 Let's look at an example of how we use this assignment method to assign tasks to different resources. This can, let's say, first printing incorporated company wants to find the minimum total cost assignment of three jobs to three typesetters, three resources. This data is given in it. R34 job is 66, T50, and resources are ABC and their relative cost, we have provided the data of the cost. Now, the method of this is basically, as I told you, step one is to subtract the minimum cost value from the total value of the row. Similarly, if you want to do it in a row, then we have this figure. If we subtract it, then we have a zero here and a zero is here. Now, in the next step, we have to look at column-wise to see which values are column-wise minimum. In column A, this is minimum, zero. In column B, this two is minimum. And in column C, zero is again minimum. So now we have to subtract the lowest value from which we have to subtract the rest of the values. So if we subtract zero from zero and two from five, then the value will remain the same. Similarly, column C will remain the same because there are two zeros and we will subtract zero from three. But in column B, the lowest is two. So when we use and subtract this, then if we subtract it from eight, then it will be six. If we subtract it from two, then it will be zero. And if we subtract it from five, then it will be three. So the new matrix that we have, that will be, now this is the new matrix. It has changed the value of six zero three. Now the second step is that we have to draw the minimum number of lines that cover the zeros. So what is this? One line comes, the other line comes. So basically we have two lines coming here. So we are covering two lines whereas we have three columns and three rows. At least three lines should come to cover zeros. So in this, we will go to basically the third step. So in this, the second step is that we have drawn these lines. And as I said, because these are two lines, then we will go to the next step, number three. In this, this is intersection. So this is important. Here our value is this three. So in this third step, the uncovered numbers are four. We have to subtract it from these smallest numbers. So instead of two it will be zero, instead of three it will be one, instead of six it will be four, and instead of five it will be three. But another step is that the number on the intersection is three, we will add this two to it. So what do we have now? Three, zero, zero. Because two minus two is zero. So that is four. If two comes out of six, then four is zero one. And that is zero. If two is added in three, then that is zero. Five is zero. So we have reached this level. Now we have to see how many lines we can cover if we cover these zeroes. So if we can see, it takes minimum three lines to cover these zeroes. And in this we have a line here, we have covered the second zeroes and we have covered the third zeroes. Basically even if you go by column wise, there should be three lines because this will cover three, the second line, the third line. If we go daily, then we don't need three. So now we have to assign tasks to different resources. We have three tasks and three resources. We have to assign them here. So the way is that let's say we go row wise and we see which row has only one place, location is zero. R34 has only one zero. And zero, as I said, is part of the optimum solution. That is, this location will be in the optimum solution. That says that this task should be provided to C. We will give C this task. So on that basis, this is our assignment. Because this task cannot be assigned to anyone else because there is no other zero. So we cross out rows and columns which intersect with zero. So we have crossed it. Now we have seen that we have three zeroes. Again, we will look at uncovered locations and in T50 there is only one zero location and below B there is one. So that means we can assign A to B because there is no competition. Whereas in S66, A is zero and B is zero. There will be competition in which to assign. So it is better to assign the competition first and the solution will come out. So we have assigned A to T50 and this will also come out. And then this will also come out. So we have only B to be left. So in the second step, we have assigned T50 to A and now what we have left is one task and one resource. So we have assigned S66 to B. So this will give us the minimum cost task assignments which these workers will perform. And if we look at our original table, cost table, then on that basis A cost was $9. B cost $10 and C cost $6. And that gives us the minimum lowest cost of $25 by assigning these tasks to these resources.