Handbook Of Pollution Control And Waste Minimization - Chapter 18 docx

If the decision alternatives are sented by the different continuous values of a decision vector, then the decision repre-problem is called continuous.. In the final problem formulation i

Elements of Multicriteria Decision Making

Abdollah Eskandari and

deci-is, we have to take multiple criteria into account This chapter will give a briefintroduction into modeling and solving such engineering decision problems

2 MODELING DECISION PROBLEMS

Every decision-making problem is based on choice We need to have options toselect from Options, or in other words, decision alternatives, may be the types of

cars available on the market, or the different technology variants to treat a region

or to build a water treatment plant The first step in modeling decision problems

is to see which set of alternatives we may choose from This set can be finite or

infinite If the number of decision alternatives is finite, then the decision problem

is called discrete For example, if finitely many technical variants are available

for an engineering project, then the decision problem is discrete In this case wehave to prepare a list of the alternatives If the number of alternatives is infinite,then no such list is possible In such cases decision variables have to beintroduced, and the decision alternatives are identified by the different values of

the decision variables If x denotes the capacity of a wastewater treatment

plant, then the different values of this variable represent the different variantsfor the capacity The decision variables are usually collected in the compo-

nents of a decision vector If x1, x2, , x m are the decision variables, then

x= (x1,x2, , x m) is the decision vector If the decision alternatives are sented by the different continuous values of a decision vector, then the decision

repre-problem is called continuous Sometimes discrete repre-problems with a large number

of alternatives are approximated by continuous models

The second step of the modeling process is known as the feasibility check.

There might be many different reasons why some alternatives should be droppedfrom the list For example, our budget poses a constraint on which cars could bepurchased In the case of a discrete problem, the infeasible alternatives have to

be dropped from the list The result of the feasibility check is therefore a reducedlist of alternatives In the case of a continuous problem, we are unable to checkall options one by one, since there are usually too many of them or we might evenhave infinitely many possibilities In such cases feasibility has to be expressed bycertain constraints, which model the additional requirements that determine

feasibility As an example, assume that a function g o (x1, , x m) represents the

cost of a project, and we cannot spend more than an amount of Q o dollars Thenthis budgetary constraint can be formulated as the inequality

In most practical problems, several similar constraints should be satisfiedfor feasibility These constraints are either <=, or >=, or = types Without restrictinggenerality, we may assume mathematically that all constraints are given byinequalities of >= type If an inequality is defined by a <= relation, then it must havethe form

which is equivalent to the relation

which is a >= type of inequality If a constraint is given by an equality, then it can

be rewritten as a pair of >= type inequalities in the following way Assume that

Therefore in our further discussions we will always assume that the feasible

We will use the notation X for the set of the feasible decision alternatives

of continuous problems By using the above notation we may write

shows that the result of the feasibility check of continuous problems is the

formulation of the system of constraint (7) or (9), and the definition of the feasibleset (10) It is worthwhile to mention that even discrete problems can be demon-

strated in this way If there are r alternatives to select from, then X can be defined

as the discrete set X = {1,2, , r}.

The third step of modeling decision problems is the formulation of the

evaluation criteria Each criterion evaluates the goodness of the different

alterna-tives from a specific point of view If only one decision maker is present, then thecriteria reflect his or her complicated evaluation system In the case of multipledecision makers, the evaluation systems of all decision makers are combined andsummarized in the criteria In the first case a suitable alternative is determinedthat gives an acceptable trade-off among the criteria, and in the second case anappropriate compromise solution is determined that is acceptable by all partiesinvolved in the decision-making process

Some criteria are easy to quantify For example, cost, profit, volume,capacity, etc., have specific values for each decision alternative Some othercriteria might not be easily quantifiable For example, the esthetic consequences

of a forestry treatment technology can be judged only subjectively Subjectivejudgments can be made verbally, such as very good, good, fair, bad, unacceptable,

or in any other similar way These subjective measures can be then quantified.For example, on a [0, 100] scale, the above evaluation categories can be identified

by 100, 75, 50, 25, and 0, respectively In the case of a discrete problem, a payoff

matrix is hence constructed, where the rows correspond to the decision

alterna-tives, the columns correspond to the evaluation criteria, and the (i, j) element a ij

of this matrix is the evaluation of alternative i with respect to criterion j as shown

in Figure 1

After the payoff matrix is constructed, the decision problem is considered

to be mathematically well defined In the case of a continuous problem, let

ƒ1 , ƒ2, , ƒ s denote the evaluation criteria as earlier, let x be the decision vector,

and assume that the feasibility of the decision vector is given by the system ofinequalities (7) or (9) Then the decision problem is mathematically well defined

by the multicriteria optimization problem

as cleaning wastewater, providing certain products to the public, etc At the sametime, we want to avoid certain negative effects such as worsening air quality,water quality, paying too much for the project, etc After the goals and thenegative effects have been identified verbally, we have to find those quantitieswhich can be used to characterize the goodness of the different alternatives withrespect to the goals and additional consequences These quantities should measurehow the goals are satisfied if the different alternatives are selected and carriedout, how certain it is that the project will be successful under the differentalternatives, and how severe the additional negative consequences will be Thesemeasurable or at least subjectively quantifiable quantities are usually selected as

the evaluation criteria Without restricting generality, we may assume that allcriteria are maximized If smaller values are better for a criterion, then bymultiplying this criterion by (–1) we obtain a maximizing criterion If a targetvalue has to be reached in the case of a certain criterion, then the distance betweenthe actual and target values has to be minimized, or the negative of this distancehas to be maximized In the final problem formulation in both the discrete andcontinuous cases, we will always assume therefore that all criteria are to bemaximized.

In the decision science literature problem (11) or (12) and the one given inFigure 1 are called multiple–criteria decision problems, or multiobjective pro-gramming problems In the next section the solution of such problems will bedefined and examined, and in the later parts of this chapter we will give thefundamentals of the most popular solution algorithms A comprehensive summaryand additional details are presented in Ref 1

As a simple illustration, consider the problem of designing a wastewater treatmentplant Assume that there are two design alternatives which cost $2,000,000 and

m3/day, respectively It is also assumed that both versions result in the sameoutput quality Since lower cost but higher capacity is better, the first criterion has

to be transformed into a maximizing criterion by multiplying it by (–1) Theresulting payoff matrix is given in Table 1 Notice that this problem is a discretemultiobjective programming problem

As a simple continuous example, consider the following problem Assumethat a combination of three alternative technologies can be used in a wastewater

treatment plant Let x1,x2 denote the proportion (in percent) of the application oftechnology variants 1 and 2; then 1 − x1− x2 is the proportion of the third

technology Therefore we have two decision variables, x1 and x2, and they have

to satisfy the conditions x1 >= 0, x2 >= 0, and x1 + x2 <= 1 The feasible set isillustrated in Figure 2

T ABLE 1 Payoff Matrix for Wastewater

Treatment Plan

Alternatives/Criteria –Cost (–$)

Capacity (m 3 /day)

Assume that there are two major pollutants that have to be removed fromthe wastewater, and the three technology variants remove 3, 2, 1 mg/m3, respec-tively, of the first kind of the pollutant, and 2, 3, 1 mg/m3 of the second kind ofpollutant Then the total removed amount of the two pollutants can be obtained

as [3x1+ 2x2+ 1(1 − x1− x2 )]V and [2x1+ 3x2+ 1(1 − x1− x2)]V, where V is the

total amount of treated wastewater Maximizing those objectives is equivalent to

maximizing 2x1+ x2 and x1+ 2x2 Therefore we have the following tive programming problem:

x1 + x2 <= 1Notice that (13) is a continuous problem with two decision variables and twoobjective functions

In many decision problems we are also interested in knowing the set of all

feasible objective values For each x∈ X, the vector

ƒ (x) = (ƒ1(x1, , x m ), ƒ2(x1, , x m ), , ƒ s (x1, , x m))

is called the objective vector at x, and the set of all feasible objective vectors,

is called the objective space If there are s objectives, then H ⊂ R s

In our discrete problem we have two objectives, therefore the objectivespace is two-dimensional Since there are only two alternatives, the objectivespace has only two points: (–2,000,000, 1,500) and (–2,500,000, 2,000) In every

discrete problem, every point of the objective space corresponds to a decisionalternative, but different alternatives may have the same point, if the objective

points when the first objective is given on the horizontal axis and the second

objective is identified by the vertical axis Point A1 shows the simultaneous objective function values for the first alternative, and point A2 shows the same

values for the second alternative

The question that arises now is the choice among the two alternatives,

which is the same as selecting the “better” point among A1 and A2 If cost is our only concern, then point A1 (that is, alternative 1) must be our choice If capacity

is our only concern, then point A2 with alternative 2 is our choice The dilemma

we face here is the conflict between the two objectives, since higher capacity can

be obtained by selecting A2 but then the cost becomes worse If both objectives

are important to us, then neither of the alternatives is better than the other andhence either alternative seems to be a reasonable choice With decision scienceterminology we might say that no alternative is dominated by another alternative

and therefore both alternatives are nondominated In the economic literature, nondominated alternatives are called Pareto optimal Let us define these terms mathematically In a discrete problem, let p and q be two alternatives, and let

ƒ (p) and ƒ(q) denote the corresponding objective vectors We say that alternatives

p and q are equivalent if ƒ (p) = ƒ(q) Here the equality of vectors is defined by

component-wise equality That is, two alternatives are equivalent to each other ifthey result in the same values in all objective functions Similarly, we say that

alternative p dominates alternative q, if ƒ (p)>=ƒ (q) and there is a strict inequality

in at least one of the objectives Here the notation ƒ (p)>=ƒ (q) means that each

Objective 2 A2

2,000 A1

1,500

-2,500,000 -2,000,000

F IGURE 3 Objective space for a discrete problem.

component of vector ƒ (p) is greater than or equal to the corresponding component

of ƒ (q) In our example, ƒ(1) = (–2,000,000, 1,500) and ƒ(2) = (–2,500,000, 2,000)

and none of the alternatives dominates the other one If no other alternatives

dominate an alternative p, then it is called nondominated In our example,

therefore, both alternatives 1 and 2 are nondominated

In the case of continuous problems, the objective space can be constructed

by representing the set H = {u|u = (u1, , u s )|u j = ƒ j (x) with some x ∈ X for

j = 1,2, , s} in the s-dimensional space In our continuous example

2u2 − u1

3 >= 0 2u1 − u2

3 >= 0and

2u2 − u1

3 + 2u13− u2 >= 1These inequalities can be simplified as

u2 >= u1

2 u2 <= 2u1 u1 + u2 <= 3 (14)The set of feasible payoff vectors (u1,u2) satisfying these conditions are illustrated

in Figure 4. The vertices A, B, C of the triangle can be obtained as the intercepts

of the lines u2= u1/2 and u2= 2u1, u2= 2u1 and u1+ u2= 3, u2= u1/2 and

u1+ u2= 3, respectively Simple calculation shows that A = (0, 0), B = (1, 2) and

C = (2, 1) We will next show that the nondominated objective vectors form the

linear segment connecting points B and C including the endpoints.

All points between A and B including point A are dominated by the objective vector B, since both coordinates of B are larger Similarly, all points

between A and C are dominated by C Let now u be a point inside the triangle The continuation of the linear segment connecting points A and u intercepts the

linear segment between B and C in a point D Since the slope of the line

dominates u Hence u is dominated If D is any point of the linear segment

connecting B and C (including the endpoints), then it is impossible to increase

any objective value without worsening the value of the other objective function

Thus, the points of the linear segment between B and C are the nondominated

objective vectors

Any nondominated point and the corresponding decision alternative is areasonable choice The nondominated decision alternatives can be obtained in thefollowing way:

where H o is the set of nondominated objective vectors In the case of our example,

the decision vector x= (x1,x2) is nondominated if and only if

1 <= 2x1 + x2 <= 2 and (2x1 + x2 ) + (x1 + 2x2) = 3These relations can be rewritten as

In the case of our discrete example both alternatives are nondominated, and

in the case of the continuous example there are infinitely many nondominatedalternatives Notice that the different nondominated solutions give differentobjective function values, contrary to single-objective optimization problems,where all alternative optimal solutions give the same optimal objective functionvalue In other words, in single-objective optimization problems the differentoptimal solutions are equivalent to each other, but in the presence of multiple

F IGURE 4 Objective space for a continuous problem.

objectives this is not true anymore If we consider two different nondominatedobjective vectors, then the first is better in one objective and the other is better inanother objective Therefore we cannot say which objective vector is better thanthe other In decision problems, however, we have to choose one alternative thatcould be considered as the best one overall In order to answer the question ofwhether an alternative is better than another alternative overall, we need addi-tional preference or trade-off information from the decision makers and/or fromthe stakeholders Depending on the nature of this additional information, differentsolution algorithms are available In the next sections the most popular methodswill be outlined In order to guarantee the existence of the solution, we willalways assume that the continuous problems have continuous bounded objectivefunctions and that the feasible set is closed.

important, the second is the second most important, and so on, and the sth

criterion is the least important The solution algorithm is then the following.Identify first the alternatives which give the highest value of the first objectivefunction If a unique alternative is found, then it is the solution of the problem.Otherwise, delete all nonoptimal alternatives, and optimize the second mostimportant objective function on the set of the remaining alternatives If a uniquealternatives is found, it is accepted as the solution of the problem; otherwise,delete again all nonoptimal alternatives, and optimize the third objective on theremaining feasible set Continue the process until a unique alternative is found oruntil the least important objective is being optimized It can be proved that thismethod always results in nondominated solutions, and if there are multiplesolutions, then they are equivalent to each other in the sense that they result inthe same objective function values

The above procedure can be summarized mathematically in the following

way Let X denote the (discrete or continuous) feasible alternative set and let

ƒ1 ,ƒ2, , ƒ s denote the objectives

If the optimal solution is unique, then it is accepted as the solution of the problem.Otherwise, go to the next step.

X j = {x|x ∈ X j−1, ƒ j−1(x) = optimal value},

and go back to optimization

In the case of a discrete problem, X has finitely many elements, X = 1,2, ,r, and the objective function values are represented by the a ij elements of the payoffmatrix (see Figure 1) The optimization step in this case can be reformulated as

max

i ∈X j

a ij

and the adjustment step is

X j = {i|i ∈ X j−1, a i,j−1 = optimal}

In the case of continuous problems the optimization step is always a

single-objective optimization problem The single-objective function is the jth single-objective of the

original problem and at each optimization a new constraint has to be added tothe constraints of the previous optimization step This new constraint requires the

optimality of the previous objective function Hence the jth optimization problem

can be written as

Maximize ƒ j (x1, , x m)subject to g l (x1, , x m) >= Q l

.

ƒ j−1(x1, , x m) = optimal valuewhere we assumed that the original problem is given as it was presented in theform (11) It is worthwhile to mention that if all objectives and original constraintsare linear, then each optimization step requires the solution of a linear program-

ming problem If j is increased by 1, then an additional constraint is added to the

constraints of the previous optimization problem This makes the solution ofthe new linear programming problem easy based on the basic optimal solution

of the previous optimization problem, since in creating a new basic solution wehave to eliminate some matrix elements only from the last constraint

Consider first the discrete problem given earlier in Table 1 If the firstobjective is more important than the second, then the alternative giving the largestnegative cost has to be selected In this case the first alternative is the choice Ifcapacity is more important than cost, then the second objective is maximized,

leading to the choice of the second alternative In both cases a unique optimalalternative is found.

Consider next the continuous problem (13) Assume first that the firstobjective function is the most important; then it has to be first maximized Thus

we have to solve the single-objective optimization problem

x1 + x2 <= 1The application of the bi-variable simplex method, or elementary graphicalapproach, gives the unique optimal solution

x1∗ = 1 and x2∗ = 0Because of uniqueness, the procedure terminates Assume next that the secondobjective is more important Then we solve the single-objective optimizationproblem

x1 + x2 <= 1There is again a unique solution,

x1∗∗ = 0 and x2 ∗∗ = 1and the algorithm terminates

Notice that in the case of the discrete problem both nondominated tives could be obtained as solutions, but in the case of the continuous problemonly the two endpoints of the nondominated alternative set could be obtained Themethod selection has excluded all other points between (1,0) and (0,1) (see Figure2) from the possibility to be obtained as a solution In order to overcome thisproblem, the method can be modified in the following way After maximizing themost important objective function, in later steps we do not require that all moreimportant objectives are at their maximal levels The optimality constraints arerelaxed by requiring that in optimizing any later objective, the more importantobjective functions are only in the neighborhood of their optimal values Forexample, we may require that their values are at least 90% of the optimal values

alterna-In the case of our continuous problem (13), assume again that the first objective

is the more important and in maximizing the second objective only at least (90%)

of the optimal value of the first objective is required From the earlier discussions

we know that the maximal value of the first objective is 2, which occurs at

x1∗= 1 and x2 ∗= 0 Therefore in the second step we have the following problem:

Maximize x1 + 2x2

Subject to x1 >= 0,x2 >= 0

2x1 + x2 >= 1.8The feasible set of this problem is shown in Figure 5, and the optimal solution is

given as x1∗= 0.8 and x2 ∗= 0.2 It can be shown that in the case of a uniquesolution, it is always nondominated; and if multiple solutions are present, thenthere is at least one nondominated alternative among the optimal solutions

In the application of the sequential optimization method we may have a uniqueoptimal alternative in an early step This was in fact the case in both the discreteand continuous examples Since the procedure terminates, no later objective istaken into account, therefore they might have very unfavorable values In order

to avoid very “bad” values for some less important objectives, we might use the

ε-constraint method We need to know which objective is the most important, and

in addition, we have to be supplied with minimal acceptable levels for all otherobjectives Without losing generality, we may assume that the first objective is

the most important and ƒ2∗, ƒ3∗, , ƒ s∗ are the minimal acceptable levels for theother objectives The method simply maximizes the first objective functionsubject to the original feasibility constraint and the additional conditions that thevalues of all other objectives must be at least as high as the minimal acceptablelevels It can be proven that in the case of a unique optimal alternative, thesolution is nondominated; and in the case of multiple optimal solutions, there is

at least one nondominated alternative among the optimal solutions

In the case of the discrete problem given in Table 1, this method can beformulated as

Maximize

where

X∗ = {i|i ∈ {1,2, , r},a ij >= ƒ j∗ for j = 2,3, , s} (22)

In the case of the continuous problem (11), the ε-constraint method is the solution

of the single-objective optimization problem:

Maximize ƒ1(x1, , x m)Subject to g1(x1, , x m) >= Q1

Consider first the discrete problem of Table 1, and assume that the cost isthe more important objective Therefore the additional constraint has to be given

to the capacity If the minimal capacity requirement is 1,500 or less, then bothalternatives are feasible and the less expensive alternative is selected, which isthe first alternative If the minimal capacity requirement is more than 1,500, thenthis additional constraint excludes the first alternative and hence the only andtherefore the optimal alternative is the second one

In the case of the continuous example (13), assume that the first objective

is more important than the other, and we require that the value of the secondobjective is at least 1.5 (which is 75% of its maximal value shown in Figure 4).Then problem (23) can be written now as

Subject to x1 >= 0,x2 >= 0

2x1 + x2 >= 1.5

and the optimal solution is x1∗= 0.5 and x2 ∗= 0.5 Notice that this solution gives

equal proportions to the first two technologies without using the third technology

at all

In the case of the weighting method, it is assumed that the 100% overall

importance of the objectives is divided up among the different objectives That

is, a positive weight c j is assigned to each objective j ( j = 1,2, , s) such that

c1+ c2+ + c s= 1 The problem of multiobjective optimization is then reduced

to optimize the weighted average of the objectives, where weights are selected as

presented in Figure 1), we have to compute first the weighted averages,

g n (x1, , x m) >= Q n

is solved, and the optimal solution is accepted as the solution of the multiobjectiveoptimization problem If all of the weights are possible, then the optimal solution

of problem (26) is always nondominated

Consider first the discrete problem of Table 1 and assume that the objectives

(cost and capacity) are equally important: c1= c2= 0.5 Since

Assume next that the two objectives are also equally important in the case

of the continuous problem (13) Then the optimization problem (26) can berewritten as

Maximize 0.5(2x1 + x2) + 0.5(x1 + 2x2) = 1.5x1 + 1.5x2

x1 + x2 <= 1

It is easy to see that there are infinitely many optimal solutions: x1∗∈ [0,1] is

arbitrary, and x2∗= 1 − x1 ∗ Assume next that the first objective is twice as

import-ant as the second one, then we many chose c1=2⁄3 and c2=1⁄3 Then problem(27) has to be modified is the following way:

Maximize 2⁄3(2x1 + x2) + 1⁄3(x1 + 2x2) = 5⁄3x1 + 4⁄3x2

x1 + x2 <= 1

solution is x1∗= 1 and x2 ∗= 0 It can be proven that this is the solution if c1> c2,

and the solution is x1∗= 0 and x2 ∗= 1 if c1< c2 The case of c1= c2 has beenexamined earlier

The weighting method is easy to apply If all objectives and constraints ofthe original problem are linear, then problem (26) becomes a linear programmingproblem which can be solved by the simplex method Notice in addition that byselecting a different set of weights, only the objective function changes Thereforethe solution procedure becomes very easy if an optimal basic solution is knownfor a different weight set The weighting method has a large disadvantage,however, since the solution might change if the objective functions are given in

different units As an illustration, consider again the discrete problem given inTable 1 and assume now that the cost is given in $1,000 and the capacity ispresented in dm3/day Then the payoff matrix has to be changed accordingly Thenew matrix is given in Table 2.

If equal weights are assumed again, then

maximizing the objectives If ƒ j∗ and ƒ j∗ denote the lowest and highest values for

objective j, then the jth normalized objective is obtained by simple transformation: