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

RNSengupta,IME Dept.,IIT Kanpur,INDIA Technique for Order Preference by Similarity to Ideal Solution (TOPSIS) Technique for Order Preference by Similarity to Ideal Solution (TOPSIS) method was developed for the INTEGRated Human Exploration Mission Simulation FaciliTY (INTEGRITY) project in the Johnson Space Centre to assess the priority of a set of human spaceflight mission simulators TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS (contd…) One assumes utility function is monotonic, in the sense the more/less you get more/less you want The basic premise based on which TOPSIS works is the fact that selected alternatives should have the shortest distance from the positive-ideal solution, and the farthest distance from the negative-ideal solution TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS (contd…) Choose positive ideal solution (PIS) of the original ranking problem Choose negative ideal solution (NIS) of the original ranking problem Find distances from each decisions/alternatives, 𝐴 𝑖 , 𝑖= 1,⋯,𝑚 to PIS, which is given as 𝑑 𝐴 𝑖 ,𝑃𝐼𝑆 Find distances from each decisions/alternatives, 𝐴 𝑖 , 𝑖= 1,⋯,𝑚 to NIS which is given as 𝑑 𝐴 𝑖 ,𝑁𝐼𝑆 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS (contd…) Euclidean distance measure is used and we ensure our main motivation is to minimize the dispersion Calculate 𝑟 𝑖 = 𝑑 𝐴 𝑖 ,𝑁𝐼𝑆 𝑑 𝐴 𝑖 ,𝑁𝐼𝑆 +𝑑 𝐴 𝑖 ,𝑃𝐼𝑆 = 𝐴 𝑖 −𝑁𝐼𝑆 2 𝐴 𝑖 −𝑁𝐼𝑆 2 + 𝐴 𝑖 −𝑃𝐼𝑆 2 The basic premise being Euclidean distance portrays the concept of utility function, 𝑈 𝑊 which is quadratic TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS (contd…) Minimizing 𝑈 𝑊 ensures minimizing dispersion, i.e., minimization of 𝑉𝑎𝑟 𝑋 One ranks the ratios, 𝑟 𝑖 , to get the best alternative TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA Algorithm for TOPSIS 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 TOPSIS TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

Algorithm for TOPSIS (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;𝑩 (benefit matrix);𝑪 (cost matrix); 𝑟 𝑖,𝑗 = 𝑥 𝑖,𝑗 𝑖=1 𝑚 𝑥 𝑖,𝑗 2 ; 𝐴𝐼𝑆= 𝑣 1 + ,⋯, 𝑣 𝑚 + = max ∀𝑖 𝑣 𝑖,𝑗 𝑗∈𝑩 , min ∀𝑖 𝑣 𝑖,𝑗 𝑗∈𝑪 (negative ideal solution);𝑃𝐼𝑆= 𝑣 1 − ,⋯, 𝑣 𝑚 − = min ∀𝑖 𝑣 𝑖,𝑗 𝑗∈𝑩 , max ∀𝑖 𝑣 𝑖,𝑗 𝑗∈𝑪 (positive ideal solution); 𝑆 𝑖 + = 𝑗=1 𝑛 𝑣 𝑖,𝑗 − 𝑣 𝑗 + 2 ; 𝑆 𝑖 − = 𝑗=1 𝑛 𝑣 𝑖,𝑗 − 𝑣 𝑗 − 2 ; 𝑇 𝑖 = 𝑆 𝑖 − 𝑆 𝑖 + + 𝑆 𝑖 − (relative closeness);𝑀= 𝑆 𝑖 + , 𝑆 𝑖 − (separation measure). Here 𝑖=1,⋯,𝑚 and 𝑗=1,⋯,𝑛 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;𝑩 (benefit matrix);𝑪 (cost matrix). Here 𝑖=1,⋯,𝑚 and 𝑗=1,⋯,𝑛 3: START if: 𝑖=1:𝑚 4: START if: 𝑗=1:𝑛 5: CALCULATE: 𝑟 𝑖,𝑗 = 𝑥 𝑖,𝑗 𝑖=1 𝑚 𝑥 𝑖,𝑗 2 ; 𝑣 𝑖,𝑗 = 𝑤 𝑗 𝑟 𝑖,𝑗 where 𝑖=1,⋯,𝑚 and 𝑗=1,⋯,𝑛 6: END if 7: END if 8: CALCULATE: 𝑣 𝑖 + ; 𝑣 𝑖 − ;𝐴𝐼𝑆;𝑃𝐼𝑆; 𝑆 𝑖 + ; 𝑆 𝑖 − ; 𝑀; 𝑇 𝑖 9: REPORT: 𝐴𝐼𝑆;𝑃𝐼𝑆;𝑀; 𝑇 𝑖 10: END TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS (contd…): Distance measure The Euclidean distance between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯, 𝑦 𝑛 is 𝑖=1 𝑛 𝑥 𝑖 − 𝑦 𝑖 2 1 2 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 Mahalanobis distance between random vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯, 𝑦 𝑛 is 𝒙−𝒚 𝑇 𝑺 −1 𝒙−𝒚 , where 𝑺 is the covariance matrix TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS (contd…): Distance measure The Hamming distance between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯, 𝑦 𝑛 is the number of positions at which the corresponding values are different The 𝐿 ∞ norm between vector/points 𝒙= 𝑥 1 ,⋯, 𝑥 𝑛 and 𝒚= 𝑦 1 ,⋯,𝑦 is max ∀𝑖 𝑦 𝑖 − 𝑦 𝑖 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 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

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

TOPSIS: Step # 01 (Construct the normalized decision matrix) (contd..) Assume, 𝑿= 10 05 15 20 25 15 25 30 25 35 40 30 30 40 45 35 15 10 10 05 Scale the values using normalization concept, i.e., 𝑟 𝑖𝑗 = 𝑥 𝑖𝑗 𝑘=1 𝑚 𝑥 𝑘𝑗 2 (you can use any other concept of utility also) 𝑹= 10 100+25+225+400 5 100+25+225+400 15 100+25+225+400 20 100+25+225+400 625 625+225+625+900 225 625+225+625+900 625 625+225+625+900 900 625+225+625+900 625 625+1225+1600+900 1225 625+1225+1600+900 1600 625+1225+1600+900 900 625+1225+1600+900 900 900+1600+2025+1225 1600 900+1600+2025+1225 2025 900+1600+2025+1225 1225 900+1600+2025+1225 225 225+100+100+25 100 225+100+100+25 100 225+100+100+25 25 225+100+100+25 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Step # 01 (Construct the normalized decision matrix) (contd..) 𝑹= 0.133333 0.033333 0.300000 0.533333 0.263158 0.094737 0.263158 0.378947 0.143678 0.281609 0.367816 0.206897 0.156522 0.278261 0.352174 0.213043 0.500000 0.222222 0.222222 0.055556 Check each column adds up to 1 as it should be TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: 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 , such that 𝑗=1 𝑛 𝑤 = 1 Consider, 𝑾= 0.20 0.00 0.00 0.00 0.00 0.00 0.10 0.00 0.00 0.00 0.00 0.00 0.15 0.00 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.00 0.30 Calculate 𝑽=𝑹𝑾 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 02 (Construct the weighted normalized decision matrix) (contd..) Thus 𝑽=𝑹𝑾= 0.133333 0.033333 0.300000 0.533333 0.263158 0.094737 0.263158 0.378947 0.143678 0.281609 0.367816 0.206897 0.156522 0.278261 0.352174 0.213043 0.500000 0.222222 0.222222 0.055556 × 0.20 0.00 0.00 0.00 0.00 0.00 0.10 0.00 0.00 0.00 0.00 0.00 0.15 0.00 0.00 0.00 0.00 0.00 0.25 0.00 0.00 0.00 0.00 0.00 0.30 Hence 𝑽=𝑹𝑾= 𝑣 11 𝑣 12 𝑣 13 𝑣 14 𝑣 21 𝑣 22 𝑣 23 𝑣 24 𝑣 31 𝑣 32 𝑣 33 𝑣 34 𝑣 41 𝑣 42 𝑣 43 𝑣 44 𝑣 51 𝑣 52 𝑣 53 𝑣 54 = 0.026667 0.006667 0.060000 0.106667 0.026316 0.009474 0.026316 0.037895 0.021552 0.042241 0.055172 0.031034 0.039130 0.069565 0.088043 0.053261 0.150000 0.066667 0.066667 0.016667 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 03 (Determine the most positive-ideal, most negative-ideal solutions) Calculate 𝑽 + which is most positive ideal solution Where 𝑽 + = max 𝑖 𝑣 𝑖𝑗 𝑗∈𝐽 ,𝑖=1,⋯,𝑚 = 𝑣 1 + , 𝑣 2 + ,⋯, 𝑣 𝑛 + TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 03 (Determine the most positive-ideal, most negative-ideal solutions) Calculate 𝑽 − which is most negative ideal solution Where 𝑽 − = min 𝑖 𝑣 𝑖𝑗 𝑗∈𝐽 ,𝑖=1,⋯,𝑚 = 𝑣 1 − , 𝑣 2 − ,⋯, 𝑣 𝑛 − TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 04 (Calculate the distance based on most positive-ideal, most negative-ideal solutions) Calculate 𝑆 𝑖 + based on most positive ideal solution Where 𝑆 𝑖 + = 𝑗=1 𝑛 𝑣 𝑖𝑗 − 𝑣 𝑗 + 2 , 𝑖= 1,⋯,𝑚 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 04 (Calculate the distance based on most positive-ideal, most negative-ideal solutions) Calculate 𝑆 𝑖 − based on most negative ideal solution Where 𝑆 𝑖 − = 𝑗=1 𝑛 𝑣 𝑖𝑗 − 𝑣 𝑗 − 2 , 𝑖= 1,⋯,𝑚 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Step # 05 (Calculate the relative proximity based on ideal solution) Calculate 𝑐 𝑖 = 𝑆 𝑖 − 𝑆 𝑖 − + 𝑆 𝑖 + , 𝑖= 1,⋯,𝑚 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

RNSengupta,IME Dept.,IIT Kanpur,INDIA TOPSIS: Example Consider the problem related to buying a house/apartment among four (04) choices, where the decision to buy the house/apartment is based on eleven (11) different parameters/criterion which are City Price Loan availability/conditions Location Number of rooms Safety Proximity to markets Proximity to schools Proximity to hospitals Facilities available Resale condition TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) 𝑿= 45.5 50.0 40.0 41.7 40.0 33.3 42.9 25.0 66.7 83.3 62.5 77.8 09.2 08.5 06.8 10.6 1.8 1.5 1.3 2.3 0.0 1.3 0.8 1.8 1.3 0.0 0.0 1.9 2.3 2.4 1.2 3.0 1.6 2.4 1.8 3.2 4.8 3.3 4.2 3.8 3.1 1.6 2.2 1.3 Use the normalization formulae as , 𝑟 𝑖𝑗 = 𝑥 𝑖𝑗 𝑘=1 𝑛 𝑥 𝑖𝑘 2 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) 𝑹= 0.5116 0.5622 0.4498 0.4689 0.5561 0.4629 0.5964 0.3475 0.4565 0.5701 0.4278 0.5325 0.5180 0.4785 0.3828 0.5968 0.5097 0.4248 0.3681 0.6513 0.0000 0.5508 0.3390 0.7627 0.5647 0.0000 0.0000 0.8253 0.4961 0.5177 0.2589 0.6471 0.3427 0.5140 0.3855 0.6854 0.5908 0.4062 0.5169 0.4677 0.7169 0.3700 0.5087 0.3006 For example 𝑟 3,2 = 42.9 40.0 2 + 42.9 2 + 62.5 2 + 6.8 2 + 1.3 2 + 0.8 2 + 0.0 2 + 1.2 2 + 1.8 2 + 4.2 2 + 2.2 2 =0.5964 𝑟 4,8 = 3.0 41.7 2 + 25 2 + 77.8 2 + 10.6 2 + 2.3 2 + 1.8 2 + 1.9 2 + 3.0 2 + 3.2 2 + 3.8 2 + 1.3 2 =0.6471 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Consider 𝑾= (0.119, 0.106, 0.115, 0.090, 0.113, 0.061, 0.064, 0.113, 0.096, 0.065, 0.058), where 𝑗=1 𝑛=11 𝑤 𝑗 = 0.119+0.106+0.115+0.090+0.113+0.061+0.064+0.113+0.096+0.065+0.058 0.119+0.106+0.115+0.090+0.113+0.061+0.064+0.113+0.096+0.065+0.058 =1 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) 𝐕= 0.0609 0.0669 0.0535 0.0558 0.0589 0.0491 0.0632 0.0368 0.0525 0.0656 0.0493 0.0613 0.0466 0.0430 0.0331 0.0586 0.0576 0.0480 0.0416 0.0736 0.0000 0.0336 0.0207 0.0446 0.0361 0.0000 0.0000 0.0528 0.0545 0.0569 0.0427 0.0758 0.0358 0.0493 0.0370 0.0658 0.0385 0.0265 0.0337 0.0304 0.0417 0.0215 0.0295 0.0117 For example v 2,11 = 0.5622 0.4629 0.5701 0.4785 0.4248 0.5508 0.0000 0.5177 0.5140 0.4062 0.3700 × 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.058 =0.0215 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Calculate V+=(0.0669, 0.0632, 0.0656, 0.0586, 0.0736, 0.0446, 0.0528, 0.0758, 0.0658, 0.0385, 0.0417), where 𝐕 + = max i v ij j∈J ,i=1,⋯,m = v 1 + , v 2 + ,⋯, v n + v 1 + =max 0.0609,0.0669,0.0535,0.0558 =0.0669 v 2 + =max 0.0589,0.0491,0.0632,0.0368 =0.0632 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Calculate V-=(0.0535, 0.0368 , 0.0493, 0.0331, 0.0416, 0.0000, 0.0000, 0.0427, 0.0358, 0.0265, 0.0117), where 𝐕 − = min 𝑖 𝑣 𝑖𝑗 𝑗∈𝐽 ,𝑖=1,⋯,𝑚 = 𝑣 1 − , 𝑣 2 − ,⋯, 𝑣 𝑛 − v 1 − =min 0.0609,0.0669,0.0535,0.0558 =0.0535 v 2 − =min 0.0589,0.0491,0.0632,0.0368 =0.0368 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Calculate the distance of each project to most positive ideal solution using 𝑆 𝑖 + = 𝑗=1 𝑛 𝑣 𝑖𝑗 − 𝑣 𝑗 + 2 , thus 𝑆 1 + =0.1045 𝑆 2 + =0.1543 𝑆 3 + =0.0870 𝑆 4 + =0.0425 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Calculate the distance of each project to most negative ideal solution using 𝑆 𝑖 − = 𝑗=1 𝑛 𝑣 𝑖𝑗 − 𝑣 𝑗 − 2 , thus 𝑆 1 1 =0.0589 𝑆 2 − =0.0477 𝑆 3 − =0.0387 𝑆 4 − =0.0928 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Calculate the relative proximity index of each alternative (which is buying the house/apartment) to the ideal solution according to formula 𝑐 𝑖 = 𝑆 𝑖 − 𝑆 𝑖 − + 𝑆 𝑖 + , thus 𝑐 1 = 0.0589 0.0589+0.1045 =0.3605 𝑐 2 = 0.0477 0.0477+0.1543 =0.2361 𝑐 3 = 0.0387 0.0387+0.0870 =0.3078 𝑐 3 = 0.0928 0.0928+0.0425 =0.6853 TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA

TOPSIS: Example (contd..) Thus the ranking is 𝑐 4 ≻ 𝑐 1 > 𝑐 3 > 𝑐 2 Hence Alternative # 04 is the best (position # 01) choice followed by Alternative # 01 (position # 02), then by Alternative # 03 (position # 03) and finally followed by Alternative # 02 (position # 04) TOPSIS RNSengupta,IME Dept.,IIT Kanpur,INDIA