Visualizing Exact and Approximated 3D Empirical Attainment ...

Research ArticleVisualizing Exact and Approximated 3D EmpiricalAttainment Functions

Tea Tušar and Bogdan FilipiI

Department of Intelligent Systems Jozef Stefan Institute and Jozef Stefan International Postgraduate School Jamova cesta 391000 Ljubljana Slovenia

Correspondence should be addressed to Tea Tusar teatusarijssi

Received 4 June 2014 Accepted 10 August 2014 Published 17 September 2014

Academic Editor Chunlin Chen

Copyright copy 2014 T Tusar and B Filipic This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Most real-world engineering optimization problems are inherently multiobjective for example searching for trade-off solutionsof high quality and low cost As no single optimal solution exists for such problems multiobjective optimizers provide sets ofoptimal (or near-optimal) trade-off solutions to choose fromThe empirical attainment function (EAF) is a powerful tool that canbe used to analyze and compare the performance of these optimizers While the visualization of EAFs is rather straightforwardin two objectives the three-objective case presents a great challenge as we need to visualize a large number of 3D cuboids Thispaper addresses the visualization of exact as well as approximated 3D EAF values and differences in these values provided bytwo competing multiobjective optimizers We show that the exact EAFs can be visualized using slicing and maximum intensityprojection (MIP) while the approximated EAFs can be visualized using slicing MIP and direct volume rendering In addition thepaper demonstrates the use of the proposed visualization techniques on a steel casting optimization problem

1 Introduction

When solving engineering optimization problems it is usuallynot enough to find the best solution with regard to singlemeasure of quality but other objectives such as other mea-sures of quality robustness of the solution or its cost can beimportant too While traditional approaches were used tocombine all these heterogeneous objectives into a single oneand solve the resulting singleobjective optimization problemin the last two decades a lot of research has been directedinto developing algorithms able to tackle these problems intheir true multiobjective form Most notably multiobjectiveevolutionary algorithms (MOEAs) have proven to be effectivein solving such problems [1 2]

A MOEA provides a set of trade-off solutions where nosolution from the set is better than any other in all objectivesThis is called an approximation set and MOEAs typicallyreturn a different approximation set in every run If the per-formance of aMOEAneeds to be analyzed or compared to theperformance of another algorithm several differentmeasurescan be adopted [3] Nevertheless visualization of approxima-tion sets can give an important insight into the properties of

the algorithms (or the problem at hand) and is often usedin this regard While simple scatter plots can visualize 2Dand 3D approximation sets (approximation sets of higherdimensions that require more sophisticated methods to bevisualized will not be discussed in this papermdashthe interestedreader is referred to [4] for a comprehensive review of suchmethods) scatter plots of multiple approximation sets canbecome too cluttered and loose their interpretation potentialThe empirical attainment function (EAF) which assigns toeach vector in the objective space a value signifying howoften the vector was attained by the given approximation setscan be used instead [5] Visualization of 2D EAFs entailsplotting of rectangles in different colors (or shades of gray)corresponding to EAF values or differences in EAF values andis rather straightforward to do

This paper focuses on the visualization of EAFs in the 3Dcase which is more demanding than the 2D case since a largenumber of cuboids or rather unions of cuboids need to be firstcomputed and then visualized Building on some previouswork [6 7] we present how the visualization of exact EAFscuboids can be done using slicing (showing the 2D images

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 569346 18 pageshttpdxdoiorg1011552014569346

2 Mathematical Problems in Engineering

Start

ComputeEAF values

Inspecting oneor comparing

two algorithms

Compute EAFdifferences

Is exactnessimportant


Computethe voxels

Computethe cuboids

Computethe voxels

Computethe cuboids

Visualize voxelsusing slicingandor DVR

Visualize cuboidsusing slicing

Visualize voxelsusing slicing

MIP andor DVR

Visualize cuboidsusing slicingandor MIP

End

One

Two

No

Yes

No

Yes

Figure 1 Flowchart of EAF visualization depending on the preferences

obtained by slicing the cuboids at different angles) and max-imum intensity projection (MIP) [8] However if exactnessis not crucial the 3D objective space can be approximatedusing a grid of voxels that is values in a 3D grid The paperintroduces the idea of visualizing approximated EAFs usingslicing MIP and direct volume rendering (DVR) [9] Aflowchart explaining the connections among these methodsis shown in Figure 1 Finally we demonstrate the use of theproposed methods on steel casting optimization problem

The paper is organized as follows Section 2 details thebackground and related work while Section 3 introduces thebenchmark approximation sets that are used in the rest ofthe paper Visualization of exact and approximated EAFs ispresented in Sections 4 and 5 respectively and summarizedin Section 6The steel casting use case is depicted in Section 7The paper concludes with final remarks in Section 8

2 Background and Related Work

In addition to recalling some relations and concepts oftenused in multiobjective optimization this section presentsformal definitions of some new terms related to the EAFsuch as attainment anchors and opposites It also reviews theexisting methods for visualization of EAFs and concludeswith the presentation of slicing MIP and DVR

21 Multiobjective Optimization Concepts The multiobjec-tive optimization problem consists of finding the optimumof a function

f 119883 997888rarr 119865

f (1199091 119909

119899) 997891997888rarr (119891

1(1199091 119909

119899) 119891

119898(1199091 119909

119899))

(1)

where 119883 is an 119899-dimensional decision space and 119865 is an 119898-dimensional objective space (119898 ge 2) Each solution x isin 119883 iscalled a decision vector while the corresponding element z =

f(x) isin 119865 is an objective vector Without loss of generality weassume that 119865 sube R119898 and all objectives 119891

119894 119883 rarr R are to be

minimizedAs this paper deals with visualization in the objective

space which can be viewed rather independently from thedecision space the following definitions are confined to theobjective space

Definition 1 (Pareto dominance of vectors) The objectivevector z119860 = (119911

119860

1 119911

119860

119898) isin 119865 dominates the objective vector

z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z119860 ≺ z119861 if

119911119860

119894le 119911119861

119894forall119894 isin 1 119898

there exists a 119895 isin 1 119898 for which 119911119860

119895lt 119911119861

119895

(2)

Mathematical Problems in Engineering 3

Definition 2 (weak Pareto dominance of vectors) The objec-tive vector z119860 = (119911

119860

1 119911

119860

119898) isin 119865weakly dominates the objec-

tive vector z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z119860 ⪯ z119861 if

119911119860

119894le 119911119861

119894forall119894 isin 1 119898 (3)

Definition 3 (strict Pareto dominance of vectors) The objec-tive vector z119860 = (119911

119860

1 119911

119860

119898) isin 119865 strictly dominates the

objective vector z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z1198607z119861 if

119911119860

119894lt 119911119861

119894forall119894 isin 1 119898 (4)

Definition 4 (incomparability of vectors) The objective vec-tors z119860 = (119911

119860

1 119911

119860

119898) isin 119865 and z119861 = (119911

119861

1 119911

119861

119898) isin 119865 are

incomparable that is z119860 z119861 if

z119860 997410 z119861 z119861 997410 z119860 (5)

Definition 5 (approximation set) A set of objective vectors119885 sube 119865 is called an approximation set if z119860 z119861 for any twoobjective vectors z119860 z119861 isin 119885

Definition 6 (weak Pareto dominance of approximation sets)The approximation set 119885

119860sube 119865 weakly dominates the

approximation set 119885119861 sube 119865 that is 119885119860 ⪯ 119885119861 if every z119861 isin 119885

119861

is weakly dominated by at least one z119860 isin 119885119860

Definition 7 (ideal point) The ideal point zlowast = (119911lowast

1 119911

lowast

119898)

represents the minimum possible value in each objective(typically it cannot be achieved)

119911lowast

119894= min 119891

119894(x) x isin 119883 forall119894 isin 1 119898 (6)

Definition 8 (minimal element minimal set) Let 119885 sube 119865 bea set of objective vectors An objective vector m isin 119885 is aminimal element of 119885 if

forallz isin 119885 z ⪯ m 997904rArr z = m (7)

All minimal elements of 119885 constitute the minimal set of 119885denoted in this paper by 119885

min

Maximal elements and the maximal set 119885max can be

defined dually

Definition 9 (Pareto front) The set of Pareto optimal objec-tive vectors known as the Pareto front can be formally definedas the minimal set of f(119883)

119875f = f(119883)min

(8)

Definition 10 (Edgeworth-Pareto hull) Let 119885 sube 119865 be anapproximation set The Edgeworth-Pareto hull (EPH) of 119885includes all objective vectors weakly dominated by 119885

119885EPH = z isin 119865 exists z1015840 isin 119885 so that z1015840 ⪯ z (9)

22 Empirical Attainment Function Based on themultiobjec-tive concept of goal-attainment [5] we say that an objectivevector is attained when it is weakly dominated by an approx-imation set that is when it is contained in the setrsquos EPHThe surface delimiting the attained vectors from the ones thathave not been attained by an approximation set is called theattainment surface and its minimal elements are called theattainment anchors

Definition 11 (set boundary) The boundary of a set 119885 sube 119865 isthe intersection of the closure of 119885 119885 with the closure of itscomplement (119865 119885)

120597119885 = 119885 cap (119865 119885) (10)

Definition 12 (attainment surface attainment anchors) Let119885 sube 119865 be an approximation set The attainment surface of119885 is the boundary of the EPH of 119885

119878119885= 120597119885EPH (11)

Theminimal elements of 119878119885are called attainment anchors and

are equal to the original approximation set 119885

119860119885= 119878

min119885

= 119885 (12)

Now imagine that an optimization algorithm is run 119903

times producing 119903 approximation sets Then each objectivevector can be attained between 0 and 119903 times

Definition 13 (empirical attainment function) For algorithmA the empirical attainment function of the objective vectorz isin 119885 sube 119865 represents the frequency of attaining z by itsapproximation sets 119885119860

1 119885

119860

119903

120572A119903(z) = 120572 (119885

A1 119885

A119903 z) =

1

119903

119903

sum

119894=1

I (119885A119894⪯ z) (13)

where I is the indicator function defined as

I (119887) =

1 if 119887 is true0 otherwise

(14)

and ⪯ is the weak Pareto dominance relation between sets

This means that for algorithm 119860 with approximationsets 119885

119860

1 119885

119860

119903an EAF value from the set of frequencies

0 1119903 2119903 1 can be assigned to every vector in theobjective space Of course in practice the EAF cannot becomputed for every objective vector and only attainmentsurfaces with a constant EAF value (and their correspondingattainment anchors) are of interest They are called 119896-attainment surfaces (also summary attainment surfaces) inorder to distinguish them from the ldquoordinaryrdquo attainmentsurfaces of single approximation sets Vectors in the 119896-attainment surface are the tightest objective vectors that havebeen attained in at least 119896 of the runs


0

3

6

9

12

15

0 3 6 9 12 15

Z1

Z1Z1

Z1

Z2

Z2

Z2

Z2

Z4

Z4

Z4

Z4

Z3

Z3

Z3

Z3

f1

f2

S25S50S75

S100Attainment anchors

Figure 2 Four approximation sets in a two-objective optimizationproblem their summary attainment surfaces and all attainmentanchors Colors denote areas with equal EAF values (darker colorscorrespond to larger values)

Definition 14 (summary attainment surfaces) Let 1198851 119885

119903

be 119903 approximation sets of algorithm119860 and 120572119860

119903(z) its EAF Let

the 119905119903-superlevel set of 120572119860119903be defined as

119871+

119905119903(120572

A119903) = z isin 119865 120572

A119903(z) ge 119905119903 forall119905 isin 1 119903 (15)

Then the summary (or 119905119903-) attainment surface 119878119905119903

is equal tothe boundary of 119871+

119905119903(120572

A119903)

119878119905119903

= 120597119871+

119905119903(120572

A119903) forall119905 isin 1 119903 (16)

We will use 119860119905119903

to denote the attainment anchors of 119878119905119903

119860119905119903

= 119878min119905119903

forall119905 isin 1 119903 (17)

Note how the definition of an attainment surface coin-cides with the definition of a summary attainment surfaceif 119903 = 1 In general the number of all attainment anchors(including duplicates) 119899

119860= sum119903

119905=1|119860119905119903

| is much larger thanthe number of vectors contained in the approximation sets119899119885

= sum119903

119905=1|119885119905| (in the context of the EAF computation [10]

vectors from the approximation sets and attainment anchorswere called input and output points resp) It has been shownin [10] that 119899

119860isin 119874(119903

2119899119885) for three-objective optimization

problems

We illustrate these concepts in the two-objective caseFigure 2 shows an example of four approximation sets1198851 119885

4(each consisted of four objective vectors indicated

with crosses) and their summary attainment surfaces Dotsdenote all attainment anchors and colors are used to empha-size areas with equal EAF values (darker colors correspond tohigher values)

0

3

6

9

12

15

0 3 6 9 12 15

ZA1

ZA1

ZA1

ZA1ZB1

ZB1

ZB2

ZB1

ZB1

ZB2

ZB2

ZB2

ZA2

ZA2

ZA2

ZA2

f1

f2

S25 S75S50 S100

Figure 3 Four approximation sets in a two-objective optimizationproblem two produced by algorithm 119860 and two by algorithm 119861Colored areas show where algorithm 119860 outperformed algorithm 119861

(green hues) and algorithm 119861 outperformed algorithm119860 (red hues)

If two algorithms 119860 and 119861 are run 119903 times each they canbe compared by computing and visualizing their EAF values120572119860

119903and 120572

119861

119903separately However there exists a straightforward

way to directly compare the algorithms by computing EAFdifferences that is differences in their EAF values [11]

Definition 15 (EAF differences) Assume that algorithms 119860

and 119861 are run 119903 times each The EAF differences betweenalgorithms119860 and119861 are defined for each objective vector z isin 119885

as

120575119860minus119861

119903(z) = 120572

119860

119903(z) minus 120572

119861

119903(z) (18)

Note that EAF differences need to be computed forall attainment anchors of the combined 2119903 approximationsets Defined in this way the differences can adopt valuesfrom the set minus1 minus(119903 minus 1)119903 0 1119903 1 Positive EAFdifferences correspond to areas in the objective space wherethe algorithm119860 outperforms the algorithm 119861 while negativeEAF differences denote areas in the objective space wherethe algorithm 119861 outperforms the algorithm 119860 Naturallywhere the differences are zero both algorithms attain the areaequally well

Let us focus on the example from Figure 3 where algo-rithms119860 and119861 solving a two-objective optimization problemproduced two approximation sets each The lines show theoverall summary attainment surfaces The colored areasemphasize areas of the objective space where algorithm 119860

outperformed algorithm119861 (green hues) and areaswhere algo-rithm119861 outperformed algorithm119860 (red hues)We can readilynotice that algorithm 119860 is more successful in optimizing thefirst objective while algorithm119861 attains better lower values ofthe second objective This small example nevertheless showsthat visualization of 2D EAF differences can be very helpfulwhen analyzing and comparing the results of two algorithms


0

3

6

9

12

15

0 3 6 9 12 15f1

f2

S25S50A25

A50

Opposite of A50

r1

r2

Figure 4 The area between the 2D 25- and 50-attainmentsurfaces from the example in Figure 2 Vectors from the opposite ofthe attainment anchors of the 50-attainment surface are denotedby squares

We can safely assume that we are interested in a limitedportion of the objective space for example the one enclosedby reference vectors r1 and r2 where r17r2 Then areaswith equal EAF valuesdifferences correspond to unions ofrectangles in 2D and unions of cuboids in 3D [7] If weallow overlapping then each rectangle or cuboid betweenthe 119905119903- and (119905 + 1)119903-attainment surfaces where 119905 isin

1 119903minus1 can be defined by two vertices one from the 119905119903-attainment surface and the other from the (119905+1)119903-attainmentsurface (Figure 4 shows 2D areas between 25- and 50-attainment surfaces) While the lower vertices correspondto the attainment anchors of the 119905119903-attainment surface theupper vertices are not the attainment anchors of the (119905+1)119903-attainment surface but their opposite The opposite can bedefined for an arbitrary approximation set as follows

Definition 16 (approximation set opposite) Let referencevectors r1 and r2 where r17r2 define the boundaries of theobserved objective space

119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (19)

Let 119885 sube 119877 be an approximation set enclosed in this spacewith the attainment surface 119878

119885 The opposite 119874

119877

119885of the

approximation set 119885 within the observed objective space 119877

is the maximal set of 119878119885cap 119877

119874119877

119885= (119878119885cap 119877)

max (20)

Finding the opposites of attainment anchors in 2D israther trivial [7] The computation of opposites (and the cor-responding cuboids) for the 3D case is more demanding andis presented in detail in Section 41

Finally let us recall the formal definition of the hypervol-ume indicator [12] which is often used to measure the qualityof approximation sets

Definition 17 (hypervolume indicator) Let again referencevectors r1 and r2 where r17r2 define the boundaries of theobserved objective space

119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (21)

The hypervolume indicator 119868119867of an approximation set 119885 sube 119877

can be formulated via the empirical attainment function of119885as

119868119867(119885) = int

zisin119877120572 (119885 z) 119889z (22)

23 Visualization of EAFs 2D EAFs are most often visual-ized through best median and worst summary attainmentsurfaces corresponding to 1119903 sim50 and 100-attainmentsurfaces respectively The first attempt to visualize 3D sum-mary attainment surfaces is documented in [13] At the timean efficient algorithm for computing 3D attainment anchorswas not yet available so separate summary attainment sur-faces were approximated by discretizing the objective spaceusing a grid Assume the 119905119903-attainment surface needs tobe visualized The algorithm examines the objective spacefor each objective separately First it finds the intersectionsbetween all approximation sets and the lines stemming fromthe 2D grid of the remaining two objectives Then theintersections on each of these lines are counted and if theyamount to 119905 a marker is drawn at the corresponding heightCombining these markers by considering all objectives yieldsinformative visualization of the chosen 3D summary attain-ment surface Years later an efficient algorithm for computingattainment anchors and their EAF values for the 3D case wasfinally provided [10 14] The attainment function tools from[14] were used to compute all EAF values in this paper

While [10] offers an estimation of the number of attain-ment anchors also for the general 119898D case where 119898 gt 3no algorithm for computing 4D EAFs exists yet As a conse-quence we do not discuss visualization of EAFs beyond 3D

Recently the idea of visualizing EAF differences in addi-tion to EAF values was introduced in [11] When visuallycomparing algorithms 119860 and 119861 four plots can be producedeach corresponding to one of the values of 120572

119860 120572119861 120575119860minus119861and 120575

119861minus119860 All attainment anchors are plotted using gray-shaded dots where the shade corresponds to the value of thechosen function (120572119860 120572119861 120575119860minus119861 or 120575119861minus119860) On top of this thebest median andworst overall summary attainment surfacesare drawn Examples from [11 15] show that this kind ofvisualization can be very useful in exploratory analysis

Building on this work we have performed some initialresearch on visualizing 3D EAFs using slicing and maximumintensity projection in [6 7]These twomethods followed bydirect volume rendering are introduced next

24 Slicing This is a simple method where spatial data issliced using a plane Then the intersection of the plane withthe data is visualized in 2D In its most general form slicingcan be applied to any kind of spatial data that is the data doesnot have to be represented by voxels


While the plane used in slicing is most often axis-alignedthe nature of multiobjective optimization problems yields theneed for a plane that always contains one axis and the originof the objective space (which usually corresponds to theideal point) and intersects all summary attainment surfacesTherefore (similarly to the prosection plane used in [4 16])the slicing plane cuts the objective space at an angle 120593 isin

(0∘ 90∘)

While not used for visualizing attainment surfaces theinteractive decision maps (IDMs) are related to slicing usingaxis-aligned planes [17] They visualize a number of axis-aligned sampling surfaces of the EPH and use different colorsto represent each slice As the surface of 119885EPH is exactly theattainment surface of 119885 it should be rather straightforwardto apply this method for visualizing attainment surfaces

25 Maximum Intensity Projection Maximum intensity pro-jection (MIP) is a volume rendering method for spatial datarepresented by voxels The method inspects voxels in direc-tion of parallel rays traced from the viewpoint to the projec-tion plane and takes the maximum value encountered in thevoxels along a ray as the projection value for the ray

The method originally called maximum activity projec-tion (MAP) [8] was proposed for 3D image rendering innuclear medicine and tested in tomographic studies It waslater accepted not only in medical imaging but also in scien-tific data visualization in general

The advantages of MIP are its simplicity and efficiencyand the ability of achieving high contrast which arises fromthe fact that maximum voxel values are projected On theother hand as a limitation the resulting projections lack thesense of depth of the original data (see [18 19] for attemptsto remedy this issue) Moreover the viewer cannot distin-guish between left and right and front and back As animprovement animations are usually provided consisting ofa sequence of MIP renderings at slightly different viewpointswhich results in the illusion of rotation

26 Direct Volume Rendering Direct volume rendering(DVR) [20] is another volume rendering method based onthe voxel representation of data that is often used for visual-ization in medicine and science It can generate very appeal-ing results by employing advanced shading techniques andcan achieve real-time interactive rendering DVR is able toproduce images of volumetric datawithout the need to explic-itly extract geometry or surfaces (hence the name direct)

First each voxel value is assigned color and opacity (theRGBA value) by means of a chosen transfer function Thetransfer function can be a simple ramp a piecewise linearfunction or an arbitrary table (the problem of specifying agood transfer function is a research area on its own [21 22])Then one of the following techniques is used to project theRGBA values to the screen ray casting (for only ray castingis used in this paper any mention of DVR can be understoodto refer to ray-casting DVR from now on) splatting shear-warp factorization or texture-based volume rendering Inray casting [9] viewing rays are traced from the viewpointthrough the volume accumulating color and opacity values

002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical

Figure 5 The linear and spherical approximation sets used in allvisualization examples in Sections 4 and 5

at each sample position along the ray The final RGB valuesshown in the image are computed from the accumulatedRGBA values using the volume rendering integral [23]

3 Benchmark Approximation Sets

In order to show the outcome of different visualizationmethods we need some approximation sets to visualize Thetwo sets of approximation sets used in this paper are notproducts of optimization algorithms but rather benchmarkapproximation sets with known properties used for visualiza-tion purposes [7] They are shown in Figure 5

The first set consists of five linear approximation sets witha uniform distribution of vectors that satisfy the constraint

3

sum

119894=1

119911119894= 119888119871 (23)

In order to simulate the behavior of a stochastic algorithmthe values 119888

119871follow the normal distribution 119873(1 005) The

second set contains five spherical approximation sets with anonuniform distribution of vectors where only few vectorsare located in themiddle of the approximation set whilemostof them are near its corners The vectors from the sphericalapproximation sets satisfy the constraint

3

sum

119894=1

1199112

119894= 1198882

119878 (24)

where 119888119878sim 119873(075 005) Each individual approximation set

contains 100 objective vectorsThe values of 119888

119871and 119888

119878were chosen so that the sets

are intertwined in one region the linear approximation setsdominate the spherical ones while in others the sphericalsets dominate the linear ones This assures that there willalways be visible EAF differences between the two sets ofapproximation sets


Input Sets of attainment anchors 1198601 119860

119903and 119886

reference vector r2 for which 1198601⪯ sdot sdot sdot ⪯ 119860

119903

and z7 r2 for all z isin 119860119894 119894 = 1 119903

Output The set of cuboids 119862119860119903+1

larr r2r1 larr zlowastforeach adjacent pair of sets 119860

119894and 119860

119894+1 119894 isin 1 119903

do119874 larr opposite(119860

119894+1 r1 r2)

foreach z isin 119860119894do

1198741015840larr o isin 119874 z7 o

foreach o isin 1198741015840 do

119862 larr 119862cup cuboid(z o value(z))end

endend

Algorithm 1 Computing the cuboids

We will use 1205721198715and 120572

119878

5to denote EAF values of the linear

and spherical approximation sets respectively In addition120575119871minus119878

5and 120575

119878minus119871

5will denote differences between those values

4 Visualizing Exact EAFs

Exact EAFs can be represented as unions of cuboids withvalues corresponding to either EAF values of one algorithmor EAF differences between two algorithmsThis section willfirst show the procedure for computing the cuboids from thegiven attainment surfaces and then present visualization ofthese cuboids using slicing and MIP

41 Computing the Cuboids The algorithm for computingEAFs [10] takes as input a series of approximation sets1198851 119885

119903and returns as output a series of sets containing

attainment anchors1198601 119860

119903that define the corresponding

summary attainment surfaces and for which the followingholds 119860

119894⪯ 119860119894+1

for 119894 isin 1 2 119903 minus 1 From these sets ofattainment anchors and a reference vector r2 that limits theobserved objective space the cuboids can be computed Wepropose a very simple algorithm for this purpose that uses thefirst set of attainment anchors and the opposite of the secondset of attainment anchors to compute overlapping cuboids(see Algorithm 1)

First a set containing only the given reference vector r2 isadded at the end of the input set of attainment anchors so thatany cuboids ranging from the last set of attainment anchorsto r2 can also be computed Then the reference vector r1needed for computation of the opposites is set to be equalto the ideal point Next the main loop iterates through alladjacent pairs of sets 119860

119894and 119860

119894+1and computes the opposite

of119860119894+1

usingAlgorithm 2 For each attainment anchor z isin 119860119894

all dominated vectors from the opposite are stored in1198741015840 Each

pair (z o) where o isin 1198741015840 represents the two vertices required

to define the cuboid In addition the cuboid is given a valueeither the EAF value or the EAF difference in z

Input Approximation set 119860 and reference vectors r1 andr2 for which r1 ⪯ z7 r2 for all z isin 119860

119894 119894 = 1 119903

Output The opposite 119874119874 larr r2foreach z isin 119860 do

1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)

for 119894 larr 1 to 3 doif not redundant (o119899119894 r1 119874) then

119874 larr 119874 cup o119899119894 end

endend

end

Algorithm 2 Computing the opposite of a 3D approximation set

Note that there might be multiple cuboids stemmingfrom the same attainment anchor (see the analog exampleof multiple rectangles in Figure 4)mdashwe call this a union ofcuboids This is not a problem since all such overlappingcuboids have the same value (if they did not have the samevalue another attainment anchor would have already splitthe cuboid) The result of Algorithm 1 is a set of cuboids (orunions of cuboids)

Next we show with the help of Figure 6 how to find theopposite of a 3D approximation set Imagine a single cuboiddefined by the reference vectors r1 and r2 The opposite isinitialized to r2 Every time a vector from 119885 is ldquocut intordquothe existing cuboid the vectors strictly dominated by it areremoved from the opposite and new vectors are added to theopposite Assume we have already performed this step forvectors z1 and z2 (see Figure 6(a)) and vector z3 is next in line(see Figure 6(b)) First we delete from the opposite all vectorsthat are strictly dominated by the current vector z (thismeansthat in our example we delete vector o4 but not o5) Nextfor every deleted vector o we create three new vectors in thefollowing way

o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)

Finally each of these vectors is added to the opposite if itis not redundant This means that it has to contribute to theopposite (it cannot be coplanar with r1 or collinear with anyof the vectors from the existing opposite) In our example weadd to the opposite vectors o6 and o7 but not o8 = (2 65 5)

(not pictured in the figure) as it is collinear with o5 and thusredundantWhen these steps have been taken for every vectorfrom 119885 the resulting set 119874 represents the opposite of 119885 SeeAlgorithm 2 for the algorithmic notation of the describedprocedure


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2

(a) Opposite of z1 z2

0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3

(b) Opposite of z1 z2 z3

Figure 6 Two steps in the construction of the opposite of the approximation set 119885

As we cannot visualize the entire (infinite) objectivespace Algorithms 1 and 2 use two reference vectors r1 and r2to limit it In order to preserve all information they need tobe set in such a way that r1 ⪯ z7r2 for all attainment anchorsz While the ideal point is the most sensible choice for r1 wecan choose any objective vector dominated by all attainmentanchors for r2

The presented approach for computing the cuboids isvery simple and easy to implement but comes at highcomputational cost Computing the cuboids between two setsof 119899 attainment anchors has the worst-case computationalcomplexity of119874(119899

3) which means that all cuboids between 119903

pairs of attainment surfaces can be computed in119874(1199031198993) time

However in practice the loop foreach o isin 1198741015840 do in Algo-

rithm 2 is executed only a few times for each z isin 119860 meaningthat in practice the computational complexity is quadratic inthe attainment anchors set size (the check for redundancy stilltakes linear time)

Another possible approach would be to deal only withnonoverlapping cuboids which is somewhat similar to theproblem of computing the hypervolume indicator in 3DIn future work we wish to investigate if a more efficientalgorithm for our problem could be found by for examplemodifying the space-sweep algorithm from [24] whichcomputes the hypervolume indicator in the 3D case with thecomputational complexity of only 119874(119899 log 119899) If this wouldbe possible the total cost of computing the cuboids wouldbe 119874(119903119899 log 119899) However time complexity is not the onlyimportant aspect in the computation of cuboids We wouldalso like to study which method would yield a lower numberof cuboids as this considerably affects the tractability of theirvisualization

42 Visualizing Cuboids Using Slicing In this paper theslicing plane is aligned with one of the three axes (assume fornow that this is 119891

3) includes the origin r1 and cuts through

the underlying plane (11989111198912) at some angle 120593 isin (0

∘ 90∘) In

this way each slice always captures all attainment surfaces

f1

f2

f3

r1

oo

998400

zz 998400

120593

Figure 7 A cuboid sliced using the plane aligned with 1198913that slices

the11989111198912plane under angleThe cuboid vertices z and o are projected

to 3D rectangle vertexes z1015840and o1015840 respectively

Slicing through the objective space containing a largenumber of cuboids means that only those cuboids thatintersect the slicing plane are visualized When a cuboid issliced the result is a rectangle in 3D (see the example inFigure 7) Two steps are needed to compute this rectangle in2D First we need to calculate the projected cuboid verticesz1015840 and o1015840 yielding the rectangle in 3D and second we need torotate the vertices by angle minus120593 to get z10158401015840 and o10158401015840 Dependingon the angles

120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)

this is done in the following way

z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘

Figure 8 Slices of the exact 3D EAF values and differences for the benchmark approximation sets under two angles Darker colors representhigher EAF valuesdifferences

o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)

When slicing through a union of cuboids stemming fromthe same vertex z some of the projected vertices o10158401015840 can beredundant In order to decrease the number of total rectanglesto plot it is therefore sensible to keep only nonredundantvertices o10158401015840 (those that are nondominated with regard to theinverted weak dominance relation ⪰)

The cuboids can be visualized by slicing them at differentangles thus showing different parts of the objective spaceWepresent visualization using slicing of our benchmark approx-imation sets at angles 120593 = 5

∘ and 120593 = 45∘ The EAF values 120572119871

5

and 120572119878

5are shown separately while the EAF differences 120575119871minus119878

5

and 120575119878minus119871

5are presented on the same plot (see Figure 8)

From the plots of EAF values it is easy to distinguishlinear sets (green hues) from the spherical ones (red hues)as the shape of the sets is readily visible We can also see thatthe two best spherical approximation sets are better than theremaining three by a solidmarginWhile these plots provide alot of information by themselves the comparison between thesets is best visualized using EAF differencesThey nicely showregions in the objective space where linear sets outperformthe spherical ones and vice versa

Note that in order to fully explore the entire objectivespace slicing planes at several different angles should beperformed Also one might want to consider slicing planesaligned to the axis 119891

1or 1198912 too However as our benchmark

approximation sets are rather symmetrical there is no needto do this here

43 Visualizing Cuboids Using MIP As explained in Sec-tion 25 MIP shows only the maximum value encounteredon rays from the viewpoint to the projection plane This


(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715

Figure 9 MIP of exact 3D EAF differences between the benchmark approximation sets

means that it is not sensible to use this method for visualizingEAF values as most of the objective space has the maximumvalue and the resulting visualization would not be very infor-mative However MIP seems a good choice for visualizingEAF differences where the highest values are of most inter-est as they represent the largest differences between the algo-rithms

MIP could be used to visualize cuboids of different valuesby sorting the cuboids so that those with higher values wouldbe put on top of those with lower values However in generalthe 3D plotting tools capable of visualizing cuboids renderthem in a sequence that tries to maintain some notion ofdepth (cuboids near the viewpoint are shown in front of thecuboids further away) and do not allow for custom sorting ofthe cuboids according to their values Therefore this sortingcan only be done after the viewpoint has been set and thecuboids are already projected to 2D At that time we canchoose to visualize them in the order of ascending valueswhich (although a bit cumbersome) effectively achieves theMIP visualization of the cuboids

Another difficulty in usingMIP for visualizing exact EAFdifferences is that we need to visualize a large number ofcuboids which can be challenging to doWhilemany of themare completely covered by cuboids with larger values andcould therefore be spared without altering the final imageremoval of such cuboids is not trivial as it depends also onthe position of the viewpoint

Nevertheless we present visualization of exact EAF dif-ferences usingMIP in Figure 9The differences 120575119871minus119878

5and 120575119878minus119871

5

need to be visualized separately in order to avoid overlappingof cuboids The MIP visualizations are very informative wecan easily see which parts of the objective space are betterattained by the linear approximation sets and which by thespherical approximation sets As is typical with MIP whilewe are able to ldquosee through the cuboidsrdquo we loose the senseof depth Although it is inevitable to lose some informationwhen projecting 3D structures onto 2D this can be amendedby combining two visualization techniques MIP and slicingTogether they give a good idea of the 3D ldquocloudrdquo of cuboids

5 Visualizing Approximated EAFs

If we wish to visually compare the outcome of two algorithmsand are not interested in the details of such a visualization we

can use approximated instead of exact EAFs Approximationmeans that the objective space is discretized into a grid ofvoxels (similarly to what was proposed in [13]) This sectionfirst presents how such discretization is performed and thenillustrates visualization of approximated EAFs using slicingMIP and DVR

51 Discretization into Voxels A voxel is a single point on aregularly spaced 3D grid of V

1times V2times V3voxels Recall that

based on the reference vectors r1 and r2 the 3D observedobjective space 119877 is defined as

119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)

If 119877 is a cube V1= V2= V3 otherwise some care must be

taken to ensure that the grid of voxels is truly regular Either119877must be normalized prior to approximation or the numberof voxels in each dimension V

119894must be set to be proportional

to 1199032

119894minus 1199031

119894for all 119894 isin 1 2 3 Note that a voxel represents only

a single point not a volume How this missing information isreconstructed depends on the visualization method

From the observed objective space 119877 the grid of V1times V2times

V3voxels is constructed so that the voxel at grid position

(1198961 1198962 1198963) has the following coordinates

(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3

Computing voxel values when the cuboids have alreadybeen computed is rather straightforward Iterating over allcuboids the voxels that are ldquocoveredrdquo by the cuboid adopt itsvalue If we are not interested in visualizing the exact EAFs(and therefore have not computed the cuboids) we can com-pute the voxel values directly from the set of approximationsets The value of each voxel is set to either EAF value or EAFdifference by counting how many approximation sets weaklydominate it


0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1

(b) Using 643 voxels

0

02

04

06

08

1

0 02 04 06 08 1

(c) Using 1283 voxels

0

02

04

06

08

1

0 02 04 06 08 1

(d) Using 2563 voxels

Figure 10 Slices of exact and approximated 120575119871minus119878

5and 120575

119878minus119871

5at angle 120593 = 5∘ showing the approximation error

52 Visualizing Voxels Using Slicing For visualization usingslicing we assume that the whole volume of the voxel has thesame value as its centerTherefore slicing of voxels is done inthe same way as slicing of cuboids (see Section 42) In orderto avoid showing plots that are very similar to those presentedin Figure 8 Figure 10 shows only the slices of 120575119871minus119878

5and 120575119878minus119871

5at

angle 120593 = 5∘ produced using different discretizations

We can see that by refining the discretization we getresults that are increasingly similar to the exact EAFs As webelieve that for the purpose of this study the discretizationinto 1283 voxels suffices we will use this discretization in theremainder of the paper

53 Visualizing Voxels Using MIP Visualizing voxels usingMIP is trivial with a tool supporting such visualization asthere are no additional parameters to set We use a volumerendering engine called Voreen [25 26] to produce the MIPimages from Figure 11 and all DVR images

Figure 11 shows the MIP for 120575119871minus119878

5and 120575

119878minus119871

5and we can

see that although these are approximations of the imagespresented in Figure 9 the results are quite similar

54 Visualizing Voxels Using DVR Visualization usingDVR is a bit trickier as we need to define the transfer func-tion that assigns color and opacity to each voxel valueIn our case voxel values are discrete as they equal either theEAF values or the EAF differences which makes this taskeasier

Figures 12 and 13 show visualization using DVR ofthe EAF differences between the two benchmark approxi-mation sets 120575

119871minus119878

5and 120575

119878minus119871

5 respectively The first five plots

in both figures show the voxels with a single value from15 55 The transfer functions used to obtain theseplots are simple piecewise linear functions that set the opacityof voxels of the desired value to 1 and the opacity of allother voxels to 0 In this way we are able to visualize each


(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715

Figure 11 MIP of approximated 3D EAF differences between the benchmark approximation sets

(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785

= 55 (f) DVR of 120575119871minus1198785isin 15 55

Figure 12 DVR of approximated 3D EAF differences between the linear and spherical benchmark approximation sets 120575119871minus1198785

of the values separately which is reflected in the nice andinformative plots in Figures 12 and 13

The biggest advantage of DVR is that by setting thetransfer function ldquothe right wayrdquo it is possible to see inside thevisualized volume This was attempted on the plots shown inFigures 12(f) and 13(f) which show completely opaque voxelswith value 55 and almost transparent voxels with value 15

(both plots are rotated to give a nicer view of inner voxels)In contrast to MIP DVR can be used for visualizing EAF

values too Figure 14 shows an example of such a visualizationfor the spherical approximation sets where only the values120575119878

5isin 15 45 are shown (opacity is set to 03 for these two

values and to 0 for all other values)

6 Discussion

Let us summarize the properties of all presented visualiza-tion methods (see also Table 1) Slicing is a rather simplemethod that enables visualization of exact EAF values anddifferences Its biggest advantage is its accuracy it enables adetailed analysis of the approximation sets in the same waythat can be done for 2D EAFs Its biggest drawback is thatit is able to visualize only one slice at a time Generally theapproximation sets are not as symmetrical as in our bench-mark case therefore multiple slices are needed to sufficientlyinspect the EAFs Also note that the meaning of the angle 120593changes if the observed objective space has different ranges


Table 1 Summarized properties of the presented visualization methods

Slicing MIP DVR

Exact EAF values

+ Enables detailed analysis of theapproximation setsminus Visualizes only one slice at atime

minus Not sensible to use sinceusually a large portion of theobjective space has themaximum EAF value

minus Not possible to use

Exact EAFdifferences

+ Enables detailed analysis of theapproximation sets+ Capable of simultaneouslyvisualizing positive and negativedifferences without overlappingminus Visualizes only one slice at atime

+ No parameters to set+ Shows all values in a singleimageminus No sense of depthminus Feasible only for a limitednumber of cuboids (impractical)


Approximated EAFvalues

minus No need to use it on theapproximation as it works wellfor exact EAFs


+ Nice visualizations+ Enables looking through thevolume and preserves the senseof depthminus Requires definition of thetransfer function

Approximated EAFdifferences


+ No parameters to set+ Shows all values in a singleimageminus No sense of depth


(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715


Figure 13 DVR of approximated 3D EAF differences between the spherical and linear benchmark approximation sets 120575119878minus1198715

(if 119877 is not a cube it is not cut exactly in half by the plane atangle 120593 = 45

∘) While slicing can be used also for visualizingapproximated EAFs there is no need to do so as it works wellfor exact EAFs

It is not reasonable to use MIP for visualizing EAF valuessince usually a large portion of the objective space has themaximumEAF value and such plots are not very informativeWhile MIP can be used for visualizing exact EAF differences


Figure 14 DVR of approximated 3D EAF values of the sphericalbenchmark approximation sets for 120572119878

5isin 15 45

it works much better on the approximated case for which itwas conceived Visualizing 3D cuboids is rather impractical(especially for a very large number of cuboids) and requiressorting of cuboids with regard to their value in order to pro-duceMIP imageHoweverMIP can easily be used to visualizeapproximated EAF differences as it has no parameters to setIts biggest advantage is that it combines all values in a singleimage giving precedence to largest differences (which arethe most important when visualizing EAF differences) whileits the biggest disadvantage is the lost sense of depth

Finally DVR can be used to visualize approximated EAFsIt produces nice and informative visualizations of both theEAF values and differences With an insightful setting of thetransfer function it is even possible to ldquolook throughrdquo thecloud of cuboids The need to define the transfer functionis at the same time the biggest disadvantage of DVR as itmight be demanding for a user not familiar with the methodNevertheless it is much easier to find the right transferfunction for EAFs than for example for some medical dataas EAFs take only discrete values

7 Steel Casting Use Case

This section shows how the described visualization methodscan be used on a real-world optimization problem with threeobjectives from a previous study [27]

71 The Steel Casting Problem Continuous casting of steel isa complex metallurgical process where molten steel is cooledand shaped into semimanufactures of desired dimensionsCooling is done using water in the primary and secondarycooling subsystems Primary cooling is performed in themold while secondary cooling comprehends wreath coolingat the exist from the mold and spray cooling of the strand

The goal is to set the parameters of this process (castingspeedmold outlet coolant temperature andwreath and spraycoolant flows (see Table 2)) in such a way that the qualityof the cast steel is as high as possible The quality of steel isdefined by the following three objectives distance fromdesired metallurgical length (the length of the liquid core inthe strand) distance from desired shell thickness (thicknessof the solid shell at mold exit) and distance from desired

Table 2 Steel casting process parameters

Parameter Lowerbound

Upperbound Disc step

Casting speed (mmin) 15 20 001Mold outlet coolant temp (∘C) 33 35 1Wreath coolant flow (m3h) 10 40 5Spray coolant flow (m3h) 25 65 5

Table 3 Variables defining the optimization objectives

Variable Lowerbound

Upperbound

Desiredvalue

Metallurgical length (m) 10 11 10Shell thickness (mm) 11 15 13Surface temperature (∘C) 1115 1130 11225

surface temperature at a predefined point in the strandTable 3 details the variables defining these optimizationobjectivesTheir bounds and desired values were determinedby domain experts

If a parameter setting yields values outside the boundaryconstraints presented in Table 3 it is deemed infeasible Theobjectives of the feasible solutions are computed by

119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)

where 119910119896is the value of the observed variable and 119910

lowast

119896is its

desired valueAs real-world experimentation with parameter settings is

expensive and time-consuming and could also be dangerouswe have simulated it using a numerical model of steel castingbased on a meshless technique for diffusive heat transport[28] One simulation of the steel casting process takesapproximately 2 minutes on a standard desktop computer

Two instances of this optimization problem were studieddiscrete and continuousThe discrete problem instancewas setby domain experts as presented in Table 2 in such a way thatall possible solutions (9639 in total) could be exploredWhilethe discrete problem instance enables the rough explorationof the objective space the continuous problem instancewhereany value within the variable bounds could be chosen is theone we wish to solve

72 Results of Optimization Algorithms Thediscrete probleminstance was solved using the exhaustive search (ES) algo-rithm These solutions form the Pareto front of the discreteproblem instance and we wanted to see whether an algorithmtackling the continuous problem instance could come close tothis Pareto front

We chose the algorithm DEMO (differential evolutionfor multiobjective optimization) [29] which is a MOEAalgorithm that uses differential evolution [30] to explore thedecision space for the continuous problem instance DEMOwas run five times each time exploring 3200 solutions Moredetailed information on the employed experimental setup canbe found in [27]


001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES

Distance from desired

metallurgical length (m)Distance from desired

shell thickness (mm)

Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)

Figure 15 The best solutions found by ES on the discrete probleminstance and by DEMO on the continuous problem instance (all fiveruns shown)

Themajority of the explored solutionswere infeasible Forexample out of 9639 solutions found by ES only 1242 werefeasible and of those only 72 were mutually nondominatedDEMO was solving an extended problem and thereforefound more nondominated solutions on average 645 perrun (although their number varied significantly over differentruns)

First we visualize all final (feasible and nondominated)solutions found by both algorithms in Figure 15 We cansee that ES found two distinct subsets of solutions A moredetailed analysis of results revealed that they correspondto two of the three mold outlet coolant temperatures (theremaining one always produces infeasible solutions) SinceDEMO was not bound by this discretization it was able tofind solutions also around these subsets Both algorithmsfound a few solutions with low distances from desired shellthickness that are rather ldquodetachedrdquo from the rest Theyactually lie on a larger disconnected region of the Pareto frontof which just this minor part is feasible

This visualization is able to show that DEMO is able toreach the Pareto front of the discrete problem and is thereforea good choice for solving this problem However there aretwo aspectswe are interested in that are hard to infer from thisvisualization alone (they could of course be computed fromthe solutions) First we wish to see whether different runs ofDEMO produce similar results Visualizing all five approxi-mation sets together with differentmarkers or visualizing oneapproximation set at a time does not provide a good meansfor comparison Second although the approximation setsfound by DEMO look linear they are in fact slightly convexmaking it very hard to see (even by rotating the objectivespace) whether results by ES are in fact dominated by thoseby DEMO We will try to find the answers to these questionsusing the proposed visualization methods

In order to use the proposed visualization methods tovisualize the results of ES and DEMO we need to addresstwo issues The first issue is the uneven ranges of the attain-ment surfaces which are important when computing voxels

(a) DVR of 120575DEMOminusES5

(b) DVR of 120575ESminusDEMO5

Figure 16 DVR of approximated 3D EAF differences between thetwo algorithms for all values in 15 55

As we can infer from Table 3 the observed objective space isequal to [0 1] times [0 2] times [0 75] Moreover the feasible resultsfound by the two algorithms actually lie in an even smallercuboid contained in [0 06] times [05 2] times [0 75] If the numberof voxels was proportional to the objective ranges the firsttwo objectives would be discretized too roughly Thereforewe choose to normalize the objective vectors with regard to[0 06] times [05 2] times [0 75] The second issue is the unevennumber of runs of both algorithms (ES was run once) Toleave the meaning of the EAFs differences intact we copy theresults by ES to get five runs in total This seems a reasonableapproach as the deterministic ES would actually produce fiveequal results if it was run five times

Now visualization methods as described in Sections 4and 5 can be used on these results Figure 16 presents theDVR of approximated EAF differences between the twoalgorithms 120575DEMOminusES

5and 120575

ESminusDEMO5

which gives us a generalidea of their performance However this does not answerour two questions To check the repeatability of DEMO weneed to inspect the difference between the best and worst


(a) DVR of 120572DEMO5isin 15 45

(b) DVR of 120572DEMO5isin 15 55

Figure 17 DVR of approximated 3D EAF values of DEM

summary attainment surfaces The narrow layer between120572DEMO5

= 15 and 120572DEMO5

= 45 as well as 120572DEMO5

= 15 and120572DEMO5

= 55 (see Figure 17) suggests that although DEMOrsquosapproximation sets were of considerably different cardinalitythey attain a similar portion of the objective space This isfurther confirmed by amore detailed inspection using slicingat angles 120593 = 25

∘ and 45∘ (see Figure 18) which shows that

only a small border of the attained objective space is attainedless than five times (denoted by light green hues)

Finally MIP can be used to check whether DEMO everfinds better solutions than those found by ES If this was notthe case each solution found by ES would have exactly oneunion of cuboids (albeit small) for which 120575

ESminusDEMO5

= 55As accuracy is important in this case (small cuboids can beomitted if the discretization into voxels is not fine enough)we useMIP on exact EAF differences Figure 19 clearly shows

0

02

04

06

08

1

0 02 04 06 08 1

(a) Slice of 120572DEMO5

at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121

(b) Slice of 120572DEMO5

at 120593 = 45∘

Figure 18 Slices of exact 3D EAF values of DEMO at two angles

that although 120575ESminusDEMO5

= 55 for some solutions this doesnot hold for all of them meaning that DEMO was actuallyable to find solutions that dominate those by ES This was ofcourse possible only because DEMOwas solving an extendedinstance of the problem

8 Conclusions

When comparing the results of multiobjective optimizationalgorithms it is important to be able to visualize them asthis can provide new information regarding the algorithmsor the given problem If the algorithms are stochastic theEAF can be used to describe how well the algorithms attainthe objective space with their multiple approximation setsVisualization of EAFs is rather straightforward in 2D butpresents a challenge in 3D as multiple cuboids need to bevisualized


Figure 19MIP of exact 3DEAF differences between ES andDEMO120575ESminusDEMO5

This paper presented how these cuboids can be computedand visualized using slicing and MIP If accuracy of visual-ization is not crucial the EAF values and differences can beapproximated by discretizing the objective space into a gridof voxels In this way visualization can be performed usingslicingMIP andDVRWehave shownhow all thesemethodsperform on two sets of benchmark approximation sets anddiscussed their advantages and disadvantages

In addition we demonstrated the use of slicing MIPand DVR on a real-world optimization problem solved byexhaustive search andMOEA algorithmWe have shown thatthese powerful visualization methods are able to give a newinsight regarding the performance of the algorithms that can-not be otherwise seen using solely ldquostandardrdquo visualization ofapproximation sets

In the future we wish to find amore efficient way to com-pute the cuboids either by improving the provided algorithmor by adjusting existing efficient algorithms for hypervolumecalculation to suit our needs

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors are grateful to Robert Vertnik for making thecasting process simulator available and Miha Mlakar forproviding the results of DEMO and ES The work presentedin this paper was carried out as part of research Program P2-0209 and research Projects L2-3651 and J2-4120 all funded bythe Slovenian Research Agency

References

[1] K DebMulti-Objective Optimization Using Evolution ary Algo-rithms John Wiley amp Sons Chichester UK 2001

[2] K Deb ldquoAdvances in evolutionary multi-objective optimiza-tionrdquo in Search Based Software Engineering G Fraser and J Tde Souza Eds vol 7515 of Lecture Notes in Computer Sciencepp 1ndash26 Springer New York NY USA 2012

[3] E Zitzler L Thiele M Laumanns C M Fonseca and V G DaFonseca ldquoPerformance assessment of multiobjective optimiz-ers an analysis and reviewrdquo IEEE Transactions on EvolutionaryComputation vol 7 no 2 pp 117ndash132 2003

[4] T Tusar and B Filipic ldquoVisualization of Pareto front approxi-mations in evolutionary multiobjective optimization a criticalreview and the prosection methodrdquo IEEE Transactions onEvolutionary Computation 2014

[5] V D G da Fonseca C M Fonseca and A O Hall ldquoInferentialperformance assessment of stochastic optimisers and the attain-ment functionrdquo inProceedings of the 1st International Conferenceon Evolutionary Multi-Criterion Optimization (EMO rsquo01) vol1993 ofLectureNotes in Computer Science pp 213ndash225 Springer2001

[6] T Tusar and B Filipic ldquoAn approach to visualizing the 3Dempirical attainment functionrdquo in Proceeding of the 15th AnnualConference on Genetic and Evolutionary Computation (GECCO13) pp 1367ndash1372 New York NY USA July 2013

[7] T Tusar and B Filipic ldquoInitial experiments in visualizationof empirical attainment function differences using maximumintensity projectionrdquo in Proceedings of the 16th Annual Con-ference Companion on Genetic and Evolutionary Computation(GECCO rsquo14) pp 1099ndash1106 Vancouver Canada 2014

[8] JWWallis T RMiller C A Lerner and E C Kleerup ldquoThree-dimensional display in nuclear medicinerdquo IEEE Transactions onMedical Imaging vol 8 no 4 pp 297ndash303 1989

[9] M Levoy ldquoDisplay of surfaces from volume datardquo IEEE Com-puter Graphics and Applications vol 8 no 3 pp 29ndash37 1988

[10] C M Fonseca A P Guerreiro M Lopez-Ibanez and LPaquete ldquoOn the computation of the empirical attainmentfunctionrdquo in Proceedings of the 6th International Conference onEvolutionary Multi-Criterion Optimization (EMO rsquo11) vol 6576of Lecture Notes in Computer Science pp 106ndash120 SpringerNew York NY USA 2011

[11] M Lopez-Ibanez L Paquete andT Stiitzle ldquoExploratory analy-sis of stochastic local search algorithms in biobjective optimiza-tionrdquo in Experimental Methods for the Analysis of OptimizationAlgorithms T Bartz-BeielsteinM Chiarandini L Paquete andM Preuss Eds pp 209ndash222 Springer 2010

[12] E Zitzler D Brockhoff and L Thiele ldquoThe hypervolume indi-cator revisited on the design of Pareto-compliant indicatorsvia weighted integrationrdquo in Proceedings of the 4th InternationalConference on EvolutionaryMulti-CriterionOptimization (EMOrsquo07) vol 4403 of Lecture Notes in Computer Science pp 862ndash876 Springer 2001

[13] J Knowles ldquoA summary-attainment-surface plotting methodfor visualizing the performance of stochastic multiobjectiveoptimizersrdquo in Proceedings of the 5th International Conferenceon Intelligent Systems Design and Applications (ISDA rsquo05) pp552ndash557 September 2005

[14] ldquoAttainment function toolsrdquo httpedendeiucptsimcmfonsecsoftwarehtml

[15] M Lopez-Ibanez and T Stutzle ldquoAn experimental analysis ofdesign choices of multi-objective ant colony optimization algo-rithmsrdquo Swarm Intelligence vol 6 no 3 pp 207ndash232 2012


[16] T Tusar and B Filipic ldquoVisualizing 4D approximation sets ofmultiobjective optimizers with prosectionsrdquo in Proceedings ofthe 13th Annual Genetic and Evolutionary Computation Confer-ence (GECCO rsquo11) pp 737ndash744 July 2011

[17] A V Lotov and K Miettinen ldquoVisualizing the Pareto frontierrdquoinMultiobjective Optimization J Branke K Deb K Miettinenand R Slowinski Eds vol 5252 of Lecture Notes in ComputerScience pp 213ndash243 Springer 2008

[18] W Heidrich MMcCool and J Stevens ldquoInteractive maximumprojection volume renderingrdquo in Proceedings of the IEEE Con-ference on Visualization pp 11ndash18 1995

[19] J Diaz and P Vazquez ldquoDepth-enhanced maximum intensityprojectionrdquo in Proceedings of the 8th IEEEEG InternationalConference on Volume Graphics (VG rsquo10) pp 93ndash100 2010

[20] K Engel M Hadwiger J M Kniss C Rezk-Salama and DWeiskopf Real-time Volume Graphics A K Peters NatickMass USA 2006

[21] C P Botha and F H Post ldquoNew technique for transfer functionspecification in direct volume rendering using real-time visualfeedbackrdquo in Proceeding of the Medical Imaging VisualizationImage-Guided Procedures and Display pp 349ndash356 San DiegoCalif USA February 2002

[22] P Sereda A Vilanova and F A Gerritsen ldquoAutomating transferfunction design for volume rendering using hierarchical clus-tering of material boundariesrdquo in Proceedings of the 8th JointEurographicsIEEE VGTC Conference on Visualization (EURO-VIS rsquo06) pp 243ndash250 2006

[23] N Max ldquoOptical models for direct volume renderingrdquo IEEETransactions on Visualization and Computer Graphics vol 1 no2 pp 99ndash108 1995

[24] N Beume C M Fonseca M Lopez-Ibanez L Paquete and JVahrenhold ldquoOn the complexity of computing the hypervol-ume indicatorrdquo IEEE Transactions on Evolutionary Computa-tion vol 13 no 5 pp 1075ndash1082 2009

[25] J Meyer-Spradow T Ropinski J Mensmann and K HinrichsldquoVoreen a rapid-prototyping environment for ray-casting-based volume visualizationsrdquo IEEE Computer Graphics andApplications vol 29 no 6 pp 6ndash13 2009

[26] ldquoVoreen volume rendering enginerdquo httpvoreenuni-muen-sterde

[27] M Mlakar T Tusar and B Filipic ldquoDiscrete vs continuousmultiobjective optimization of continuous casting of steelrdquoin Proceeding of the 14th International Conference on Geneticand Evolutionary Computation (GECCO 12) pp 587ndash590 NewYork NY US July 2012

[28] R Vertnik and B Sarler ldquoSimulation of continuous casting ofsteel by a meshless techniquerdquo International Journal of CastMetals Research vol 22 no 1ndash4 pp 311ndash313 2009

[29] T Robic and B Filipic ldquoDEMO differential evolution formultiobjective optimizationrdquo in Proceedings of the 3rd Interna-tional Conference on EvolutionaryMulti-CriterionOptimizationvol 3410 of Lecture Notes in Computer Science pp 520ndash533Springer 2005

[30] K V Price and R Storn ldquoDifferential evolutionmdasha simpleevolution strategy for fast optimizationrdquoDr Dobbrsquos Journal vol22 no 4 pp 18ndash24 1997

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Start

ComputeEAF values

Inspecting oneor comparing

two algorithms

Compute EAFdifferences



Computethe voxels

Computethe cuboids

Computethe voxels

Computethe cuboids

Visualize voxelsusing slicingandor DVR

Visualize cuboidsusing slicing

Visualize voxelsusing slicing

MIP andor DVR

Visualize cuboidsusing slicingandor MIP

End

One

Two

No

Yes

No

Yes

Figure 1 Flowchart of EAF visualization depending on the preferences

obtained by slicing the cuboids at different angles) and max-imum intensity projection (MIP) [8] However if exactnessis not crucial the 3D objective space can be approximatedusing a grid of voxels that is values in a 3D grid The paperintroduces the idea of visualizing approximated EAFs usingslicing MIP and direct volume rendering (DVR) [9] Aflowchart explaining the connections among these methodsis shown in Figure 1 Finally we demonstrate the use of theproposed methods on steel casting optimization problem

The paper is organized as follows Section 2 details thebackground and related work while Section 3 introduces thebenchmark approximation sets that are used in the rest ofthe paper Visualization of exact and approximated EAFs ispresented in Sections 4 and 5 respectively and summarizedin Section 6The steel casting use case is depicted in Section 7The paper concludes with final remarks in Section 8

2 Background and Related Work

In addition to recalling some relations and concepts oftenused in multiobjective optimization this section presentsformal definitions of some new terms related to the EAFsuch as attainment anchors and opposites It also reviews theexisting methods for visualization of EAFs and concludeswith the presentation of slicing MIP and DVR

21 Multiobjective Optimization Concepts The multiobjec-tive optimization problem consists of finding the optimumof a function

f 119883 997888rarr 119865

f (1199091 119909

119899) 997891997888rarr (119891

1(1199091 119909

119899) 119891

119898(1199091 119909

119899))

(1)

where 119883 is an 119899-dimensional decision space and 119865 is an 119898-dimensional objective space (119898 ge 2) Each solution x isin 119883 iscalled a decision vector while the corresponding element z =

f(x) isin 119865 is an objective vector Without loss of generality weassume that 119865 sube R119898 and all objectives 119891

119894 119883 rarr R are to be

minimizedAs this paper deals with visualization in the objective

space which can be viewed rather independently from thedecision space the following definitions are confined to theobjective space

Definition 1 (Pareto dominance of vectors) The objectivevector z119860 = (119911

119860

1 119911

119860

119898) isin 119865 dominates the objective vector

z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z119860 ≺ z119861 if

119911119860

119894le 119911119861

119894forall119894 isin 1 119898

there exists a 119895 isin 1 119898 for which 119911119860

119895lt 119911119861

119895

(2)



119860

1 119911

119860


tive vector z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z119860 ⪯ z119861 if

119911119860

119894le 119911119861

119894forall119894 isin 1 119898 (3)


119860

1 119911

119860



1 119911

119861

119898) isin 119865 that is z1198607z119861 if

119911119860

119894lt 119911119861

119894forall119894 isin 1 119898 (4)


119860

1 119911

119860

119898) isin 119865 and z119861 = (119911

119861

1 119911

119861

119898) isin 119865 are


z119860 997410 z119861 z119861 997410 z119860 (5)





119861



1 119911

lowast

119898)


119911lowast

119894= min 119891





min


defined dually


119875f = f(119883)min

(8)





120597119885 = 119885 cap (119865 119885) (10)


119878119885= 120597119885EPH (11)



119860119885= 119878

min119885

= 119885 (12)




1 119885

119860

119903

120572A119903(z) = 120572 (119885

A1 119885

A119903 z) =

1

119903

119903

sum

119894=1

I (119885A119894⪯ z) (13)


I (119887) =


(14)



119860

1 119885

119860




0

3

6

9

12

15

0 3 6 9 12 15

Z1

Z1Z1

Z1

Z2

Z2

Z2

Z2

Z4

Z4

Z4

Z4

Z3

Z3

Z3

Z3

f1

f2

S25S50S75




119903




119871+

119905119903(120572

A119903) = z isin 119865 120572

A119903(z) ge 119905119903 forall119905 isin 1 119903 (15)



119905119903(120572

A119903)

119878119905119903

= 120597119871+

119905119903(120572

A119903) forall119905 isin 1 119903 (16)

We will use 119860119905119903


119860119905119903

= 119878min119905119903

forall119905 isin 1 119903 (17)


119860= sum119903

119905=1|119860119905119903


= sum119903



119860isin 119874(119903


problems




0

3

6

9

12

15

0 3 6 9 12 15

ZA1

ZA1

ZA1

ZA1ZB1

ZB1

ZB2

ZB1

ZB1

ZB2

ZB2

ZB2

ZA2

ZA2

ZA2

ZA2

f1

f2

S25 S75S50 S100




119903and 120572

119861





as

120575119860minus119861

119903(z) = 120572

119860

119903(z) minus 120572

119861

119903(z) (18)





0

3

6

9

12

15

0 3 6 9 12 15f1

f2

S25S50A25

A50

Opposite of A50

r1

r2





119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (19)



119877

119885of the



119874119877

119885= (119878119885cap 119877)

max (20)




119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (21)



119868119867(119885) = int

zisin119877120572 (119885 z) 119889z (22)




119860 120572119861 120575119860minus119861and 120575






(0∘ 90∘)







002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical






3

sum

119894=1

119911119894= 119888119871 (23)




3

sum

119894=1

1199112

119894= 1198882

119878 (24)



119871and 119888





119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119860

1 119911

119860


tive vector z119861 = (119911119861

1 119911

119861

119898) isin 119865 that is z119860 ⪯ z119861 if

119911119860

119894le 119911119861

119894forall119894 isin 1 119898 (3)


119860

1 119911

119860



1 119911

119861

119898) isin 119865 that is z1198607z119861 if

119911119860

119894lt 119911119861

119894forall119894 isin 1 119898 (4)


119860

1 119911

119860

119898) isin 119865 and z119861 = (119911

119861

1 119911

119861

119898) isin 119865 are


z119860 997410 z119861 z119861 997410 z119860 (5)





119861



1 119911

lowast

119898)


119911lowast

119894= min 119891





min


defined dually


119875f = f(119883)min

(8)





120597119885 = 119885 cap (119865 119885) (10)


119878119885= 120597119885EPH (11)



119860119885= 119878

min119885

= 119885 (12)




1 119885

119860

119903

120572A119903(z) = 120572 (119885

A1 119885

A119903 z) =

1

119903

119903

sum

119894=1

I (119885A119894⪯ z) (13)


I (119887) =


(14)



119860

1 119885

119860




0

3

6

9

12

15

0 3 6 9 12 15

Z1

Z1Z1

Z1

Z2

Z2

Z2

Z2

Z4

Z4

Z4

Z4

Z3

Z3

Z3

Z3

f1

f2

S25S50S75




119903




119871+

119905119903(120572

A119903) = z isin 119865 120572

A119903(z) ge 119905119903 forall119905 isin 1 119903 (15)



119905119903(120572

A119903)

119878119905119903

= 120597119871+

119905119903(120572

A119903) forall119905 isin 1 119903 (16)

We will use 119860119905119903


119860119905119903

= 119878min119905119903

forall119905 isin 1 119903 (17)


119860= sum119903

119905=1|119860119905119903


= sum119903



119860isin 119874(119903


problems




0

3

6

9

12

15

0 3 6 9 12 15

ZA1

ZA1

ZA1

ZA1ZB1

ZB1

ZB2

ZB1

ZB1

ZB2

ZB2

ZB2

ZA2

ZA2

ZA2

ZA2

f1

f2

S25 S75S50 S100




119903and 120572

119861





as

120575119860minus119861

119903(z) = 120572

119860

119903(z) minus 120572

119861

119903(z) (18)





0

3

6

9

12

15

0 3 6 9 12 15f1

f2

S25S50A25

A50

Opposite of A50

r1

r2





119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (19)



119877

119885of the



119874119877

119885= (119878119885cap 119877)

max (20)




119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (21)



119868119867(119885) = int

zisin119877120572 (119885 z) 119889z (22)




119860 120572119861 120575119860minus119861and 120575






(0∘ 90∘)







002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical






3

sum

119894=1

119911119894= 119888119871 (23)




3

sum

119894=1

1199112

119894= 1198882

119878 (24)



119871and 119888





119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

3

6

9

12

15

0 3 6 9 12 15

Z1

Z1Z1

Z1

Z2

Z2

Z2

Z2

Z4

Z4

Z4

Z4

Z3

Z3

Z3

Z3

f1

f2

S25S50S75




119903




119871+

119905119903(120572

A119903) = z isin 119865 120572

A119903(z) ge 119905119903 forall119905 isin 1 119903 (15)



119905119903(120572

A119903)

119878119905119903

= 120597119871+

119905119903(120572

A119903) forall119905 isin 1 119903 (16)

We will use 119860119905119903


119860119905119903

= 119878min119905119903

forall119905 isin 1 119903 (17)


119860= sum119903

119905=1|119860119905119903


= sum119903



119860isin 119874(119903


problems




0

3

6

9

12

15

0 3 6 9 12 15

ZA1

ZA1

ZA1

ZA1ZB1

ZB1

ZB2

ZB1

ZB1

ZB2

ZB2

ZB2

ZA2

ZA2

ZA2

ZA2

f1

f2

S25 S75S50 S100




119903and 120572

119861





as

120575119860minus119861

119903(z) = 120572

119860

119903(z) minus 120572

119861

119903(z) (18)





0

3

6

9

12

15

0 3 6 9 12 15f1

f2

S25S50A25

A50

Opposite of A50

r1

r2





119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (19)



119877

119885of the



119874119877

119885= (119878119885cap 119877)

max (20)




119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (21)



119868119867(119885) = int

zisin119877120572 (119885 z) 119889z (22)




119860 120572119861 120575119860minus119861and 120575






(0∘ 90∘)







002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical






3

sum

119894=1

119911119894= 119888119871 (23)




3

sum

119894=1

1199112

119894= 1198882

119878 (24)



119871and 119888





119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

3

6

9

12

15

0 3 6 9 12 15f1

f2

S25S50A25

A50

Opposite of A50

r1

r2





119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (19)



119877

119885of the



119874119877

119885= (119878119885cap 119877)

max (20)




119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 119898 (21)



119868119867(119885) = int

zisin119877120572 (119885 z) 119889z (22)




119860 120572119861 120575119860minus119861and 120575






(0∘ 90∘)







002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical






3

sum

119894=1

119911119894= 119888119871 (23)




3

sum

119894=1

1199112

119894= 1198882

119878 (24)



119871and 119888





119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(0∘ 90∘)







002

0406

081 0

0204

0608

10

02

04

06

08

1

LinearSpherical






3

sum

119894=1

119911119894= 119888119871 (23)




3

sum

119894=1

1199112

119894= 1198882

119878 (24)



119871and 119888





119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119903and 119886


119903




119894and 119860

119894+1 119894 isin 1 119903


119894+1 r1 r2)


1198741015840larr o isin 119874 z7 o



endend


We will use 1205721198715and 120572

119878



5and 120575

119878minus119871









119894⪯ 119860119894+1



119894and 119860


of119860119894+1






119894 119894 = 1 119903


1198741015840larr o isin 119874 z7 o

119874 larr 119874 minus 1198741015840


o1198991 larr (1199111 1199002 1199003)

o1198992 larr (1199001 1199112 1199003)

o1198993 larr (1199001 1199002 1199113)


119874 larr 119874 cup o119899119894 end

endend

end




o1198991 = (1199111 1199002 1199003)

o1198992 = (1199001 1199112 1199003)

o1198993 = (1199001 1199002 1199113)

(25)




0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o2

o3o4 o5

z1 z2


0246810

0 2 4 6 8 10

0

1

2

3

4

5

o1

o6

o2

o3o7

o5

z1 z2

z3













∘ 90∘) In


f1

f2

f3

r1

oo

998400

zz 998400

120593





120593z = arctan(

1199112minus 1199031

2

1199111minus 1199031

1

)

120593o = arctan(

1199002minus 1199031

2

1199001minus 1199031

1

)

(26)


z1015840 =

(1199031

1+

1199112minus 1199031

2

tan120593

1199112 1199113) if 120593z ge 120593

(1199111 1199031

2+ (1199111minus 1199031

1) tan120593 119911

3) if 120593z lt 120593


0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

02

04

06

08

1

12

0 02 04 06 08 1 12

(a) Slice of 1205721198715at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(b) Slice of 1205721198785at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(c) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 5∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(d) Slice of 1205721198715at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(e) Slice of 1205721198785at 120593= 45∘

0

02

04

06

08

1

12

0 02 04 06 08 1 12

(f) Slice of 120575119871minus1198785

and 120575119878minus1198715

at 120593= 45∘


o1015840 =

(1199001 1199031

2+ (1199001minus 1199031

1) tan120593 119900

3) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

tan120593

1199002 1199003) if 120593o lt 120593

z10158401015840 =

(1199031

1+

1199112minus 1199031

2

sin120593

1199113) if 120593z ge 120593

(1199031

1+

1199111minus 1199031

1

cos120593 1199113) if 120593z lt 120593

o10158401015840 =

(1199031

1+

1199001minus 1199031

1

cos120593 1199003) if 120593o ge 120593

(1199031

1+

1199002minus 1199031

2

sin120593

1199003) if 120593o lt 120593

(27)




5

and 120572119878


5

and 120575119878minus119871








(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715






5and 120575119878minus119871

5








119877 = z isin 119865 119911119894isin [1199031

119894 1199032

119894] forall119894 isin 1 2 3 (28)




to 1199032

119894minus 1199031






(

(1199032

1minus 1199031

1) (21198961minus 1)

2V1

(1199032

2minus 1199031

2) (21198962minus 1)

2V2

(1199032

3minus 1199031

3) (21198963minus 1)

2V3

)

(29)

where 119896119894isin 1 V

119894 for 119894 isin 1 2 3



0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

02

04

06

08

1

0 02 04 06 08 1

(a) Using cuboids

0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1


0

02

04

06

08

1

0 02 04 06 08 1



5and 120575

119878minus119871



5and 120575119878minus119871

5at





5and 120575

119878minus119871

5and we can




119871minus119878

5and 120575

119878minus119871




(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










(a) MIP of 120575119871minus1198785

(b) MIP of 120575119878minus1198715


(a) DVR of 120575119871minus1198785

= 15 (b) DVR of 120575119871minus1198785

= 25 (c) DVR of 120575119871minus1198785

= 35

(d) DVR of 120575119871minus1198785

= 45 (e) DVR of 120575119871minus1198785









6 Discussion




Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Slicing MIP DVR

Exact EAF values
















(a) DVR of 120575119878minus1198715

= 15 (b) DVR of 120575119878minus1198715

= 25 (c) DVR of 120575119878minus1198715

= 35

(d) DVR of 120575119878minus1198715

= 45 (e) DVR of 120575119878minus1198715








5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











5isin 15 45












Variable Lowerbound

Upperbound

Desiredvalue




119891119896=1003816100381610038161003816119910119896minus 119910lowast

119896

1003816100381610038161003816

forall119894 isin 1 2 3 (30)


lowast

119896is its







001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










001

0203

0405

06

0408

1216

2

0

2

4

6

8

DEMOES




Dist

ance

from

desir

edsu

rfac

e

tem

pera

ture

(∘C

)











5and 120575

ESminusDEMO5







= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














= 15 and 120572DEMO5


= 15 and120572DEMO5





ESminusDEMO5


0

02

04

06

08

1

0 02 04 06 08 1


at 120593 = 25∘

0

02

04

06

08

1

02 04 06 08 121


at 120593 = 45∘




8 Conclusions









Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Acknowledgments


References







































Volume 2014




Journal of











Journal of


Function Spaces






Algebra
































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Visualizing Exact and Approximated 3D Empirical Attainment ...

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Transcript of Visualizing Exact and Approximated 3D Empirical Attainment ...