1. Professor: Chu, Ta Chung Student: Nguyen Quang Tung Student’s ID: M977Z235 Fuzzy multiobjective linear model for supplier selection in a supply chain

2. An integrated multi-objective supplier selection Notation definition: x i : the number of units purchased from the ith supplier P i : price of the ith supplier D: demand over the period C i : capacity of the ith supplier F i : percentage of items delivered late for the ith supplier S i : percentage of rejected units for the ith supplier n: number of suppliers

3. An integrated multi-objective supplier selection Objective functions : Constraints

4. A fuzzy multi-objective programming The model is proposed by Zimmermann (1978) with the purpose of finding a vector x T = [x 1, x 2, …, x n ] to satisfy Subject to For fuzzy constraints For deterministic constraints (7) (8) (9)

5. The crisp model of fuzzy optimization problem Objective function Subject to (10) (11) (12) (13) (14) (15)

6. Model Algorithm The following steps are to solve a multi-objective supplier selection problem: Step 1: Construct the supplier selection model according to the criteria and constraints of the buyer and suppliers. Step 2: Solve the multi-objective supplier selection problem as a single-objective supplier selection problem using each time only one objective. This value is the best value for this objective as other objectives are absent. Step 3: From the results of step 2 determine the corresponding values for every objective at each solution derived.

7. Model Algorithm (cont’) Step 4: From step 3, for each objective function find a lower bound and an upper bound corresponding to the set of solutions for each objective. Let and denote the lower bound and upper bound for the kth objective (Z k ) Step 5: For the objective functions and fuzzy constraints find the membership function. Step 6: From step 5 and DM’s preferences, based on fuzzy convex decision-making formulate the equivalent crisp model of the fuzzy optimization problem. Step 7: Find the optimal solution vector x*, where x* is the efficient solution of the original multi-objective supplier selection problem with the DM’s preferences.

8. Numerical Example Suppose the demand for material of a company is predicted to be about 1000 units. There are 3 suppliers supplying material for this company. The information on these suppliers are given as follows: SupplierPrice% rejected Items % late Deliveries Capacity S1315%25%500 S2220%10%600 S355%15%550

9. Numerical Example The supplier selection model (the linear model): Objective functions: Min Z 1 = 3x 1 + 2x 2 + 5x 3 (net cost) Min Z 2 = 0.15x 1 + 0.2x 2 + 0.05x 3 (rejected items) Min Z 3 = 0.25x 1 + 0.1x 2 + 0.15x 3 (late delivered items) Subject to x 1 + x 2 + x 3 = 1000(demand constraint) 0 ≤ x 1 ≤ 500(capacity constraint) 0 ≤ x 2 ≤ 600(capacity constraint) 0 ≤ x 3 ≤ 550(capacity constraint)

10. By solving the multi-objective supplier selection problem as a single-objective supplier selection problem using each time only one objective we get the following data: μ = 0μ = 1μ = 0 Z 1 (net cost)41002400- Z 2 (rejected items)18095- Z 3 (late delivered item)200120- Demand95010001100

11. Membership functions Net cost 2400 4100 1 0 Rejected items 1 0 95 180 1 0 120 200 Late delivery items Demand 1 0 950 1000 1100

12. The fuzzy multi-objective model is formulated to find x T = (x 1,x 2,x 3 ) to satisfy: Z 1 = 3x 1 + 2x 2 + 5x 3 ≤ ~ Z 2 = 0.15x 1 + 0.2x 2 + 0.05x 3 ≤ ~ Z 3 = 0.25x 1 + 0.1x 2 + 0.15x 3 ≤ ~ Subject to: x 1 + x 2 + x 3 ≅ 1000 0 ≤ x 1 ≤ 500 0 ≤ x 2 ≤ 600 0 ≤ x 3 ≤ 550

13. Suppose the weight of fuzzy goals and fuzzy constraints are given as: w 1 = 0.15, w 2 = 0.5, w 3 = 0.25, β = 0.1. Then the model can be formulated as follows: Max 0.15 λ 1 + 0.5 λ 2 + 0.25 λ 3 + 0.1 γ Subject to

14. 0 ≤ x 1 ≤ 500 0 ≤ x 2 ≤ 600 0 ≤ x 3 ≤ 550 Using Lingo 8.0 to solve this problem, the optimal solution is obtained as follows: x 1 = 0; x 2 = 450; x 3 = 550 Z 1 = 3650; Z 2 = 118; Z 3 = 128

15. Lingo programming model MAX = 0.15 * A1 + 0.5 * A2 + 0.25 * A3 + 0.1 * B; A1 * 1700 <= 4100 - 3 * X1 - 2 * X2 - 5 * X3; A2 * 85 <= 180 - 0.15 * X1 - 0.2 * X2 - 0.05 * X3; A3 * 80 <= 200 - 0.25 * X1 - 0.1 * X2 - 0.15 * X3; B * 100 <= 1100 - X1 - X2 - X3; B * 50 <= X1 + X2 + X3 - 950; A1 >= 0; A1 <= 1; A2 >= 0; A2 <= 1; A3 >= 0; A3 <= 1; B >= 0; B <= 1; @GIN( X1); @GIN( X2); @GIN( X3); X1 <= 500; X2 <= 600; X3 <= 550; X1 >= 0; X2 >= 0; X3 >= 0;

16. Optimal Solution Global optimal solution found at iteration: Objective value: 0.7339154 Variable Value Reduced Cost A1 0.2647059 0.000000 A2 0.7352941 0.000000 A3 0.9062500 0.000000 B 1.000000 0.000000 X1 0.000000 0.2928309E-02 X2 450.0000 0.2665441E-02 X3 550.0000 0.2204044E-02

17. THANK YOU FOR YOUR ATTENTION!