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Incomparable, what now ?IV: An (unexpected) modelling
challenge
R. Bruggemann(1) and L. Carlsen(2)
(1): Leibniz-Institute of Freshwater Ecology and Inland Fisheries, Berlin, Germany
(2): Awareness Center, Roskilde, Denmark
Flor_incomp25_anim.ppt 27.2.2015 – 4.4.2015
Ranking I
• Most often there is no measure for theranking aim
• Hence a multi-indicatorsystem (MIS) is neededas a proxy for the ranking aim. Examples:
Poverty Sustainability Child-well being
Ranking I (cont‘d)
• MCDA-methods, so far a ranking is intended, are constructing a ranking index CI.
• The order due to CI is by construction a weak(or better a linear) order.
• Hence, there are no incomparabilities. Everything seems to be on its best way!
Really??
Consequences
• By construction a linear order is intended:
– No ambiguity
– Most often no ties
• A metric is available
• Modelling of the knowledge ofstakeholders/decision makers
However….
• Conflicts are hidden from the very beginning
• There may be robustness problems
• Problems to get the needed parameters (forexample the weights) of the MCDA-method
Ranking II
• HDT-equation: x y: qi(x) qi(y) for all qi of MIS• The HDT-equation is very strict: Incomparabilities are
arising:– Minute numerical differences i: = abs(qi(x) – qi(y)) – No care how many indicator pairs taken from MIS are
contributing to x || y– No care, whether or not an incomparability is induced by
contextual similar qi of the MIS
• Regarding knowledge to eliminate someincomparabilities: How to model within the frameworkof HDT?
Our talk here in Florence:
1. An empirical data set (taken fromenvironmental chemistry)
2. Weight intervals
3. Towards a controlling law
4. Results
5. Discussion, Future tasks
Pesticides in the environment
DDT Aldrin (ALD)
… and nine other pesticides.How do they affect the environment? There is no single measure. Hence a multi-indicator System:Persistence (Pers), Bioaccumulation (BioA) and Toxicity (Tox)
Hexachlorobenzene (HCB)
Hasse diagram (HD) of 12 Pesticides
There are many incomparabilities (such as DDD || HCL), so there is a need for modelling.
Pesticides:DDTALD CHLDDEDDDHCL HCBMECDIE PCNPCPLINCharacterized by a MIS {Pers, BioA, Tox}
U = |{(xi,xj)XxX, xi||xj, with i<j}| = 31
Modelling by weight intervals
1. Basic paper: Match – Commun. Math. Comput. Chem. 2013, 69, 413-432
2. Idea: Let(x) = wi qi(x) the value of a composite indicator , one ofthe simplest constructions to get a linear orweak order.
Modelling by weight intervals (cont‘d)
1. The problem is the selection of weights, which causes
– subjectivity and
– hides incomparabilities expressing major conflictsin the data
2. Hence: A relaxation by weight-intervals
U = 5
0 1Range of weights
PersistenceBioaccumulation
Toxicity
In our example:
Many weak orders
Concatenationof these ordersAlgebraically:Intersection ofthese orders.
CI1 CI2 …
Questions
1. What can be said about stability? Could ithappen that another MC – run changes theresult?
2. How can we judge the role of weight-intervals?
– Effect of lengths of the intervals
– Effect of location of the intervals
Idea• Let Wi be the ith weight interval (located within the
span of [0,1])
• U = f(Wi) i = 1,…,m ; m number of indicators (1)
• Equation (1) by far too detailled
• Hence: Introduction of an artificial parameter:
Vr = (realized wi,max – realized wi,min)for all (realized wi,max – realized wi,min ) 0
ExpectationU, incomparability
Vr = 0Certainty aboutweights, i.e. exactlyone weight for eachindicator:
U = 0
0 < Vr < 1Some uncertaintyHence intervals withLengths < 1
0 < U < Umax
Vr = 1Completeweight‘s span.No knowledge.Original poset
U = Umax
0
U = f(Vr)
• Having m indicators, m* may be the numberof indicators, where Vr 0. Then:
• U = f(Vr, m*)
• Focus on the length of the intervals
• Disregarding the position of the intervals
Hypothesis
• U = Umax* Vrs, s = 1/m*,
• Umax = U(Vr= 1),
• Vr = 1: all intervals have length 1, maximal uncertainty, original poset (without anyweights)
• The values U(Vr) may be considered asdescribing a kind of normal behaviour.
• Realistic assumption?
Results: Pesticides data
0
5
10
15
20
25
30
35
0 0.2 0.4 0.6 0.8 1 1.2
Urealmin
Urealmax
U
Vr
0
5
10
15
20
25
30
35
0 0.2 0.4 0.6 0.8 1 1.2
U(V)
U(V)
Umax = 31
Results/Summary
• Modelling stakeholders knowledge within theframework of partial order theory: weightintervals
• Need of a function to get an overviewU = f(Vr).
• A power law seems to be a suitable approach
• Other modelling concepts (not shown here): power law seems to describe pretty well U = f(p), i.e. U = Umax *ps, p methods parameter.
Check of the ideas with anotherdataset
• Polluted sites in South-Westgermany
• Four indicators, 59 regions
• Power law seems to fail!
Results: Pollution in a South-Western region of Germany
0
200
400
600
800
1000
1200
1400
1600
0 0.2 0.4 0.6 0.8 1 1.2
Ucalc
Ureal
The values obtained by the MC simulations seem to belarger then the values obtained by the power law U = 1386 * V(1/4)
The problem: In contrast to the pesticides data, the single indicatorsinduces orders with large equivalence classes. I.e. the degree of degeneracy is high.
Umax = 1386
Pollution dataPlaying with ideas…..
K(qi) describes the degeneracy with respect to indicator qi.
K = Ni*(Ni – 1) , Ni = |ith equivalence class|
The degree of degeneracy k(qi) related to each indicator:
)1(
)1(:)(
nn
NNqk
ii
i
1/m is valid, only if no degeneracy appears, becausethen each indicator contributes its own linear order to the poset.
Idea: (1/m) eff(ectiv) = f(m, k(qi))
n: number of objects
Approach:))(1(
1:
1
,..,1
mi
i
eff
qkmm
Pesticide data : k(qi) = 0 for all i. No correction neededPollution data: k(q1) = 0.074 , k(q2) = 0.036, k(q3) = 0.037, k(q4) = 0.016
(1/m)eff = 0.21. UcalcP is calculated following (1/m)eff:
0
200
400
600
800
1000
1200
1400
1600
0 0.2 0.4 0.6 0.8 1 1.2
Ureal
UcalcP
U
Vr
Eq. 2
-100
-50
0
50
100
150
200
250
300
350
400
0 0.2 0.4 0.6 0.8 1 1.2
delta1
delta2
Vr
Future Work
• Explain, why a power law is the correct law!
• Find out, whether Vr is a good selection
• Find methods for a correct interpretation
• Are there other more reasonable controllingparameter?
• Is s generally well described as a funktion of1/m and k(qi)? Could we do it better?
Thank you for attention!!
Background
Remarks
1) „Differential“ view for effect of V for CI‘s2) Linear orders w.r.t. to the indicators or any CI imply equal distance to
the original poset
3) In linear orders the number of 1 in the -matrix equals (n*(n-1)/2) and is independent of the special CI All 1 in for the original poset are realized in all (CI).Therefore the distances of all CI to the original poset are the same, namely (in the case of the pesticides) 31
4) If not a linear but weak orders with nontrivial equivalence classesappear then irregularities appear as in case of the second data set
njiotherwise
jiifji ,...,1,
0
1,
Example wi (i=1,2,3 and wi = 1)
w3 w2
w1
w*(1)
w*(2)
w*(3)
Three tuples w* selected
Example continued
Pollution data: 59 objects , 4 indicators)
orig Pb cd zn S CIrelatorig 0 1513 1448 1449 1414 1386Pb 1513 0 1851 1884 1455 1359cd 1448 1851 0 1183 1922 742zn 1449 1884 1183 0 1703 1185S 1414 1455 1922 1703 0 1534CIrelat 1386 1359 742 1185 1534 0
Orig (poset) Max distance: 1513,• The attributes Pb, Cd, Zn and S induce weak orders with• Pretty large nontrivial equivalence classes.
Degeneracy (K) : CI 0; Pb 254; Cd 124; Zn 126; S 56• Assumption s = ¼ may not a good one, because some
indicators are insufficient in differentiating the order.
Check of the exponent s in U = Umax*Vs
Pollution data
0
200
400
600
800
1000
1200
1400
1600
0 0.2 0.4 0.6 0.8 1 1.2
Ureal
Ucalc3
Ucalc4
Ucalc3:s = 0.1Ucalc4: s = 0.2
Hypothesis: s = f(m*,K)
U
V
Discussion
• Selection of interval length: – Seems to describe crudely the behavior of U
– Position of intervals in dependence of any indicator isconsidered as „fine tuning“
– There is no need to find a law which exactlymeets the points (U,Vr).
• Main effect at the very beginning of the Vr-scale: Vr = 0 Vr = – Sensitivity?
– Distances (orders due to single indicators)
V 0
•V = 0 means, we start from a weak order ,•V = means, there is an influence from other indicators
Orig (partial order)
Pers Tox
BioA
5816
50
3131
31
46 16
12
31
A composite indicatordue to equal weights
The behaviour U = f(Vr) could be asfollows:
Range a) due to mc-runsandb) how differentweight intervals areassociated with theindicators
Vr
U
Umax
U = 0
Vr = 0 Vr = 1
Constructed test data set
7 objects, three indicators. By construction a high degree ofdegeneracy: K(q1) = 8, K(q2) = 8, K(q3) = 14. i.e. k(q1) = 0.19, k(q2) = 0.19, k(q3) = 0.333
0
2
4
6
8
10
12
14
0 0.2 0.4 0.6 0.8 1 1.2
U
Ucalc
Ucalcwithoutdeg
Calculatedindividually
Calculated, by eq. 2
CalculatedWithout regardingdegeneracy
Balance equation
a, b d, e, f
c
V = |{(c,a),(c,b),(c,d),(c,e),(c,f),(a,b),(b,a),(d,e),(e,d),(d,f),(f,d),(e,f),(f,e)}|U = |{(a,d),(a,e),(a,f),(b,d),(b,e),(b,f)}|K = = |{(a,b),(b,a)} {(d,e),(e,d),(d,f),(f,d),(e,f),(f,,e)}|
2*U + 2*V – K = n*(n-1)
Discussion
• Assume U = f(p) can be established, then• The control is concerned with stability, not…• …at which values of p the „best“ partial order
will be found (see with respect to fuzzymodelling: Annoni et al., 2008, De Baets et al., 2011)
• The distance graph will help us to decide theeffect of mixing the indicators due to their weightintervals
• As to how far Vinput can be seen as a betterleading quantity is a task for the future