Algorithmic improvements on dynamic programming for the bi-objective {0,1} knapsack problem

Abstract

This paper presents several methodological and algorithmic improvements over a state-of-the-art dynamic programming algorithm for solving the bi-objective {0,1} knapsack problem. The variants proposed make use of new definitions of lower and upper bounds, which allow a large number of states to be discarded. The computation of these bounds are based on the application of dichotomic search, definition of new bound sets, and bi-objective simplex algorithms to solve the relaxed problem. Although these new techniques are not of a common application for dynamic programming, we show that the best variants tested in this work can lead to an average improvement of 10 to 30 % in CPU-time and significant less memory usage than the original approach in a wide benchmark set of instances, even for the most difficult ones in the literature.

Appendices

Appendix A: Simplex algorithm

The continuous 0-1 bi-objective knapsack problem can be formalized as follows (see Sect. 1 for the notation and Eq. (1) for the discrete variant):

$$ \begin{array}{l@{\quad}l} \max & f(x) = \bigl(f_1(x), f_2(x) \bigr)\\[5pt] \mbox{subject to:}& w^\mathrm{T}x \leq W,\quad x \in [0,1]^n\\ \end{array} $$

(4)

The simplex algorithm computes the set of all nondominated vectors of problem (4). It starts from one lexicographically-optimal solution. Then, it proceeds with an efficient pivoting process until an efficient solution of optimal value for objective f ₂ is obtained. In order to obtain the lexicographically-optimal solution, the objects are sorted in non-increasing order of the profit-to-weight ratio $v_{k}^{1}/w_{k}$. In case of ties, the objects with the same profit-to-weight ratio are sorted in non-increasing order of the profit-to-weight ratio for $v_{k}^{2}/w_{k}$.

Remark 1

If two objects j and ℓ have the same profit-to-weight ratio on the first and the second objective, then consider only one object of weights w _j +w _ℓ and of profit $(v^{1}_{j} + v^{1}_{\ell}, v^{2}_{j} + v^{2}_{\ell})$.

The lexicographical-optimal solution x ^∗ is obtained by using a greedy procedure that includes sorted objects until the capacity is achieved. Let

$$c = \min \Biggl\{j: \sum_{k=1}^j w_j>W \Biggr\} $$

be the critical object; then $x^{*} = (x^{*}_{1},\ldots,x^{*}_{n})$ is such that:

$$x^*_j = \begin{cases} 1 & \mbox{for } j= 1,\ldots,c-1,\\[5pt] (W- \sum_{k=1}^{c-1}w_k )/w_c, & \mbox{for } j = c,\\[5pt] 0 & \mbox{for } j= c+1,\ldots,n.\\ \end{cases} $$

Given that $x^{*}_{c}$ is not equal to zero (that case will be considered later on), it constitutes an efficient basic variable. Note that an efficient basis is constituted by only one efficient basic variable.

The algorithm uses a pivoting procedure that ensures the transition from one efficient solution to an adjacent one that decreases the value on the first objective. This procedure, which starts with solution x ^∗ and one of its associated efficient basis $\{x^{*}_{c}\}$, is described as follows for the general case considering an efficient solution x and one of its associated efficient basis {x _c }.

• 1. Given that variable x _c is basic, compute the reduced costs associated to variable x _j on the objective i that is $V^{i}_{(c,j)}= v_{j}^{i} - v_{c}^{i} \cdot w_{j}/w_{c}$, i=1,2 and j=1,…,n. The indexes of candidates variables to enter the basis are:

  $$J = \left \{j \in \{1,\ldots,n\}\setminus c \mbox{ such that: } \begin{array}{l} V^1_{(c,j)}<0 \mbox{ and } V^2_{(c,j)}>0, \mbox{ if } x_j = 0\\[7pt] V^1_{(c,j)}>0 \mbox{ and } V^2_{(c,j)}<0, \mbox{ if } x_j = 1\\ \end{array} \right \} $$

  If J is an empty set, the algorithm terminates since it is not possible to increase the value of solution x on the second objective. Hence, x is an efficient solution that is optimal for the second objective. Otherwise, the algorithm proceeds with an efficient pivoting between x _c and variable $x_{j^{*}}$ (j ^∗∈J), which provides a new efficient solution such that the slope of the line through this new solution and the previous solution x is the smallest. Then, j ^∗ is such that

  $$\frac{V^2_{(c,j^*)}}{V^1_{(c,j^*)}} = \min_{j \in J} \frac{V^2_{(c,j)}}{V^1_{(c,j)}} $$

Remark 2

Since the objects of the same profit-to-weight ratio for all j∈{1,…,n}∖c were added, it does not hold that $V^{1}_{(c,j)} = 0$ and $V^{2}_{(c,j)} = 0$.

Remark 3

However, it is possible to obtain variables such that $V^{1}_{(c,j)} = 0$ and $V^{2}_{(c,j)} < 0$ or $V^{1}_{(c,j)} < 0$ and $V^{2}_{(c,j)} = 0$, for x _j =0, as well as variables such that $V^{1}_{(c,j)} = 0$ and $V^{2}_{(c,j)} > 0$ or $V^{1}_{(c,j)} > 0$ and $V^{2}_{(c,j)} = 0$, for x _j =1. Note that these variables are not candidates to enter the basis since they cannot lead to an efficient pivoting and to a new efficient solution.

Remark 4

Index j ^∗ is the index in J that maximizes $-V^{2}_{(c,j)}/ (V^{1}_{(c,j)}-V^{2}_{(c,j)} )$.

• 2. Let δ=w _c /w _j . The values of variables x _c and x _j are updated as follows.

  $$\mbox{if } x_j = 0 \begin{cases} \mbox{if } \delta x_c < 1 \qquad \begin{cases} x_j := \delta x_c\\ x_c := 0\\ x_j \mbox{ enters the basis in substitution of $x_{c}$} \end{cases} \\[20pt] \mbox{if } \delta x_c > 1 \qquad \begin{cases} x_j := 1\\ x_c := x_c - \delta\\ x_c \mbox{ remains in the basis} \end{cases} \\[20pt] \mbox{if } \delta x_c = 1 \qquad \begin{cases} x_j := 1\\ x_c := 0\\ \mbox{The solution is integer} \end{cases} \end{cases} $$
  $$\mbox{if } x_j = 1 \begin{cases} \mbox{if } \delta (1-x_c) < 1 \begin{cases} x_j := 1 - \delta(1 - x_c)\\ x_c := 1\\ x_j \mbox{ enters the basis in substitution of $x_{c}$} \end{cases} \\[20pt] \mbox{if } \delta (1-x_c) > 1 \begin{cases} x_j := 0\\ x_c := x_c + \delta\\ x_c \mbox{ remains in the basis} \end{cases} \\[20pt] \mbox{if } \delta (1-x_c) = 1 \begin{cases} x_j := 0\\ x_c := 1\\ \mbox{The solution is integer} \end{cases} \\ \end{cases} $$

The final solution is efficient. The pivoting process is repeated with the new basic solution. In case of degeneration, i.e., when we obtain an integer solution, the pivoting procedure is slightly different. We consider all possible increasing pivots between x _i and x _j for i=1,…,n and j=i+1,…,n satisfying that

(5)

Among all these possible pivots satisfying (5), the algorithm pivots between x _c and $x_{j^{*}}$ such that

$$\frac{V^2_{(c,j^*)}}{V^1_{(c,j^*)}} = \min \frac{V^2_{(c,j)}}{V^1_{(c,j)}} $$

Hence, the algorithm updates the values of x _c and $x_{j^{*}}$ using Step 2 described above.

Appendix B: Computational results

Table 1 Average CPU-time of several approaches for the benchmark set of instances; see text for more details

Table 2 Maximum amount of states in memory of several approaches for the benchmark set of instances; B-DP1,3a_∗ denotes variants B-DP1, B-DP3a_.95, B-DP3a_.90 and B-DP3a_.85; see text for more details