IME634: Management Decision Analysis Raghu Nandan Sengupta Industrial & Management Department Indian Institute of Technology Kanpur VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (VIseKriterijumska Optimizacija I Kompromisno Resenje) The idea of VIKOR (VIseKriterijumska Optimizacija I Kompromisno Resenje) i.e., Multicriteria Optimization and Compromise Solution), a MCDM technique, was developed by Serafim Opricovic during his Ph.D. work VIKOR method was introduced as one applicable technique to be implemented within MCDM problem and it as developed as a multi attribute decision making method to solve a discrete decision making problem with non-commensurable (different units) and conflicting criteria VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR (contd..) In this method the decision maker likes a solution that is closest to the ideal, and hence the decisions/alternatives are evaluated/compared/ranked accordingly While ranking, the decisions/alternatives, rather than the best solution, is the target as finding out the ideal solution is not always feasible, but rather the closest to the ideal is what is practically possible. VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR (contd..) Two of the MCDM methods, i.e., VIKOR and TOPSIS are based on an aggregating function which represents the concept of closeness of the solution to the ideal solution In VIKOR we follow linear normalization while in TOPSIS it is vector normalization VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR (contd..) The use of normalization is used to eliminate the units of criterion functions and thus ensure a level playing field for different criterion In VIKOR we determine a maximum group utility for the majority and a minimum of an individual regret for the opponent In TOPSIS a solution with the shortest distance to the ideal solution and the greatest distance from the negative-ideal solution is required to be found. While doing this we do not consider the relative importance of these distances. VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR (contd..) Assume you have 𝑚 decisions/alternatives, 𝐴 𝑖 , 𝑖=1,⋯,𝑚 and 𝑛 attributes/decision criteria/goals 𝐶 𝑗 , 𝑗=1,⋯,𝑛 Consider 𝐶 𝑗 𝐴 𝑖 as the value of the 𝑗 𝑡ℎ attributes/decision criteria/goals for the 𝑖 𝑡ℎ alternative such that 𝐿 𝑝,𝑖 = 𝑗=1 𝑛 𝑤 𝑗 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 𝑖 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 − 𝑝 1 𝑝 , 𝑖=1,⋯,𝑚; 0≤𝑝≤∞ Here: 𝐶 𝑗 𝐴 + = max 𝑖 𝐶 𝑗 𝐴 𝑖 and 𝐶 𝑗 𝐴 − = min 𝑖 𝐶 𝑗 𝐴 𝑖 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR (contd..) Remember 𝐿 𝑝,𝑖 is used to formulate the ranking measure, where 𝑝=1,2,⋯, is integer and denotes the distance measure norm used for 𝑖= 1,⋯,𝑚 𝑝=1 signifies the Manhattan norm while 𝑝= ∞ denotes the infinity norm VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (contd..): Distance The 𝐿 1 norm or Manhattan distance between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯, 𝑦 𝑛 is 𝑖=1 𝑛 𝑥 𝑖 − 𝑦 𝑖 . The name relates to the distance a taxi has to drive in a rectangular street grid The 𝐿 𝑝 norm between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯, 𝑦 𝑛 is 𝑖=1 𝑛 𝑥 𝑖 − 𝑦 𝑖 𝑝 1 𝑝 The 𝐿 ∞ norm between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯,𝑦 is max ∀𝑖 𝑦 𝑖 − 𝑦 𝑖 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (contd..): Distance (Example) 𝒙= 2,−5,20 and 𝒚= −12,15,0 𝐿 1 = 𝑖=1 3 𝑥 𝑖 − 𝑦 𝑖 = 2− −12 + −5 −15 + 20−0 =14+ 20+20=54 𝐿 2 = 𝑖=1 3 𝑥 𝑖 − 𝑦 𝑖 2 1 2 = 2− −12 2 + −5 −15 2 + 20−0 2 1 2 2− −12 2 + −5 −15 2 + 20−0 2 1 2 = 196+400+400 1 2 = 996 1 2 =31.55 𝐿 ∞ = max ∀𝑖 𝑥 𝑖 − 𝑦 𝑖 =𝑚𝑎𝑥 2− −12 , −5 −15 , 20−0 = 𝑚𝑎𝑥 14,20,20 =20 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (contd..): Distance (Example) RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (contd..): Distance RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR (contd..): Distance The green line (L2-norm) is the unique shortest path, while the red, blue, yellow (L1-norm) are all same length (=12) for the same route This is why L2-norm has unique solution while L1- norm does not have any unique solution One can generalize this to n-dimension case VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR Algorithm Assume decisions/alternatives as 𝐴 𝑖 , 𝑖= 1,⋯,𝑚 Assume attributes/decision criteria/goals are 𝐶 𝑗 , 𝑗=1,⋯,𝑛 We state the pseudo-codes for the working principle of VIKOR VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR Algorithm (contd..) 1: DEFINE: 𝑿 𝑚×𝑛 (matrix consisting of priority scores assigned to decisions/alternatives), 𝐴 𝑖 ,based on attributes/decision criteria/goals, 𝐶 𝑗 ; 𝑤 𝑗 (weight for the attributes/decision criteria/goals) such that 𝑗=1 𝑛 𝑤 𝑗 =1; 𝐶 𝑗 𝐴 𝑖 (function relationship between attributes/decision criteria/goals for each decisions/alternatives); 𝐶 𝑗 𝐴 + = max 𝑖 𝐶 𝑗 𝐴 𝑖 ; 𝐶 𝑗 𝐴 − = min 𝑖 𝐶 𝑗 𝐴 𝑖 ; 𝐿 𝑝,𝑖 = 𝑗=1 𝑛 𝑤 𝑗 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 𝑖 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 − 𝑝 1 𝑝 . Here 𝑖=1,⋯,𝑚;𝑗=1,⋯,𝑛 and 𝑝=1,2,..,∞ (distance norm) 2: INPUT: 𝑿 𝑚×𝑛 (matrix consisting of priority scores assigned to decisions/alternatives), 𝐴 𝑖 ,based on attributes/decision criteria/goals, 𝐶 𝑗 ; 𝑤 𝑗 (weight for the attributes/decision criteria/goals) such that 𝑗=1 𝑛 𝑤 𝑗 =1; 𝐶 𝑗 𝐴 𝑖 (function relationship between attributes/decision criteria/goals for each decisions/alternatives). Here 𝑖=1,⋯,𝑚 and 𝑗=1,⋯,𝑛. 3: START if: 𝑖=1:𝑚 4: START if: 𝑗=1:𝑛 5: CALCULATE: 𝐶 𝑗 𝐴 𝑖 ; 𝐶 𝑗 𝐴 + = max 𝑖 𝐶 𝑗 𝐴 𝑖 ; 𝐶 𝑗 𝐴 − = min 𝑖 𝐶 𝑗 𝐴 𝑖 ; 𝐿 𝑝,𝑖 = 𝑗=1 𝑛 𝑤 𝑗 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 𝑖 𝐶 𝑗 𝐴 + − 𝐶 𝑗 𝐴 − 𝑝 1 𝑝 where 𝑖=1,⋯,𝑚;𝑗=1,⋯,𝑛 and 𝑝=1,2,..,∞ (distance norm) 6: END if 7: END if 8: CALCULATE: 𝐶 𝑗 𝐴 + ; 𝐶 𝑗 𝐴 − ; 𝐿 𝑝,𝑖 9: REPORT: 𝐶 𝑗 𝐴 + ; 𝐶 𝑗 𝐴 − ; 𝐿 𝑝,𝑖 10: END VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Steps # 01 (Construct the normalized decision matrix) Assume the decision matrix, 𝑿= 𝑥 11 ⋯ 𝑥 1𝑛 ⋮ ⋱ ⋮ 𝑥 𝑚1 ⋯ 𝑥 𝑚𝑛 𝑚×𝑛 , m is number of Alternatives and n is number of Criterion VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Steps # 01 (Construct the normalized decision matrix) 𝑿= 30 40 50 20 90 100 120 110 190 150 0.9 0.3 0.5 0.5 0.7 9 7 3 4 5 5×4 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Steps # 01 (Construct the normalized decision matrix) Convert the entries in X into scaled normalized values, where 𝑟 𝑖𝑗 = 𝑥 𝑖𝑗 2 𝑘=1 𝑚 𝑥 𝑘𝑗 2 , which has no dimension Thus 𝑹= 𝑟 11 ⋯ 𝑟 1𝑛 ⋮ ⋱ ⋮ 𝑟 𝑚1 ⋯ 𝑟 𝑚𝑛 𝑚×𝑛 = 𝑥 11 2 𝑘=1 𝑚 𝑥 𝑘1 2 ⋯ 𝑥 1𝑛 2 𝑘=1 𝑚 𝑥 𝑘𝑛 2 ⋮ ⋱ ⋮ 𝑥 𝑚1 2 𝑘=1 𝑚 𝑥 𝑘1 2 ⋯ 𝑥 𝑚𝑛 2 𝑘=1 𝑚 𝑥 𝑘𝑛 2 𝑚×𝑛 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Steps # 01 (Construct the normalized decision matrix) 𝑹= 30 30+100+0.9+9 40 40+120+0.3+7 50 50+110+0.5+3 20 20+190+0.5+4 90 90+150+0.7+5 100 30+100+0.9+9 120 40+120+0.3+7 110 50+110+0.5+3 190 20+190+0.5+4 150 90+150+0.7+5 0.9 30+100+0.9+9 0.3 40+120+0.3+7 0.5 50+110+0.5+3 0.5 20+190+0.5+4 0.7 90+150+0.7+5 9 30+100+0.9+9 7 40+120+0.3+7 3 50+110+0.5+3 4 20+190+0.5+4 5 90+150+0.7+5 5×4 , where 𝑟 𝑖𝑗 = 𝑥 𝑖,𝑗 𝑘=1 𝑚 𝑥 𝑘,𝑗 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Steps # 01 (Construct the normalized decision matrix) 𝑹= 0.214439 0.239091 0.305810 0.093240 0.366300 0.714796 0.717274 0.672783 0.885781 0.610501 0.006433 0.001793 0.003058 0.002331 0.002849 0.064332 0.041841 0.018349 0.018648 0.020350 5×4 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 02 (Construct the weighted normalized decision matrix) If the decision maker decides on the set of weights, depending on his/her preference, then the weight, 𝑾= 𝑤 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑤 𝑛 𝑛×𝑛 , such that 𝑗=1 𝑛 𝑤 =1 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 02 (Construct the weighted normalized decision matrix) 𝑾= 0.25 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.25 4×4 , such that 𝑗=1 𝑛 𝑤 =1 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 02 (Construct the weighted normalized decision matrix) Calculate 𝑭=𝑹𝑾= 𝑓 1,1 ⋯ 𝑓 1,𝑛 ⋮ ⋱ ⋮ 𝑓 𝑚,1 ⋯ 𝑓 𝑚,𝑛 𝑚×𝑛 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 02 (Construct the weighted normalized decision matrix) 𝑭=𝑹𝑾= 0.214439 0.239091 0.305810 0.093240 0.366300 0.714796 0.717274 0.672783 0.885781 0.610501 0.006433 0.001793 0.003058 0.002331 0.002849 0.064332 0.041841 0.018349 0.018648 0.020350 5×4 0.25 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.25 4×4 = 0.0536 0.0598 0.0765 0.0233 0.0916 0.1787 0.1793 0.1682 0.2214 0.1526 0.0016 0.0004 0.0008 0.0006 0.0007 0.0161 0.0105 0.0046 0.0047 0.0051 5×4 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 03 (Determine the maximum/best from the criterion values) Determine the maximum/best: 𝑓 𝑗 ∗ = max ∀𝑖 𝑓 𝑖,𝑗 , 𝑗=1,⋯,𝑛 𝑓 1 ∗ =𝑚𝑎𝑥 𝑓 1,1 , 𝑓 2,1 ,⋯, 𝑓 𝑚−1,1 , 𝑓 𝑚,1 𝑓 2 ∗ =𝑚𝑎𝑥 𝑓 1,2 , 𝑓 2,2 ,⋯, 𝑓 𝑚−1,2 , 𝑓 𝑚,2 . 𝑓 𝑛−1 ∗ =𝑚𝑎𝑥 𝑓 1,𝑛−1 , 𝑓 2,𝑛−1 ,⋯, 𝑓 𝑚−1,𝑛−1 , 𝑓 𝑚,𝑛−1 𝑓 𝑛 ∗ =𝑚𝑎𝑥 𝑓 1,𝑛 , 𝑓 2,𝑛 ,⋯, 𝑓 𝑚−1,𝑛 , 𝑓 𝑚,𝑛 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 03 (Determine the maximum/best from the criterion values) 𝑓 1 ∗ =𝑚𝑎𝑥 0.0536,0.0598,0.0765,0.0233,0.0916 = 0.0916 𝑓 2 ∗ =𝑚𝑎𝑥 0.1787,0.1793,0.1682,0.2214,0.1526 = 0.2214 𝑓 3 ∗ =𝑚𝑎𝑥 0.0016,0.0004,0.0008,0.0006,0.0007 = 0.0016 𝑓 4 ∗ =𝑚𝑎𝑥 0.0161,0.0105,0.0046,0.0047,0.0051 = 0.0161 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 03 (Determine the minimum/worst from the criterion values) Determine the minimum/worst 𝑓 𝑗 − = min ∀𝑖 𝑓 𝑖,𝑗 , 𝑗=1,⋯,𝑛 𝑓 1 − =𝑚𝑖𝑛 𝑓 1,1 , 𝑓 2,1 ,⋯, 𝑓 𝑚−1,1 , 𝑓 𝑚,1 𝑓 2 − =𝑚𝑖𝑛 𝑓 1,2 , 𝑓 2,2 ,⋯, 𝑓 𝑚−1,2 , 𝑓 𝑚,2 . 𝑓 𝑛−1 − =𝑚𝑖𝑛 𝑓 1,𝑛−1 , 𝑓 2,𝑛−1 ,⋯, 𝑓 𝑚−1,𝑛−1 , 𝑓 𝑚,𝑛−1 𝑓 𝑛 − =𝑚𝑖𝑛 𝑓 1,𝑛 , 𝑓 2,𝑛 ,⋯, 𝑓 𝑚−1,𝑛 , 𝑓 𝑚,𝑛 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 03 (Determine the minimum/worst from the criterion values) 𝑓 1 − =𝑚𝑖𝑛 0.0536,0.0598,0.0765,0.0233,0.0916 = 0.0233 𝑓 2 − =𝑚𝑖𝑛 0.1787,0.1793,0.1682,0.2214,0.1526 = 0.1526 𝑓 3 − =𝑚𝑖𝑛 0.0016,0.0004,0.0008,0.0006,0.0007 = 0.0004 𝑓 4 − =𝑚𝑖𝑛 0.0161,0.0105,0.0046,0.0047,0.0051 = 0.0046 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 03 (Graphical representation of maximum/best and minimum/worst for each criterion) VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 04 (Determine the relative ratios based on 𝑳 𝟏 norm utilizing maximum/best and minimum/worst ratios) Compute: 𝐿 1,𝑖 = 𝑗=1 𝑛 𝑤 𝑗 𝑓 𝑗 ∗ − 𝑓 𝑖,𝑗 𝑓 𝑗 ∗ − 𝑓 𝑗 − , 𝑖=1,⋯,𝑚 𝐿 1,1 = 𝑤 1 𝑓 1 ∗ − 𝑓 1,1 𝑓 1 ∗ − 𝑓 1 − + 𝑤 2 𝑓 2 ∗ − 𝑓 1,2 𝑓 2 ∗ − 𝑓 2 − +⋯+ 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 1,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − + 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 1,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − 𝐿 1,2 = 𝑤 1 𝑓 1 ∗ − 𝑓 2,1 𝑓 1 ∗ − 𝑓 1 − + 𝑤 2 𝑓 2 ∗ − 𝑓 2,2 𝑓 2 ∗ − 𝑓 2 − +⋯+ 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 2,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − + 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 2,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − . 𝐿 1,𝑚−1 = 𝑤 1 𝑓 1 ∗ − 𝑓 𝑚−1,1 𝑓 1 ∗ − 𝑓 1 − + 𝑤 2 𝑓 2 ∗ − 𝑓 𝑚−1,2 𝑓 2 ∗ − 𝑓 2 − +⋯+ 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑚−1,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − + 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 𝑚−1,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − 𝐿 1,𝑚 = 𝑤 1 𝑓 1 ∗ − 𝑓 𝑚,1 𝑓 1 ∗ − 𝑓 1 − + 𝑤 2 𝑓 2 ∗ − 𝑓 𝑚,2 𝑓 2 ∗ − 𝑓 2 − +⋯+ 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑚,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − + 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 𝑚,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 04 (Determine the relative ratios based on 𝑳 𝟏 norm utilizing maximum/best and minimum/worst ratios) 𝐿 1,1 =0.25× 0.0916−0.0536 0.0916−0.0233 +0.25× 0.2214−0.1787 0.2214−0.1526 +0.25× 0.0016−0.0016 0.0016−0.0004 +0.25× 0.0161−0.0161 0.0161−0.0046 = 0.2943 𝐿 1,2 =0.25× 0.0916−0.0598 0.0916−0.0233 +0.25× 0.2214−0.1793 0.2214−0.1526 +0.25× 0.0016−0.0004 0.0016−0.0004 +0.25× 0.0161−0.0105 0.0161−0.0046 = 0.6418 𝐿 1,3 =0.25× 0.0916−0.0765 0.0916−0.0233 +0.25× 0.2214−0.1682 0.2214−0.1526 +0.25× 0.0016−0.0008 0.0016−0.0004 +0.25× 0.0161−0.0046 0.0161−0.0046 = 0.6807 𝐿 1,4 =0.25× 0.0916−0.0233 0.0916−0.0233 +0.25× 0.2214−0.2214 0.2214−0.1526 +0.25× 0.0016−0.0006 0.0016−0.0004 +0.25× 0.0161−0.0047 0.0161−0.0046 = 0.7194 𝐿 1,5 =0.25× 0.0916−0.0916 0.0916−0.0233 +0.25× 0.2214−0.1526 0.2214−0.1526 +0.25× 0.0016−0.0007 0.0016−0.0004 +0.25× 0.0161−0.0051 0.0161−0.0046 = 0.6822 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 04 (Determine the relative ratios based on 𝑳 ∞ norm utilizing maximum/best and minimum/worst ratios) Compute: 𝐿 ∞,𝑖 = max ∀𝑗 𝑗=1 𝑛 𝑤 𝑗 𝑓 𝑗 ∗ − 𝑓 𝑖,𝑗 𝑓 𝑗 ∗ − 𝑓 𝑗 − , 𝑖=1,⋯,𝑚 𝐿 ∞,1 =𝑚𝑎𝑥 𝑤 1 𝑓 1 ∗ − 𝑓 1,1 𝑓 1 ∗ − 𝑓 1 − , 𝑤 2 𝑓 2 ∗ − 𝑓 1,2 𝑓 2 ∗ − 𝑓 2 − ,⋯, 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 1,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − , 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 1,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − 𝐿 ∞,2 =𝑚𝑎𝑥 𝑤 1 𝑓 1 ∗ − 𝑓 2,1 𝑓 1 ∗ − 𝑓 1 − , 𝑤 2 𝑓 2 ∗ − 𝑓 2,2 𝑓 2 ∗ − 𝑓 2 − ,⋯, 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 2,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − , 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 2,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − . 𝐿 ∞,𝑚−1 =𝑚𝑎𝑥 𝑤 1 𝑓 1 ∗ − 𝑓 𝑚−1,1 𝑓 1 ∗ − 𝑓 1 − , 𝑤 2 𝑓 2 ∗ − 𝑓 𝑚−1,2 𝑓 2 ∗ − 𝑓 2 − ,⋯, 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑚−1,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − , 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 𝑚−1,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − 𝐿 ∞,𝑚 =𝑚𝑎𝑥 𝑤 1 𝑓 1 ∗ − 𝑓 𝑚,1 𝑓 1 ∗ − 𝑓 1 − , 𝑤 2 𝑓 2 ∗ − 𝑓 𝑚,2 𝑓 2 ∗ − 𝑓 2 − ,⋯, 𝑤 𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑚,𝑛−1 𝑓 𝑛−1 ∗ − 𝑓 𝑛−1 − , 𝑤 𝑛 𝑓 𝑛 ∗ − 𝑓 𝑚,𝑛 𝑓 𝑛 ∗ − 𝑓 𝑛 − VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA VIKOR: Step # 04 (Determine the relative ratios based on 𝑳 ∞ norm utilizing maximum/best and minimum/worst ratios) 𝐿 ∞,1 =𝑚𝑎𝑥 0.25× 0.0916−0.0536 0.0916−0.0233 ,0.25× 0.2214−0.1787 0.2214−0.1526 ,0.25× 0.0016−0.0016 0.0016−0.0004 ,0.25× 0.0161−0.0161 0.0161−0.0046 = 0.1553 𝐿 ∞,2 =𝑚𝑎𝑥 0.25× 0.0916−0.0598 0.0916−0.0233 ,0.25× 0.2214−0.1793 0.2214−0.1526 ,0.25× 0.0016−0.0004 0.0016−0.0004 ,0.25× 0.0161−0.0105 0.0161−0.0046 = 0.2500 𝐿 ∞,3 =𝑚𝑎𝑥 0.25× 0.0916−0.0765 0.0916−0.0233 ,0.25× 0.2214−0.1682 0.2214−0.1526 ,0.25× 0.0016−0.0008 0.0016−0.0004 ,0.25× 0.0161−0.0046 0.0161−0.0046 = 0.2500 𝐿 ∞,4 =𝑚𝑎𝑥 0.25× 0.0916−0.0233 0.0916−0.0233 ,0.25× 0.2214−0.2214 0.2214−0.1526 ,0.25× 0.0016−0.0006 0.0016−0.0004 ,0.25× 0.0161−0.0047 0.0161−0.0046 = 0.2500 𝐿 ∞,5 =𝑚𝑎𝑥 0.25× 0.0916−0.0916 0.0916−0.0233 ,0.25× 0.2214−0.1526 0.2214−0.1526 ,0.25× 0.0016−0.0007 0.0016−0.0004 ,0.25× 0.0161−0.0051 0.0161−0.0046 = 0.2500 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 05 (Determine the maximum/minimum based on 𝑳 𝟏 norm) Find: 𝐿 1,∀𝑖 𝑚𝑎𝑥 = max ∀𝑖 𝐿 1,𝑖 , i.e., 𝐿 1,∀𝑖 𝑚𝑎𝑥 = max 𝐿 1,1 , 𝐿 1,2 ,⋯, 𝐿 1,𝑚−1 , 𝐿 1,𝑚 Find: 𝐿 1,∀𝑖 𝑚𝑖𝑛 = min ∀𝑖 𝐿 1,𝑖 , 𝑖=1,⋯,𝑚, i.e., 𝐿 1,∀𝑖 𝑚𝑖𝑛 =min 𝐿 1,1 , 𝐿 1,2 ,⋯, 𝐿 1,𝑚−1 , 𝐿 1,𝑚 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 05 (Determine the maximum/minimum based on 𝑳 𝟏 norm) RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 05 (Determine the maximum/minimum based on 𝑳 ∞ norm) Find: 𝐿 ∞,∀𝑖 𝑚𝑎𝑥 = max ∀𝑖 𝐿 ∞,𝑖 , 𝑖=1,⋯,𝑚, i.e., 𝐿 ∞,∀𝑖 𝑚𝑎𝑥 =max 𝐿 ∞,1 , 𝐿 ∞,2 ,⋯, 𝐿 ∞,𝑚−1 , 𝐿 ∞,𝑚 Find: 𝐿 ∞,∀𝑖 𝑚𝑖𝑛 = min ∀𝑖 𝐿 ∞,𝑖 , 𝑖=1,⋯,𝑚, i.e., 𝐿 ∞,∀𝑖 𝑚𝑖𝑛 =min 𝐿 ∞,1 , 𝐿 ∞,2 ,⋯, 𝐿 ∞,𝑚−1 , 𝐿 ∞,𝑚 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 05 (Determine the maximum/minimum based on 𝑳 ∞ norm) RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 06 (Calculate relative ranking) Compute: 𝑄 𝑖 =𝑣 𝐿 1,𝑖 − 𝐿 1,𝑖 𝑚𝑖𝑛 𝐿 1,𝑖 𝑚𝑎𝑥 − 𝐿 1,𝑖 𝑚𝑖𝑛 + 1−𝑣 𝐿 ∞,𝑖 − 𝐿 ∞,𝑖 𝑚𝑖𝑛 𝐿 ∞,𝑖 𝑚𝑎𝑥 − 𝐿 ∞,𝑖 𝑚𝑖𝑛 , 𝑖=1,⋯,𝑚 𝑄 1 =𝑣 𝐿 1,1 − 𝐿 1,1 𝑚𝑖𝑛 𝐿 1,1 𝑚𝑎𝑥 − 𝐿 1,1 𝑚𝑖𝑛 + 1−𝑣 𝐿 ∞,1 − 𝐿 ∞,1 𝑚𝑖𝑛 𝐿 ∞,1 𝑚𝑎𝑥 − 𝐿 ∞,1 𝑚𝑖𝑛 𝑄 2 =𝑣 𝐿 1,2 − 𝐿 1,2 𝑚𝑖𝑛 𝐿 1,2 𝑚𝑎𝑥 − 𝐿 1,2 𝑚𝑖𝑛 + 1−𝑣 𝐿 ∞,2 − 𝐿 ∞,2 𝑚𝑖𝑛 𝐿 ∞,2 𝑚𝑎𝑥 − 𝐿 ∞,2 𝑚𝑖𝑛 . 𝑄 𝑚−1 =𝑣 𝐿 1,𝑚−1 − 𝐿 1,𝑚−1 𝑚𝑖𝑛 𝐿 1,𝑚−1 𝑚𝑎𝑥 − 𝐿 1,𝑚−1 𝑚𝑖𝑛 + 1−𝑣 𝐿 ∞,𝑚−1 − 𝐿 ∞,𝑚−1 𝑚𝑖𝑛 𝐿 ∞,𝑚−1 𝑚𝑎𝑥 − 𝐿 ∞,𝑚−1 𝑚𝑖𝑛 𝑄 𝑚 =𝑣 𝐿 1,𝑚 − 𝐿 1,𝑚 𝑚𝑖𝑛 𝐿 1,𝑚 𝑚𝑎𝑥 − 𝐿 1,𝑚 𝑚𝑖𝑛 + 1−𝑣 𝐿 ∞,𝑚 − 𝐿 ∞,𝑚 𝑚𝑖𝑛 𝐿 ∞,𝑚 𝑚𝑎𝑥 − 𝐿 ∞,𝑚 𝑚𝑖𝑛 VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA

VIKOR: Step # 06 (Calculate relative ranking) 𝑣 is introduced as weight of the strategy of ‘‘the majority of criteria” (or ‘‘the maximum group utility”), here one can consider 𝑣=0.5 Rank the alternatives, sorting by the values 𝐿 1,𝑖 , 𝐿 ∞,𝑖 and 𝑄 𝑖 in decreasing order The results are three ranking lists. Propose as a compromise solution the alternatives obtained above VIKOR RNSengupta,IME Dept.,IIT Kanpur,INDIA