An exact scalarization method with multiple reference points for bi-objective integer linear optimization problems

Abstract

This paper presents an exact scalarization method to solve bi-objective integer linear optimization problems. This method uses diverse reference points in the iterations, and it is free from any kind of a priori chosen weighting factors. In addition, two new adapted scalarization methods from literature and the modified Tchebycheff method are studied. Each one of them results in different ways to obtain the Pareto frontier. Computational experiments were performed with random real size instances of two special problems related to the manufacturing industry, which involve lot sizing and cutting stock problems. Extensive tests confirmed the very good performance of the new scalarization method with respect to the computational effort, the number of achieved solutions, the ability to achieve different solutions, and the spreading and spacing of solutions at the Pareto frontier.

Appendices

Appendices

Appendixes A.1 and A.2 consider the following illustrative instance for an integer linear bi-objective problem.

$$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} z(x)=(c^T x,d^Tx)\\ \text {subject to}&{} Ax=b\\ &{} x \in {\mathbb {Z}}_+^6, \end{array} \end{aligned}$$

(12)

where $c^T=[-5 \,\,\, 2]$, $d^T=[1 \,\, -4]$, $A=[{\bar{A}}\, \, |\, \, I_4]$, $b=[3,6,8,4]^T$, $I_4$ is the identity matrix of dimension 4, and ${\bar{A}}^T=\left[ \begin{array}{cccc} -1&{} 1&{} 1&{} 0 \\ 1&{} 0&{} 1&{} 1 \end{array} \right]$. Let $X=\{ x \in {\mathbb {Z}}_+^6| Ax=b\}$ be the feasible set of this problem.

1.1 Numerical example for MRV method

Example 1

Consider $\ell =100$. This means that we expect to obtain at maximum 100 non-dominated vectors.

• Calculation of lexicographic vectors: First, we obtain the lexicographic vectors $z^{\text {lex}_1}=A=(-30,6)^T$ and $z^{\text {lex}_2}=B=(3,-15)^T$ individually optimizing the objectives $z_1(x)$ and $z_2(x)$, respectively, and cheeking they if are dominated vectors.

  We calculate $\delta _1$ and $\delta _2$ as follows:

  $$\begin{aligned} \delta _1 = \frac{B_1-A_1}{100}=0.33 \quad \text {and} \quad \delta _2=\frac{B_2-A_2}{100}=0.21. \end{aligned}$$

  Let $t=2$ and go to the next step.

• Iteration 1: first, $R'=(-30,-15)^T+(0.33,0.21)^T=(-29.67,-14.79)^T$ and $N'=(3,6)^T-(0.33,0.21)^T=(2.67,5.79)$.

  We solve Problem (3) given by:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -29.67 \leqq z_1(x) \leqq 2.67\\ &{} -14.79 \leqq z_2(x) \leqq 5.79 \\ &{} z_1(x)-(-29.79) \leqq u \\ &{} z_2(x)-(-14.67) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  whose optimal solution is a Pareto optimal solution corresponding to the objective vector $C=(-19,-7)^T$. Let $t=3$.

• Iteration 2: We organize the non-dominated vectors already determined, that is, $A=(-30,6)^T$, $B=(-19,-7)^T$, $C=(3,-15)^T$. We consider the consecutive vectors A and B, hence $R'=(-30,-7)^T+(0.33,0.21)^T=(-29.67,-6.79)^T$ and $N'=(-19,6)^T-(0.33,0.21)^T=(-19.33,5.79)^T$ and solve the subproblem:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -29.67 \leqq z_1(x) \leqq -19.33\\ &{} -6.79 \leqq z_2(x) \leqq 5.79 \\ &{} z_1(x)-(-29.67) \leqq u \\ &{} z_2(x)-(-6.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  whose solution corresponds to the objective vector $H=(-26,-2)^T$. Considering the consecutive non-dominated vectors B and C, let $A \leftarrow B$ and $B \leftarrow C$, we have $R=(-19,-15)^T$ and $N=(3,-7)^T$ and so $R'=(-18.67,-14.79)^T$ and $N'=(2.67,-7.21)^T$. A new subproblem must be solved:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -18.67 \leqq z_1(x) \leqq 2.67\\ &{} -14.79 \leqq z_2(x) \leqq -7.21 \\ &{} z_1(x)-(-18.67) \leqq u \\ &{} z_2(x)-(-14.79) \leqq u \\ &{} x \in X, \, u \geqq 0. \end{array} \end{aligned}$$

  Its solution provides the objective vector $C=(-7,-13)^T$, and there are $t=5$ non-dominated vectors.

• Iteration 3: Let us order the non-dominated vectors determined, that is, $A=(-30,6)^T$, $B=(-26,-2)^T$, $C=(-19,-7)^T$, $D=(-7,-13)^T$ and $E=(3,-15)^T$. Hence, considering the first pair of consecutive vectors, we have $R'=(-29.67,-1.79)^T$, $N'=(-26.33,6.21)^T$, and the new subproblem to be solved is the following:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -29.67 \leqq z_1(x) \leqq -26.33\\ &{} -1.79 \leqq z_2(x) \leqq 6.21 \\ &{} z_1(x)-(-29.67) \leqq u \\ &{} z_2(x)-(-26.33) \leqq u \\ &{} x \in X, \, u \geqq 0. \end{array} \end{aligned}$$

  It provides a solution whose objective vector is $F=(-28,2)^T$. Now, $A \leftarrow B$ and $B \leftarrow C$ and thereby $R'=(-25.67,-6.79)^T$ and $N'=(-19.33,1.79)^T$. We solve the subproblem:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -25.67 \leqq z_1(x) \leqq -19.33\\ &{} -6.79 \leqq z_2(x) \leqq 1.79 \\ &{} z_1(x)-(-25.67) \leqq u \\ &{} z_2(x)-(-6.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  whose optimal solution corresponds to the objective vector $G=(-21,-3)^T$. Consequently, $A \leftarrow C$ and $B \leftarrow D$ and the new reference vectors are $R'=(-18.67,-12.79)^T$ and $N'=(-7.33,-7.21)^T$. The new subproblem is given by:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -18.67 \leqq z_1(x) \leqq -7.33\\ &{} -12.79 \leqq z_2(x) \leqq -7.21 \\ &{} z_1(x)-(-18.67) \leqq u \\ &{} z_2(x)-(-12.79) \leqq u \\ &{} x \in X, \, u \geqq 0. \end{array} \end{aligned}$$

  This subproblem has an optimal solution whose associated objective vector is $H=(-14,-8)^T$. Let $A \leftarrow D$ and $B \leftarrow E$ and consequently, $R'=(-6.67,-14.79)^T$ and $N'=(2.67,-13.21)^T$, thus originating the following subproblem:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -6.67 \leqq z_1(x) \leqq 2.67\\ &{} -14.79 \leqq z_2(x) \leqq -13.21 \\ &{} z_1(x)-(-6.67) \leqq u \\ &{} z_2(x)-(-14.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  which provides the objective vector $I=(-2,-14)^T$. Now, we have $t=9$ non-dominated vectors.

• Iteration 4: Let us sort all the vectors already determined, that is, $A=(-30,6)^T$, $B=(-28,2)^T$, $C=(-26,-2)^T$, $D=(-21,-3)^T$, $E=(-19,-7)^T$, $F=(-14,-8)^T$, $G=(-7,-13)^T$, $H=(-2,-14)^T$ and $I=(3,-15)^T$. We have as reference vectors $R'=(-29.67,2.21)^T$ and $N'=(-28.33,5.79)^T$, thus resulting the following subproblem:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -29.67 \leqq z_1(x) \leqq -28.33\\ &{} 2.21 \leqq z_2(x) \leqq 5.79 \\ &{} z_1(x)-(-29.67) \leqq u \\ &{} z_2(x)-(2.21) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  which is infeasible. The next four subproblems are infeasible. For the fifth problem, we have $R'=(-13.67,-12.79)^T$ and $N'=(-7.33,-8.21)^T$. Now, the subproblem to be solved is:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -13.67 \leqq z_1(x) \leqq -7.33\\ &{} -12.79 \leqq z_2(x) \leqq -8.21 \\ &{} z_1(x)-(-13.67) \leqq u \\ &{} z_2(x)-(-12.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  It provides a Pareto optimal solution whose associated objective vector is $J=(-12,-12)^T$. The other subproblems obtained from the consecutive vectors already found, are infeasible. Then, we update $t=10$.

• Iteration 5: We sort all the non-dominated vectors determined: $A=(-30,6)^T$, $B=(-28,2)^T$, $C=(-26,-2)^T$, $D=(-21,-3)^T$, $E=(-19,-7)^T$, $F=(-14,-8)^T$, $G=(-12,-12)^T$, $H=(-7,-13)^T$, $I=(-2,-14)^T$ and $J=(3,-15)^T$. The first five subproblems determined were already checked in the previous iteration and are infeasible. It remains to analyze the sixth subproblem, hence $A \leftarrow F$ and $B \leftarrow G$, whose modified reference vectors are $R'=(-13.67,-11.79)^T$ and $N'=(-12.33,-8.21)^T$. Now we must solve the subproblem:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -13.67 \leqq z_1(x) \leqq -12.33\\ &{} -11.79 \leqq z_2(x) \leqq -8.21 \\ &{} z_1(x)-(-13.67) \leqq u \\ &{} z_2(x)-(-11.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  which is infeasible. For the seventh subproblem, $A \leftarrow G$ and $B \leftarrow H$, and, consequently, $R'=(-11.67,-12.79)^T$ and $N'=(-13.33,-12.21)^T$. The respective subproblem is the following:

  $$\begin{aligned} \begin{array}{lll} \text {Minimize} &{} u \\ \text {subject to}&{} -11.67 \leqq z_1(x) \leqq -13.33\\ &{} -12.79 \leqq z_2(x) \leqq -12.21 \\ &{} z_1(x)-(-11.67) \leqq u \\ &{} z_2(x)-(-12.79) \leqq u \\ &{} x \in X, \, u \geqq 0, \end{array} \end{aligned}$$

  which is also infeasible. In the previous iteration, it was verified that the nine subproblems are infeasible, thus attaining the stopping criterion for this procedure and with all the non-dominated vectors determined. The set of all non-dominated vectors for the problem instance given in this example is

  $$\begin{aligned} \begin{array}{lll} \displaystyle Z^*&{}=&{}\left\{ \displaystyle \left[ \begin{array}{c} -30 \\ 6 \end{array} \right] , \left[ \begin{array}{c} -28 \\ 2 \end{array} \right] ,\left[ \begin{array}{c} -26 \\ -2 \end{array} \right] ,\left[ \begin{array}{c} -21 \\ -3 \end{array} \right] , \right. \\ &{} &{}\displaystyle \left[ \begin{array}{c} -19 \\ -7 \end{array} \right] ,\left[ \begin{array}{c} -14 \\ -8 \end{array} \right] ,\left[ \begin{array}{c} -12 \\ -12 \end{array} \right] ,\left[ \begin{array}{c} -7 \\ -13 \end{array} \right] , \\ &{} &{} \left. \left[ \begin{array}{c} -2 \\ -14 \end{array} \right] , \left[ \begin{array}{c} 3 \\ -15 \end{array} \right] \right\} . \end{array} \end{aligned}$$

  (13)

It is important to note the following aspect: in spite of using $\ell =100$, thus expecting a maximum 100 iterations, the procedure only solved 10 subproblems up to optimality and tested nine for feasibility.

1.2 Numerical example for MB method

Example 2

The lexicographic vectors of Problem (12) are given by $A=z^{\text {lex}_1}=(3,-15)^T$ and $B=z^{\text {lex}_2}=(-30,6)^T$. Let $\ell =21$, and, consequently, $\delta _1=1.57$ and $\delta _2=1.00$.

• Iteration 1: For $k=1$, $z^0=(1.43,-14)$ and the following subproblem is solved:

  $$\begin{aligned} \begin{array}{lll} \text {Maximize} &{} l_1\\ \text {subject to}&{} 1.43 - z_1(x) = l_1 \\ &{} -14 - z_2(x) = l_2 \\ &{} x \in X, \ l_1 \geqq 0,\ l_2 \geqq 0, \end{array} \end{aligned}$$

  thus giving a new Pareto optimal solution, whose associated objective vector is ${z^1}=(-2,-14)^T$.

• Iteration 2: $k=2$, $z^{0k}=(-3.57,-13)^T$, we have $z^0=(1.43,-14)$ and the following subproblem is solved:

  $$\begin{aligned} \begin{array}{lll} \text {Maximize} &{} l_1\\ \text {subject to}&{} -3.57 - z_1(x) = l_1 \\ &{} -13 - z_2(x) = l_2 \\ &{} x \in X,\ l_1 \geqq 0,\ l_2 \geqq 0, \end{array} \end{aligned}$$

  thus determining a new Pareto optimal solution, whose associated objective vector is ${z^2}=(-7,-13)^T$.

By following the same steps, until the value of the second coordinate of $z^0$ researches 6, all the non-dominated vectors for Problem (12) are determined, as presented in (13).