ANALYTICAL HEIRARCHY PROCESS Theory with a Case AHP Process is explained with two methods namely Normalization and Eigen vector Method and error rectification is explained for global weights calculations. Same is explained with an example as well.
DEEPAK AGGARWAL 1/29/2018 Analytical Hierarchy ProcessINTRODUCTIONThe AHP process was first developedand introduced by Thomas L Saaty as early as in 1970 which afterwards was usedand researched exhaustively.As per Thomas L Saaty, theAnalytical Hierarchy process can be defined as is “Structured technique fororganizing and analysing complex decisions, based on mathematics andpsychology.”Basically it is a decision makingmethod. It can be used at number of areas like government, purchasing goods orservices or making decisions based on perception like going for a college.AHP need not necessarily give thecorrect or accurate decision but can be made the most consistent one with thedata based on ratings or perception.
It helps individuals or organisations toset priorities where data based on ideas, feelings etc. can be quantified.AHP PROCESSAHP can be used in number of steps:A.
Organizingthe parameters of criteria and alternative: 1. Setting of Goal orObjective: In order to approach the solutionof the problem, the first step includes freezing of agenda or output intendedto become the outcome like selection of college or purchasing a house etc.2. Setting of Criterion orCriteria: The next step involves setting ofthe criteria based on ideas, feelings or ratio data collected throughquestionnaire3. Setting of Sub Criterion or Criteria: Theabove step can be further de-clustered into Sub criterion for the objective setto give in depth analysis of the decision making.4.
Alternatives: Thefinal Process is to evaluate the alternatives available among which the choiceshave to be made.5. Tree Structure:Finally arrange all the above steps into a tree like structure to evaluate theglobal preferences of all the alternatives. B. Collectdata for comparison matrix:1. Onlytwo alternatives / criterion can be compared at a time.
2. Thepair can be compared using fundamental scale (1 to 9)3. Itcan measure both tangible and intangible criterions.4. Thecriterions can be scaled in following manner:5: Extremely favourable à For Alternative “A”4: Very Strongly Favourable3: Strongly Favourable2: Slightly Favourable1: Equally Favourable _______________________________________2: Slightly Favourable3: Strongly Favourable4: Very Strongly Favourable5: Extremely favourable à For Alternative “B”From the above example if choose let’s say 4 for Balternative then preference of B over A is 4 OR Ratio of B: A = 4. In thesimilar manner other data can be collected through questionnaire or any othermethod.C. Developthe comparison matrix:While preparing comparison matrix,arrange all alternatives both on Rows and columns and while calculatingpreference Numerator to be kept as Row Alternative and Denominator to be keptas column alternative.
For example if the weight-age ofeach alternative is S1, S2……Sn, then the Matrix cell will look like “S1/S2” for1st row and 2nd column and so on. The resultant will be asquare matrix as per the following details: Alternative 1 Alternative 2 Alternative 3 Alternative 1 S1/S1 S1/S2 S1/S3 Alternative 2 S2/S1 S2/S2 S2/S3 Alternative 3 S3/S1 S3/S2 S3/S3 OR Alternative 1 Alternative 2 Alternative 3 Alternative 1 1 S1/S2 S1/S3 Alternative 2 S2/S1 1 S2/S3 Alternative 3 S3/S1 S3/S2 1 Withweight-age of Alternative 1 à S1 Alternative2 à S2 Alternative 3 à S3Note:1. If there are nalternatives then comparison matrix can be made using nC2 formula which givesnumber of comparisons required for it.2. Also it can be noticedthat diagonal elements are always unity as it’s the ratio of same alternatives.3. It can also be noticedthat elements which are diagonally opposite to each other are inverse of eachother i.
e.aij= (1/aji)Continuing toit,S1/S2= 1 / (S2/S1)S1/S3= 1 / (S3/S1)S2/S3= 1 / (S3/S2)D. ProblemSolving:There are two ways inwhich these problems can be solved to prioritise the decisions:a. NormalizationMethodb. EigenVector Method a. Normalizationmethod:In this method theabove square matrix is first normalized column wise using the followingformula: After normalization ofthe matrix, the sum of column elements will be unity.
Now if all the columnsare identical then the normalized matrix is consistent without any error andthe weights of each alternative are individual row value for that particularalternative. For example, if normalized matrix comes out to be: Alternative 1 Alternative 2 Alternative 3 Weights Alternative 1 0.5 0.
5 0.5 0.5 Alternative 2 0.4 0.4 0.4 0.4 Alternative 3 0.
1 0.1 0.1 0.1 However, if the columnsare not identical then average of each column element of a row is taken to getthe weight of that particular row or alternative. And consistency error has tobe calculated using the following formulas:For ?max calculationsuse the following formula:Now if the consistency ratio Cr< 0.1, then the weights are consistent else correction is required which wewill see in Eigen vector method.
b. EigenVector Method:Inthis method, instead of normalising the square matrix, we create a unity matrixand multiply it with some value ? (keep it blank initially, we will solve andtry to find its value) to create a matrix “A-?I” and we have to make itsdeterminant zero to solve for the value of ? using the following formula:To calculate the determinant of theabove equation, use “Mdeterm” in excel, it will give some value of determinantof above equation.Now for solving the above equation,· First set ? to anyhigher value say 20, and then run the If analysis with goal seek.· To solve, set the goalof making determinant to value zero by changing the values of ?.· Then run it.
Wewill be able to get the value of ? which will be “?max” forcalculation of consistency ratio. Now, to solve for the weights of thealternatives make n equations using following formulas:Andmake objective function as: Using solver for above n equations and an objective function find individualweights of all alternatives.E.
Error Calculation and rectification:Ifthe matrix and weights are consistent and there is no error then the followingequation will hold true: If not, then following equation will holdtrue:Where,?ijis the error associated with it. If we solve the above equation we get:So we have to take transpose of square matrix and multiply it with / matrix. In the result following analysis to bedone:· In this matrix if checkfor the maximum value and if it lies above the unity diagonal then find outcorresponding wi / wj valueand substitute it the original square matrix in the same/corresponding cell andagain run the process for calculation of consistency ratio till it becomes lessthan 0.1· If the maximum valuelies below the unity diagonal, then take corresponding mirror value (acrossdiagonal) which will be the minimum value of matrix and find out correspondingwi / wj value andsubstitute it the original square matrix in the same/corresponding cell andagain run the process for calculation of consistency ratio till it becomes lessthan 0.
1This methodology gives you theweights of all the corresponding alternatives. And to select the globalpriority, multiply each corresponding weight of criterion and sub criterion andsum for different products. This is explained further in detail with example.ProblemRam has to buya new house for himself and his dependent family. He came across threedifferent choices of houses which he thinks are good enough to be bought. And whileshort listing these three houses he chose following criteria:a. Locationb. Amenitiesc.
Cost/resaled. Safety/SecurityHe asked tohis friend for advice who was studying analytical hierarchy process in hiscollege. He agreed to help him and gave him following questionnaire:Rate circle, the following with ascore from 1-5 with 5 as most favourable and 1 as least favourable:Q1: Location Amenities Q2: Location Cost Q3: Location Security Q4: Amenities Cost Q5: Amenities Security Q6: Cost Security Q7: Location A LocationB Q8: Location A LocationC Q9: Location B LocationC Q10: AmenitiesA Amenities B Q11: AmenitiesA Amenities C Q12: AmenitiesB Amenities C Q13: Cost A CostB Q14: Cost A CostC Q15: Cost B CostC Q17: SecurityA SecurityB Q18: SecurityA SecurityC Q19: SecurityB SecurityC The responseto all of the questions were collected and used for analysis purpose usingAnalytical Hierarchy Process.SOLUTIONThe followingmatrixes were obtained from above questionnaire: X = Location Amenities Cost Security Location 1 4 2 1 Amenities 0.25 1 0.25 1 Cost 0.
5 4 1 0.5 Security 1 1 2 1 XL = A B C A 1 0.5 0.25 B 2 1 0.
5 C 4 2 1 XA = A B C A 1 2 4 B 0.5 1 2 C 0.25 0.5 1 XC = A B C A 1 0.
5 2 B 2 1 4 C 0.5 0.25 1 XS = A B C A 1 4 2 B 0.25 1 0.5 C 0.
5 2 1 Now usingNormalisation method 1st we get following matrixes as: X = Location Amenities Cost Security Weights Location 0.36 0.40 0.38 0.29 0.36 Amenities 0.09 0.
10 0.05 0.29 0.13 Cost 0.18 0.40 0.19 0.
14 0.23 Security 0.36 0.10 0.
38 0.29 0.28 Here, thecolumns are not identical and there is some inconsistency. The consistencyratio comes out to be: n= 4.
00 CI= 0.16 ?MAX = 4.48 RI= 0.99 CR= 0.16 Which isgreater than 0.1. Hence Eigen Matrix will be used for error rectification. Wewill do it in later part of the solution.
Now other matrixes: XL = A B C Weights A 0.14 0.14 0.14 0.14 B 0.
29 0.29 0.29 0.29 C 0.
57 0.57 0.57 0.57 LocationPreferences of A, B & C and it is consistent so no need of errorcalculation. XA = A B C Weights A 0.
57 0.57 0.57 0.57 B 0.
29 0.29 0.29 0.29 C 0.
14 0.14 0.14 0.14 AmenitiesPreferences of A, B & C and it is consistent so no need of errorcalculation. XC = A B C Weights A 0.
29 0.29 0.29 0.29 B 0.
57 0.57 0.57 0.57 C 0.14 0.14 0.14 0.14 CostPreferences of A, B & C and it is consistent so no need of errorcalculation.
XS = A B C Weights A 0.57 0.57 0.57 0.57 B 0.14 0.
14 0.14 0.14 C 0.
29 0.29 0.29 0.29 SecurityPreferences of A, B & C and it is consistent so no need of errorcalculation.Hence weightsof A,B & C with respect to Location, Amenities, Cost and Security iscalculated using normalization method and it is consistent since all thecolumns are identical in each matrix. Therefore in these weights there is noneed to calculate error as it is nil.However, whilecalculating weights of Location, Amenities, Cost and Security, the columns arenot identical and there is consistence error of 0.
16 which is greater than 0.1.Hence, error rectification is required in it for further process.
To rectifyerror, we need to first calculate weights using Eigen vector method: X = Location Amenities Cost Security Location 1.00 4.00 2.00 1.00 Amenities 0.25 1.00 0.25 1.
00 Cost 0.50 4.00 1.00 0.
50 Security 1.00 1.00 2.00 1.00 Form a unitmatrix for further calculation, I = 1.
00 0.00 0.00 0.00 0.00 1.
00 0.00 0.00 0.00 0.00 1.00 0.00 0.
00 0.00 0.00 1.00 Now usingblank values of ?= Determinant = Make followingmatrix: X-?I Location Amenities Cost Security Location 1.00 4.00 2.00 1.00 Amenities 0.
25 1.00 0.25 1.00 Cost 0.50 4.00 1.00 0.50 Security 1.
00 1.00 2.00 1.00 Substitute a value of ?= 20 let’s say and runwhat if analysis with goal seek of determinantsetting to zero by changing ?. We get, ?= 4.
45 Determinant = 0.00 And, X-?I Location Amenities Cost Security Location -3.45 4.00 2.00 1.00 Amenities 0.25 -3.
45 0.25 1.00 Cost 0.
50 4.00 -3.45 0.50 Security 1.00 1.00 2.00 -3.45 For weightscalculations, first make empty weights cells like: Location Amenities Cost Security And make 4equations/constraints using formula:And set theobjective function as per the following formula:Now, run thesolver with these constraints and objective with linear function and nonnegative variables.
We will get following results: Location Amenities Cost Security Objective 0.36 0.12 0.24 0.28 1.00 Now calculatethe consistency ratio using following formulas:We get thefollowing results: n= 4.
00 CI= 0.15 ?MAX = 4.45 RI= 0.99 CR= 0.15 The consistencyratio is 0.15 which is greater than 0.1.
Hence error rectification is required.For errorrectification we first take transpose of original matrix: XTRANSPOSE 1.00 0.25 0.
50 1.00 4.00 1.00 4.00 1.00 2.
00 0.25 1.00 2.00 1.00 1.00 0.50 1.
00 Now using thefollowing formula we calculate the error matrix:It comes outto be: 0.36 0.12 0.
24 0.28 0.36 1.
00 0.73 0.76 1.30 0.
12 1.37 1.00 2.10 0.45 0.
24 1.31 0.48 1.00 1.70 0.
28 0.77 2.24 0.59 1.00 Here, the valueof 2.24 is highest. But it is below the diagonal.
Hence the corresponding mirror value across diagonalis taken which is 0.45Now the Wi/Wjvalue of that cell is 0.45 which is needed to be replaced in original matrixwhich comes out to be: X = Location Amenities Cost Security Location 1.00 4.00 2.00 1.00 Amenities 0.
25 1.00 0.25 0.45 Cost 0.50 4.00 1.
00 0.50 Security 1.00 1.00 2.
00 1.00 If we againsolve it with Goal Seek and then solver and calculate the final consistencyratio it comes out to be following values: ?= 4.07 Determinant = 0.00 n= 4.00 CI= 0.02 ?MAX = 4.07 RI= 0.99 CR= 0.
02 Location Amenities Cost Security Objective 0.37 0.09 0.23 0.30 1.00 Now the finalconsistency ratio is 0.02.
Hence it is consistent with above weights.For the finalpreference we have to prepare the chart once again: 0.37 0.09 0.23 0.3 0.29 0.57 0.29 0.57 0.14 0.57 0.14 0.29 0.14 0.57 0.29 0.14 Now the globalweights of each of the house A, B and C is calculated by multiplyingalternative weight of category and category weight and summing all theseproducts to get the total weight-age of that particular alternative.WA= (0.14)*(0.37) + (0.57)*(0.09) + (0.29)*(0.23) + (0.57)*(0.3) = 0.341Similarly,WB= (0.29)*(0.37) + (0.29)*(0.09) + (0.57)*(0.23) + (0.14)*(0.3) = 0.307And,WC= (0.57)*(0.37) + (0.14)*(0.09) + (0.14)*(0.23) + (0.29)*(0.3) = 0.343Hence weight-age of house C ishighest. So the friend recommend ram to go for house C as it is the best suitedaccording to his preferences with an error of 2% approximately.