Fuzzy multiobjective linear model for
supplier selection in a supply chain
Professor: Chu, Ta Chung
Student: Nguyen Quang Tung
Student’s ID: M977Z235
An integrated multi-objective supplier
selection
 Notation definition:
• xi: the number of units purchased from the ith supplier
• Pi: price of the ith supplier
• D: demand over the period
• Ci: capacity of the ith supplier
• Fi: percentage of items delivered late for the ith supplier
• Si: percentage of rejected units for the ith supplier
• n: number of suppliers
An integrated multi-objective supplier selection
 Objective functions:
 Constraints
A fuzzy multi-objective programming
 The model is proposed by Zimmermann (1978) with the purpose of
finding a vector xT = [x1, x2, …, xn] to satisfy
(7)
Subject to
(8)
For fuzzy constraints
(9)
For deterministic constraints
The crisp model of fuzzy optimization problem
Objective function
(10)
Subject to
(11)
(12)
(13)
(14)
(15)
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.
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 (Zk)
 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.
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:
Supplier
Price
% rejected % late
Items
Deliveries
Capacity
S1
3
15%
25%
500
S2
2
20%
10%
600
S3
5
5%
15%
550
Numerical Example
 The supplier selection model (the linear model):
Objective functions:
Min Z1 = 3x1 + 2x2 + 5x3 (net cost)
Min Z2 = 0.15x1 + 0.2x2 + 0.05x3 (rejected items)
Min Z3 = 0.25x1 + 0.1x2 + 0.15x3 (late delivered items)
Subject to
x1 + x2 + x3 = 1000 (demand constraint)
0 ≤ x1 ≤ 500
(capacity constraint)
0 ≤ x2 ≤ 600
(capacity constraint)
0 ≤ x3 ≤ 550
(capacity constraint)
 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=0
μ=1
μ=0
Z1 (net cost)
4100
2400
-
Z2 (rejected items)
180
95
-
Z3 (late delivered item)
200
120
-
Demand
950
1000
1100
 Membership functions
Net cost
Rejected items
1
1
0
0
2400
4100
95
Demand
Late delivery items
1
1
0
0
120
200
180
950 1000
1100
 The fuzzy multi-objective model is formulated to find
xT= (x1,x2,x3) to satisfy:
Z1 = 3x1 + 2x2 + 5x3 ≤ ~
Z2 = 0.15x1 + 0.2x2 + 0.05x3 ≤ ~
Z3 = 0.25x1 + 0.1x2 + 0.15x3 ≤ ~
Subject to:
x1 + x2 + x3 ≅ 1000
0 ≤ x1 ≤ 500
0 ≤ x2 ≤ 600
0 ≤ x3 ≤ 550
 Suppose the weight of fuzzy goals and fuzzy constraints are
given as: w1 = 0.15, w2 = 0.5, w3 = 0.25, β = 0.1. Then the
model can be formulated as follows:
Max 0.15λ1 + 0.5λ2 + 0.25λ3 + 0.1γ
Subject to
0 ≤ x1 ≤ 500
0 ≤ x2 ≤ 600
0 ≤ x3 ≤ 550
Using Lingo 8.0 to solve this problem, the optimal
solution is obtained as follows:
x1 = 0; x2 = 450; x3 = 550
Z1 = 3650; Z2 = 118; Z3 = 128
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;
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
THANK YOU FOR YOUR ATTENTION!