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Assignment Problem - 3 Hungarian Assignment Method - HAM - 1
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Solution of an Assignment Problem through Hungarian Assignment Method (HAM) made easy explaining in simple way
#OR #Math #OperationsManagement #Statistics #Hungarian #Assignment #Method #Problem #Solution #FreeStudy
Hungarian Method - Steps:
Step 1 Develop the cost table from the given problem. If the number of rows is not equal to the number of columns and vice versa, a dummy row or dummy column must be added. The assignment costs for dummy cells are always zero.
Step 2 Find the opportunity cost table (a) Locate the smallest element in each row of the given cost table and then subtract that from each element of that row, and (b) In the reduced matrix obtained from 2(a), locate the smallest element in each column and then subtract that from each element of that column. Each row and column now have at least one zero value.
Step 3 Make assignments in the opportunity cost matrix. The procedure of making assignments is as follows:
a) Examine rows successively until a row with exactly one unmarked zero is obtained. Make an assignment to this single zero by making a square around it.
b) For each zero value that become assigned, eliminate (strike off) all other zeros in the same row and/or column.
c) Repeat Steps 3(a) and 3(b) for each column also with exactly single zero value cells that have not been assigned.
d) If a row and/or column have two or more unmarked zeros and one cannot be chosen by inspection, then choose the assigned zero cell arbitrarily.
e) Continue this process until all zeros in rows/columns are either enclosed(assigned) or struck off(×)
Step 4 Optimality criterion If the number of assigned cells is equal to the number of rows/columns, then it is an optimal solution. The total cost associated with this solution is obtained by adding original cost figures in the occupied cells. If a zero cell was chosen arbitrarily in Step 3, there exists an alternative optimal solution. But if no optimal solution is found, then go to Step 5.
Step 5 Revise the opportunity cost table Draw a set of horizontal and vertical lines to cover all the zeros in the revised cost table obtained from Step 3, by using the following procedure:
a) For each row in which no assignment was made, mark a tick (√)
b) Examine the marked rows. If any zero cells occur in those rows, mark to the respective columns that contain those zeros.
c) Examine marked columns. If any assigned zero occurs in those columns, tick the respective rows that contain those assigned zeros.
d) Repeat this process until no more rows or columns can be marked.
e) Draw a straight line through each marked column and each unmarked row. If the number of lines drawn (or total assignments) is equal to the number of rows (or columns), the current solution is the optimal solution, otherwise go to Step 6.
Step 6 Develop the new revised opportunity cost table –
a) From among the cells not covered by any line, choose the smallest element. Call this value k.
b) Subtract k from every element in the cell not covered by a line.
c) Add k to every element in the cell covered by the two lines, i.e. intersection of two lines.
d) Elements in cells covered by one line remain unchanged. Step 7 Repeat Steps 3 to 6 until an optimal solution is obtained.
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#OR #Math #OperationsManagement #Statistics #Hungarian #Assignment #Method #Problem #Solution #FreeStudy
Hungarian Method - Steps:
Step 1 Develop the cost table from the given problem. If the number of rows is not equal to the number of columns and vice versa, a dummy row or dummy column must be added. The assignment costs for dummy cells are always zero.
Step 2 Find the opportunity cost table (a) Locate the smallest element in each row of the given cost table and then subtract that from each element of that row, and (b) In the reduced matrix obtained from 2(a), locate the smallest element in each column and then subtract that from each element of that column. Each row and column now have at least one zero value.
Step 3 Make assignments in the opportunity cost matrix. The procedure of making assignments is as follows:
a) Examine rows successively until a row with exactly one unmarked zero is obtained. Make an assignment to this single zero by making a square around it.
b) For each zero value that become assigned, eliminate (strike off) all other zeros in the same row and/or column.
c) Repeat Steps 3(a) and 3(b) for each column also with exactly single zero value cells that have not been assigned.
d) If a row and/or column have two or more unmarked zeros and one cannot be chosen by inspection, then choose the assigned zero cell arbitrarily.
e) Continue this process until all zeros in rows/columns are either enclosed(assigned) or struck off(×)
Step 4 Optimality criterion If the number of assigned cells is equal to the number of rows/columns, then it is an optimal solution. The total cost associated with this solution is obtained by adding original cost figures in the occupied cells. If a zero cell was chosen arbitrarily in Step 3, there exists an alternative optimal solution. But if no optimal solution is found, then go to Step 5.
Step 5 Revise the opportunity cost table Draw a set of horizontal and vertical lines to cover all the zeros in the revised cost table obtained from Step 3, by using the following procedure:
a) For each row in which no assignment was made, mark a tick (√)
b) Examine the marked rows. If any zero cells occur in those rows, mark to the respective columns that contain those zeros.
c) Examine marked columns. If any assigned zero occurs in those columns, tick the respective rows that contain those assigned zeros.
d) Repeat this process until no more rows or columns can be marked.
e) Draw a straight line through each marked column and each unmarked row. If the number of lines drawn (or total assignments) is equal to the number of rows (or columns), the current solution is the optimal solution, otherwise go to Step 6.
Step 6 Develop the new revised opportunity cost table –
a) From among the cells not covered by any line, choose the smallest element. Call this value k.
b) Subtract k from every element in the cell not covered by a line.
c) Add k to every element in the cell covered by the two lines, i.e. intersection of two lines.
d) Elements in cells covered by one line remain unchanged. Step 7 Repeat Steps 3 to 6 until an optimal solution is obtained.
OR, Operations Management, Math, Transportation, HAM, Hungarian Assignment Method, Tie, Optimal Solution, Statistics, OM, Operations Management, MBA, MCA, CA, CS, CPA, CMA, CWA, BBA, BCA, BCom, MCom, GRE, GMAT, Grade 11, Grade 12, Class 11, Class 12, IAS, CAIIB, FIII, IBPS, BANK PO, UPSC, Probability, standard deviation, binomial distribution, normal distribution, poisson distribution, probability distribution, assignment problem in operations research, transportation problem in operations research, queuing theory, cpm, pert, binomial distribution in probability, chi square distribution, conditional probability, poisson distribution probability, job sequencing problem, assignment problem, northwest corner method, hungarian method for solving assignment problem, time series analysis in statistics, queuing theory in operations research, replacement problems in operations research, degeneracy in transportation problem, simulation problem in operational research, inventory model in operations research, fitting of binomial distribution, fitting of poisson distribution, pchart, cchart, npchart, mean chart, range chart, fit a binomial distribution, fit a poisson distribution, fitting of poisson distribution, fitting of binomial distribution, dual of transportation problem, duality, dual of linear programming problem, linear programming, simplex method, graphic method, graphical method, graph, odds, data science, data analysis, data mining, data management, dbms, data interpretation, reasoning, logic, coding, python, java, c, c++, c#, exponential distribution, uniform distribution, zvalue, zscore, index number,
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