noc19-mg15 Lec 38-VIKOR

Welcome back, my dear friends, a very good morning, good afternoon, good evening to all of you, wherever you are in this part of this world, whether in India or abroad. And as you know, this is the DADM2, which is Data Analysis and Decision Making 2 course under the NPTEL MOOC series. This course total duration is for 12 weeks, which is 30 hours and 30 hours basically being split into 60 lectures, each lecture is for half an hour. And we have, as I said, it is 12 weeks, so each week we have about 5 lectures, each being for half an hour and after each week we have assignments. So, as you can see from the slide, we are in the lecture number 38, which is the 8th week. That means you already have taken 7 assignments and we are in the verge of trying to finish this 8th week and you will basically have the 8th assignment for assignment number 8. And my good name is Raghunandan Sengupta from the IME department at IIT Kanpur. So, if you remember, we were discussing about the VICOR method, the steps. So, I did discuss the initial part of this pseudo code and I will try to wrap up the pseudo code and discuss because I did not finish the whole part. So, I will add it here and so it will be easy for all of us to do it accordingly. So, as we were discussing, so you will basically have the ranking system based on a normalized rank. That means in the LP norm for each and every ith alternative will be based on a scaling system where the denominator is the difference between the maximum and the minimum. And in the numerator, you will have basically the difference between the maximum and the actual distance which you have or the weightage which you have, the priority which you have, which is C j a i. And this factor can be multiplied by the weights depending on the level of importance you want to put for each and every alternative. So, if the alternative weights are basically 1 by n depending on the number of alternatives which is the criteria 1 by n. So, each would basically get the same weightages and then you will basically multiply, find out the pth power, then sum them up, find out the 1 by pth power corresponding to the norms which you have. And we have already used the concept of the norms to illustrate that as p increases how the difference between the norms for L 30, L 31, L 32 so on and so forth decreases. I am giving 30, 31, 32 as an example. So, you will start for each. So, once this ranking system is known to you, how will rank them based on the distance concept, you will take for each i is equal to 1 to m, you will take each and every criteria. For i is equal to 1, you will take the criteria 1 to n, find out those corresponding L p 1 norm, put it those values such that you can rank them. Then change i is equal to 2, find out for j is equal to 1 to n, again the L p 2 norm, put it L p remains fixed, remember that 2 means the second alternative, continue doing so till we have L p 1 to L p m and based on that we will proceed. So, you will calculate the values of C j a plus and C j a minus and use them for the ranking. So, I will start going through slides and do that in an excel sheet also. So, assume the, you will basically construct the normalized decision matrix. So, assume the decision matrix is x, size is m cross n. This is same which I have been repeating time and again, but I will still do it and without being to repetitive in order to make you understand that you are trying to basically compare and find out the overall priority weights or priority values for each and every criteria and each every alternatives. So, the size as I said is m cross n. So, the first row would be x 1 1 till x 1 n, where x 1 1 means the so called actual non-normalized priority values for the first criteria on the first alternative and while x 1 n basically being the priority weight or the value for the nth criteria on the first alternative. Similarly, x 2 1, x 2 2 till x 2 n would be the corresponding values of the priority scores. Say for example, or values whatever you say for the first, second, third, fourth till the nth criteria for the second alternative only. Similarly, the last row x m 1 till x m n is basically the criteria values being given for the mth alternatives corresponding to the fact that I am considering the first criteria till the last criteria. Now, you can basically use the normalization concept. The normalization concept again I am considering the very simple, this is a very simple concept which I which I consider where you take the values corresponding to the fact that you are for trying to find out the normalized scores as the sum is 1. So, whether you use just cube, square whatever it does not matter. I will show you an example which may be a little bit variant, but you will understand that. Then once you find out the normalized values you have the r and then these and again this is of size m cross n which is the normalized scores or the weights or the values corresponding to the jth criteria for the ith alternative. So, let us basically consider a value and I go a little bit slows in order to make you understand. So, this I will mention as x, so I will consider a 1, I am writing it, so that is everything is easy for us to understand, I will take four alternatives and I will take a little bit more number of criteria. So, even though it will take time, but I am sure you will understand. So, I have five criteria, they are values are given, but the corresponding units may be different which I said that they are attributes, so the units may be different. So, we are able to consider that in the Y code, I am not going to the detailed discussion again I will just solve the problem. So, consider I will go column wise, so it will be easy for us to understand, so consider these are in on a say for example, on a unit of two units places 30, 40, 70, 40, 30, 50, 70, 40 and the units and the 10 values are there, others I will consider this decimal, this I will consider, these values I will consider on a scale of say for example, I guess I am consider, this I am taking arbitrary values in order to make you understand. So, these are for each and every criteria and the weight, so I will mark these criteria with a color yellow, these as, so this is done. Now weights I would not take now, so let us go one by one, so I am basically finding the row, consider it is basically row wise. So, let me use the first concept of normalization, normalization number one, so I am using different normalization, which you can change accordingly, but let us proceed. So, if I use the simple, let us consider I am using the L3, so let us proceed by L3, so this is done. Now I will basically freeze the values and at each cell, so they do not want to be any change as I copy the values, so let us consider, so sum should be 1, this is 1. So, now I normalize the second value, so it will be now C24, D24, E24, F24, G24. So, this value is done, again 1, then I go to values, this is 25, this is 25, 25, 25 and the last row would be 20. So, let us, I will double check that, do not worry, and when I come to 26, 26, 26, 26, let us check whether the values are same, I could have done the column wise also. Now, once it is done, what does the, so these part is done, so I have normalized them. So, once normalized, I go to the next step, which is basically find out the weights, give the weights. So, let us go and give some weights, so let us, so this will be of 5 cross 5, so I will write the value as, so it will be, let us consider them of equal weightage, so it is one fifth, so these are the, so these are the overall matrix, which I have. So, if you want, I can shift it and write the corresponding, so these are the criteria weights, which I have. So, this becomes 2, this becomes 3, this becomes 4, this becomes 5. So, this you have the weights. Now, once the weights are done, what does it give, I have to find out the multiplied value. So, I will reduce the view and multiply them. So, I need to find out f. So, let me put it f, the normalization step 1, I can change. So, this is of size 4 cross 5, this is 5 cross 5, 4 cross 5 into 5 cross 5 would give you 4 cross 5 matrix. So, this would be normalized, so called normalized values for the 4 alternatives and 5. So, this is the f value. So, this is MMULT of the first row and the first column. So, obviously this row will get fixed here. So, I, so freeze it, then I cap it for all others. Let me check whether I am right. So, I am considering all the different columns with respect to the first row. Then I multiply, this is the second row, which is for A2 corresponding to C1. So, again I freeze the values. So, let us check whether they are the same. So, this is, the columns are changing, the rows is the second one. Then I go to the third row, which is for A3, alternate 3 multiplied by corresponding C1, C2, C3, C4 for each cell. So, let me freeze the values. Let us see they are frozen. So, it is basically the row remains the same, the columns are changing. Finally, it is the last row multiplied by each corresponding column. Last row is basically the fourth alternate. Let us freeze the values. So, it is basically. Now, these f values are obtained, which is multiplication of each and every normalized vector, which was basically R into weights, which is W. So, normalized vector, let me put the name also. So, it is easy. Now, let us do one thing. Let us change the normalization and note these values. Let me, I will come to that later on, later on. Normalization, which I am using concept is, let me mention it. So, it will be L3. So, we are taking PS3. Now, we have found out f 1 1 till f 1 n, which is the overall normalized weights, weighted criteria for the first criteria, for the first alternative. Similarly, f 1 n would basically be the weighted normalized priority for the n-th criteria for the first alternative. Similarly, the last row, which is, you can see f, I should, which is f m 1 till f m n is basically normalized weighted, so-called points or scores for the first criteria on the m-th alternative. Similarly, f m n, so this m and r in the suffix m, n are the weighted normalized scores for the n-th criteria for the m-th alternative. So, these being done, now we proceed. So, we need to basically find out f j star and f j minus, which are basically for each and every row or column, depending on how we have mentioned them, we will find the best and the worst. Now, remember f j star, that means for j is equal to 1, you will take along each and every alternatives. So, f j is there. So, let us go to, so we will put it at f j plus, so f star 1. So, this we can find out for each criteria. So, we will take, so this will be for the second criteria, this will be for the third criteria, this will be for the fourth criteria, this will be for the fifth criteria. Similarly, I will have f j minus, let me write it. So, minus value would goes constant to the, I will put the minus value as this. So, this is minus for the third criteria and this is for the third, this is for the third, then I come to the fourth, then I come to the fifth, done. Now, I need to find out the maximum, here the maximum and here the minimum. So, for all the i's, so remember for, I will take for all the i's. So, it is going to max for this. So, I take max, so this is the max values. Now, I take the min for the worst criteria. So, the means are calculated. So, this is the max value and the min value. Now, remember here, the max and min would be utilized depending on the weights, which you have already considered. So, giving the max and the min, we want to find out, now the weights are already given. So, you want to find out for f j minus, each and every alternate multiply the weights. So, let us go. So, let us consider the normalized denominator. So, the f j plus star minus f j minus. So, this is what I have. So, this would be f star comma 1 minus f minus. So, these are the values. So, let me, this is the second, this is the third, this is the fourth, this is the fifth. So, here I have the first criteria, this is for the second, this is for the third and this is for the fourth, this is for the fifth, for each and every criteria, we have already done that. So, let us find out, this will come in the denominator. So, this will be, now I go to the ratios, that means, f j plus minus that ith one. So, for each and every j is equal to 1 to n, I will sum them up. So, I either sum up for at 1. So, I am going to multiply the weights. So, it will be the weights multiplied. So, I will put weight multiplied by max value minus the 1, which is the max value of this minus each and every cell value plus, I am taking equal weightages. So, max value minus the second plus the weights multiplied, sorry, I should put the multiplication sign or else it would not come. So, this would be max minus the third cell bracket closed plus the fourth weight W 4 multiplied by the max minus the 35, 36, 37, 38. So, you will basically have, let me check, you are going to go by the, sorry. So, I have just the j, then I am changing j is equal to 1, 2. So, weights minus the, this would be 40 D, D 41 minus j is equal to 1. So, it will be second cell value, which will be D 35. So, it will be, this will become E and this will become F. So, F finally, you will basically have the last cell multiplied by, I have not done it. So, it will be F, it will be G, G. So, I have basically taken all the values along for the first one. Similarly, I need to find out the corresponding. So, the values are being calculated. So, I need to copy them. So, once I copy them, I need to basically, so, the weights are to be fixed because they are not changing. So, I again, the max values are also to be fixed. So, if I go there, I will just do one, one of them and it will become much easier for us to solve. So, once I have this, again I go for the, there is some change, weight, weight, weight. So, I have these values for each and every, you know, I can find out for the second one. So, these values we shall get for the maximum and minimum. I will do it accordingly for the F minus of 1, 2, 3, 4, 5. I will solve this example and also show, continue this in the last two classes and try to basically give the overall picture that how you will basically rank in the concept of Euclid method. With this, I will end this class and continue the discussion about the solving step by step. So, it will be easier for you to understand. Have a nice day and thank you very much.