 Dear participants, welcome to the course on supply chain digitization. It is jointly being taught by Professor Piyanka Burma, Professor Sushmita Narayana and Professor Devav Brathath Das from IIM, Mumbai. So, in today's lecture, I am going to focus on network optimization and how supply chain to digital twin can be useful to do this. So, first of all we will take a case study and then we will explain this concept of supply chain digital twin specifically from network optimization point of view. The supply chain planning head of the western region estimated that the demand would be doubled in the next 3 years. So, demand is going to increase. Therefore, after deliberation in the boardroom, the senior management of the company decided to set up a new factory in this region, new factory. The prospective locations are Nasik and Orangabad. The lease for the DC in Mumbai as found in the Greenfield analysis. So, when we did Greenfield analysis, we have found out that the optimal location of the DC was Mumbai, but that lease will be expiring. So, the senior management is deliberating where to have the distribution center. The prospective locations are V1D and RP. So, they have two prospective locations for factory, they have two prospective locations for distribution centers. So, what should be the optimized supply chain network for their business if demand becomes doubled in the next 3 years. So, this is the data of demand. I have 4 markets Pune, Mumbai, Ahmedabad and Surat. Early demand has become doubled compared to the earlier scenario and I also have their latitude and longitude. So, I know where these markets are exactly located. I know their yearly demand as well. So, my objective is to find out what should be the optimized supply chain network for their business. So, that means, I need to find out where should they locate the factory, should they locate in Orangabad, should they locate in Nasik or should they locate two factories, one in Orangabad another one is Nasik. Where should they locate the distribution center, should they locate in V1D or should they locate in Bapi or should they locate distribution centers both in V1D as well as Bapi. So, first they have to decide where should they locate the factory, where should they locate the distribution centers. Suppose they decide that then they also need to find out which distribution center will be serving to which market, which distribution center will be serving to which market all of this decision has to be taken and how much quantity should they be sending to this market. That means, from which DC to which market how much quantity will be transported so that demand at Mumbai, Pune, Surat and Ahmedabad is satisfied. So, this is the decision that the manager has to take. So, I need to find out what would be the optimized supply chain network, where should they locate the factory, where should they locate the distribution centers and which factory will be serving to which distribution center, which distribution center will be serving to which customers and how the movement of products will happen from factory to DC and DC to customers that is what they need to decide. So, for that I need data. So, first of all I need the location of Nasik that is latitude and longitude of Nasik factory, latitude and longitude of Orangabad factory. Then I also need to get the fixed cost per year. For Nasik it is this value, for Orangabad it is this value. I also need production cost per unit in both the location in Nasik it is little costly 7 dollar per unit in Orangabad it is 5 dollar per unit. I also need to find out the distance from both the factory to both the DC. So, Nasik to Bapi distance is 144 kilometer, Nasik to V1D distance is 129 kilometer, Orangabad to Bapi distance is 322 kilometer, Orangabad to V1D distance is 297 kilometer. I also need the data from DC to markets. So, I have 2 DC Bapi and V1D, Bapi latitude and longitude is given. For V1D latitude and longitude is also given. I also have the fixed cost per year for both the location. I also have inbound processing cost which is 0.25 dollar per unit, outbound processing cost is 0.25 per unit at Bapi whereas, inbound processing cost in V1D 0.5 dollar per unit, outbound processing cost in V1D 0.5 dollar per unit. Then I also have the distance from DC to market. So, 169 is the distance from Bapi to Mumbai, Bapi to Pune distance is 298 kilometer, Bapi to Surat distance is 102 kilometer, Bapi to Amadabha distance is 346 kilometer and so on. So, V1D to these 4 markets distance is also given. And I also have the data of transportation cost which is 0.001 per kilometer per unit and the revenue is dollar 15 per unit. So, these are the data given to us. I have to find out what should be the optimized supply chain network for their business so that overall cost is minimized. Now, if you look into this map, I have put the same data in the map. We have 2 factory, one located in Rangabad, one located in Nasik. We have 2 prospective distribution center, one is Bapi, another one is V1D and we have the location of the customers. So, now I need to find out where should they locate the factory, where should they locate the distribution center DC and how the movement of products will happen from factory to DC and DC to customer centers. So, with that objective in mind we developed an optimization model. So, optimization model has various parameters. I will explain slowly, slowly there is no need to worry. So, the first decision variable is this one, YI. YI is a binary variable. It will take value 1 if ith factory is open, 0 otherwise. If factory 1 is open, Y1 will be 1. If factory 2 is open, Y2 will be 1. If factory 1 is not opened, Y1 will be 0. If factory 2 is not open, Y2 will be 0. Then I have another decision variable ZJ. So, ZJ will be 1. If jth DC is open, 0 otherwise. So, these 2 are my decision variables which tells me which factory will open, which DC will open. Then I need the movement of product from factory to DC and DC to customers. So, XIJ represents how much quantity to be shift from ith factory to jth DC. QJK determines how much quantity to be shift from jth DC to kth customer. So, these are my decision variables. These values I will get after doing the optimization. Then I need some parameters. The revenue per unit is represent at REV. Then I have fixed cost of operating factory I. I have fixed cost of operating DCJ. I have per unit production cost at factory I. I have per unit inbound cost at DCJ. Per unit outbound cost at DCJ. Per unit transportation cost from factory I to DCJ. Per unit transportation cost from DCJ to customer k. And I have demand at customer location k. So, all the parameters as well as decision variables are listed in this table. And M is a very big number. Large value we will explain where should I use this value. So, our objective is to maximize the profit. So, profit is nothing but revenue minus cost. So, what is revenue? Revenue per unit is REV. I am multiplying this with QJK. That means, quantity shift from jth DC to kth customer. So, this value I need to maximize. This is the revenue. Now, there are some cost. What are the cost? First is fixed cost of operating the factory. Fixed cost of operating the DC. Now, if we look into this term, I have YI multiplied over here. So, that means, if YI equal to 1, this fixed cost will be deducted. If YI equal to 0, that means, if ith factor is not open, then I will not deduct any cost. The cost will be 0. Similarly, for ZJ also, if ZJ equal to 1, that jth distribution center will be open and my cost will be subtracted. If J distribution center is not open, fixed DCJ into ZJ will be 0. Then we have production cost. Then we also have inbound cost. I also have outbound cost. So, inbound and outbound cost is at the DC location. So, inbound processing cost at DC, outbound processing cost at DC. Then we have transportation cost from ith factory to jth distribution center. And if I sum it up with I equal to 1 to 2, j equal to 1 to 2, I will get total transportation cost from factory to DC. I will also get total transportation cost from DC to customers. So, I have revenue which is represented in this term and then I have cost. Fixed cost, production cost, inbound cost, outbound cost, transportation cost. So, revenue minus cost will give me the profit which I need to maximize. Then I have few constraints. So, first of all, my demand has to be satisfied. So, I have 4 market k equal to 1, 2, 3, 4 and at each market my demand should be satisfied. That means, summation q j k, j equal to 1, 2, 2 should be demand k. So, that means, how much I am sending from various distribution centers to demand points? It should match. So, demand should be same as how much I am getting at kth customers location. Then this is very interesting. If you look into this, y i will take 1 or 0. If y i equal to 1 that means, the i th factory is open and in this specific problem I am assuming that factory has unlimited capacity. That is why I am taking a very large value. So, m is a very large value. If y i equal to 1, the right hand side will be a very large value and then this much summation x i j is the quantity which is going out from i th factory. How much capacity it has? So, I have to make sure that the amount I am sending out from i th factory should be less than or equal to its capacity. Now, if y i equal to 1, my capacity will be m. If y i equal to 0, my capacity will come down to 0. That means, if i th factory is not open, I would not be able to send anything from that factory. Similarly, I have m j j, j j can take value 1 or 0. If j j equal to 1, j distribution center is open. If j j equal to 0, j distribution center is not open, then my capacity of that d c will be 0. If j distribution center is open, the capacity of the d c will be m and in this case, I am assuming initially that capacity is unlimited, but going forward we will relax this and then we will be able to get the exact capacity in the model. So, as of now, I am assuming that capacity is unlimited. That is very large, but if you want to put a fixed capacity, you replace m with that fixed capacity. Now, I also have to make balance. So, this is balancing constant. So, whatever is coming to d c j, I am sending the exact amount to the customer's location. So, I am not keeping any inventory in my d c. Whatever is flowing in, I am sending it out to the customer's and then we have decision variables y i can take value either 0 or 1, j j again binary variable can take value between 0 to 1 and x i j and q j k are continuous variable which can take value more than equal to 0. So, now, this is my optimization model. If I solve this optimization model for given demand and other value of the parameters, we will get an optimum solution. So, first I have solved this using excel solver because that is easy to visualize, easy to understand it, then we will move into analytics. So, we have to optimize the location of d c, location of d c, location of factory and so on. So, first thing if you see my decision is where should I open my factory. So, Nasik the facility open is 0 1. So, I can take if I press 1 over here. So, it will tell me that my Nasik factory should be open. If I put 0 over here, the model will tell me I should not open the Nasik factory. So, this yellow region is my decision variable. So, the first decision is should I open a factory in Nasik or not? Should I get the value 0 1 1? If I get 0, that means the factory should not open. If I get 1, it will represent factory should open. So, initially I am putting this value 0 and 1, but the model will give you the optimum value. So, right now I am assuming this is 0, this is 1, then I also have to find out how much quantity should I send from Nasik to Bapi, how much quantity should I send from Nasik to V1D, how much should I send from Orangabad to Bapi, Orangabad to V1D. So, these are my decision variables. So, initially just to start the optimization model, I am putting 0 0 because Nasik factory is not open. So, therefore, this value has to 0. Since it is not open, it cannot send anything out and then Orangabad I am putting 80. So, from Orangabad to Bapi, 80 units would be transported, Orangabad to V1D is 0. And I also have fixed cost per year, I also have production cost per unit, these are my input variables. So, now, since Nasik factory is not open, I am not sending anything out of it. So, there is a constraint for that. So, initially I have taken the value of m is so big, this is m very large number. If the facility is not open, the capacity will come down to 0, can you see? It has become 0. So, this multiplied by this. That means, if facility is not open, your capacity will automatically be restricted to 0. If the facility is open, then the capacity will be the big value which you have taken it over here. But in reality, you can replace this big M with your actual capacity. So, now, for Orangabad factory, I have a capacity of very large value because the factory is open and taking value 1. Now, let us see how much quantity is coming from Nasik to Vapi, Orangabad to Vapi. Nasik to Vapi, nothing will come because the factory is not open. Orangabad to Vapi, I am getting 8 units. So, inflow to DC, Vapi is 8 units, inflow to DC V1D is 0. Why V1D? Nothing is going, because if you see here, this is a variable, V1D facility is not open. Since facility is not open, I cannot take any inbound material over here. So, therefore, this is 0. So, now facility, Vapi facility is open, V1D is not open. So, the distribution center Vapi is open, distribution center V1D is not open. This is the initial value. Again and again I am saying it is not the optimum value. This is the random value which I have put in. Once you optimize, you will get the optimum value. But suppose this is correct, then understand the results. So, I got at Vapi 8 units. So, inflow to DC is 80 which is same as this. Then I need to see how much I am sending out. From Vapi to Mumbai, I am sending out 20, Vapi to Pune 20, Surat 20, I am whatever 20. This is the random data I have generated. So, 80 units has come, 80 units are going out. Now I have to make sure that demand is satisfied. Now the demand is this and I am only sending 20. So, therefore, it is not a feasible solution. My demand is not satisfied. So, in this model, we are assuming that all demand has to be satisfied. These values will be changed and I will get the exact value of demand over here. So, now how much is coming into Mumbai? This much is coming into Mumbai. Vapi to Mumbai 20 is V1D to Mumbai 0 because V1D is not open. So, I have to make sure that this value, this 20 is same as this because demand is 359, 160 which I need to satisfy. Vapi to Pune 20 units, V1D to Pune 0, total to Pune 20, this value should match with this. Vapi to Surat 20, V1D to Surat 0 and total inflow at Surat is 20. This should match with 87600 and so on. So, now this is telling me that demand is satisfied. Then I have to also see inflow is equal to outflow. So, inflow is 80 units, outflow is 80 units, inflow is 0 at V1D, outflow is also 0. So, my this constant is also satisfied. Now, I have another constant, this and this. So, this says if a factor is not open, it capacity should be 0 and that is what we are writing it over here. If a facility is not open, the actual capacity will be 0. If a facility is open, actual capacity will be m, big m which I have taken a very large value. Similarly, facility this is opened, my capacity will be how much m, big large value. If this is not open, then my capacity will come down to 0. So, if you see this is 0 into this value. So, my model is taking care of all the constants. Now, we have to see the objective function value. So, for objective function value, we need this input parameters, distance from factory to DC, distance from DC to customers. We also need the revenue, we also need the revenue data which is 15 dollar per unit. We also need the transportation cost data 0.001 per kilometer per unit. So, now, how can I calculate revenue? So, revenue is nothing but how much I am sending to customers. So, I am sending these many units to customers multiplied by revenue. So, if you see these many units, I am sending out multiplied by 15. So, 15 is the revenue per unit. So, 15 into sum of this is my total revenue minus cost. So, what are the cost? Sum of L 3 to Q 3. So, these are my cost. Fixed cost is this much, production cost is this, inbound cost is this, outbound cost is this, factory to DC transportation cost is this, factory to customer transportation cost is this. So, I have revenue, this multiplied by revenue per unit, this is my total revenue and then I need to subtract the cost. So, now, let us spend time to understand how this cost is calculated. So, I have fixed cost, fixed cost is if I open this DC, then only I will incur the cost. So, for Nashik, I will not incur any cost. So, this multiplied by 0 plus this multiplied by 1. So, for Orangabad factory, I will incur a cost of 1825000 and for DC, for VAPI, I have to incur a cost 1825000. So, if you add these two, you will get the value 365000. Now, what is the production cost? The production cost per unit is 7 at Nashik, 5 at Orangabad and I am not producing anything at Nashik. I am producing only at Orangabad. So, 5 into how much I am producing? 80 units. So, 5 into 80 will be 400, then inbound cost. So, inbound cost to DC. So, from factory to DC, how much I am sending? 80 units I am sending from Orangabad to VAPI and I am sending 0 units. So, total 80 units are coming to VAPI and per unit inbound cost is 0.25. So, 80 into 0.25 plus 0 units coming to V1D 0 into 0.5 which will be 20. Outbound cost again, how much is my outbound cost? How much I am sending out? 80 units from VAPI, 0 units from V1D. So, 80 into 0.25 plus 0 into 0.5, it will be 20. Then I need to calculate the transportation cost. So, how much I am sending? This much I am sending from Nashik to VAPI, Nashik to V1D, Orangabad to VAPI, Orangabad to V1D. So, these values should be multiplied by the distance. So, 0 into 144, 0 into 129, 80 into 132 plus 0 into 297. So, this is my product into kilometer and this has to be multiplied by 0.001 because that is the cost per kilometer per unit. So, if you do that, you will get the value 26. Similarly, I need to calculate from DC to customers. These are my movement of products from DC to customers. These are my kilometer. So, I need to apply some product of this comma this. So, this will get 20 into 169 plus 0 into 35, 20 into 298 plus 0 into 151, 20 into 102 plus 0 into 254, 20 into 346 plus 0 into 487. That will be my product into kilometer. Then I need to multiply the value 0.001, which is cost per kilometer per unit. If I do that, I will get 18. Now, these are my total fixed cost, which I need to subtract from my revenue. So, that is what we have done in total profit calculation. So, this first term is giving me total revenue. Second term is my cost. So, this is not optimum value. This is not feasible also because if you see this, it is not matching with the executive value. So, now I need to go to solver and optimize this. So, before I go for optimization, I need to explain the solver also. So, first thing I am trying to maximize E 19, then by changing variable cells. So, what are the cells you are going to change? D 3, D 4. So, D 3 to D 4, then I am changing D 13 to D 14. So, D 3, D 4, D 13, D 14. These two are binary variables. So, that is why I have also written D 13, D 14 is binary, D 3, D 4 is binary. Then I am also changing E 3 to F 4 and E 13 to H 14. So, these value are my movement of products from factory to DC. This is from DC to customers. So, these are all my decision variables. Whatever cells are highlighted in yellow, these are my decision variables. Then I need to add constant E 5 to F 5. If you see E 5 to F 5 should be less than equal to E 9 to F 9. So, this is restricting the capacity of distribution center. Then I have E 15 to H 15, E 15 to H 15 is my actual cells what I am doing. That should be same as my demand. So, E 15 to H 15 should be E 17 to H 17. This value should be same as this value. Then I have H 3 to H 4 should be less than equal to K 3 to K 4. This is restricting the capacity of the factory. Then I have I 13 to I 14 that is these two value should be same as K 13 to K 14. That is whatever is coming in that should go out and this is called the balancing constant. Then since this model is linear, I am solving using simplex LB. If you solve, you will get the optimal solution. So, now optimum solution is I need to open factory at Orangabad and this is the movement of the product from Orangabad to Bapi, Orangabad to V1D no movement. So, out of two factory, I am opening a factory at Orangabad. Out of two DC, I am opening one DC at Bapi. So, Orangabad factory is opened, Bapi DC is opened. So, from Orangabad factory to Bapi DC, I am sending this much quantity. From Bapi, I am sending Mumbai 359160, which is matching the demand of Mumbai. From Bapi to Pune, I am sending 315360, which is also matching the demand of Pune. Bapi to Surat, I am sending this. Bapi to Amalabha, I am sending this, which are also matching with their corresponding demand. And the total profit is coming out to be 5927495. So, this is the optimum value of the decision variables and this is the optimum value of the objective functions. So, I am sure you understood how this excel is working. So, let us move to the PPT slide. We have already done this hands on and this was my initial proposed supply chain network. Now, after optimizing, we are getting the solution that my factories should be located at Orangabad. My DCs should be located in Bapi. So, although I had two options of factory, but only one factory is open. I had two options for distribution center, but only one DC at Bapi is open. So, from factory to Bapi, Finnish good will move. Then from Bapi to Mumbai, Bapi to Pune, Bapi to Surat, Bapi to Amalabha, products will move to meet the demand of these customers location. The same thing we have plotted in map also. So, from Orangabad product will move to Bapi DC. From Bapi DC it will go to Surat, it will go to Amalabha, it will go to Mumbai, it will go to Pune. That is how your demand will be satisfied and this is the optimized supply chain network, because we got the results using optimization model. So, now for small scale problem, the excel solver is fine. It is giving me the solution in a reasonable time and I am also able to get where my factory should be located, where my DC should be located and how the movements of products will happen from factory to DC and DC to various customers location. Now imagine, if I have 4000 customers, 40,000 customers, multiple DC, multiple factory, can I solve it using excel? Of course, no, because excel does not have that capability to handle so many decision variable plus I cannot visualize it better way, because if I have 40,000 customers located across the country, 15 DC, 45 factory is difficult to visualize it and difficult to see how the movements of products are happening from factory to DC, DC to customers. So, therefore, we need better techniques and tool. So, for that any logistics will be a very useful software. As you have done it for green field analysis, same thing we will do it for network optimization. So, now let us make the problem little complex and then we will solve it using green using any logistics. So, in earlier case I had only one product in Pune, but now I have 4 SKUs, I have 4 SKUs in Mumbai, I have 4 SKUs in Surat, I need to meet the demand of 4 SKUs in Ahmedabad also. So, Pune location has demand for 4 SKUs, Mumbai location has demand for 4 SKUs, Surat 4 SKUs, Ahmedabad 4 SKUs. So, instead of one product, now I have 4 products. So, complexity has increased from 1 to 4 products in place of yearly demand, now I have daily demand. So, we have purposefully increased the complexity of the model little bit. Now if I want to solve this, excel cannot do this, because number of variables would be more and excel will not have the capability to solve this. So, therefore, I need a tool which is more user friendly, which is easy to operate and I can get the solution by clicking a button. So, for that we are using any logistics. So, we have used network optimization model of unit logistics and got the same result. You can see the Orangabad factory is opened, then VAPDC is opened and this is how the movements of products are happening from DC to various customers locations. And from factory to DC, this is the way products would move. So, it is not only telling me where the DC should be located, where the factory should be located. The energy software is also telling me actual road route, which the trucks would follow. So, that I can reach to the customers at a quicker time and total transportation cost is minimized. Now let us go to analytics software and see how network optimization can be done. Now if you see here, I will go to network optimization. Now this is the data. So, I have already entered the data the same way as we have entered the data for gain fill analysis. So, the same way we have to enter the data over here. So, quickly I will show you some of the data points. So, I have 4 customers location Pune, Mumbai, Amalabad and Surat. Then I have DCs and factories. Orangabad is factory, V1 is DC, Asikis factory, Babi is DC. Then I can see the demand of 4 SKUs in Pune is given. If you click here, it will tell you what is the demand per day in Pune for SKU 1. Then I have revenue data also, 15 dollar per unit for SKU. Then I have facility expense also. I can see the expenses. Orangabad, this is the fixed cost per day because we are entering per day. So, 7500 per day, 7500 into 365. If you do, you will get the fixed cost per year. Then we have group 3. This is 2 group V1, V1, V2 factories. Orangabad, Asik, 4 customers, 4 locations, Mumbai, Surat, Amalabad and Pune. I can see the location, exact location is also given over here. So, if you have many locations, you can auto fill coordinates, but make sure that your spelling is correct. And then we have objective member. So, revenue minus total fixed cost, inbound cost, outbound cost. In this case, I have not taken penalty. So, penalty will be 0. Production cost, storage cost also we have not taken. Total supply cost, transportation cost. So, if you take this cost, this cost will be deducted. If you do not take this cost, this cost will not be deducted. Then we have run it for your first January 2021 to 31st December 2021. You can make it live by putting the today's date to next one year. Depending upon what would be your period, you can change this. So, all of this information is given here. You can enter it manually based on your products. If you do not want to enter it manually, there is Excel template, which has been uploaded in the toolkit, which I have shown you the link. If you go to the link, you will be able to get the Excel template also, because many time problem becomes large and you are not able to solve it. You are not able to enter the data manually. So, you can take the help of those templates. Now, if I run this network optimization model, you will see this running. And can you see, I am getting the result. Total profit is 5927494.8. That is 5927495. It is matching exactly with the value, which we got in Excel solver. So, therefore, if I solve it using analog sticks, the results are coming same. But the advantage is, it is giving me a good visualization, how the products movements are happening, which route it is taking, what is the exact location. All of this can be visualized. Not only that, I can also see how the movements are happening. If you click over flow details, you will be able to see how the movements are happening. From Orangobar to Vapi, you can see SK1313900 units are being transported. I can also see what is the distance. I can also see transportation cost. I can also see processing cost, flow cost. So, for each node, from each node to another node, that is from factory to DC, DC to customer location, I can find out how many units are moving and what are their corresponding costs. So, everything can be visualized in a nice manner if you have analogistic software. So, these are the advantages and this is how you can form a supply chain digital twin for network optimization. So, I am sure you got the idea, how a digital twin can be formed. If you run this using analog stick software, you will get the similar results as we have done. So, after you run the analog sticks over there, you will get the similar output as we have shown in the analog stick software. So, we will stop today's class over here. So, thank you so much. Look forward to seeing you in the next class. Thank you.