 In today's session we are going to see the decision making under certainty at the end of the session the students is going to formulate the business problem or industry problem in the linear programming for the optimal assignment of the jobs or the resource along with that is going to determine the schedule of job assignment with the minimum cost or the time involved in the process and the maximizing with the maximizing the profit or the sale of the assignment problem here this problem consists of the four labors or you can say the four workers to whom the particular four jobs should be assigned here the question is what there is a decision making regarding that the particular ABCD workers which job job number one job number two job number three and job number four means for person or the worker a either the first job should be assigned or job two should be assigned or three or four here the below there is a particular table which is defining the a minutes required by the particular worker to complete is respective job for example if I say if the worker a is assigned job number one means 45 minutes are required to complete the particular job one by the particular worker a and forty minutes are required by the particular worker a to complete the job number two and fifty one minutes are required to complete the particular job three by worker a in this way for each worker each job there is a payoff table which is defining the minutes required to complete the particular job by the is respective worker or the person now as I've done we are discussing earlier session further we have to come from the first step is what to come from whether the given problem or the given assignment problem is a balanced or unbalanced type of problem here the number of workers are four and the jobs are four means what four rows and four column means a number of rows are equal to number of columns that's why the given assignment problem is a balanced type of problem means there is no need to convert the particular problem in the balanced type which because already it is a balanced yeah that this particular problem we are going to solve it with the hangarians assignment method as per the hangarians assignment method the first step is the row reduction yeah in the original table the first table yeah we have to select in each row we have to select the smallest element or you can say smallest value in first that is a worker a that is 45 40 51 and 67 the smallest is 40 in second row or for worker be out of 50 for us or 57 42 63 55 the smallest is 42 as a highlighted or I have highlighted by the showing the arrows in each row we have shown it by the red color the row reduction means yeah we as we have selected 40 we have to separate it with the respective row to each p of element which are one of the example I have shown yeah 40 my sorry 45 minus 45 40 minus 40 it will get 0 51 minus 40 we are 11 and 67 minus 40 27 this is been done for the respective row that is a first row for the BCD same way 57 minus 42 42 minus 42 63 minus 42 55 minus 42 in same way we have to carry for the each rows this is a obtained table after the row reduction as per the rule the hangarians method there is at least one zero in each row at least one zero there may be more than one zero in each row but there should be at least one zero in each row second step is what after getting the row reduction table this is a row reduction table in that the second step we have to see the minimum value in each column that is of job one two three four yeah out of this first column 5 15 1 0 the minimum is 0 in second column 0 0 4 4 there is 0 is a minimum in third 11 21 0 and 19 the minimum is 0 and 27 13 16 14 there is minimum is 13 again similarly the selected minimum value should be subtracted with the each respective column elements after subtracting will get the table this is a table which is showing the subtraction I have shown the fourth column because when I have seen the row reduced table in each column I have getting at least one zero when I might subtract it with the each element of that respective row the value will be same but in the fourth column there is no any zero the minimum is 13 that's why the particulars in this table I have shown the only calculation of the fourth table that is 27 minus 13 13 minus 13 16 minus 13 14 minus 13 this is the table finally after the column reduction in this when we observe the row reduction and the column reduction here each row there is one zero in and in each column there is at least one zero yeah after this the particularly we have to draw the horizontal line and the vertical line as per the requirement and we have to bracket the zero the process is what we have to select the row first of all we will go with the row selection this is the table after the row and column reduction which is shown in this table we have to carry out this step number three yeah in each row we are first we are solving it with the row here in each row we have to see first row is a year only one zero is there so the rule is what select that row which is having only one zero now first of only we are getting zero that's why I will mark it by the bracket and I will hide it or I will particularly draw the vertical line as I shown here similarly after hiding this there will be a reduction of the means reduction of the second column of that is job number but to only again we have to see the next step again 15 21 and 0 is been left yeah I have not considered this second job because already it has been hired idea here there is 15 21 0 there is one again one once more there is a zero again I will hide similarly I will cover it by the vertical line as shown is here after that the second and the four job is been hired only first and third job are been non-hired again I will start with the first 5 and 11 there is no any 0 15 and 21 there is no any 0 1 and 0 there is 0 at least in third row that is C workup again I will make a bracket 0 bracket bracket for that particular 0 again I will hide in this step 2 3 and 4 columns are been hired it only one column is left that is a first and I start again with the first 5 15 1 0 at the last row there is 0 I will again mark it by the bracket again I will hide the obtain table will be in such a fashion here the 0 which is been marked by the prox are 1 2 3 4 as per the Hangarians rule when the 0 which is marked by the box if it is equal to the n that is number of column or row then the optimal solution is been obtained when I compare this particular table with the original table here I have shown the original table for the first step we have considered for the first step in the first and we have started the problem I marked here are highlighted first or original table here the 0 box here the value is 40 in second row where the 0 been marked by the box it is 55 similarly for the third and fourth here this values 40 41 and 48 and 55 these are the values means this table is saying what the particular worker a should be assigned job number 2 particular and worker b should be assigned job number 4 worker C the job number 3 should be assigned and worker d job number 1 should be assigned which is again I have shown in the next table or slide this is a worker a b c d job as I have told worker a is assigned here worker a is assigned job number 2 I have shown here worker a as assigned number job number 2 b 4 c 3 3rd one job and the d first job and the these are overall minutes required to complete the particular jobs here all the four jobs are been scheduled to each respective or you can all four jobs are been scheduled to each workers at a time we can assign only one job to one worker or one person which has been shown here also and the total minutes required is 184 this is the questions let us select the correct answer for the particular questions hope so you have selected this answers princess for the today's session