Multi criteria decision making

1. MULTIPLE OBJECTIVE
DECISION MAKING (MODM)
Ha Thi Xuan Chi, PhD
1

2. Contents
Introduction
Classification of MODM methods
Modeling
The optimal solutions and efficient
solutions.
The ideal solutions
2

3. Introduction
Multiple Objective Decision Making (MODM) is
not completely a minimization or maximization
problem.
MODM is a special combination among single
optimization problems, usually these objectives are
conflicting. Called “Satisficing”: satisfying and
sacrificing.
Mathematical Programming played important role
in modeling .
3

4. Classification of MODM methods

5. MODELING
When k objectives are considered
simultaneously, the MODM problem can be
modeled as follows:
5
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍𝑖 = 𝑓𝑖 𝑋 ; 𝑖 = 1, 2, . . . , 𝑘
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜:
𝑔𝑗 𝑋 ≤ 𝑏𝑗; 𝑗 = 1, 2, . . . , 𝑚
Where:
𝑋 = (𝑥1, 𝑥2, …, 𝑥𝑛)
𝑓𝑖 𝑋 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑡ℎ
𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
𝑔𝑗 𝑋 ≤ 𝑏𝑗 𝑖𝑠 𝑡ℎ𝑒 𝑗𝑡ℎ
𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡.

6. MODELING
The matrix form:
maximize Z = CX
Subject to: AX  B;
X  0
Where:
Z: matrix with dimension kx1;
X: matrix with dimension nx1;
B: matrix with dimension mx1;
A: matrix with dimension mxn and
C: matrix with dimension kxn.
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7. Modeling
Example 1:
max z1 = x1 + x2
max z2 = x2–x1
Subject to:
x1  3
x2  3
x1 , x2  0
There is no optimal solution.

8. Modeling
Example 2:
max z1 = x2
max z2 = x2–x1
Subject to:
x1  3
x2  3
x1 , x2  0
The optimal solution is (0,3)

9. MODELING
Notes:
 It is impossible to satisfy all conflicting objectives.
 Optimality is replaced by concept “satisfying and
sacrificing” or “the best compromise solutions”. This is
is based on Decision Maker preferences.
Assumptions:
 The solutions were exist.
 Objectives might not be modified and compacting to
to simpler or less objectives than the original one.
9

10. Solutions
The optimal solution,
The efficient solution
The ideal solution.

11. Optimal solutions
The optimal solution: is the maximum value which achieving
from all objectives simultaneously.
Solution x* is optimal if and only if:
x* S and fj(x*)≥ fj (x)
j and x  S;
S is set of feasible solutions.
Notes:
- In general, there is no feasible optimal solution.
- If objectives are not conflicting, than one can find out optimal
solution. However, by nature, objectives are conflicting, so we
consider only efficient solutions.
11

12. Dominance
x1 dominates x2, if
Solution x1 is no worse than x2 in all
objectives
Solution x1 is strictly better than x2 in at least
least one objective
x1 dominates x2 <=>x2 is dominated by x1

13. Efficient solutions
Efficient solution (Pareto solution, frontier
solutions, non-dominant solution)
 A solution that no increase can be obtained at
any objective without simultaneously decrease at
least one of objectives.
 Solution x* is efficient if and only if there does
not exist any x  S sao cho:
fj (x)  fl (x*), for All j
fj (x) > fj (x*),for at least one value of j.
The solution is not unique.
13

14. Efficient solutions 14
Z1=f1(x)
Z2=f2(x)
a
b
Taä
p caù
c ñieå
m hieä
u quaû
0
Set of efficient solutions
b
A
Z = f
2
(x)
Z = f
1
(x)
Fig. 2.1 Illustration on concept of Efficient solutions
0
Feasible solutions

15. THE BEST EFFICIENT SOLUTIONS
In most of the cases, we need a “best”
among efficient solutions.
Therefore, some additional criteria will be
introduced to select the “best”.
The term “best compromise solution” will be
used.
15

16. THE BEST EFFICIENT SOLUTIONS
Example 3 Consider following bi-criterion problem. Find
the efficient solutions.
maximize z1 = –2x1 + x2;
maximize z2 = 2x1 + x2
Sub. to: - x1 + x2  1
x1 + x2  7
x1  5
x2  3
x1, x2  0
16

17. THE BEST EFFICIENT SOLUTIONS 17
Z = f 2 (x)
2
Z = f 1 (x)
1
x2
3
(0, 1)
0 (5,0)
(2, 3) (4, 3)
x1
Fig. 2.2 Efficient solutions of the problems
(5,2)

18. THE IDEAL SOLUTION
Each objective has individual optimum
solution. Corresponding with the optimum of
one objective, the other has not reached
optimum.
18

19. THE IDEAL SOLUTION
The payoff matrix:
square matrix, the diagonal of matrix is optimal
values for each individual objective.
xh* is the optimum solution for individual
objective h and this value was used to compute
values of the other objective.

20. THE IDEAL SOLUTION
Payoff Matrix:
20
x
z
Z1 f1(x
1*
) f1(x
2*
) ... f1(x
h*
) ... f1(x
k*
)
Z2 f2(x
1*
) f2(x
2*
) ... f2(x
h*
) ... f2(x
k*
)
zl fl(x
1*
) fl(x
2*
) ... fl(x
h*
) ... fl(x
k*
)
zk fk(x
1*
) fk(x
2*
) ... fk(x
h*
) ... fk(x
k*
)
x1*
x2*
... xh*
... xk*
    
    

21. THE IDEAL SOLUTION
Example 4
maximize z1 = –x1 + 3x2
maximize z2 = 2x1 + x2
maximize z3 = –2x1 + x2
Subject to:
–x1 + x2  1
x1 + x2  7
x1  5
x2  3
x1, x2  0
21

22. THE IDEAL SOLUTION 22

23. THE IDEAL SOLUTION
Individual Optimum values::
x1
1* = 2; x2
1* = 3; z1* = 7
x1
2* = 5; x2
2* = 2; z2* = 12
x1
3* = 0; x2
3* = 1; z3* = 1
The payoff matrix:
23
Z1 7 1 3
Z2 7 12 1
Z3 -1 -8 1
x1*
x2*
x3*

24. TECHNIQUES
•The method of global criterion
•Goal Programming
•Epsilon-constraint
•Denovo programming
24

25. The method of global criterion
MODM problem :
𝑀𝑎𝑥. 𝑓𝑗 𝑋 ; j = 1,2, . . . , 𝐾
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
Transform to the method of global criterion for the vector maximum
problem
𝑀𝑖𝑛. 𝑍 = 𝑗=1
𝐾 𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗(𝑥)
𝑝
, 𝑗 = 1,2, . . . , 𝐾,
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
The ideal solutions, 𝑓𝑗
∗
(x)
25

26. The method of global criterion
Case 1: p =1: f(x) and g(x) are linear functions
When all the objectives, i =1, 2, .., k, and all the constraint
constraint functions, i = 1, ... , m, are linear functions,
𝑀𝑖𝑛. 𝑍 = 𝑗=1
𝐾 𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗
(𝑥)
𝑝
, 𝑗 = 1,2, . . . , 𝐾,
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
becomes a linear programming problem
The simplex method for L.P. can solve the problem.

27. The method of global criterion
Case 2: p = 2: f(x) and g(x) are linear functions
When both f(x) and g(x) are linear functions
𝑀𝑖𝑛. 𝑍 = 𝑗=1
𝐾 𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗(𝑥)
𝑝
, 𝑗 = 1,2, . . . , 𝐾,
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
becomes a convex programming problem; more
precisely, a quadratic programming problem.
27

28. The method of global criterion
Case f(x) and g(x) are non-linear functions
If anyone or all of the objectives,
 𝑓𝑗 𝑋 , j = 1, 2, ••• ,k,
and 𝑔𝑖 𝑋 , i =1, 2, •••, m, are nonlinear functions,
𝑀𝑖𝑛. 𝑍 = 𝑗=1
𝐾 𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗(𝑥)
𝑝
, 𝑗 = 1,2, . . . , 𝐾,
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
is a nonlinear programming problem (for both p =1 or p =2).

29. Case p=∞:
Become :
Minmax d∞ =
𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗(𝑥)
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚
𝑀𝑖𝑛. 𝑍 = 𝑗=1
𝐾 𝑓𝑗
∗
(𝑥)−𝑓𝑗(𝑥)
𝑓𝑗
∗(𝑥)
𝑝
, 𝑗 = 1,2, . . . , 𝐾,
𝑆𝑢𝑏. 𝑡𝑜: 𝑔𝑖 𝑋 ≤ 𝑏𝑖 ; 𝑖 = 1,2, . . . , 𝑚

30. Ex: Production scheduling of the
Hardee toy company
The Hardee toy company makes two kinds of toy dolls.
Doll A is a high quality toy and Doll B is of lower quality.
The respective profits are $0.40 and $0.30 per doll. Each
Doll A requires twice as much time as a Doll B, and if all
dolls were of type B, the company could make 500 per
day. The supply of material is sufficient for only 400 dolls
per day (both A and B combined). The problem assumes
that all the dolls for type A and type B the factory can
make could be sold, and that the best customer of the
company wishes to have as many as possible of type A
doll. The manager realizes that two objectives:
 (1) the maximization of profit, and
 (2) the maximum production of Doll A, should be considered in
in scheduling the production.

31. Formulate the problem

32. Solution
x1 and x2 are the number of Doll A and that
of Doll B produced
The method of global criterion follows three
steps:
Step 1. Obtain the ideal solution;
Step 2. Construct a pay-off table;
Step 3. Obtain the preferred solution.

33. Step 1: Obtain the ideal solution

The solution is: x1 =100, x2= 200, f1 (x*) = 130

34. Step 1: Obtain the ideal solution
The optimal solution: x1 = 250, x2 = 0, f2(x*) =250

36. Step 2: Construct a pay-off table

37. Step 2: Construct a pay-off table
Row 1 gives the solution vector:
x*= (100, 300) , f1(x*) = 130; f2 (100, 300)
=100 is the value taken on by the objective
f2 when f1 reaches its maximum.
Row 2 gives the solution vector:
 x* = (250, 0) , f2 (x*) = 250; f1 (250, 0) = 100
is the value taken on by f1 when f2 reaches
its maximum.

38. Step 3. Obtain the preferred
solution
 A preferred solution is a non-dominated
solution which is a point on the segment of
straight line BC

40. Step 3. Obtain the preferred
solution
Case1. P=1
x1= 250,
x2= 0
f1= 100,
f2=250
The solution is given by
which is point C

41. Step 3. Obtain the preferred
solution
Case 2. p = 2
The solution is :
x1= 230.7;
x2= 38.6
f1=103.9,
f2=230.7
which is point D