Chapter 4 ELECTRE METHOD

1
Chapter 4 ELECTRE METHOD
2
OVERVIEW

• ELECTRE is an MADM technique developed by Prof.
  B.Roy and associates
• Basic concept choose those alternatives which
  are preferred for most of the criteria and yet do
  not cause an unacceptable level of discontent for
  any one criterion.
• Corollary concepts concordance, discordance,
  threshold values
• Concordance - C(k,l) for any two alternatives k
  and l, this is a weighted measure of the number
  of criteria for which alternative k is preferred
  over alternative l (denoted by k P l) or for
  which k is equal to l (denoted by k E l)
• Discordance D(k,l) It handles the set of
  criteria for which k is not preferred over l and
  gives a measure of the degree of discomfort or
  discontent as a result of preferring k to l
• Threshold values (p, g) defined by the DM to
  quantify the degree of concordance. He/she wants
  (value of p) and the amount of discordance he/she
  can tolerate ( value of q), 0lt (p, q) lt 1

3
ELECTRE I

• Step 1 the weights of the criteria, w(i) for i
  1, 2, , n, are given a priori by the DM.
• Step 2 Determine concordance index
• Where the set i k P l ? k E l is the union of
  the set of the criteria for which k is preferred
  over l and the set of criteria for which k is
  equal to l
• c1 c2 c3 c4
• (0.5) (0.3) (0.1) (0.1)
• a1
• a2
• C(a1, a2) (0.3 0.1)/1 0.4
• C(a2, a1) (0.5 0.1 0.1)/1 0.7

• 20 30 5
• 20 10 30 10

4
ELECTRE I

• Step 3 Determine discordance index (k ? l)
• Where
• fi(xk) evaluation of alternative xk with
  respect to criterion fi
• fi(xl) evaluation of alternative xl with
  respect to criterion fi
• R largest range of the n criteria scales

5
ELECTRE I

• Computation
• Define an interval scale common to all criteria
• Define a range for the interval scale which may
  be different for each criterion with the best
  rating having the highest value of the range and
  the worst rating having the lowest value of the
  range
• Evaluate for each criterion of the alternatives
  and assign the corresponding rating fi(xj) or
  range value as defined in the interval scale for
  the criterion.
• c1 c2 c3 c4
• w 0.5 0.3 0.1 0.1
• a1 10 20 30 5
• a2 20 10 30 10
• D(a1, a2) Max(20-10), (10-5)/100 0.1
• D(a2, a1) Max(20 10)/100 0.1

6
ELECTRE I

• Step 4 DM specifies values of p and q
• Step 5 for each pair (k ? l) check the
  inequalities
• C(k,l) p and D(k,l) q
• If both conditions hold, then in the partial
  ordering k gt l. Otherwise, there is no preference
  order between k and l.
• Step 6 construct the preference graph where the
  nodes are the alternatives and the directed arcs
  show the preferences
• meaning 1 gt 2 gt 3
• Step 7 ask the DM whether he/she is satisfied
  with this partial ordering, i.e. whether a
  decision can be made with this partial ordering.
  If yes, the algorithm terminates. If no, go back
  to step 4 where the DM must relax the values of p
  and q.
• The result of ELECTRE I is a preference graph
  representing a partial ordering of alternatives

7
Example

• Problem evaluate three diverse energy programs
  of a government
• Alternatives
• Energy development and demonstration program
  (EDD)
• Renewable energy financial incentive (REFI)
• Wood capitalization (WC)
• Attributes
• Net cost of energy provided to user (/m.BTU)
• Potential long-term direct job creation (jobs)
• Potential maximum energy savings (trillion
  BTU/yr)
• Program cost to government (million)

8
Example

• From feasibility study
• DM1 w1 0.40, w2 0.12, w3 0.31, w4 0.17
• DM2 w1 0.28, w2 0.27, w3 0.19, w4 0.26

9
Concordance Matrix

• l
• 1 2 3
• 1 - 0.6 0.29
• DM1 k 2 0.4 - 0.17
• 3 0.71 0.83 -
• l
• 1 2 3
• 1 - 0.72 0.53
• DM2 k 2 0.28 - 0.26
• 3 0.47 0.74 -

10
Discordance Matrix

• Choose interval scale 0 10
• Assign worst and best case for each
  criterion
• Criteria Worst(0) Best (10)
• 1 25 5
• 2 0 400
• 3 0 1.0
• 4 4 0
• Ratings (by linear interpolation)
• Alternatives
• 1 2 3
• Criteria EDD REFI WC
• 1 4.5 5.5 7.5
• 2 6.9 0.12 3.1
• 3 7.1 2.4 10
• 4 9.9 7.5 5

11
Discordance Matrix

• Discordance
• l
• 1 2 3
• 1 - 0.1 0.3
• k 2 0.68 - 0.76
• 3 0.49 0.25 -
• Given DMs provide threshold values p 0.65, q
  0.4.
• Ranking
• DM1 DM2

1
2
3
12
Discordance Matrix

• DM1 C(k,l) D(k,l)
• l l
• 1 2 3 1 2 3
• 1 - 0.6 0.29 1 - 0.1 0.3
• k 2 0.4 - 0.17 2 0.68 - 0.76
• 3 0.71 0.83 - 3 0.49 0.25 -
• For p 0. 65, q 0.4
• Ranking
• DM2 C(k,l) D(k,l)
• 1 2 3 same as above
• 1 - 0.72 0.53 for p 0.65, q 0.4
• 2 0.28 - 0.26 Ranking
• 3 0.47 0.74 -

13
Discordance Matrix

• For DM1 p 0.65, q 0.5 p 0.6, q
  0.5
• complete
  ranking
• 3 gt 1 gt 2
• For DM2 p 0.53, q 0.4
• complete ranking
• 1 gt 3 gt 2

14
Conclusion

• DM1 alternative 3 preferred over alternative 2
  but is unable to give any preference ranking
  between alt. 1 and alt. 3 alt. 1 and alt. 2
• If DM raises his discordance threshold value to
  0.5, alt. 3 will be preferred over alt.1.
• If he lowers his concordance to 0.6, then alt. 1
  will be preferred over alt.2 and a complete
  ranking will be obtained 3 gt 1 gt 2 (WC gt EDD gt
  REFI)
• DM2 alts 1 and 3 preferred over alt. 2. Alt. 1
  will be preferred over alt.3 if the DM lowers his
  concordance threshold value to 0.52, thus
  enabling a complete ranking 1 gt 3 gt 2 (EDD gt WC
  gt REFI)

15
ELECTRE II

• In ELECTRE I the necessary and sufficient
  conditions for ordering is
• C(k,l) p and D(k,l) q, for all k ? l
• This may result in just partial ordering
• In ELECTRE II complete ordering is accomplished
  through the introduction of the concepts of
  strong and weak ranking relationship and the
  delineation of high, average and low concordance,
  and high, average, and low discordance

16
ELECTRE II

• Relationship strong and weak
• Concordance high, average, and low
• Discordance high, average, and low
• Concordance condition (see Figure 1)
• (note in all cases (2) should hold)
• w sum of weights for which k gt l
• w sum of weights for which k l
• w- sum of weights for which k lt l

17
ELECTRE II
C
0 p- p0 p
1
high

average low
C(k,l) ? p p gt C(k,l) ? p0 p0 gt C(k,l) ?
p-
Figure 1 Concordance Condition
18
ELECTRE II

• Discordance condition (see Figure 2)
• Relationship
• (a) Strong (SF) k strongly outranks l if
• C(k,l) is high and D(k,l) is at most average, or
• (2) C(k,l) is at least average and D(k,l) is at
  most low
• (b) Weak (Sf) k weakly outranks l if
• (1) both C(k,l) and D(k,l) are low, or
• (2) both C(k,l) and D(k,l) are average, or
• (3) combinations of (1) and (2)
• These two binary relations, (a) and (b), define
  two graphs.
• Let all loops be eliminated from the graph
  associated with SF,
• defining a reduced graph GF (Y, UF)
• Y set of nodes alternatives
• UF set of arcs - relationship

19
ELECTRE II
D
0 q0 q
1 low
average high

D(k,l) ? q0 q0 lt D(k,l) ? q
q lt D(k,l)
Figure 2 Discordance Condition
20
ELECTRE II
C(k,l) D(k,l) 1
1 High High
SF Average SF Sf
SF Average Low Sf Low
0
0
21
ELECTRE II
Gf

• Check the recycling to reduce inconsistency

22
ELECTRE II

• Corresponding to the weak relationship (b) above,
  Gf is defined from Sf. Gf (Y,Uf).
• Ranking procedure forward, reverse, average.
• Forward ranking r
• Start with k 1 and let Y(1) GF
• From GF select all nodes in Y(k) not having a
  precedent. Denote this set of nondominated
  alternatives by Z(k)
• Use Gf to remove as many ties as possible between
  alts. in Z(k). For this purpose, look for the set
  of arcs in Uf with both extremities in Z(k) call
  this set Xf. Construct the graph (Z(k), Xf)
• Select all nodes of (Z(k), Xf) not having a
  precedent. Denote this set by A(k). The set A(k)
  consists of all nodes having no precedent in
  either GF or Gf
• Rank x (the alt.) as r(x) k for every x
  ?A(k), i.e. every element of the set A(k) will
  have rank k
• Determine Y(k1) Y(k) A(k) and delete all
  arcs emanating from A(k). If Y(k1) is an empty
  set, then all elements in GF have been ranked
  otherwise set k k1 and go to Step 2

23
ELECTRE II

• Reverse Ranking, r
• Reverse the directions of the arcs UF of GF and
  Uf of Gf to obtain a mirror image of the direct
  outranking relationship
• Obtain a ranking ?(x) on these new graphs by
  using forward ranking steps
• Re-establish the correct ranking order by
  setting r(x) 1 max?(x) - ?(x)
• Average Ranking, r is considered the final
  ranking r(x) r(x) r(x)/2

24
ELECTRE II

• Forward ranking
• r(k) 1 ? 2 ? 3
• Reverse Ranking
• r(1) 1 max?(x) - ?(1)
• 1 3 3
  1
• r(2) 1 3 2 2
• r(3) 1 3 1 3
• r(x) 1 ? 2 ? 3

1
2
3
25
Example

• Consider the strong and weak relationship graphs
  in Fig 1a and 1b. There are no loops hence GF
  SF and Gf Sf
• Forward ranking First iteration
• k 1, Y(1) GF
• Z(1) 1, 2, 3
• Xf (2,3)
• A(1) 1,2
• Ranking r(1) r(2) 1
• Y(k1) Y(2) Y(1) A(1)
• 3,4,5,6,78,9?0
• Hence continue, k k1 2
• And Y(2) 3,4,5,6,7,8,9

Fig.1a
Fig. 1b.
26
Example

• Second iteration
• Z(2) 3,4,5
• Xf (4,3)
• A(2)4,5
• Ranking r(4) r(5) 2
• Y(3) Y(2) A(2)
• 3,6,7,8,9? 0
• k k1 3

27
Example

• Third iteration
• Z(3) 3,6,7
• Xf (3,7), (6,7)
• A(3) 3,6
• Ranking r(3) r(6) 3
• Y(4) Y(3) A(3)
• 7,8,9?0
• k k1 4

28
Example

• Fourth iteration
• Z(4) 7,8,9
• Xf (8,9)
• A(4) 7,8
• Ranking r(7) r(8) 4
• Y(5) Y(4) A(4) 9? 0
• k k1 5
• Fifth iteration yields simply r(9) 5
• Forward ranking (1,2) ?(4,5)?(3,6)?(7,8)?9

29
Example

• Reverse ranking
• Reverse arcs UF of GF and Uf of Gf to obtain the
  respective graphs shown below.

30
Example

• Obtain ?(x)
• node1 2 3 4 5 6 7 8 9
• ?(x) 4 5 2 3 4 3 1 2 1
• Reverse ranking
• Node1 2 3 4 5 6 7 8 9
• r(x)2 1 4 3 2 3 5 4 5
• Average ranking
• Node1 2 3 4 5 6 7 8 9
• r(x) 1 1 3 2 2 3 4 4 5
• r(x)2 1 4 3 2 3 5 4 5
• r(x)1.5 1 3.5 2.5 2 3 4.5 4 5
• Final ranking the final ranking in decreasing
  order of preference is 2, 1, 5, 4, 6, 3, 8, 7, 9

31
Example Energy Problem same as in ELECTRE I

• Additional data p 0.8, p0 0.65, p- 0.5,
  q 0.6, q0 0.4
• DM1 GF SF Gf Sf
• Ranking Node 1 2 3
• r(x) 2 3 1
• r(x) 2 3 1
• r(x) 2 3 1
• Final ranking 3 1 2

1
2
3
32
Example Energy Problem same as in ELECTRE I

• DM2 GF SF Gf Sf
• Ranking Node 1 2 3
• r(x) 1 3 2
• r(x) 1 3 2
• r(x) 1 3 2
• Final ranking 1 3 2

1
2
3