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Скачать или смотреть Assignment Problem - 4 Hungarian Assignment Method - HAM - 2

  • PUAAR Academy
  • 2016-05-12
  • 116343
Assignment Problem - 4 Hungarian Assignment Method - HAM - 2
OROperations ResearchAssignment ProblemHAMHungarian Assignment MethodMBAMCACACSCWACPACFACMABBABComMComCAIIBFIIIProbabilitystandard deviationbinomial distributionnormal distributionpoisson distributionprobability distributionassignment problemtransportation problemcpmpertchi square distributionconditional probabilityqueuingoddsdata sciencedata analysisdata miningdata managementreasoninglogiccodingpythonjavac++c#
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Assignment Problem

Hungarian Assignment Method - HAM

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 the previous Step , there exists an alternative optimal solution. But if no optimal solution is found, then go to the next following Step .

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 the next Step.

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.

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