(v) Now draw straight lines which pass through all the un marked rows and marked columns.
(v) Now draw straight lines which pass through all the un marked rows and marked columns.It can also be noticed that in an n x n matrix, always less than ‘n’ lines will cover all the zeros if there is no solution among them. In step 4, if the number of lines drawn are equal to n or the number of rows, then it is the optimum solution if not, then go to step 6. Select the smallest element among all the uncovered elements. Repeat the procedure from step (3) until the number of assignments becomes equal to the number of rows or number of columns.Tags: Importance Of Sports In Education EssaysFuel Cells Research PapersWrite An Analysis EssayEssay On Can Computers Replace BooksAssignment And Assumption Of LeaseEssay Stories With Moral ValuesProblem Solving First Grade
(iii) Step 3, (i) and 3 (ii) are repeated till all the zeros are either marked or crossed out.
Now, if the number of marked zeros or the assignments made are equal to number of rows or columns, optimum solution has been achieved. At this stage, draw the minimum number of lines (horizontal and vertical) necessary to cover all zeros in the matrix obtained in step 3, Following procedure is adopted: (i) Tick mark () all rows that do not have any assignment.
Note that this permutation is not the optimal solution.
[More information on computational complexity and applications to come.] Here we present the Koopmans-Beckmann formulation of the QAP.
The objective of the problem is to assign a set of facilities to a set of locations in such a way as to minimize the total assignment cost.
The assignment cost for a pair of facilities is a function of the flow between the facilities and the distance between the locations of the facilities.Following steps are involved in solving this Assignment problem, 1.Locate the smallest cost element in each row of the given cost table starting with the first row.Now, this smallest element is subtracted form each element of that row.So, we will be getting at least one zero in each row of this new table. Having constructed the table (as by step-1) take the columns of the table.Now, this element is subtracted from all the uncovered elements and added to the element which lies at the intersection of two lines. Summary: The objective of the Quadratic Assignment Problem (QAP) is to assign \(n\) facilities to \(n\) locations in such a way as to minimize the assignment cost.Having performed the step 1 and step 2, we will be getting at least one zero in each column in the reduced cost table. Now, the assignments are made for the reduced table in following manner.(i) Rows are examined successively, until the row with exactly single (one) zero is found.Any basic feasible solution of an Assignment problem consists (2n – 1) variables of which the (n – 1) variables are zero, n is number of jobs or number of facilities.Due to this high degeneracy, if we solve the problem by usual transportation method, it will be a complex and time consuming work. Before going to the absolute method it is very important to formulate the problem. Now as the problem forms one to one basis or one job is to be assigned to one facility or machine.