) + s (cid:18) δ + 2 s + 1 (cid:19) = s ( δ + 1) (cid:18) δ + 1 s (cid:19) − ( sδ + s + 1) (cid:18) δ + 1 s (cid:19) + s δ + 2 s + 1 (cid:18) δ + 1 s (cid:19) = sδ + s − s + 1 (cid:18) δ + 1 s (cid:19) . Here, the second equality uses the identity (17) in the first and third summand, while the second equality isthe mere algebraic sum of the previous line. Finally, a direct simplification of terms yields the identity (27b)and the result is complete.
Theorem 10.
Let S and Z G def = S − P i = 1 P ( i ) be the split item and the greedy algorithm profit random variables forthe 0-1RKP (22) . Then, E (cid:0) Z G (cid:12)(cid:12) S = s (cid:1) = 2 δ − s + 44 δs + s − s + 1 , for all s = 2 , . . . , µ, (28a) E ( Z G ) = − ( δ + 1) δ (cid:0) δ (cid:1) δ − + (2 δ + 3)( δ + 2)( δ + 1)4 δ n(cid:0) δ (cid:1) δ − o − ( δ + 2) n(cid:0) δ (cid:1) δ +1 − δ + 1 δ o + 2 δ + 52 δ n(cid:0) δ (cid:1) δ +2 − δ + 7 δ + 22 δ o , (28b)19 ith µ = δ + 1 .Proof. We compute the identity (28a) directly, using the definition of P ( i ) introduced in Equation (21) E (cid:0) Z G (cid:12)(cid:12) S = s (cid:1) = s − X j = 1 E (cid:0) P ( j ) (cid:12)(cid:12) S = s (cid:1) = s − X j = 1 E (cid:0) W ( j ) G ( j ) (cid:12)(cid:12) S = s (cid:1) = s − X j = 1 E (cid:16) W ( j ) µ X t = j T ( t ) (cid:12)(cid:12) S = s (cid:17) . Recalling that the variables (cid:0) W ( i ) (cid:1) µi =1 and (cid:0) T ( i ) (cid:1) µi =1 are independent, we have E (cid:0) Z G (cid:12)(cid:12) S = s (cid:1) = s − X j = 1 E (cid:0) W ( j ) (cid:12)(cid:12) S = s (cid:1) µ X t = j E (cid:0) T ( t ) (cid:12)(cid:12) S = s (cid:1) = s − X j = 1 δs + s − s − µ − j + 12= ( s − µ − s + 2)4 δs + s − s − . Here, the second equality holds due to the identity (27b) and the distribution of the increments ( T ( i )) µi =1 introduced in Hypothesis 2. Simplifying the expression above, the Equation (28a) follows.Next, we compute the expectation of Z G conditioning on the possible values of S and combining the withthe identities (28a), (26a); this gives E ( Z G ) = δ +1 X s = 2 E (cid:0) Z G (cid:12)(cid:12) S = s (cid:1) P (cid:0) S = s (cid:1) = δ +1 X s = 2 δ − s + 44 δs + s − s + 1 s − δ s (cid:18) δ + 1 s (cid:19) = 14( δ + 2) δ +2 X m = 3 ( m − δ − m + 5) (cid:0) ( δ + 1) m − δ − (cid:1) δ m − (cid:18) δ + 2 m (cid:19) = 14( µ + 1) µ +1 X m = 3 ( m − µ − m + 3) (cid:0) µm − µ − (cid:1) δ m − (cid:18) µ + 1 m (cid:19) . The third equality in the expression above is a convenient reindexing of the sum while the last equality followsfrom the substitution µ = δ + 1. Next, consider the polynomial identity( m − µ − m + 3)( µm − µ −
1) = − µm ( m − m −
2) + (2 µ + 1)( µ + 1) m ( m − − µ + 1) m + (4 µ + 6)( µ + 1) , and combine it with the expression above. We get E ( Z G ) = − δ µµ + 1 µ +1 X m = 3 m ( m − m − δ m − (cid:18) µ + 1 m (cid:19) + 2 µ + 14 δ µ +1 X m = 3 m ( m − δ m − (cid:18) µ + 1 m (cid:19) − ( µ + 1) µ +1 X m = 3 mδ m − (cid:18) µ + 1 m (cid:19) + δ µ + 64 µ +1 X m = 3 δ m (cid:18) µ + 1 m (cid:19) = − δ µµ + 1 ( µ + 1) µ ( µ − (cid:0) δ (cid:1) µ − + 2 µ + 14 δ n ( µ + 1) µ (cid:0) δ (cid:1) µ − − ( µ + 1) µ o − ( µ + 1) n ( µ + 1) (cid:0) δ (cid:1) µ − ( µ + 1) − µ ( µ + 1) δ o + δ µ + 64 n(cid:0) δ (cid:1) µ +1 − − µ + 1 δ − ( µ + 1) µ δ o . δ = µ −
1, the equality (28b) follows.Next we find the distribution, conditional expectation with respect to S and expectation for the slack K . Theorem 11.
The slack random variable K , introduced in Definition P (cid:0) K = k (cid:1) = δ − kδ (cid:0) δ (cid:1) δ − k − , for all k = 0 , . . . , δ. (29a) E (cid:0) K (cid:12)(cid:12) S = s (cid:1) = δ + 1 − ss + 1 , for all s = 2 , . . . , µ. (29b) E (cid:0) K (cid:1) = − ( δ + 1) δ n(cid:0) δ (cid:1) δ − o + ( δ + 3) n(cid:0) δ (cid:1) δ +1 − δ + 1 δ o − δ n(cid:0) δ (cid:1) δ +2 − δ + 7 δ + 22 δ o . (29c) Proof.
Revisiting the cornerstone lemma 7, observe that for K = k , the range of the split index s is { , . . . , δ − k + 1 } . Hence, P ( K = k ) = δ − k +1 X s = 2 P ( K = k, S = s ) = δ − k +1 X s = 2 s − δ s (cid:18) δ − ks − (cid:19) = 1 δ δ − k X j = 1 jδ j − (cid:18) δ − kj (cid:19) . Here, the second equality holds due to the cornerstone identity (25), the third equality is a convenient reindexingand association of terms. Finally applying the derivative of the Newton’s binomial expansion, the identity (29a)follows.Next, we show the equality (29b). From the cornerstone lemma 7 observe that if S = s the range of theslack K is { , . . . , δ − s + 1 } . Hence, recalling the conditional expectation identity (20a) we get E (cid:0) K (cid:12)(cid:12) S = s (cid:1) = δ − s +1 X k = 0 k P (cid:0) K = k (cid:12)(cid:12) S = s ) = 1 P ( S = s ) δ − s +1 X k = 0 k P (cid:0) K = k, S = s ) . Now, appealing to the basic identiy (25), we have E (cid:0) K (cid:12)(cid:12) S = s (cid:1) = 1 P ( S = s ) 1 δ s δ − s +1 X k = 0 ( δ − k ) k (cid:18) δ − k − s − (cid:19) = 1 P ( S = s ) 1 δ s δ X j = s − j ( δ − j ) (cid:18) j − s − (cid:19) = 1 P ( S = s ) δ + 1 δ s δ X j = s − j (cid:18) j − s − (cid:19) − P ( S = s ) 1 δ s δ X j = s − ( j + 1) j (cid:18) j − s − (cid:19) . Here, the second equality follows from reindexing j = δ − k , while the third equality is a mere convenientassociation of summands. Next from the identity (17), we get the equalities js − (cid:0) j − s − (cid:1) = (cid:0) js − (cid:1) , j ( j +1) s ( s − (cid:0) j − s − (cid:1) = (cid:0) j +1 s (cid:1) for the first and second summands respectively. From here, proceeding as in the proofs of Lemma 9 andTheorem 10, the identity (29b) follows.Finally, we pursue a closed form for E ( K ); to that end we apply the identity (20c) and get E (cid:0) K (cid:1) = δ +1 X s = 2 E (cid:0) K (cid:12)(cid:12) S = s (cid:1) P (cid:0) S = s (cid:1) = δ +1 X s = 2 δ + 1 − ss + 1 s − δ s (cid:18) δ + 1 s (cid:19) = 1 δ + 2 δ +1 X s = 2 ( δ + 1 − s )( s − δ s (cid:18) δ + 2 s + 1 (cid:19) = 1 δ + 2 δ +2 X j = 3 ( δ + 2 − j )( j − δ j − (cid:18) δ + 2 j (cid:19) . j = s + 1. Next, we replace thepolynomial identity ( δ + 2 − j )( j −
2) = − j ( j + 1) + ( δ + 3) j − δ + 2)in the expression above and get( δ + 2) E (cid:0) K (cid:1) = − δ δ +2 X j = 3 j ( j − δ j − (cid:18) δ + 2 j (cid:19) + ( δ + 3) δ +2 X j = 3 jδ j − (cid:18) δ + 2 j (cid:19) − δ ( δ + 2) δ +2 X j = 3 δ j (cid:18) δ + 2 j (cid:19) = − δ n ( δ + 2)( δ + 1) (cid:0) δ (cid:1) δ − ( δ + 2)( δ + 1) o + ( δ + 3) n ( δ + 2) (cid:0) δ (cid:1) δ +1 − ( δ + 2) − ( δ + 2)( δ + 1) δ o − δ ( δ + 2) n(cid:0) δ (cid:1) δ +2 − − δ + 2 δ − ( δ + 2)( δ + 1)2 δ o . Here, the second equality uses the Newton’s binomial expansion, together with its first and second derivatives.Finally, simplifying the latter expression the equality in (29c) follows.
Theorem 12.
Let Z LP be the optimal profit value given by the linear relaxation of the 0-1RKP (22) . Then,its expected value is given by E ( Z LP ) = E ( Z G ) + δ + 12 δ (cid:0) δ (cid:1) δ − − ( δ + 2)( δ + 1) δ n (1 + 1 δ (cid:1) δ − o + ( δ + 5)( δ + 2)2 n(cid:0) δ (cid:1) δ − − δ + 1 δ o − ( δ + 3) δ n(cid:0) δ (cid:1) δ +2 − δ + 7 δ + 22 δ o = − ( δ + 1)( δ − δ (cid:0) δ (cid:1) δ − + (2 δ − δ + 2)( δ + 1)4 δ n(cid:0) δ (cid:1) δ − o − ( δ + 2)( δ − n(cid:0) δ (cid:1) δ +1 − δ + 1 δ o − δ n(cid:0) δ (cid:1) δ +2 − δ + 7 δ + 22 δ o . (30) Here, Z G is the profit of the solution furnished by the greedy algorithm, whose expectation E ( Z G ) is given bythe identity (28b) .Proof. Due to Theorem 1 (ii), equation (9b), we know that Z LP = S − X j = 1 P ( j ) + 1 W ( S ) (cid:16) δ − S − X j = 1 W ( j ) (cid:17) P ( S ) = Z G + K G ( S ) = Z G + K µ X ℓ = S T ( ℓ ) . (31)22ence, conditioning on the range of S and recalling the equalities (29b), (26a), we have E ( Z LP ) = E ( Z G ) + µ X s = 2 E (cid:16) K µ X ℓ = S T ( ℓ ) (cid:12)(cid:12)(cid:12) S = s (cid:17) P ( S = s )= E ( Z G ) + µ X s = 2 E (cid:0) K (cid:12)(cid:12) S = s (cid:1) E (cid:16) µ X ℓ = S T ( ℓ ) (cid:12)(cid:12)(cid:12) S = s (cid:17) P ( S = s )= E ( Z G ) + µ X s = 2 δ + 1 − ss + 1 µ − s + 12 s − δ s (cid:18) δ + 1 s (cid:19) = E ( Z G ) + 12 1 δ + 2 µ X s = 2 ( µ − s )( µ + 1 − s )( s − δ s (cid:18) δ + 2 s + 1 (cid:19) = E ( Z G ) + 12 1 µ + 1 µ +1 X m = 3 ( µ + 1 − m )( µ + 2 − m )( m − δ m − (cid:18) µ + 1 m (cid:19)| {z } def =Σ In the expression above, the fourth equality uses the identities µ = δ + 1 and (17). The fifth equality followsfrom reindexing m = s + 1, here we also denote the second summand term by Σ. Next, we focus on derivinga closed form for Σ, to that end, we appeal to the polynomial identity( µ + 1 − m )( µ + 2 − m )( m −
2) = m ( m − m − − µ + 1) m ( m − µ + 4)( µ + 1) m − µ +2)( µ +1) . Replacing the latter in the second summand Σ, it transforms inΣ ≡ δ µ + 1 µ +1 X m = 3 m ( m − m − δ m − (cid:18) µ + 1 m (cid:19) − δ µ +1 X m = 3 m ( m − δ m − (cid:18) µ + 1 m (cid:19) + µ + 42 µ +1 X m = 3 mδ m − (cid:18) µ + 1 m (cid:19) − δ ( µ + 2) µ +1 X m = 3 δ m (cid:18) µ + 1 m (cid:19) = µ δ (cid:0) δ (cid:1) µ − − ( µ + 1) µδ n (1 + 1 δ (cid:1) µ − − o + ( µ + 4)( µ + 1)2 n(cid:0) δ (cid:1) µ − − µδ o − δ ( µ + 2) n(cid:0) δ (cid:1) µ +1 − − µ + 1 δ − ( µ + 1) µ δ o = δ + 12 δ (cid:0) δ (cid:1) δ − − ( δ + 2)( δ + 1) δ n (1 + 1 δ (cid:1) δ − o + ( δ + 5)( δ + 2)2 n(cid:0) δ (cid:1) δ − − − δ + 1 δ o − δ ( δ + 3) n(cid:0) δ (cid:1) δ +2 − − δ + 2 δ − ( δ + 2)( δ + 1)2 δ o . Here, the second equality was attained using Newton’s binomial identity, together with its first three derivatives.The last equality was attained by replacing δ = µ −
1. Performing further simplifications we get the first equalityin the identity (30) and replacing (28b) in it, we obtain the second equality.
Definition 7.
Define the post-greedy profit random variable, associated with the 0-1RLPK, as Y LP def = K G ( S ) . (32) Theorem 13 (Asymptotic Relations) . Let S , K , Z G and Z LP be the random variables defined so far, then the ollowing limits hold lim δ → ∞ E (cid:0) S (cid:1) = e, (33a)lim δ → ∞ V ar (cid:0) S (cid:1) = e (cid:0) e − e (cid:1) , (33b)lim δ → ∞ E (cid:0) K (cid:1) δ = 3 − e, (33c)lim δ → ∞ E ( Z G ) + δ − E ( S ) + 12 E ( K ) E (cid:0) Z LP (cid:1) = 1 , (33d)lim δ → ∞ E (cid:0) Z G (cid:1) E (cid:0) Z LP (cid:1) = e − , (33e)lim δ → ∞ δ − E ( S ) + 12 E ( K ) E (cid:0) Z LP (cid:1) = 3 − e. (33f) Sketch of the proof.
An elementary calculation of limits on the corresponding closed formulas developed abovegives all the desired results.
Remark 7.
Observe that if we approximate E ( K P µt = S T ( t )) with δ − E (S)+12 E ( K ) then, the expression E ( Z G )+ δ − E (S)+12 E ( K ) is an approximation of E ( Z LP ) as the equation (31) shows. Hence, the statement (33d) provesthat this is a good approximation. We close this section presenting the computation of the eligible-first algorithm expectation E ( Z eF ). Giventhat the proofs are remarkably similar to those presented in the previous section, we only present sketches ofthe proofs with some important highlights. Definition 8.
Let K and S be the random variables introduced in Definition 6(i) Let E be the set defined in (7a). We say an item i ∈ [ µ ] is eligible-fist eF if it is the least element of theset E i.e., if it is the first eligible item, once the greedy algorithm has stopped packing items.(ii) For the eligible-first algorithm, we define its corresponding post-greedy profit random variable as follows Y eF = ( P ( i ) i is eF , , E = ∅ . (34) Lemma 14.
With the definitions above we have P (cid:0) i is eF , K = k, S = s (cid:1) = δ − kδ s +1 k (cid:0) − kδ (cid:1) i − s − (cid:18) δ − k − s − (cid:19) , (35a) for i = s + 1 , . . . , µ , k = 0 , . . . , δ − s − , s = 2 , . . . , µ . E (cid:0) Y eF (cid:12)(cid:12) K = k, S = s (cid:1) = k δ − s + 1) n − (cid:0) − kδ (cid:1) δ − s +1 o − δ k (cid:0) − kδ (cid:1)n − (cid:0) δ − sδ k (cid:1)(cid:0) − kδ (cid:1) δ − s o . (35b) Sketch of the proof.
In order to prove (35a) first notice that P (cid:0) i is eF (cid:12)(cid:12) K = k, S = s (cid:1) = kδ (cid:0) − kδ (cid:1) i − s − , W ( j ) must be bigger than k for j = s + 1 , . . . , i − W ( i ) must be less or equal than k . Each of theformer events has probability 1 − kδ , which must take place i − s − i − − ( s + 1) + 1 times, while the latterevent has probability kδ . Recalling that P (cid:0) i is eF , K = k, S = s (cid:1) = P (cid:0) i is eF (cid:12)(cid:12) K = k, S = s (cid:1) P (cid:0) K = k, S = s (cid:1) together with the cornerstone identity (25), the equation (35a) follows. It is also important to stress that i iseF only if i > s .Finally, in order to prove the identity (35b) observe that E (cid:0) W ( i ) (cid:12)(cid:12) i is eF , K = k, S = s (cid:1) = k because theevent [ i is eF ] implies the event [ W ( i ) ≤ k ]. Hence, E (cid:0) P ( i ) (cid:12)(cid:12) i is eF , K = k, S = s (cid:1) = E (cid:0) G ( i ) W ( i ) (cid:12)(cid:12) i is eF , K = k, S = s (cid:1) == E (cid:0) G ( i ) (cid:12)(cid:12) i is eF , K = k, S = s (cid:1) E (cid:0) W ( i ) (cid:12)(cid:12) i is eF , K = k, S = s (cid:1) = µ − i + 12 k . From here, we get the identity (35b) using the same reasoning of the previous.
Theorem 15 (Expected values of Z eF ) . With the definitions above, the following expectation holds E ( Z eF ) = E ( Z G ) + µ X s = 2 µ − s +1 X k = 0 k δ − s + 1) n − (cid:0) − kδ (cid:1) δ − s +1 o δ − kδ s (cid:18) δ − k − s − (cid:19) − µ X s = 2 µ − s +1 X k = 0 δ k (cid:0) − kδ (cid:1)n − (cid:0) δ − sδ k (cid:1)(cid:0) − kδ (cid:1) δ − s o δ − kδ s (cid:18) δ − k − s − (cid:19) . (36) Here Z eF is the value of the objective function furnished by the eligible-first algorithm, introduced in Definition
Recalling E ( Y eF ) = µ X s = 2 µ − ( s − X k = 1 E (cid:0) Y eF (cid:12)(cid:12) K = k, S = s (cid:1) P (cid:0) K = k, S = s (cid:1) , together with the fundamental identity (25), the equation (36) follows. Corollary 16 (Approximation of E ( Z eF )) . With the definitions above, the following estimate holds E ( Z eF ) ∼ eF( δ ) def = E ( Z G ) + E ( K )4 ( δ − E ( S ) + 1) n − (cid:0) − E ( K ) δ (cid:1) δ − E (S)+1 o − δ E ( K ) (cid:0) − E ( K ) δ (cid:1)n − (cid:0) δ − E ( S ) δ E ( K ) (cid:1)(cid:0) − E ( K ) δ (cid:1) δ − E (S) o . (37) Proof.
Let k = (cid:4) E ( K ) (cid:5) and s = (cid:6) E ( S ) (cid:7) and notice the approximation E ( Y eF ) ∼ E (cid:0) Y eF (cid:12)(cid:12) K = k , S = s (cid:1) ∼ k δ − s + 1) n − (cid:0) − k δ (cid:1) δ − s +1 o − δ k (cid:0) − k δ (cid:1)n − (cid:0) δ − s δ k (cid:1)(cid:0) − k δ (cid:1) δ − s o ∼ E ( K )4 ( δ − E ( S ) + 1) n − (cid:0) − E ( K ) δ (cid:1) δ − E (S)+1 o − δ E ( K ) (cid:0) − E ( K ) δ (cid:1)n − (cid:0) δ − E ( S ) δ E ( K ) (cid:1)(cid:0) − E ( K ) δ (cid:1) δ − E (S) o . Here, the first approximation follows by assuming that K , S are constant and equal to k , s respectively. Thesecond line follows from the equality (35b) and the third line follows by merely replacing back k , s by thecorresponding expected values E ( K ) and E ( S ) respectively. Next, recalling that E ( Z eF ) = E ( Z G ) + E ( Y eF )and using the approximation above, the estimate (37) follows.25 emark 8. Observe that we denote the approximation eF( δ ), as a function depending only on the capacity δ . This is a correct statement because E ( S ) and E ( K ) are both functions, exclusively depending on δ as theequations (26b) and (29c) show.
4. Probabilistic Analysis of a D&C Pair
With the current probabilistic setting it is not possible to get exact expressions for the expected value of Z ∗ (not to mention closed formulas), because it is not possible to give explicit expressions for the optimalsolution z ∗ as we were able to attain for z G in (5) and z LP in (9b). Furthermore, it is not possible to give suchexplicit descriptions for z fG or even z eG , therefore we use the greedy algorithm and the eligible-first algorithmintroduced in Definition 1, to estimate the expected performance of the Divide-and-Conquer method. Π left and Π right random subproblems For the analysis of the Divide-and-Conquer method, the induced problems Π left and Π right must be analyzedindependently. To that end we introduce the random setting for each of these problems
Definition 9.
Define the following elements introduced by one iteration of the Divide-and-Conquer method(i) The left and right capacity random variables are given by C left def = S − X i odd W ( i ) + l K m , C right def = S − X i even W ( i ) + j K k . (38)(ii) The left and right subproblems are defined byΠ left def = (cid:10) C left , ( P ( i )) i ∈ V left , ( W ( i )) i ∈ V left (cid:11) , Π right def = (cid:10) C right , ( P ( i )) i ∈ V right , ( W ( i )) i ∈ V right (cid:11) , (39)with V left def = { i ∈ [ µ ] : i is odd } and V right def = { i ∈ [ µ ] : i is even } .(iii) We denote by Z algleft , Z algright , the corresponding objective function values to Π left , Π right furnished by thealgorithms alg = ∗ , G , fF , eF , LP. (Recall that the case alg = ∗ , stands for the optimal solution, i.e., theoptimal value generated by an exact algorithm, e.g., dynamic programming.)(iv) We denote by Y eFleft , Y eFright , the corresponding post-greedy profit random variables of Π left , Π right respectively,furnished by the algorithms alg = eF , LP and according to the definitions 8 (ii) and 7respectively.(v) Denote by S left ( S right ), K left ( K right ) be the split item and the slack of the Π left (Π right ) problem.Before proceeding to the next results, in order to reduce the cases of analysis, we adopt the following hypothesis Hypothesis 3.
From not on it will be assumed that µ = 2 λ , i.e., the quantity of eligible items is even. Inparticular, each subproblem Π left and Π right has λ eligible items. Theorem 17.
Let C left , C right be the random variables introduced in Definition s is an odd number, then E (cid:0) C left (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k (cid:6) k (cid:7) , (40a) E (cid:0) C right (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k (cid:4) k (cid:5) . (40b)26 ii) If s is an even number, then E (cid:0) C left (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k (cid:6) k (cid:7) + 14 δ − ks − , (41a) E (cid:0) C right (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k (cid:4) k (cid:5) − δ − ks − . (41b) Proof.
Recall that it K = k and S = s then P s − i = 1 W ( i ) = δ − k and W ( s ) > δ − k , then (cid:0) W ( i ) (cid:1) s − i =1 is acomposition of δ − k in s − s is odd, then s − E (cid:0) S − X i odd W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) = E (cid:0) S − X i even W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) . Recalling that E (cid:0) P S − i = 1 W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k , the result follows.(ii) If s is even, then s − ℓ + 1 is odd and due to Theorem 5 (ii) about compositions, it follows that E (cid:0) S − X i odd W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) = E (cid:0) S − X i even W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) + 12 δ − k ℓ + 1= E (cid:0) S − X i even W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) + 12 δ − ks − . The second equality is a mere replacement of s = 2 ℓ + 2. Hence, recalling that E (cid:0) P S − i = 1 W ( i ) (cid:12)(cid:12) K = k, S = s (cid:1) = δ − k and solving the 2 × Lemma 18.
Let K be the slack variable introduced in Definition E (cid:16)l K m(cid:17) = E ( K )2 + 12 δ δ X k even k (cid:0) δ (cid:1) k − , (42a) E (cid:16)j K k(cid:17) = E ( K )2 − δ δ X k even k (cid:0) δ (cid:1) k − . (42b) Proof.
We prove the statement using the definition E ( ⌈ K2 ⌉ ) = P δk = 0 ⌈ k ⌉ P ( K = k ). Hence, separating evenand odd indexes we get E (cid:16)l K m(cid:17) = λ − X ℓ = 0 l ℓ m P ( K = 2 ℓ ) + λ − X ℓ = 0 l ℓ + 12 m P ( K = 2 ℓ + 1)= λ − X ℓ = 0 ℓ P ( K = 2 ℓ ) + λ − X ℓ = 0 ( ℓ + 1) P ( K = 2 ℓ + 1)= λ − X ℓ = 0 ℓ P ( K = 2 ℓ ) + λ − X ℓ = 0 ℓ + 12 P ( K = 2 ℓ + 1) + 12 λ − X ℓ = 0 P ( K = 2 ℓ + 1)= E ( K )2 + 12 λ − X ℓ = 0 P ( K = 2 ℓ + 1) . ⌈·⌉ , the third equality is a convenientassociation of terms and the fourth equality merely recovers the expectation of the slack random variable K .Next we focus in the last sum,12 λ − X ℓ = 0 P ( K = 2 ℓ + 1) = 12 δ X k odd δ − kδ (cid:0) δ (cid:1) δ − k − = 12 δ δ X m even m (cid:0) δ (cid:1) m − . Here, the first equality comes from the identity (29a). The second equality is the reindexing m = δ − k andrecalling that δ and k are odd, it follows that m is even. Combining with the previous expression, the identity(42a) follows.In order to prove the identity (42b), it suffices to note that ⌊ K2 ⌋ = K − ⌈ K2 ⌉ and use (42a) to conclude theresult. Theorem 19.
The random variable capacities of the left and right problems have the following expectations E ( C left ) = δ δ δ X k even k (cid:0) δ (cid:1) k − + µ n(cid:0) δ (cid:1) µ + (cid:0) − δ (cid:1) µ o − δ n(cid:0) δ (cid:1) µ +1 − (cid:0) − δ (cid:1) µ +1 o . (43a) E ( C right ) = δ − δ δ X k even k (cid:0) δ (cid:1) k − − µ n(cid:0) δ (cid:1) µ + (cid:0) − δ (cid:1) µ o + δ n(cid:0) δ (cid:1) µ +1 − (cid:0) − δ (cid:1) µ +1 o . (43b) Proof.
We focus on the calculation of E ( C left ) using the definition, i.e., E ( C left ) = X s X k E (cid:0) C left (cid:12)(cid:12) K = k, S = s (cid:1) P (cid:0) K = k, S = s (cid:1) . According to the expressions (40a) and (41a), there are two paramount parts: the “head” δ − k + (cid:6) k (cid:7) , presentin both cases and the “tail” δ − ks − , present only in the case s even. We compute these separately, for the“head” we recall the cornerstone identity (25) and get δ X k = 0 δ − k +1 X s = 2 n δ − k (cid:6) k (cid:7)o δ − kδ s (cid:18) δ − k − s − (cid:19) = δ X k = 0 n δ − k (cid:6) k (cid:7)o δ − k +1 X s = 2 δ − kδ s (cid:18) δ − k − s − (cid:19) = δ X k = 0 n δ − k (cid:6) k (cid:7)o P ( K = k )= δ − E ( K )2 + E (cid:16)l K m(cid:17) = δ δ δ X k even k (cid:0) δ (cid:1) k − . (44)Here, the first equality is direct, the second holds by definition of P ( K = k ) (see the proof of (29a) in Lemma11 ), the third equality holds by definition of expectation and the fourth equality is obtained combining thelatter with (42a). Next we compute the “tail”, recalling identities (25) and (17), we have X s even δ − s +1 X k = 0 δ − ks − δ − kδ s (cid:18) δ − k − s − (cid:19) = X s even δ s δ − s +1 X k = 0 ( δ − k ) (cid:18) δ − ks − (cid:19) .
28e focus on the internal sum δ − s +1 X k = 0 ( δ − k ) (cid:18) δ − ks − (cid:19) = s δ − s +1 X k = 0 (cid:18) δ − k + 1 s (cid:19) − δ − s +1 X k = 0 (cid:18) δ − ks − (cid:19) = s (cid:18) δ + 2 s + 1 (cid:19) − (cid:18) δ + 1 s (cid:19) = sµ − s + 1 (cid:18) µs (cid:19) . Then, back to the “tail” term we have µ X s even δ s sµ − s + 1 (cid:18) µs (cid:19) = 14( µ + 1) µ X s even sµ − δ s (cid:18) µ + 1 s + 1 (cid:19) = 14( µ + 1) µ +1 X s even sµ − δ s (cid:18) µ + 1 s + 1 (cid:19) = µ µ + 1) µ +1 X s even s + 1 δ s (cid:18) µ + 1 s + 1 (cid:19) − δ µ +1 X s even δ s +1 (cid:18) µ + 1 s + 1 (cid:19) . = µ µ + 1) µ +1 X ℓ odd ℓδ ℓ − (cid:18) µ + 1 ℓ (cid:19) − δ µ +1 X ℓ odd δ ℓ (cid:18) µ + 1 ℓ (cid:19) . In the expression above, the first equality is the adjustment of the binomial coefficient using the identity (17).The second equality extends the upper limit sum from µ to µ + 1, which can be done without picking up newsummands, because we have assumed that µ is even and we are adding over s even. The third equality is aconvenient association of terms. Next, recall that F ( x ) def = n X ℓ odd (cid:18) nℓ (cid:19) x ℓ = (1 + x ) n − (1 − x ) n . Hence, using the function F ( · ) and its first derivative, the tail term gives X s even δ s sµ − s + 1 (cid:18) µs (cid:19) = µ µ + 1) n ( µ + 1) (cid:0) δ (cid:1) µ + ( µ + 1) (cid:0) − δ (cid:1) µ o − δ n(cid:0) δ (cid:1) µ +1 − (cid:0) − δ (cid:1) µ +1 o = µ n(cid:0) δ (cid:1) µ + (cid:0) − δ (cid:1) µ o − δ n(cid:0) δ (cid:1) µ +1 − (cid:0) − δ (cid:1) µ +1 o . (45)Putting together the “head” of the sum (44) and the “tail” (45), the identity (43a) follows.In order to compute the expectation of C right given by the expression (43b), we repeat the same procedure,but, keeping in mind that the “tail” (45) has to be subtracted rather than added.In oder to ease future calculations, we will use the following estimates Lemma 20.
Let S left , S right be the splitting item random variable defined above for the problems Π left , Π right then, their expectations satisfy E (cid:0) S right (cid:12)(cid:12) C right = c (cid:1) = E (cid:0) S left (cid:12)(cid:12) C left = c (cid:1) = (cid:0) c (cid:1) c , (46a) E (cid:0) S left (cid:1) ∼ (cid:0) E ( C left ) (cid:1) E (C left ) , (46b) E (cid:0) S right (cid:1) ∼ (cid:0) C right (cid:1) C right . (46c)29 ketch of the proof. The proof of equation (46a) is analogous to the proof of Lemma 8, because once C left is known/fixed, the conditional expectations depend strictly on the capacity of the particular 0-1KP, as well asthe weight random variables (cid:0) W (2 i − (cid:1) λi =1 , (cid:0) W (2 i ) (cid:1) λi =1 for Π left and Π right respectively, whose distribution isuniform and independent from each other.The estimates (46b) and (46c) follow directly using the same reasoning of Corollary 16. Lemma 21.
The slack random variables K left , K right , introduced in Definition E (cid:0) K right (cid:12)(cid:12) C right = c (cid:1) = E (cid:0) K left (cid:12)(cid:12) C left = c (cid:1) = − ( c + 1) c n(cid:0) c (cid:1) c − o + ( c + 3) n(cid:0) c (cid:1) c +1 − c + 1 c o − c n(cid:0) c (cid:1) c +2 − c + 7 c + 22 c o . (47a) E (cid:0) K left (cid:1) ∼ − ( E ( C left ) + 1) E ( C left ) n(cid:0) E ( C left ) (cid:1) E (C left ) − o + ( E ( C left ) + 3) n(cid:0) E ( C left ) (cid:1) E (C left )+1 − E ( C left ) + 1 E ( C left ) o − E ( C left ) n(cid:0) E ( C left ) (cid:1) E (C left )+2 − E ( C left ) + 7 E ( C left ) + 22 E ( C left ) o . (47b) E (cid:0) K right (cid:1) ∼ − ( E ( C right ) + 1) E ( C right ) n(cid:0) E ( C right ) (cid:1) E (C right ) − o + ( E ( C right ) + 3) n(cid:0) E ( C right ) (cid:1) E (C right )+1 − E ( C right ) + 1 E ( C right ) o − E ( C right ) n(cid:0) E ( C right ) (cid:1) E (C right )+2 − E ( C right ) + 7 E ( C right ) + 22 E ( C right ) o . (47c) Sketch of the proof.
The proof of the equation (47a) is analogous to that of Lemma 9, adjusting the argumentspresented in the proof of Lemma 20.The estimates (47b) and (47c) follow directly applying the same reasoning of Corollary 16. Π left and Π right subproblems In the present section we compute the conditional expectation of Z Gleft , Y eFleft and Z Gright , Y eFright with respectto C left and C right respectively. Theorem 22.
Let Π left , Π right be the left and right subproblems introduced in Definition Z Gleft , Z Gright e their corresponding solutions furnished by the greedy algorithm. Then, E (cid:0) Z Gleft (cid:12)(cid:12) C left = c, S left = s (cid:1) = µ − s + 22 sc + s − s + 1 , for all s = 2 , . . . , µ, (48a) E (cid:0) Z Gleft (cid:12)(cid:12) C left = c (cid:1) = − c (cid:0) c (cid:1) c +1 − ( µ + 3)( c + 2)2 n(cid:0) c (cid:1) c +1 − c + 1 c o + 2 µc + 6 c + 3 µ + 102+ c ( µ + 3) n(cid:0) c (cid:1) c +2 − − c + 2 c − ( c + 1)( c + 2)2 c o , (48b) E (cid:0) Z Gright (cid:12)(cid:12) C left = c, S left = s (cid:1) = µ − s + 12 sc + s − s + 1 , for all s = 2 , . . . , µ, (48c) E (cid:0) Z Gright (cid:12)(cid:12) C left = c (cid:1) = − c (cid:0) c (cid:1) c +1 − ( µ + 2)( c + 2)2 n(cid:0) c (cid:1) c +1 − c + 1 c o + 2 µc + 4 c + 3 µ + 72+ c ( µ + 2) n(cid:0) c (cid:1) c +2 − − c + 2 c − ( c + 1)( c + 2)2 c o , (48d) with µ = δ + 1 .Proof. Recall that V left has only odd indexes, then E (cid:0) Z Gleft (cid:12)(cid:12) C left = c, S left = s (cid:1) = s − X j = 1 E (cid:0) P (2 j − (cid:12)(cid:12) C left = c, S left = s (cid:1) = s − X j = 1 E (cid:0) W (2 j − G (2 j − (cid:12)(cid:12) C left = c, S left = s (cid:1) = s − X j = 1 E (cid:16) W (2 j − µ X t = 2 j − T ( t ) (cid:12)(cid:12) C left = c, S left = s (cid:17) . Recalling that the variables (cid:0) W ( i ) (cid:1) µi =1 and (cid:0) T ( i ) (cid:1) µi =1 are independent, we have E (cid:0) Z Gleft (cid:12)(cid:12) C left = c, S left = s (cid:1) = s − X j = 1 E (cid:0) W (2 j − (cid:12)(cid:12) C left = c, S left = s (cid:1) µ X t = 2 j − E (cid:0) T ( t ) (cid:12)(cid:12) C left = c, S left = s (cid:1) = s − X j = 1 cs + s − s − µ − (2 j −
1) + 12= ( s − µ − s + 2)2 cs + s − s − . Here, the second equality holds due to the identity (27b), while the third is its sum. Simplifying the expressionabove, the Equation (48a) follows.Next, in order to prove (48b), observe that due to the expression (20c) we get E (cid:0) Z Gleft (cid:12)(cid:12) C left = c (cid:1) = c +1 X s = 2 E (cid:0) Z Gleft (cid:12)(cid:12) C left = c, S left = s (cid:1) P (cid:0) S left = s (cid:12)(cid:12) C left = d (cid:1) . Combining the equation (48a) with the cornerstone identity (25) in the space of conditioning, gives E (cid:0) Z Gleft (cid:12)(cid:12) C left = c (cid:1) = c +1 X s = 2 µ − s + 22 cs + s − s + 1 s − c s (cid:18) c + 1 s (cid:19) . Z Gleft , the equations(48c) and (48d), involving Z Gright are attained and the result is complete.Observe that Theorem 22 computes only the conditional expectations. In order to find the expectation weshould compute, E (cid:0) Z Gleft (cid:1) = X c E (cid:0) Z Gleft (cid:12)(cid:12) C left = c (cid:1) P ( C left = c ) , (49a) E (cid:0) Z Gright (cid:1) = X c E (cid:0) Z Gright (cid:12)(cid:12) C right = c (cid:1) P ( C right = c ) . (49b)However, as it has been shown above, the random variables C left and C right are really wild to be used in thiscalculation (see the proof of Theorem 19). Hence, we adopt the following estimate Corollary 23.
Let Π left , Π right be the left and right subproblems introduced in Definition Z Gleft , Z Gright be their corresponding solutions furnished by the greedy algorithm. Then, the following estimates hold E (cid:0) Z Gleft (cid:1) ∼ − E ( C left )2 (cid:16) E ( C left ) (cid:17) E (C left )+1 − ( µ + 3)( E ( C left ) + 2)2 n(cid:16) E ( C left ) (cid:17) E (C left )+1 − E ( C left ) + 1 E ( C left ) o + 2 µ E ( C left ) + 6 E ( C left ) + 3 µ + 102+ E ( C left )( µ + 3) n(cid:0) E ( C left ) (cid:1) E (C left )+2 − − E ( C left ) + 2 d − ( E ( C left ) + 1)( E ( C left ) + 2)2 E ( C left ) o , (50a) E (cid:0) Z Gright (cid:1) ∼ − E ( C right )2 (cid:16) E ( C right ) (cid:17) E (C right )+1 − ( µ + 2)( E ( C right ) + 2)2 n(cid:16) E ( C right ) (cid:17) E (C right )+1 − E ( C right ) + 1 E ( C right ) o + 2 µ E ( C right ) + 4 E ( C right ) + 3 µ + 72+ E ( C right )( µ + 2) n(cid:16) E ( C right ) (cid:17) E (C right )+2 − − E ( C right ) + 2 E ( C right ) − ( E ( C right ) + 1)( E ( C right ) + 2)2 E ( C right ) o , (50b) with µ = δ + 1 .Proof. It follows directly from Theorem 22.Next we compute some convenient conditional expectations of the post-greedy profit random variables Y left and Y right . Theorem 24.
With the definitions above, we have P (cid:0) i − is left eF , K left = k, S left = s (cid:12)(cid:12) C left = c (cid:1) = c − kc s +1 k (cid:0) − kc (cid:1) i − s − (cid:18) c − k − s − (cid:19) , (51a) P (cid:0) i is right eF , K right = k, S right = s (cid:12)(cid:12) C right = c (cid:1) = c − kc s +1 k (cid:0) − kc (cid:1) i − s − (cid:18) c − k − s − (cid:19) , (51b)32 or i = s + 1 , . . . , λ , k = 0 , . . . , δ − s − , s = 2 , . . . , λ . E (cid:0) Y eFleft (cid:12)(cid:12) K left = k, S left = s, C left = c (cid:1) = k µ − s ) n − (cid:0) − kc (cid:1) λ − s o − c k (cid:0) − kc (cid:1)n − (cid:0) λ − s − c k (cid:1)(cid:0) − kc (cid:1) λ − s − o (51c) E (cid:0) Y eFright (cid:12)(cid:12) K right = k, S right = s, C right = c (cid:1) = k µ − s − n − (cid:0) − kc (cid:1) λ − s o − c k (cid:0) − kc (cid:1)n − (cid:0) λ − s − c k (cid:1)(cid:0) − kc (cid:1) λ − s − o . (51d) Here Y eFleft and Y eFright are the post-greedy profit random variables introduced in Definition
The result is attained adjusting the procedure used in the proof of Lemma 14. Theidentities (51a) and identity (51b) follow directly. For the proof of (51c), we only provide details of thefollowing conditional expectation. Recall that due to the hypothesis 3 V left = V right = λ = µ , then E (cid:0) Y eFleft (cid:12)(cid:12) K left = k, S left = s, C left = c (cid:1) = λ X i = s +1 E (cid:0) P (2 i − (cid:12)(cid:12) i − , K left = k, S left = s, C left = c (cid:1) × P (cid:0) i − (cid:12)(cid:12) K left = k, S left = s, C left = c (cid:1) = λ X i = s +1 µ − i + 22 k kc (cid:0) − kc (cid:1) i − s − . Here, the first equality is a mere definition of conditional expectation, while the second equality computesdirectly the conditional probability of the event in the summand. From here, solving the sum with the techniquespresented in the proof of Lemma 14, the identity (51c) follows. The proof of the identity (51d) is similar.
Corollary 25 (Expected values of Z eFleft and Z eFright ) . With the definitions above, the following conditional expec-tations hold E (cid:0) Z eFleft (cid:12)(cid:12) C left = c (cid:1) = E (cid:0) Z Gleft (cid:12)(cid:12) C left = c (cid:1) + λ X s = 2 λ − s +1 X k = 1 k µ − s ) n − (cid:0) − kc (cid:1) λ − s o c − kc s +1 (cid:18) c − k − s − (cid:19) − λ X s = 2 λ − s +1 X k = 1 k (cid:0) − kc (cid:1)n − (cid:0) λ − s − c k (cid:1)(cid:0) − kc (cid:1) λ − s − o c − kc s (cid:18) c − k − s − (cid:19) , (52a) E ( Z eFright (cid:12)(cid:12) C right = c ) = E ( Z Gright (cid:12)(cid:12) C right = c )+ λ X s = 2 λ − s +1 X k = 1 k µ − s − n − (cid:0) − kc (cid:1) λ − s o c − kc s +1 (cid:18) c − k − s − (cid:19) − λ X s = 2 λ − s +1 X k = 1 k (cid:0) − kc (cid:1)n − (cid:0) λ − s − c k (cid:1)(cid:0) − kc (cid:1) λ − s − o c − kc s (cid:18) c − k − s − (cid:19) . (52b) Here Z eFleft , Z eFright are the corresponding values of the objective function, furnished by the eligible-first algorithmfor the problems Π left and Π right .Sketch of the proof. The proof is analogous to the one presented in Theorem 15.33inally, we close this section presenting an estimate for the expected performance of the eligible-first algo-rithm on the Π left and Π right subproblems.
Corollary 26 (Approximation of E ( Z eFleft ) , E ( Z eFright )) . With the definitions above, the following estimates hold E (cid:0) Z eFleft (cid:1) ∼ E (cid:0) Z Gleft (cid:1) + E ( K left )4 (cid:0) µ − E ( S left ) (cid:1)n − (cid:0) − E ( K left ) E ( C left ) (cid:1) λ − E (S left ) o − E ( C left )2 E ( K left ) (cid:0) − E ( K left ) E ( C left ) (cid:1)n − (cid:0) λ − E ( S left ) − E ( C left ) E ( K left ) (cid:1)(cid:0) − E ( K left ) E ( C left ) (cid:1) λ − E (S left ) − o (53a) E (cid:0) Z eFright (cid:1) ∼ E (cid:0) Z Gright (cid:1) + E ( K right )4 ( µ − E ( S right ) − n − (cid:0) − E ( K right ) E ( C right ) (cid:1) λ − E (S right ) o − E ( C right )2 E ( K right ) (cid:0) − E ( K right ) E ( C right ) (cid:1)n − (cid:0) λ − E ( S right ) − E ( C right ) E ( K right ) (cid:1)(cid:0) − E ( K right ) E ( C right ) (cid:1) λ − E (S right ) − o . (53b) Sketch of the proof.
Similar to the proof of Corollary 16
5. Performance Estimates for the Divide-and-Conquer Method
In the current section, we use the previous analysis to derive performance parameters, some for efficiency-reference and other as lower bound estimates for the expected (average) performance of the Divide-and-Conquermethod. We also compute with higher accuracy, the performance of the method for the Z eF and Z LP boundingalgorithm solutions, to estimate the expected performance of Divide-and-Conquer on the optimal solution Z ∗ .We begin this section evaluating numerically the aforementioined parameters for one iteration of the method. In this section, we finally apply the analytical results previously developed to estimate the performance ofone iteration of the Divide-and-Conquer method. Observe that the complexity of the analytical expressions,forces us to seek a numerical evaluation of these expressions in order to attain a tangible value (or referencelower bounds) of the method’s efficiency. It is important to stress that for most of the cases, the numericalcomputations will use the approximations introduced in the lemmas 20, 21 and the corollaries 23, 26 above.This approach is adopted because, the conditional expectations of C left and C right with respect to K and S havea very wild structure, as they heavily depend on whether the split value is even or odd (see the equations (40)and (41) in Theorem 17). This case-wise structure makes hard to use the identities (40) and (41) for furthercalculations beyond the expectations E ( C left ) and E ( C right ) (e.g., the equations (52a) and (52b)).On the other hand, it is important to observe that the approximation eF( δ ) for E ( Z eF ) given in (37) (similarto all the estimates adopted) is very accurate with respect to the exact values (36), as it can be seen in thetable 4 below. Additionally, Theorem 13 shows a convergent asymptotic behavior for the paramount randomvariables of the 0-1RKP (equation (22)). Furthermore, the statement (33d) in Theorem 13 shows analytically,that the upper bound E ( Z LP ) can be accurately approximated, as pointed out in Remark 7, in an analogousway to our approximation E ( Z eF ) ∼ eF( δ ). Accuracy δ
10 20 30 40 50 60 70 80 90 100 110 120100 × eF( δ ) E ( Z eF ) 2.66 0.75 0.45 0.34 0.28 0.24 0.21 0.19 0.17 0.15 0.14 0.13 Table 4: Accuracy of the approximation eF( δ ). We present the relative accuracy of the approximation in percentage terms forseveral values of the capacity δ . Definition 10.
Let Π be an instance of the 0-1RKP introduced in Definition 5 and let Π left and Π right bethe problems induced by one iteration of the Divide-and-Conquer method (see Definition 3). Let E ( Z ∗ ) E ( Z eF ) and E ( Z LP ) be the expected objective function values for the optimal, eligigle-first and linear relaxationrespectively; moreover the analogous notation holds when the subindex makes reference to the Π left or Π right random subproblems.(i) Define the following efficiency-reference parameters ρ def = E (cid:0) Z ∗ left (cid:1) + E (cid:0) Z ∗ right (cid:1) E (cid:0) Z ∗ (cid:1) × , ρ left def = E (cid:0) Z ∗ left (cid:1) E (cid:0) Z ∗ (cid:1) × , ρ right def = E (cid:0) Z ∗ right (cid:1) E (cid:0) Z ∗ (cid:1) × , (54a) ρ eF def = E (cid:0) Z eFleft (cid:1) + E (cid:0) Z eFright (cid:1) E (cid:0) Z eF (cid:1) × , ρ eFleft def = E (cid:0) Z eFleft (cid:1) E (cid:0) Z eF (cid:1) × , ρ eFright def = E (cid:0) Z eFright (cid:1) E (cid:0) Z eF (cid:1) × , (54b) ρ LP def = E (cid:0) Z LPleft (cid:1) + E (cid:0) Z LPright (cid:1) E (cid:0) Z LP (cid:1) × , ρ LPleft def = E (cid:0) Z LPleft (cid:1) E (cid:0) Z LP (cid:1) × , ρ LPright def = E (cid:0) Z LPright (cid:1) E (cid:0) Z LP (cid:1) × . (54c)(ii) Define the following lower bound parameterslb G def = E (cid:0) Z Gleft (cid:1) + E (cid:0) Z Gright (cid:1) E (cid:0) Z LP (cid:1) × , lb Gleft def = E (cid:0) Z Gleft (cid:1) E (cid:0) Z LP (cid:1) × , lb Gright def = E (cid:0) Z Gright (cid:1) E (cid:0) Z LP (cid:1) × , (55a)lb eF def = E (cid:0) Z eFleft (cid:1) + E (cid:0) Z eFright (cid:1) E (cid:0) Z LP (cid:1) × , lb eFleft def = E (cid:0) Z eFleft (cid:1) E (cid:0) Z LP (cid:1) × , lb eFright def = E (cid:0) Z eFright (cid:1) E (cid:0) Z LP (cid:1) × . (55b)It is direct to see that the parameters of equations (54) account for the efficiency of the Divide-and-Conquermethod acting on the three solutions Z ∗ , Z eF and Z LP . The analogous holds whenever the subindex side ∈{ left , right } is present. However, we still need to show that the parameters introduced in the equations (55)are actually lower bounds. Proposition 27.
With the definitions above for the performance parameters, the following estimates hold lb G ≤ lb eF ≤ ρ, (56a)lb Gleft ≤ lb eFleft ≤ ρ left , (56b)lb Gright ≤ lb eFright ≤ ρ right . (56c) Proof.
Recall that due to the algorithms’ definition Z ∗ ≤ Z LP and Z Gside ≤ Z eFside ≤ Z ∗ side for side = left , right,for any instance of the problem. Then, E ( Z Gleft ) + E ( Z Gright ) ≤ E ( Z eFleft ) + E ( Z eFright ) ≤ E ( Z ∗ left ) + E ( Z ∗ right ),consequently E ( Z Gleft ) + E ( Z Gright ) E ( Z LP ) = lb G ≤ E ( Z eFleft ) + E ( Z eFright ) E ( Z LP ) = lb eF ≤ E ( Z ∗ left ) + E ( Z ∗ right ) E ( Z ∗ ) = ρ. The above shows the inequality (56a). The proof of the estimates (56b) and (56c) is analogous.It is clear that we are interested in computing the value of ρ, ρ left and ρ right , however, as discussed inRemark 2 above, the probabilistic analysis of Z ∗ , or even Z eG , Z fG is not tractable. Hence, we use the values of Z G , Z eF , Z LP whose probabilistic analysis has been described accurately enough in the sections 3 and 4 above.We analyze the behavior of the Divide-and-Conquer method from two points of view,35 apacity Items Z eF Z LP δ µ ρ eF ρ eFleft ρ eFright ρ LP ρ LPleft ρ LPright
49 50 99.59 68.35 31.24 91.05 63.62 27.4399 100 99.78 68.37 31.41 91.97 64.23 27.74149 150 99.85 68.38 31.48 92.28 64.44 27.84199 200 99.89 68.38 31.51 92.44 64.54 27.90249 250 99.91 68.38 31.53 92.53 64.6 27.93299 300 99.93 68.39 31.54 92.59 64.64 27.95399 400 99.94 68.39 31.56 92.67 64.69 27.98499 500 99.96 68.39 31.57 92.72 64.72 28.00599 600 99.96 68.39 31.57 92.75 64.75 28.01699 700 99.97 68.39 31.58 92.77 64.76 28.01799 800 99.97 68.39 31.58 92.79 64.77 28.02899 900 99.97 68.39 31.58 92.81 64.78 28.03999 1000 99.98 68.39 31.59 92.82 64.79 28.03
Table 5: Expected performance of Z eF and Z G .
200 400 600 800 1000Capacity30405060708090100 LPLP_LeftLP_RighteFeF_LefteF_Right
Figure 3: Expected performance of the Divide-and-Conquermethod on Z LP and Z eF . view a. We compute the efficiency of the method for Z eF and Z LP (equations (54b) and (54c)) to have an ideaof the expected performance of the method for Z ∗ (equation (54a)), see Table 5 and Figure 3 below.view b. We compute lower bounds (equations (55)) for the expected performance of the Divide-and-Conquermethod for Z ∗ , see Table 6 and Figure 4 below. Capacity Items Z G / Z LP Z eF / Z LP δ µ lb G lb Gleft lb Gright lb eF lb eFleft lb eFright
49 50 72.74 49.75 22.99 79.19 55.34 23.8599 100 72.28 49.44 22.84 79.51 55.53 23.98149 150 72.13 49.33 22.79 79.62 55.59 24.02199 200 72.05 49.28 22.77 79.67 55.62 24.04249 250 72.01 49.25 22.76 79.70 55.64 24.06299 300 71.98 49.23 22.75 79.72 55.66 24.07399 400 71.94 49.20 22.74 79.75 55.67 24.08499 500 71.92 49.19 22.73 79.76 55.68 24.08599 600 71.90 49.18 22.72 79.77 55.69 24.09699 700 71.89 49.17 22.72 79.78 55.69 24.09799 800 71.88 49.16 22.72 79.79 55.69 24.09899 900 71.88 49.16 22.72 79.79 55.70 24.09999 1000 71.87 49.16 22.72 79.79 55.70 24.10
Table 6: Lower bounds, ratios Z G / Z LP and Z eF / Z LP .
200 400 600 800 1000Capacity20304050607080 lb^Glb^G_Leftlb^G_Rightlb^eFlb^eF_Leftlb^eF_Right
Figure 4: Lower bounds for the expected performance of theDivide-and-Conquer method on Z LP and Z eF . Remark 9.
Strictly speaking we are adopting the following approximations for the performance parameters. E (cid:16) Z ∗ left + Z ∗ right Z ∗ × (cid:17) ∼ E (cid:0) Z ∗ left (cid:1) + E (cid:0) Z ∗ right (cid:1) E (cid:0) Z ∗ (cid:1) ×
100 = ρ, E (cid:16) Z Gleft + Z Gright Z lb × (cid:17) ∼ E (cid:0) Z Gleft (cid:1) + E (cid:0) Z Gright (cid:1) E (cid:0) Z LP (cid:1) ×
100 = lb G , (57)and similarly for all the efficiency-reference (equations (54)) and the lower bound (equations (55)) parametersthat we have introduced in Definition 10. However, it must be observed that these assumptions are mild astheir values are very close to the empirical results. On the other hand, finding the expectation of the left handside in the estimates (57) is significantly more complex and provides little extra accuracy. Finally, given that36e want to merely estimate the expected efficiency of the Dividie-and-Conquer method on the 0-1RKP, it issave to give up such level of precision. In this section we can finally deliver tangible values for the performance of the Divide-and-Conquer method.First for one iteration and then, we furnish a method to estimate the performance for any D&C tree (seeExample 2 below).Observe that for all the parameters introduced in the previous section, the variance is remarkably low.Therefore, we can adopt the averages as the value of the corresponding performance parameters for oneiteration of the Divide-and-Conquer method, see Table 7. Moreover, due to the low value of the variance, it issafe to assume the same performance of the method, through all the iterations of the full binary D&C tree. Z eF Z LP Z G / Z LP Z eF / Z LP ρ eF ρ eFleft ρ eFright ρ LP ρ LPleft ρ LPright lb G lb Gleft lb Gright lb eF lb eFleft lb eFright mean 99.93 68.39 31.54 92.59 64.64 27.95 71.98 49.23 22.75 79.72 55.65 24.07variance 0.01 0.00 0.00 0.12 0.05 0.01 0.03 0.01 0.00 0.01 0.00 0.00Table 7: Mean and Variance for the performance parameters defined in equations (54b), (54c), (55a) and (55b). Next, in order to give the performance associated to a D&C tree first we need to mark its vertices in aparticular way.
Definition 11.
Let T be a D&C tree(i) For every vertex Π of T we construct a marker m Π in the following way. If the vertex is different from theroot then the marker is the sequential list of left and/or right turns, that the unique path from the rootto it, needs to take. If the vertex is the root simply assign an empty list as its marker. (See Figure 5 inExample 2 below.)(ii) Given Π of T a vertex with its corresponding marker m Π , then we define the factorΦ(Π) def = 100 × | length Π | Y i = 1 m Π ( i ) , Φ ∈ { ρ eF , ρ LP , lb G , lb eF } , (58)with the convention that ϕ (root) = 1.(iii) The value of the performance parameter of the tree T is given byΦ( T ) def = max n , X L is a leave of T Φ( L ) , o Φ ∈ { ρ eF , ρ LP , lb G , lb eF } . (59) Remark 10.
We observe the following(i) Due to Definition 3 (iii), every internal vertex of a D&C tree has exactly two children: left and right.Therefore, the marking process is well-defined because, given any arbitrary vertex of the tree, all itsancestors excepting the root, are necessarily the left or right child of its parent.(ii) The marking of vertices is completely analogous to the well-know binary expansion of numbers in theinterval [0 , T ) involves a maximum between the derived algebraic expression and a 50% value.This is due to the quality certificate of 50% in the worst case scenario presented in Theorem 3 (ii). Example 2 (Continuation of Example1).
We compute the performance parameters for the D&C tree pre-sented in Example 1 above. The figure 5 depicts the marking of each of the vertices of the tree, while Table 5summarizes the values of the four efficiency parameters introduced in Section 5.1 above.
Element Efficiency Lower Bound ρ eF ρ LP lb G lb eF Π T Table 8: Performance parameters Example 2. • Π , m = ( ) • Π , m = ( l ) • Π , m = ( l, l ) • Π , m = ( l, r ) • Π , m = ( r ) Figure 5: Divide-and-Conquer labeled tree. For each vertex Π i the path from the root to the vertex is indicated as a sequence ofleft and right turns and denoted as its marking m i . (The notationm Π i of Definition 11 is omitted for visual purposes.) In the current section we describe the numerical verification of the results presented so far. First, we needto define a number of Bernoulli trials in our experiments, to that end we recall the following result on confidence
Theorem 28.
Let x be a scalar statistical variable with mean ¯ x , variance σ .(i) The number of Bernoulli trials necessary to get a 95% confidence interval is given by n def = (cid:0) . . (cid:1) σ . (60) (ii) The
95 percent confidence interval is given by I x def = " ¯ x − . r σ n , ¯ x + 1 . r σ n . (61) Proof.
The proof is based on the Central Limit Theorem, see [32] for details.Next, we summarize the guidelines for the experiments designa. The split index variable S is used to determine the number of Bernoulli trials for our numerical experiments,because we have an analytical expression for its variance given by Equation 26c.b. For simplicity, the sizes of the 0-1RKP’s for which the theoretical results are to be verified have the structure δ = 2 j −
1. These sizes, together with their corresponding number of Bernoulli trials, using the equations26c and (60) are summarized in the table 9 below.c. For simplicity, the D&C tree structures to be evaluated are the complete binary trees of the heights detailedin Table 10. 38 apacity Items Variance Trials δ µ V ar( S ) n
63 64 0.7329 1127127 128 0.7493 1152255 256 0.7575 1165511 512 0.7616 11711023 1024 0.7637 1174
Table 9: Summary of Experiments and Bernoulli Trials
Tree Height Number of T h Nodes1 1 32 2 63 3 144 4 30
Table 10: Summary of Tree Structures d. Each capacity δ of Table 9 is tested through all the D&C trees of Table 10.The table 11 displays the empirical efficiency results for the first case of Table 9: knapsack capacity δ = 63,number of items µ = 64. The remaining experiments of the table 9 yield similar efficiency results to the firstcase, presented in Table 11. The table 12 summarizes the theoretical results, computed using the approximationmethod introduced in Definition 11 and explained in Example 2. As it can be seen, the empirical results aremore favorable than the theoretical results, for all the analyzed trees. (The same holds for all the emainingexperiments of the table 9.) Tree Efficiency Lower Bound T ρ ρ eF ρ LP lb G lb eF Table 11: Empirical Tree Efficiencies, δ = 63, µ = 64,Number of Bernoulli Trials n = 1152. Tree Efficiency Lower Bound T ρ eF ρ LP lb G lb eF Table 12: Theoretical Tree Efficiency Estimates.These are constructed based on the values of Table 7.
Remark 11.
It is important to stress that the same set of experiments of Table 9 were used to verify theresults developed in this work. For all the random variables involved, its empirical expectation falls into theircorresponding confidence interval presented in Theorem 28. The correctness of the developed expressions wasverified, using the full knapsack problem for the results of Section 3 and using the basic D&C tree T = 1 ofthe table 10 (three nodes and height one), to check those presented in Section 4.
6. Conclusions and Final Discussion
The present work yields the following conclusions(i) A complete and detailed theoretical analysis for the Divide-and-Conquer method’s efficiency has beenpresented. The analysis has been done from two points of view: the worst case scenario and the expectedperformance. Before this work, the method’s efficiency was analyzed only from the empirical point ofview.(ii) For the worst case scenario, it suffices to find a control solution (see Theorem 2) which is computationallycheap. In our case furnished by the extended-greedy algorithm ( x eG and z eG ) and then split the problems,in order to assure it belongs ( x eG restricted) to all the search spaces of the D&C subproblems. This wasdone by carefully computing the knapsack capacities of the subproblems, given that the mechanism forsplitting items (even and odd indexes) was already decided as discussed in Remark 3.(iii) It is possible to use another control solution for the worst case scenario, rather than the one presentedhere. For instance, the algorithm G presented in [1], which is computationally more expensive, but it39ertifies a worst case scenario of 75%. However, for this or any other control solution, the computationof the knapsack capacities δ left , δ right detailed in Algorithm 2, needs to be adjutsed in order to satisfy thehypothesis of Theorem 2.(iv) The analysis of the expected performance is considerably harder than the previous one. A discrete proba-bilistic setting has to be established (see Hypothesis 2) and a randomized version of the problem, 0-1RKP,has to be introduced (see Definition 5).(v) The probabilistic analysis was done in two parts: Section 3 analyzes the 0-1RKP in full while Section 4analyzes the expected behavior of one single iteration of the Divide-and-Conquer method. In the firstcase, all the expectations were computed with absolute accuracy. In the second case, the same rigor waskept only for the computation of the left and right knapsack capacities but, in order to pursue furtherresults, we approximated the expression assuming independence of the slack K and split S variables. Thelatter approximation has solid grounds because of the smooth behavior of the expectations of the mainvariables of the model, as shown in Theorem 13.(vi) In Section 5 several parameters to measure the performance of the method were introduced. Here,the expressions previously attained were numerically evaluated (due to its complexity) in order to obtainconcrete, tangible values of the method’s performance; first for one single D&C iteration, then, anapproximation is given for a general D&C tree (see Definition 11). Once again, hypothesis of independencebetween random variables were adopted, in order to compute the desired values (see Remark 9). Finally,the theoretical results are verified empirically with numerical experiments statistically sound.(vii) The empirical verification of our results (displayed in Table 11), show that the theoretical approximations(summarized in the table 12) are a lower estimate for performance of Divide-and-Conquer and can be usedto evaluate the method in general terms. To this end, two pairs of parameters were introduced: ρ eF , ρ LP as a reference of and lb G , lb eF as lower bounds of the expected performance . Hence, if the first pair isused to decide, it is recomendable to use the method with at most three iterations ( T = 3). However, themore conservative view of the lower bounds’ pair, states that Divide-and-Conquer should be used with atmost two iterations ( T = 2), because beyond that height they are no higher than the worst case scenario,already furnished by the extended-greedy algorithm ( x eG and z eG ).(viii) Finally, a more daring approach would use the empirical evidence to decide the limit extension of Divide-and-Conquer trees, summarized in Table 11 (similar to all the other experiments of Table 9). From thispoint of view, the method is still highly recommendable for four iterations ( T = 4). This is consistentwith the empirical findings of [8] where six D&C iterations produced satisfactory results in average . Acknowledgments
The first Author wishes to thank Universidad Nacional de Colombia, Sede Medell´ın for supporting the productionof this work through the project Hermes 45713 as well as granting access to Gauss Server, financed by “ProyectoPlan 150x150 Fomento de la cultura de evaluaci´on continua a trav´es del apoyo a planes de mejoramiento de losprogramas curriculares”. ( gauss.medellin.unal.edu.co ), where the numerical experiments were executed.All the polynomial identities used in the proofs were verified in wolframalpha.com ; due to its level of complexityit would have not been possible to develop them without this remarkable free tool.
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