Interactive programming approach for solving multi-level multi-objective linear programming problem

Abstract

The paper presents an interactive programming approach to find the compromise optimal solution of the multi-level multi-objective linear programming problem. The solving process can be divided into analysis stage and decision-making stage. In the analysis stage, an evaluation function is constructed to express the difference between the objective and ideal values. In the decision-making stage, decision maker can compare the objective values with ideal values. If there has unsatisfied objective values, decision maker can make a concession of the satisfied values to improve the unsatisfied values. When the satisfactory degree of decision maker in upper levels have been mat, the problem of lower levels will be in solving. A characteristic of the proposed method comes from its continuous interaction with the decision maker. Finally, a numerical example is solved by this algorithm.

Keywords

Interactive programming approach multi-level multi-objective linear programming membership function compromise optimal solution

1 Introduction

Multi-level multi-objective linear programming (ML-MOLP) problem is a special mathematical programming problem which has several objectives in different levels. The main idea of the ML-MOLP is that high level decision makers feedback the result to lower level decision makers and solve the lower level on the premise that the upper policy makers make certain concessions. The results obtained ensure the maximum possible satisfaction of each level decision maker. The most important feature of this model is that the objective functions can be solved according to the important degrees. Zimmermann [9] used the fuzzy set theory [6] to solve decision problem which has several conflicting objectives. Lai [16] introduced the concept of membership function to multi-objective programming problem for satisfactory solution. Then several literatures proposed fuzzy programming approach and fuzzy goal programming approach to solve multi-level programming problem [5 , 10–14]. Zhao et al. [19] presented an evaluation function to express decision makers’ satisfactory degree to solve programming problem which has two levels. Razmi [4] proposed an effective method to obtain compromise solution of the multi-objective programming problem. Fuzzy goal programming approach of Mohamed [8] which solves MOLP problem is extend by Baky [2, 3] to solve bi-level multi-objective programming problem in 2009, then in 2010 he solved ML-MOLP problem with the same method. Shi [17, 18] proposed the model of bi-level multi-objective programming and solved it with interactive algorithm. Lachhwani [7, 15] presented methods to solve multi-level linear fractional programming problem with fuzzy goal programming approach. The method in literature 7 given some modifications to revise the method which suggested by Baky (2010). Ren [1] studied fully fuzzy linear programming problem with two levels and solved it with interactive programming approach.

The aims of this paper is to proposed an efficient method to solve ML-MOLP problem. The feature of the proposed method is that in the solving process the decision maker would have the opportunity to adjust the unsatisfactory solution. Therefore, this approach ensures that the compromise optimal solution can be obtained that satisfies the decision makers’ preferences. The structure of the article is arranged as follows: The first section mainly introduces the research status of ML-MOLP problem. In Section 2, the model of ML-MOLP problem and relevant definitions are proposed. In Section 3, using interactive programming approach to solve the proposed model to obtain the compromise optimal solution. In Section 4 numerical example is solved to illustrate the proposed method. Then some conclusions are discussed in Section 5.

2 Formulation of ML-MOLP problem

In this section a p-level multi-objective programming problem is considered as follows: $\begin{matrix} \max_{x 1} z_{1} (x) = max_{x_{1}} (z_{11} (x), z_{12} (x), \dots, z_{1 m_{1}} (x)) \\ where x_{2}, x_{3}, \dots, x_{p} solves \\ \max_{x 1} z_{2} (x) = max_{x_{2}} (z_{21} (x), z_{22} (x), \dots, z_{2 m_{2}} (x)) \\ ⋮ \\ where x_{p} solves \\ \max_{x p} z_{p} (x) = max_{x_{p}} (z_{p 1} (x), z_{p 2} (x), \dots, z_{{pm}_{p}} (x)) \\ s . t . x \in D : {Σ_{i = 1}^{p} A_{i} x_{i} \leq B, B \in R^{m}, x_{i} \geq 0} \neq \emptyset \end{matrix}$ (1) where $z_{ij} (x) = c_{1}^{ij} x_{1} + c_{2}^{ij} x_{2} + \dots + c_{p}^{ij} x_{p}, i = 1, 2, \dots, p, j = 1, 2, \dots, m_{i} .$ $c_{r}^{ij} = (c_{r 1}^{ij}, c_{r 2}^{ij}, \dots, c_{{rn}_{r}}^{ij})$ , x_r = (x_r1, x_r2, ⋯, x_{rn
_r}) ^T, r = 1, 2, ⋯, p . Problem (1) contains n decision variables and m constraints. Set this problem includes one decision maker on each level. Let x = (x₁, x₂, ⋯, x_p) ^T is the decision vector. The vector x_i is controlled by the ith level decision maker and contains n_i decision variables, that is n = n₁ + n₂ + ⋯ + n_p. A_i is the coefficient matrices with size m × n_i, i = 1, 2, ⋯, p.

In problem (1), decision maker at the lth level (l = 1, 2, ⋯, p) makes the following multi-objective maximization problem: $\begin{matrix} \max_{x 1} z_{l} (x) = \max_{x 1} (z_{l 1} (x), z_{l 2} (x), \dots, z_{{lm}_{l}} (x)) \\ s . t . x \in D \end{matrix}$ (2)

In this paper, the interactive fuzzy programming algorithm is used to solve ML-MOLP problem. In problem (1), each level contains multiple objective functions, so we can not find the optimal solution for this problem, but find the efficient solution of each level and the compromise optimal solution of the whole problem.

Definition 2.1. Set x^* ∈ D, if there exists no other x ∈ D such that z_ij (x) ≥ z_ij (x^*) for all i and j, i = 1, 2, ⋯, p, j = 1, 2, ⋯, m_i and z_ij (x) > z_ij (x^*) for at least one i and j, then x^* is an efficient solution to ML-MOLP problem.

Definition 2.2. For the ML-MOLP problem, the efficient solution is called the compromise optimal solution when it is considered to be the best solution which meet the requirements of all decision makers.

3 Interactive programming formulation

In the process of solving the multi-objective problem in each level, we can solve it by the combination of the solver and the decision maker, that is, interactive programming. The solving process is divided into analysis stage and decision-making stage. In the analysis stage, the analyst provides the objective value with the ideal value to the decision maker. In the decision-making stage, the decision makers judge whether the current efficient solution is satisfied, the satisfied goal can give some compromise to improve the unsatisfied goal. The related information was provided to decision maker to go on processing. By repeating the process, the satisfied solution for each goal will be obtained.

3.1 Construct membership function

For each objective function z_ij of the ML-MOLP problem, we can get two values $z_{ij}^{U}$ and $z_{ij}^{L}$ as its upper bounds and lower bounds. Then the linear membership function μ_{z
_ij} (z_ij (x)) of the ijth objective function can be constructed as the following form [3]: $μ_{z_{ij}} (z_{ij} (x)) = {\begin{matrix} 0, & if & z_{ij} (x) \leq z_{ij}^{L}, \\ \frac{z_{ij} (x) - z_{ij}^{L}}{z_{ij}^{U} - z_{ij}^{L}}, & if & z_{ij}^{L} \leq z_{ij} (x) \leq z_{ij}^{U}, \\ 1, & if & z_{ij} (x) \geq z_{ij}^{U} . \end{matrix}$ (3)

Assume that $x^{i^{*}} = (x_{1}^{i^{*}}, x_{2}^{i^{*}}, \dots, x_{p}^{i^{*}})^{T}$ is an efficient solution to the ith level. In the decision vector, let $t_{r}^{i_{L}}$ is the maximum negative values and $t_{r}^{i_{R}}$ is the positive tolerance values which considered by the ith level decision maker, r = 1, 2, ⋯, n_r, i = 1, 2, ⋯, p - 1. The membership function of each decision vector $x_{i}^{i^{*}} = (x_{i 1}^{*}, x_{i 2}^{*}, \dots, x_{{in}_{i}}^{*})^{T}$ is controlled by the decision makers that can be constructed as: $μ_{x_{ir}} (x_{ir}) = {\begin{matrix} \frac{(x_{ir}^{*} + t_{r}^{i_{R}}) - x_{ir}}{t_{r}^{i_{R}}}, & if & x_{ir}^{*} \leq x_{ir} \leq x_{ir}^{*} + t_{r}^{i_{R}}, \\ \frac{x_{ir} - (x_{ir}^{*} - t_{r}^{i_{L}})}{t_{r}^{i_{L}}}, & if & x_{ir}^{*} - t_{r}^{i_{L}} \leq x_{ir} \leq x_{ir}^{*}, \\ 0, & otherwise . \end{matrix}$ (4)

Noted that, the tolerances $t_{r}^{i_{L}}$ and $t_{r}^{i_{R}}$ are not necessarily same [3].

3.2 Interactive programming approach

Consider the problem (1), we can obtain the optimal solutions and optimal values by solving each objective function under the given set of constraint D. Let $x_{ij}^{*}$ is the respective optimal solution of the objective function z_ij (x). In general cases, $x_{ij}^{*}$ are different. For each objective function z_ij (x), calculate the ideal values $z_{ij}^{*} (x) = z_{ij} (x_{ij}^{*})$ and the minimum acceptable values $z_{ij}^{L} (x) = \min {z_{ij} (x_{11}^{*}), \dots, z_{ij} (x_{1 m_{1}}^{*}), \dots,$ $z_{ij} (x_{p 1}^{*}), \dots, z_{ij} (x_{{pm}_{p}}^{*})}, j = 1, \dots, m_{i}, i = 1, \dots, p .$

Consider the multi-objective programming problem of the lth level, $z_{l}^{*} = (z_{l_{1}}^{*}, z_{l_{2}}^{*}, \dots, z_{l_{m_{l}}}^{*})$ is an idea vector composed of ideal values, then solving problem (2) can be converted to solve the minimization problem as follows: $\begin{matrix} \min_{x \in D_{l}^{k}} d (z_{l}, z_{l}^{*}) \end{matrix}$ (5) Where d (·, ·) represents the distance between z_l and $z_{l}^{*}$ . Constructing a weighted maximize evaluation function

$\begin{matrix} \min_{x \in D_{l}^{k}} d (z_{l}, z_{l}^{*}) = \min_{x \in D_{l}^{k}} \max_{1 \leq j \leq m_{l}} {w_{j} | z_{lj}^{*} - z_{lj} |} \end{matrix}$ (6) Where w_j is the nonnegative weight to the jth objective function. In order to solve the problem (5), an auxiliary variable λ is introduced as follows $\begin{matrix} λ = \max_{1 \leq j \leq m_{l}} {w_{j} | z_{lj}^{*} - z_{lj} |} \end{matrix}$

Thus, the solution of the above problem (6) can be translated into solving the following problem $\begin{matrix} min λ \\ s . t . w_{lj} | z_{lj}^{*} - z_{lj} (x) | \leq λ, (j = 1, 2, \dots, m_{l}) \\ x \in D_{l}^{k} \\ λ \geq 0 \end{matrix}$ (7)

Theorem 3.1. Set (λ^*, x^*) is the optimal solution of problem (7), then it is also the optimal solution of problem (6).

Proof. Suppose that (λ^*, x^*) is not an optimal solution of problem (6), then there exits $\bar{x} \in D$ and $\bar{λ} \geq 0$ such that $w_{lj} | z_{lj}^{*} - z_{lj} (\bar{x}) | \leq \bar{λ}$ and $\bar{λ} \leq λ^{*}$ , this contradicts the fact that (x^*, λ^*) is an optimal solution of problem (7).

In problem (5), $D_{l}^{k}$ is the constraint set for solving the lth level in the kth round, when k = 1, $\begin{matrix} D_{l}^{1} = {x \in D | μ_{z_{ij}} (z_{ij} (x)) \geq α_{ij}, μ_{x_{ir}} (x_{ir}) \geq α_{x_{ir}}, \\ j = 1, \dots, m_{i}, i = 1, \dots, l - 1, r = 1, \dots, n_{i}} \end{matrix}$ μ_{z
_ij} (z_ij (x)), μ_{x
_ir} (x_ir) represent the membership function of objective functions and decision variables. $z_{ij}^{U}$ in the membership function of μ_{z
_ij} is the compromise optimal value for solving the l - 1th problem. α_ij and α_{x
_ir} are the minimum acceptable values for the objective functions and decision variables given by the decision maker, where α_ij, α_{x
_ir} ∈ [0, α^*] . α^* is the optimal solution of problem (8) which represents the maximum level of membership function can be achieved at the same time. $\begin{matrix} \max α \\ s . t . μ_{z_{ij}} (z_{ij} (x)) \geq α i = 1, \dots, l - 1, j = 1, \dots, m_{i} \\ μ_{x_{ir}} (x_{ir}) \geq α i = 1, \dots, l - 1, r = 1, \dots, n_{i} \\ x \in D \\ 0 \leq α \leq 1 \end{matrix}$ (8)w_lj is nonnegative weight to the jth objective function. Using information of linear objective function, it is noted as [20]

$w_{l j} = \frac{β_{l j}}{\sum_{j = 1}^{m_{l}} β_{l j}}$ and $β_{l j} = {\begin{matrix} \frac{z_{l j}^{*} - z_{l j}^{L}}{z_{l j}^{*}} \frac{1}{\sqrt{\sum_{r = 1}^{p} {(c_{r}^{l j})}^{2}}}, & i f z_{l j}^{L} \geq 0 \\ \frac{z_{l j}^{L} - z_{l j}^{*}}{z_{l j}^{L}} \frac{1}{\sqrt{\sum_{r = 1}^{p} {(c_{r}^{l j})}^{2}}}, & i f z_{l j}^{L} < 0 \end{matrix}$

To solve problem (7) to get the optimal solution $x_{l}^{k}$ . In order to test whether the obtained solution is satisfactory, the decision maker compares the objective value $z_{lj} (x_{l}^{k})$ with the corresponding ideal value $z_{lj}^{*} = z_{lj} (x_{lj}^{*}), j = 1, 2, \dots, m_{l}$ . If the objective value z_ls (1 ≤ s ≤ m_l) is satisfying, but z_lt (1 ≤ t ≤ m_l) is unsatisfying, then it needs further improvement on the existing basis. In order to improve the unsatisfactory z_lt, the constraint $z_{lt} \geq z_{lt} (x_{l}^{k})$ is added to the next step to ensure that the solution of the next round is not worse than $x_{l}^{k}$ . Then in the next round appropriate concession should be made for the satisfaction value, set $Δ_{ls}^{k} > 0$ is the maximum tolerance that objective function z_ls allows, then the tolerance constraint $z_{ls} \geq z_{ls} (x_{l}^{k}) - Δ_{ls}^{k}$ is added in the next round of solution. Therefore we can get the new constraint set as follows: $\begin{matrix} D_{l}^{k + 1} = {x \in D_{l}^{k} | z_{lt} \geq z_{lt} (x_{l}^{k}), z_{ls} \geq z_{ls} (x_{l}^{k}) - Δ_{ls}^{k}} \end{matrix}$

A new round of solving is carried out under the constraint set $D_{l}^{k + 1}$ . At this time, the satisfactory objective function is no longer involved in the solution, that is, the weight coefficient of this term is zero, and the solution to the objective function z_lt in this way is better than the previous round.

The solution procedure for solving ML-MOLP problem using the interactive programming approach is described below.

ALG(3.1).

Step 1: Calculate the optimal solutions $x_{ij}^{*}$ of all objective functions z_ij under the given constraint set D. If the optimal solutions are the same, then it outputs $x_{ij}^{*}$ as the optimal solution to the original ML-MOLP problem, or goes to the Step 2.

Step 2: According to the optimal solution $x_{ij}^{*}$ , the ideal values $z_{ij}^{*}$ and the minimum acceptable values $z_{ij}^{L}$ of each objective function can be obtained.

Step 3: Let $D_{1}^{1} : = D, l : = 1, k : = 1$ , that is the initial constraint set.

Step 4: Evaluate the weights w_lj, j = 1, 2, ⋯, m_l.

Step 5: Solve the problem (7) to get the optimal solution $x_{l}^{k}$ .

Step 6: To compare objective value and ideal value.

If all the objective values are satisfying, then output the satisfactory solution $x_{l}^{k}$ and the satisfactory values $z_{ij}^{U} = z_{ij} (x_{l}^{k}), i = 1, 2, \dots, l,$ j = 1, 2, ⋯, m_l in lth level, then goes to Step 8.

If there exists unsatisfying solution, then goes to Step 7.

Step 7: Let decision maker be dissatisfied with kth objective value and given the maximum tolerance $Δ_{ls}^{k} > 0$ to the satisfying objective value z_ls, then make the new constraint set $D_{l}^{k + 1}$ . Let w_ls = 0, k : = k + 1 be and turn to the Step 4.

Step 8: If l > p - 1, then go to Step 9; otherwise let l : = l + 1 and turn to the Step 4;

Step 9: Outputs satisfactory solution $x_{p}^{k}$ , that is the compromise optimal solution to the ML-MOLP problem.

4 Numerical example

In this section, a numerical example is given to illustrate the solution process of the algorithm. Considering three level multi-objective linear programming problem that is solved in literature 3. $\begin{matrix} \max_{x 1} (z_{11} = - x_{1} + x_{2} + 4 x_{3}, \\ z_{12} = x_{1} - 3 x_{2} + 4 x_{3}) \\ where x_{2}, x_{3} solves \\ \max_{x 2} (z_{21} = - 2 x_{1} + x_{2} - 2 x_{3}, \\ z_{22} = - 2 x_{1} - x_{2} + 3 x_{3}, \\ z_{23} = - 3 x_{1} + x_{2} - x_{3}) \\ where x_{3} solves \\ \max_{x 3} (z_{31} = - 7 x_{1} - 3 x_{2} + 4 x_{3}, z_{32} = - x_{1} - x_{3}) \\ s . t . x_{1} + x_{2} + x_{3} \leq 3 \\ - x_{1} + x_{2} + x_{3} \leq 1 \\ x_{1} + x_{2} + x_{3} \geq 1 \\ x_{1} + x_{2} - x_{3} \leq 1 \\ x_{1}, x_{2} \geq 0, 0 \leq x_{3} \leq 0.5 . \end{matrix}$

Table 1 summarizes the idea values and the minimum acceptable values of the ML-MOLP problem.

Table 1
Idea values and the minimum acceptable values

z ₁₁ z ₁₂ z ₂₁ z ₂₂ z ₂₃ z ₃₁ lz₃₂

$z_{ij}^{*}$ 2.5 3.5 1 1 1 0.5 0

$z_{ij}^{L}$ 0.5 -3 -4 -1.5 -5 -8.5 -2

	z ₁₁	z ₁₂	z ₂₁	z ₂₂	z ₂₃	z ₃₁	lz₃₂
$z_{ij}^{*}$	2.5	3.5	1	1	1	0.5	0
$z_{ij}^{L}$	0.5	-3	-4	-1.5	-5	-8.5	-2

Following the ALG 3.1, we can solve the problem as follows: For the first level, the weight coefficients of each objective function are w₁₁ = 0.308, w₁₂ = 0.692, then the first level to solve the following problem: $\begin{matrix} min λ \\ s . t . 0.77 + 0.308 x_{1} - 0.308 x_{2} - 1.232 x_{3} \leq λ \\ 2.422 - 0.692 x_{1} + 2.076 x_{2} - 2.768 x_{3} \leq λ \\ x \in D_{1}^{1} \\ λ \geq 0 . \end{matrix}$

To solve the above problem, the optimal solution $x_{1}^{1} = (0.884, 0, 0.5)$ and the objective function values $z_{11} (x_{1}^{1}) = 1.116,$ $z_{12} (x_{1}^{1}) = 2.884 .$ are obtained. Into the decision-making stage, to compare target value and ideal value, if the second value z₁₂ is satisfying, the first value z₁₁ is unsatisfying. Let the decision maker gives the maximum tolerance $Δ_{12}^{1} = 1$ and makes the new constraint set $\begin{matrix} D_{1}^{2} = {x \in D_{1}^{1} | - x_{1} + x_{2} + 4 x_{3} \geq 1.116, \\ x_{1} - 3 x_{2} + 4 x_{3} \geq 1.884} \end{matrix}$

Go to the analysis phase, since the target z₁₂ is satisfied, then w₁₂ = 0, so the new weight coefficients are w₁₁ = 1, w₁₂ = 0, continue to solve the following problem on the first level $\begin{matrix} min λ \\ s . t . 2.5 + x_{1} - x_{2} - 4 x_{3} \leq λ \\ x \in D_{1}^{2} \\ λ \geq 0 . \end{matrix}$

To solve the above problem, the optimal solution $x_{1}^{2} = (0.346, 0.154, 0.5),$ the objective function values are $z_{11} (x_{1}^{2}) = 1.808,$ $z_{12} (x_{1}^{2}) = 1.884 .$ Once again into the decision-making stage, the decision maker compares the current target value and the ideal value. Let the decision maker be satisfied with both objectives, the end of the solution. Put out the compromise optimal solution $x_{1}^{2} = (0.346, 0.154, 0.5)$ and the compromise optimal values $z_{11}^{U} = 1.808, z_{12}^{U} = 1.884 .$

For the second level, let the decision maker at the first level can make appropriate concessions to the solution x₁ = 0.346, set the negative tolerances $t_{1}^{1_{L}} = 0.346$ and the tolerances positive $t_{1}^{1_{R}} = 0.154 .$ Construct the membership function of the decision variable x₁ and the objective function z₁₁ (x), z₁₂ (x). Solving problem (8) to obtain the maximum satisfaction value $α_{1}^{*} = 1$ that enables the membership function to reach at the same time. When solving the second level, decision maker can make appropriate concessions to the decision variables and the compromise optimal values of the first level. Let α₁₁ = α₁₂ = 0.9, α_{x
₁} = 0.5 . The weight coefficients of the second level are w₂₁ = 0.341, w₂₂ = 0.365, w₂₃ = 0.294, then the second level to solve the following problem: $\begin{matrix} min λ \\ s . t . 0.341 + 0.682 x_{1} - 0.341 x_{2} + 0.682 x_{3} \leq λ \\ 0.365 + 0.73 x_{1} + 0.365 x_{2} - 1.095 x_{3} \leq λ \\ 0.294 + 0.882 x_{1} - 0.294 x_{2} + 0.294 x_{3} \leq λ \\ - x_{1} + x_{2} + 4 x_{3} \geq 1.677 \\ x_{1} - 3 x_{2} + 4 x_{3} \geq 1.396 \\ 0.173 \leq x_{1} \leq 0.423 \\ x \in D \\ λ \geq 0 . \end{matrix}$

To solve the above problem, the optimal solution $x_{2}^{1} = (0.224, 0.276, 0.5),$ the objective function values are $z_{11} (x_{2}^{1}) = 2.052,$ $z_{12} (x_{2}^{1}) = 1.396,$ $z_{21} (x_{2}^{1}) = - 1.172,$ $z_{22} (x_{2}^{1}) = 0.776,$ $z_{23} (x_{2}^{1}) = - 0.846 .$ Into the decision-making stage, to compare target value and ideal value, decision makers think that all the objective function values are satisfied, then output the compromise optimal solution of the second level $x_{2}^{1} = (0.224, 0.276, 0.5),$ compromise optimal values are $z_{ij}^{U} = z_{ij} (x_{2}^{1}) .$ Continue to solve the third level, let x₁ = 0.224 in $x_{2}^{1}$ with the negative and positive tolerances $t_{2}^{1_{L}} = 0.224, t_{2}^{1_{R}} = 0.276$ is given by the decision maker on the first level. x₂ = 0.276 in $x_{2}^{1}$ with the negative and positive tolerances $t_{2}^{2_{L}} = 0.276, t_{2}^{2_{R}} = 0.724$ is given by the decision maker of the second level. Construct membership function of the decision variable x₁, x₂ and the objective function z₁₁ (x), z₁₂ (x), z₂₁ (x), z₂₂ (x), z₂₃ (x). Solving problem (8) to obtain the maximum satisfaction value $α_{2}^{*} = 0.99$ that enables the membership function to reach at the same time. When solving the third level, decision maker can make appropriate concessions to the decision variables and the compromise optimal values of the first and the second levels. Set up α_ij = 0.9, i = 1, 2, α_{x
₁} = α_{x
₂} = 0.5. The weight coefficients of the third level are w₃₁ = 0.148, w₃₂ = 0.852, then the third level to solve the following problem: $\begin{matrix} min λ \\ s . t . 0.074 + 1.036 x_{1} + 0.444 x_{2} - 0.592 x_{3} \leq λ \\ 0.852 x_{1} + 0.852 x_{3} \leq λ \\ - x_{1} + x_{2} + 4 x_{3} \geq 1.897 \\ x_{1} - 3 x_{2} + 4 x_{3} \geq 0.9564 \\ 2 x_{1} - x_{2} + 2 x_{3} \leq 1.455 \\ - 2 x_{1} - x_{2} + 3 x_{3} \geq 0.548 \\ 3 x_{1} - x_{2} + x_{3} \leq 1.3064 \\ 0.112 \leq x_{1} \leq 0.362 \\ 0.138 \leq x_{2} \leq 0.638 \\ x \in D \\ λ \geq 0 . \end{matrix}$

The optimal solution of the above problem is $x_{3}^{1} = (0.114, 0.386, 0.5),$ the objective function values are $z_{11} (x_{3}^{1}) = 2.272,$ $z_{12} (x_{3}^{1}) = 0.956,$ $z_{21} (x_{3}^{1}) = - 0.842,$ $z_{22} (x_{3}^{1}) = 0.886,$ $z_{23} (x_{3}^{1}) = - 0.456,$ $z_{31} (x_{3}^{1}) = 0.044,$ $z_{32} (x_{3}^{1}) = - 0.614 .$ Into the decision-making stage, to compare target value and ideal value, decision makers think that all the objective function values are satisfied, then output the compromise optimal solution of the third level. This results are the compromise optimal solution of original problem, and we can obtain the values of membership functions μ_{z
₁₁} (z₁₁ (x)) =0.886, μ_{z
₁₂} (z₁₂ (x)) =0.609, μ_{z
₂₁} (z₂₁ (x)) =0.632, μ_{z
₂₂} (z₂₂ (x)) =0.954, μ_{z
₂₃} (z₂₃ (x)) =0.757, μ_{z
₃₁} (z₃₁ (x)) =0.949, μ_{z
₃₂} (z₃₂ (x)) =0.693 respectively.

A comparison given in Table 2 between the proposed interactive programming approach and the FGP approach given in [3] by Baky. As seen in Table 2, in addition to the z₁₁, the objective values obtained by the proposed method are superior to the objective values of reference 3.

Table 2

Comparison of the results

The FGP approach		Interactive programming approach		The idea values
z₁₁ = 2.498	μ_{z ₁₁} = 0.999	z₁₁ = 2.272	μ_{z ₁₁} = 0.886	z₁₁ = 2.5
z₁₂ = -0.494	μ_{z ₁₂} = 0.386	z₁₂ = 0.956	μ_{z ₁₂} = 0.609	z₁₂ = 3.5
z₂₁ = -1.002	μ_{z ₂₁} = 0.6	z₂₁ = -0.842	μ_{z ₂₁} = 0.632	z₂₁ = 1
z₂₂ = -0.498	μ_{z ₂₂} = 0.401	z₂₂ = 0.886	μ_{z ₂₂} = 0.954	z₂₂ = 1
z₂₃ = -1.002	μ_{z ₂₃} = 0.666	z₂₃ = -0.456	μ_{z ₂₃} = 0.757	z₂₃ = 1
z₃₁ = -4.494	μ_{z ₃₁} = 0.445	z₃₁ = 0.044	μ_{z ₃₁} = 0.949	z₃₁ = 0.5
z₃₂ = -1	μ_{z ₃₂} = 0.5	z₃₂ = -0.614	μ_{z ₃₂} = 0.693	z₃₂ = 0
x^* = (0.5, 0.998, 0.5)		x^* = (0.114, 0.386, 0.5)

5 Conclusion

This paper presents interactive programming approach to obtain the compromise optimal solution of the ML-MOLP programming problem. This kind of problem can be solved by the combination of the solver and the decision maker. In view of the numerical example, the proposed method is efficient. The advantages of the algorithm mentioned in this paper is the decision makers can identify and adjust their own decision at any time during the solution process, and on the premise of decision makers to make certain concessions to get the compromise optimal solution. In the future, We will focus on the following topics: (1) The coefficients or variables in the model are fuzzy or random. (2) The proposed method can be extended to solve multi-level multi-objective fractional programming problem. (3) In the process of solving, it is possible to use the nonlinear membership function instead of the linear membership function, and compare the results.

Footnotes

Acknowledgments

The project is supported by the Natural Science Foundation of Hebei Province of China (No. A2015209229) and Science and Technology Research Foundation of Higher Education Institutions of Hebei province of China (No. Z2017014).

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