 Dear participants, welcome to the course on Supply Chain Digitization. It is jointly being taught by Professor Piyanka Burma, Professor Sushmita Narayanan, and Professor Dev Brathadas from IIM Mumbai. So, in this lecture we will focus on the case study, which we were talking about in the last class. So, this case study is all about the efficiency measurement of 9 manufacturing facilities. So, if you see the management wants to find out the efficiency of their 9 facilities. So, for each of these facilities they collected the data related to outputs and inputs. So, they consider two major outputs production yield and overall equipment effectiveness. They consider two inputs one is cycle time another one is resource utilization. So, based on these two inputs and two outputs they want to find out out of this 9 facility, which one is performing better in terms of efficiency, which one is not performing so good. That is the first objective and second objective is if one facility is not doing good in terms of efficiency, then where they should improve should they focus on increasing output or should they focus on reducing inputs. So, that is the question which is there in the mind of senior management. So, with that objective in mind we developed one data environment analysis model and this model was developed. So, now if you see the objective function it is non-linear. So, why non-linear? Because my decision variables are there in the numerator as well as denominator. So, 92 into V 1 plus 80 into V 2 divided by 32 into U 1 plus 88 into U 2. So, it is a non-linear. Similarly, if you look into the constants these are also non-linear because numerator as well as denominator both terms have decision variables. So, it is basically kind of y by x type. So, since it is y by x type these are non-linear expression. So, now the question is can I convert non-linear terms into linear terms because non-linear optimization is it will tough to do, but we know how to solve linear optimization model. In this case specifically it will be a linear programming model. So, can I convert this non-linear programming model into a linear programming model. So, if you look into these constants very carefully it is 92 V 1 plus 80 V 2 weighted sum of output divided by weighted sum of input. So, can I convert this term linear? Yes, it is possible. So, if I do this let us say I take this denominator out from here and put it over there. So, it will be 92 into V 1 plus 80 V 2 less than equal to 32 into U 1 plus 88 U 2. So, the constant has become linear. So, it was non-linear, but I have converted it into linear term. Similarly, all these constants can be easily converted into linear term if I bring the denominator from here and multiply it with this. So, that is how all the nine constants is converted into linear term. So, all of these are now linear. So, all these are linear. Now, the constants are converted into linear. Now, if you look into the objective function it is still non-linear. So, objective function is still non-linear. So, it is having y by x type of expression. The weighted sum of output is having decision variables V 1 and V 2. Weighted sum of input which is at the denominator is also having decision variables E 1 and E 2. So, therefore, this is non-linear. Now, can I convert this term into linear? Yes or no? So, now, the one point which is important here is efficiency score. So, if you remember the last lecture, the efficiency score, maximum value of efficiency is 1. It can never go beyond 1. So, we will take that point into account and we will try to convert this objective function into linear term. So, now, if you look into this denominator. So, let us assume that this is 1. So, if I take 32 E 1 plus 88 E 2 equal to 1. If I put this as an additional constant. So, this is my additional constant. I only had 9 constant for 9 facility, but I have put 31 plus 8 to 8 E 2 equal to 1. Now, the denominator has become 1 by virtue of this constant. So, as denominator has become 1, what is the maximum value numerator can take? Maximum value numerator can take you also 1. So, therefore, if I add this constant 32 E 1 plus 88 E 2 equal to 1, I am not losing any information. I am not putting any restriction which is not part of the model. So, therefore, if I make sure that denominator is 1, the numerator by virtue of this constant, because I also have this constant. So, this constant makes sure that this term 9 to 2 E 1 plus 80 E 2 plus 32 E 1 plus 88 E 2. So, this is my objective function and if I compare this objective function with this constant. So, objective function cannot take value more than 1. It has to be less than equal to 1, because the definition of efficiency value has to be less than equal to 1. And if I make sure that this value is 1, then the maximum value numerator can achieve equal to 1 because of this constant. So, therefore, I can convert this non-linear term into linear term adding one additional constant like this and then the objective function will be maximized 92 V 1 plus 80 V 2. And this term cannot take more than 1, because if it takes more than 1, then this expression, this objective function will be more than 1, which is not satisfying this constant. So, therefore, because of this constant, because of this constant for facility 1, this objective function value will be restricted to 1 and it can be maximum 1. With this conversion, I can convert non-linear model into linear model. So, now in this case, all my constants are linear. My objective function is also linear. So, I have a linear programming model. The objective is maximized 92 V 1 plus 80 V 2 and subject to 9 constants related to the efficiency value less than equal to 1 and one more additional constant, which has come by virtue of converting non-linear objective function into linear objective function. So, now, if I solve this model, I will get the value of V 1 V 2 V 1 V 2. Now, as per solution is concerned, it has now become easier because my model is no longer non-linear. This is a linear programming model. Now, if I change the facility, facility insist of facility 1, if I put facility 2, only thing, my objective function will change. This 9 constant will remain same. So, every time, if I change the facility, my objective function will change. In the non-linear model, in the linear version of the model, objective function will change and this additional constant will change. So, now, let us see, can I put this in a generalized form because in this case, I am having 2 output, 2 inputs and 9 facilities. In reality, you might have m number of outputs, n number of input, p number of facilities and so on. So, therefore, I need to have a generalized form. So, hope you understood the concept. Now, let us see how can I put a generalized form. So, generalized data envelopment analysis model for facility p. So, now, this is objective function. What is the numerator value? It is nothing but weighted sum of outputs. So, I have weight v k. v k is nothing but weight of output k. I am multiplying it with k th output value of facility p. So, for facility p, I have the output value of like, I have k th output. So, that should be multiplied by v k and then I am summing it up all of this. So, this is nothing but weighted sum of. Now, numerator if you see, I have u j multiplied by input j p. So, u j is weight of input j and how many inputs I have j equal to 1 to m. So, I have m inputs and for each input, I have weight u 1 is the weight of input 1, u 2 is the weight of input 2 dot, dot, dot, u m is the weight of input m. So, that is being multiplied by input j p. What is input j p? For facility p, the value of input j is written over here. So, that is how I can get. So, the denominator is weighted sum of. Now, I have weighted sum of outputs divided by weighted sum of inputs that is my efficiency score, I need to maximize it. And how many outputs I have? I have s number of output. So, k equal to 1 to s. So, v 1 represent the weight of output 1, v 2 represents weight of output 2 dot, dot, dot, v s represent weight of output s. So, that is how this objective function will look like. Now, I also have to write constant. So, what is the constant? Constant is nothing but efficiency score of all of these facilities should be less than equal to 1 and I am assuming that there are n facilities each with m inputs and s outputs. So, now this is my constant. So, constant is efficiency of each facility. So, how many facility I have for i equal to 1 to n? So, for each of this facility the efficiency is less than equal to 1. So, this is my model and this is an variable sir v k greater than equal to 0 u j greater than equal to 0. So, j takes value 1 to m because I have m inputs k takes value between 1 to s because I have s outputs. So, this is my generalized DEA model for facility p. p could take any value from 1 to n. So, now, if I have to convert this into linear form I can do the similar way as we have done for the example. Now, this term is non-linear, this term is also non-linear. So, first let us see how can I focus on. So, first I am converting the constants into linear. So, this is if I put this term weighted sum of inputs and bring it in the right hand side that would be linear term. So, that is what I have written v k into output k i summation over k equal to 1 to s minus summation over j equal to 1 to m u j into input j i less than equal to 0 for all i equal to 1 to n. So, this term basically has been converted into this term and now it is linear. Now, objective function has to be linear. So, as we discussed the input value we should convert we should make it 1. So, that is what we have written u j input j p summation over j equal to 1 to m. So, that is nothing, but weighted sum of input that we are making it equal to 1. So, again this is linear. Now, the output weighted sum of output is my objective function this is also linear. So, now this is a generalized d a model after linearization for facility p. So, p could be 1 2 3 4 dot n. So, if p equal to 1 this output of facility 1 will come here if p equal to 1 input of facility 1 will come here and this value will remain same this will not change if p equal to 2 output of facility 2 will come if p equal to 2 input of facility 2 will come. So, only this term and this term only the objective function and this constraint will change based on the value of the p the rest of these constraints the n constraints will remain same. So, that is a generalized model because when you go and in a real world you might have m number of inputs you might have s number of outputs n number of facilities. So, you can use this model. So, now we have understood generalized version of d a model. Now, let us see for this specific case although I have developed the model we have converted into linear form, but we have not solved it yet. So, how can I solve it? So, for that we will use excel solver and in one of the previous lectures professor P. N. Kabarma is already taught you like how to use excel solver. So, we will also use excel solver to solve the linear programming model because now our model is no longer non-linear we have converted it into a linear programming model. So, we can easily solve it using excel solver and. So, let us see how can we solve it using excel solver. So, first we will explain the major steps in power point slide then we will take you to excel solver and solve it in front of you. So, that you can get a good hands on. So, now these were my data I had two outputs production yield and overall equipment effectiveness. I had two inputs cycle time and this was utilization for nine facilities. So, first thing you have to put this data in an excel file. So, we will provide you the excel file with these data points. Now, let us understand the steps. So, objective function after linearization was like this maximize 90 to v 1 plus 80 v 2. So, now let us see how can we put this data into excel. So, I have four decision variable v 1 v 2 u 1 u 2. So, what is v 1 is nothing, but weight of output 1 that is weight of production yield v 2 is weight of output 2 that is weight of overall equipment effectiveness. u 1 is the weight of input 1 that is weight of cycle time u 2 is the weight of input 2 that is weight of resource utilization. And initially for these four decision variables that is four weights to output weights and two input weights we are taking the value 0 0 0 0. So, to initialize the process we are taking the value 0 0 0 0 the yellow cells which you are seeing now. So, initial value of v 1 is 0, initial value of v 2 is 0, initial value of u 1 is 0, initial value of u 2 is 0. Now, the coefficient of v 1 in the objective function is 92, coefficient of v 2 in the objective function is 80. So, the green cells represents the coefficient of v 1 in objective function that is 92, this is coefficient of v 2 in the objective function. So, the objective function is 92 into v 1 plus 80 into v 2. So, now v 1 and v 2 both are 0 0 initially. So, what will be the value 92 into 0 plus 80 into 0. So, that is 0. So, this term is nothing, but 92 into v 1 plus 80 into v 2. So, this is the term. Now, initially both are 0 0. So, therefore, objective function value is also 0. So, objective function is clear. Now, let us focus on constraint 0. So, what was constraint 0? Constraint 0 was derived from objective function to make the terms linear. So, what was the constraint 0 is nothing, but weighted sum of inputs. So, 32 into u 1 plus 88 into u 2. Initially the value of u 1 is 0, value of u 2 is also 0 and 32 is nothing, but coefficient of u 1 88 is coefficient of u 2. So, 32 will be multiplied by 0, 88 will be multiplied by 0. So, this term is nothing, but 32 into 0 plus 88 into 0. But if I write it in a generalized value generalized form it will be 32 into u 1 plus 88 into u 2. So, that is how this value is coming. So, initially it is 0, but I have to make sure that this value becomes 1. So, that is the right hand side. So, although it is 0, but when I solve it I have to make sure that this value is no longer 0, it has to be equal to 1, because I need to satisfy this constraint. So, I write now 0 0 0 0 is not satisfying my constraint. So, this is not a feasible solution. So, I have to get a feasible solution to get that first I have to make sure that this 0 is not 0, it has to be 1 that is how we have put this constraint. Now, I had 9 constraints for 9 facilities. So, what was the constraint? The first constant was 92 into v 1 plus 80 v 2 less than equal to 32 into u 1 plus 88 u 2 in a generalized form. Now, this term these two terms if I take it left hand side it will be 92 into v 1 plus 80 v 2 minus 32 u 1 minus 88 u 2 less than equal to 0. So, now how to write it in excel? So, 92 into 0 92 into v 1 initially now v 1 is 0. So, 92 into 0 80 into v 2 80 into 0 80 into v 2 80 into 0 minus 32 into u 1 32 into 0 minus 88 into u 2 88 into 0. So, that is this value. So, what is this value? This is nothing, but if I just write it over here 92 into v 1 plus 80 into v 2 minus 32 into u 1 minus 88 into v 2 minus 88 into v 2 minus 92 into u 2. So, this is this term since all of them are 0 I am getting this 0 and it is has to be less than equal to 0. So, same way you can put constraint 2 constraint 3 constraint 4. So, 1 constraint 9. So, that is how this calculations are there. So, now the objective function. So, if you go to excel we will see this value objective function value is nothing, but 92 into v 1 plus 80 into v 2 this value you will see 32 into u 1 plus 88 into u 2 92 into v 1 plus 80 v 2 minus 32 u 1 minus 88 u 2. Similarly, all of these calculations are mentioned. So, the formulas are there in excel and we will see this. So, how is this? This is nothing, but this value. What is constraint 2? Constraint 2 this is nothing, but this formula. What is constraint 3? Constraint 3 is nothing, but this formula. So, that is how formulas are entered and once you go to the excel solver we will show you how to do this calculation over there. Now, all the cells are having formulas as we have shown to you. So, once all the cells are entered with the formulas relationships are there we have to enter the same thing into excel solver. So, excel solver looks like this is the screen sort. So, the set objective I am trying to maximize because my objective is maximization type. So, I will click here maximization. So, what I want to maximize 92 v 1 plus 80 v 2. What is this term? This term is nothing, but this value this cell. What is this cell? 17. So, f 17 is nothing, but 92 into v 1 plus 80 into v 2. So, this cell I want to maximize that is what we have written f 17 then by changing variable cell. These are my decision variables. What are my decision variables? v 1 v 2 v u 1 u 2. So, v 1 is v 17 initial value 0 c 17 d 17 e 17. So, v 17 to e 17. So, these are my decision variables. Initially all are 0 0 0 0 0, but once you solve it these values will be changed and we will get the actual value of the decision variables. Then what are the constant? The constant is 32 u 1 plus 88 u 2 should be 1. That means, this value this value is nothing, but 32 u 1 plus 88 u 2 should be 1. So, that means, f f 19 should be equal to h 19 f 19 should be equal to h 19 and then I have these many constant. So, what are these constant? So, this constant is nothing, but this value this value is nothing, but this one 92 v 1 plus 80 v 2 minus 80 32 u 1 minus 88 u 2 less than equal to 0. So, this value is nothing, but this third constant nothing, but this one and the ninth constant nothing, but this one. So, we have already explained. So, all of these formulas are entered here. So, all of this should be less than equal to 0 that means, f 20 f 20 to f 28 f 20 to f 28 should be less than equal to h 20 to h 28. So, that is what we have written here. So, now, we have the model which have entered into the excel solver and we are making sure that unconstrained variables are non-negative. So, all the decision variables v 1 v 2 e 1 u 2 should be non-negative greater than equal to 0 and we are using simplex LP. So, if you use all of this parameter and click on solve, you will get a solution like this. So, now, this solution is very important for us to present. So, if you look into this efficiency results. So, efficiency is nothing, but the value of the objective functions. So, when I solve for facility 1, the model which you have seen here maximize 92 v 1 plus 80 v 2, the objective function will be 0.9572. It is not efficient 95 percent efficient, 5 percent efficient I still could achieve. For facility 2, it is 87 percent efficient and so on. Now, if I look into this data, I can see factory 4 is efficient, 4 is efficient and factory 7 is efficient. So, factory 4 and 7, these are efficient 100 percent efficiency score is 100 percent whereas, other factories are not efficient. So, if we look factory 8 is having lowest efficiency score 76 percentage efficiency score. Now, I still have 24 percent to improve. So, the question is where should I focus on? If I am a manager of factory 8, where should I focus on? I can still improve 24 percent efficiency. Should I focus on increasing the outputs or should I focus on reducing the inputs? So, if you see the value of the weights that is the decision variables for factory 8, if I see as far as 2 outputs are concerned, output 2 is having high weight 0.01. Similarly, if I look into the input, the resource utilization is having high weight. Now, if I compare these 2, V 2 is having more weight V 2 is 0.01 1 whereas, U 2 is 0.010. So, V 2 is having little bit more weightage compared to U 2. If I have to increase the efficiency by putting minimum effort, I should try to increase the overall equipment effectiveness of facility 8 because weight is highest over here. So, since weight is highest over here, I should try to focus on increasing the output 2. That means, for facility 8 I should try to increase the output of overall equipment effectiveness. So, if I put my resources to increase the output of overall equipment effectiveness, my efficiency will increase at a faster rate. Let us say which factory? Let us say I am to focus on 0.92. 0.92 is the efficiency score of facility 9. So, it is not efficient enough, 8 percent efficiency I can still increase. So, if I want to increase the efficiency, where should I focus on? Should I focus on this in output? Should I focus on this output? Should I focus here? Should I focus here? I have to see the value of these output parameters. Value of the output parameters, this is the maximum 0.03. So, I can see for facility 9, if I focus on reducing the cycle time, then my efficiency will increase at a faster rate. That is how the interpretation should be. So, out of these four decision variables B 1, B 2, U 1, U 2, you have to see which decision variable is having more value. So, these are nothing but the weights. So, wherever you have more weights, my focus should be on those parameters. If more weight is on output, then I should try to increase the output value. If more weight is on input parameters, then for that specific parameter, for that specific facility, I should try to reduce the input time values. Then I will achieve efficiency at a faster rate. The efficiency score will increase at a faster rate. So, now we have solved both the problems which we stated earlier in the case. The first question was out of this nine facility, which facility is more efficient? So, I can say facility 4 and facility 7, these are the efficient because their efficiency score is 100 percent. So, 4 and 7 efficient. Now, rest all are inefficient, but among the inefficient facilities, I can see facility 8 is most inefficient because the efficiency score is 76 percent. They still can improve the efficiency by 24 percent. Then after facility 8, which is the next least efficient, 87 percent I can see. 87 percent is facility 2. So, facility 2 has still option to increase efficiency by 13 percent and so on. So, I can increase the efficiency. Now, the next question is how can they increase the efficiency? Since, 8 is having lowest efficiency score, I will see the value of waste B 1, B 2, E 1, E 2. Since, B 2 is the highest weight. So, I should try to focus on increasing the overall effectiveness, overall equipment effectiveness. That means, right now my value is 68. You see, right now my value is 68. The model suggests that you should increase it. I should try to increase it to 70, 72, 73, 74. If you see, it has maximum value, 85. Right now I am standing at 68. So, I can still improve 17 percent more from the, from my like peers. So, therefore, if I increase this value 68 to 70, 70 to 75, my efficiency will keep on increasing. Rather than focusing on all other parameters, I may focus on increasing the effectiveness of equipment. Similarly, if I focus on 9 facility, I should try to reduce this value and so on. Similarly, if I focus on facility 2, it is suggesting me that I need to increase overall equipment effectiveness. I need to increase this value 78. I have to increase it little bit. At the same time, it is suggesting me that resource utilization also you need to reduce. Right now, I am, I am increasing too much, I am utilizing too much resources, 94 percent. So, it is suggesting me, although you have to increase overall effectiveness of the equipment, you also have to reduce resource utilization. You are utilizing too much resources and output is not that great. So, therefore, it is focusing that either I should reduce resource utilization or I should increase overall effectiveness of the equipment or I can do both. Then also, I will be able to achieve efficiency. So, this data, the output of the model is not only telling me which one is efficient, which one is not efficient. It is also telling me where should I focus on out of these 4 parameters. So, they focus on output, they focus on input, they focus on combination of output and input. Like in the case of facility 2, I am focusing on combination of output as well as input. So, I am, my focus is increasing output 2 and reducing input 2. Similarly, for factory 8, my focus on increasing output 2 and reducing, you can see this value reducing the value of resource utilization. So, we got the idea, the managerial interpretation now let us see quickly how we can do this using excel. So, I already explained the excel solver excel like interpretation. We will now go to the excel and run it once again and show you how this excel works. So, if you open this excel file, then these are the data which is already there given to us. So, do not see this data now, we will explain and now this is my optimization model as we have explained in the PPT. So, initial values are 0, 0, 0 and 0. These are my initial values. So, now, if I select facility 1, I have selected facility 1, if I have selected facility 1, I am getting 92 into v 1, 80 into v 2 this value, since both of them are 0, 0 is 0 and then input values in the objective function for facility 1 is 32 and 88. Now, if I go to data solver and solve it, you will get 95 percent, 95.72 percent as the efficiency score and see the weights have changed. The initiality was 0, 0. Now, after solving weights have changed and I can see OEE is having 0.011 weight and then 0.008 for cycle time 0.008 for resource utilization. The same value I have copied and pasted it over here. Now, let us say I want to go to facility, I want to see facility 4. So, facility 4, I have 90 percent, I have 90, 78 as my output parameters, 30, 82, 30, 82 as my input parameters and I have used a V lookup function. So, automatically if you select facility 4, these 4 values will automatically be changed. Now, if I solve here, you will see a solution 100 percent efficiency. That is what we discussed in the PPT also, it is using 100 percent efficiency. So, same way let us see for factory 8. So, if I change factory 8, see automatically 74, 68 has come coefficient of outputs, coefficient of objective function. Similarly, 35, 94 has also come over here. Now, solver, solve, you are getting 76 percent efficiency. That is what we discuss in the PPT and these values are my weights and see this is having highest weight 0.011, then resource utilization and then cycle time and production. So, now, if you go through the excel file, you will see the formulas as we explained in the PPT. So, I request you please go through each of this cell and see how these formulas are calculated. Whatever you discuss in the PPT, the exactly same formulas are written over here. You can also explore on your own. So, one thing I would like to mention here, which we did not talk in next PPT is this constant. So, b 7 to e 17 is nothing but these decision variables. I am putting a very minimum value 0.000601. So, I do not want to get weight 0. So, that is why we have put a very minimum value of decision variable. So, epsilon. So, instead of writing 0 weight, I wanted to keep the value 0.000001. So, if you get 0 values, you might think that these parameters are not important. So, therefore, we have kept a very minimum value. So, that we do not get 0 for any of these weights. We should not get 0 for V 1, V 2, V 1, V 2. So, we should have at least some importance. So, therefore, I have kept a very, very low importance 0.0001. Otherwise, if I remove this, you will get 0. This parameter will be 0. This parameter will be 0. This decision variable will be 0. So, although like when you present to somebody, they might think that 0 means that parameter is having no importance, but you know in reality cycle time production yield has importance. Although in this case, it is not coming up as a top important, but it has importance. So, therefore, I have kept a very low value 0.0001. So, 601. You can, if you want, you can keep further low value also, but I suggest avoid putting 0. Then, the interpretation will be like if people might think it is of no use. Now, if you solve it, you will get like solver found a solution. All constants and optimality conditions are satisfied. So, linear programming model which you got, it is optimal and you got a solution. So, that is how you get all the solutions. So, I have to run it 9 times. If we keep on changing this, then every time you have to go solver and run it. So, if I run 9 times, I will get 9 such values of the weights and 9 efficiency score. That is how the scores are calculated and we have used this in the PPT. So, hope you could get the idea of data development analysis and the usefulness of this technique. It is very useful technique. It is used for measuring efficiency. I can compare efficiency of similar facilities, similar warehouses, similar distribution centers, similar factories. It is a very, very useful tools. Only thing we need to find out important value of outputs, important value of inputs. If we have inputs and outputs data, these models can be applied. And we can find out which one is efficient and which one is not efficient. And not only that, if a factory or facility is not efficient, like should you focus on outputs, should you focus on inputs, so efficiency increases. So, with that I would like to stop this lecture. So, thank you so much. Look forward to seeing you in the next class.