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If number of rows is not equal to number of columns, then add dummy rows or columns with cost 0, to make it a square matrix. | |
a. Identify the minimum element in each row and subtract it from each element of that row. b. Identify the minimum element in each column and subtract it from every element of that column. | |
Make assignment in the opporunity cost table a. Identify rows with exactly one unmarked 0. Make an assignmment to this single 0 by make a square ( [0] ) around it and cross off all other 0 in the same column. b. Identify columns with exactly one unmarked 0. Make an assignmment to this single 0 by make a square ( [0] ) around it and cross off all other 0 in the same rows. c. If a row and/or column has two or more unmarked 0 and one cannot be chosen by inspection, then choose the cell arbitarily. d. Continue this process until all 0 in rows/columns are either assigned or cross off( | |
(a) If the number of assigned cells = the number of rows, then an optimal assignment is found and In case you have chosen a 0 cell arbitrarily, then there may be an alternate optimal solution exists. (b) If optimal solution is not optimal, then goto Step-5. | |
Draw a set of horizontal and vertical lines to cover all the 0 a. Tick(✓) mark all the rows in which no assigned 0. b. Examine Tick(✓) marked rows, If any 0 cell occurs in that row, then tick(✓) mark that column. c. Examine Tick(✓) marked columns, If any assigned 0 exists in that columns, then tick(✓) mark that row. d. Repeat this process until no more rows or columns can be marked. e. Draw a straight line for each unmarked rows and marked columns. f. If the number of lines is equal to the number of rows then the current solution is the optimal, otherwise goto step-6 | |
Develop the new revised opportunity cost table a. Select the minimum element, say k, from the cells not covered by any line, b. Subtract k from each element not covered by a line. c. Add k to each intersection element of two lines. | |
Repeat steps 3 to 6 until an optimal solution is arrived. |
\ | I | II | III | IV | V |
A | 10 | 5 | 13 | 15 | 16 |
B | 3 | 9 | 18 | 13 | 6 |
C | 10 | 7 | 2 | 2 | 2 |
D | 7 | 11 | 9 | 7 | 12 |
E | 7 | 9 | 10 | 4 | 12 |
`I` | `II` | `III` | `IV` | `V` | ||
`A` | ||||||
`B` | ||||||
`C` | ||||||
`D` | ||||||
`E` | ||||||
`I` | `II` | `III` | `IV` | `V` | ||
`A` | `5=10-5` | `0=5-5` | `8=13-5` | `10=15-5` | `11=16-5` | Minimum element of `1^(st)` row |
`B` | `0=3-3` | `6=9-3` | `15=18-3` | `10=13-3` | `3=6-3` | Minimum element of `2^(nd)` row |
`C` | `8=10-2` | `5=7-2` | `0=2-2` | `0=2-2` | `0=2-2` | Minimum element of `3^(rd)` row |
`D` | `0=7-7` | `4=11-7` | `2=9-7` | `0=7-7` | `5=12-7` | Minimum element of `4^(th)` row |
`E` | `3=7-4` | `5=9-4` | `6=10-4` | `0=4-4` | `8=12-4` | Minimum element of `5^(th)` row |
`I` | `II` | `III` | `IV` | `V` | ||
`A` | `5=5-0` | `0=0-0` | `8=8-0` | `10=10-0` | `11=11-0` | |
`B` | `0=0-0` | `6=6-0` | `15=15-0` | `10=10-0` | `3=3-0` | |
`C` | `8=8-0` | `5=5-0` | `0=0-0` | `0=0-0` | `0=0-0` | |
`D` | `0=0-0` | `4=4-0` | `2=2-0` | `0=0-0` | `5=5-0` | |
`E` | `3=3-0` | `5=5-0` | `6=6-0` | `0=0-0` | `8=8-0` | |
Minimum element of `1^(st)` column | Minimum element of `2^(nd)` column | Minimum element of `3^(rd)` column | Minimum element of `4^(th)` column | Minimum element of `5^(th)` column |
`I` | `II` | `III` | `IV` | `V` | ||
`A` | (1) Rowwise cell `(A,II)` is assigned | |||||
`B` | (2) Rowwise cell `(B,I)` is assigned so columnwise cell `(D,I)` crossed off. | |||||
`C` | (4) Columnwise cell `(C,III)` is assigned so rowwise cell `(C,V)` crossed off. | Columnwise `(C,IV)` crossed off because (3) Rowwise cell `(D,IV)` is assigned | Rowwise `(C,V)` crossed off because (4) Columnwise cell `(C,III)` is assigned | |||
`D` | Columnwise `(D,I)` crossed off because (2) Rowwise cell `(B,I)` is assigned | (3) Rowwise cell `(D,IV)` is assigned so columnwise cell `(C,IV)`,`(E,IV)` crossed off. | ||||
`E` | Columnwise `(E,IV)` crossed off because (3) Rowwise cell `(D,IV)` is assigned | |||||
`I` | `II` | `III` | `IV` | `V` | ||
`A` | ||||||
`B` | (5) Mark(✓) row `B` since column `I` has an assignment in this row `B`. | |||||
`C` | ||||||
`D` | (3) Mark(✓) row `D` since column `IV` has an assignment in this row `D`. | |||||
`E` | (1) Mark(✓) row `E` since it has no assignment | |||||
(4) Mark(✓) column `I` since row `D` has 0 in this column | (2) Mark(✓) column `IV` since row `E` has 0 in this column |
`I` | `II` | `III` | `IV` | `V` | ||
`A` | `7=5+2` intersection cell of two lines | cell covered by a line | cell covered by a line | `12=10+2` intersection cell of two lines | cell covered by a line | |
`B` | cell covered by a line | `4=6-2` cell not covered by a line | `13=15-2` cell not covered by a line | cell covered by a line | `1=3-2` cell not covered by a line | |
`C` | `10=8+2` intersection cell of two lines | cell covered by a line | cell covered by a line | `2=0+2` intersection cell of two lines | cell covered by a line | |
`D` | cell covered by a line | `2=4-2` cell not covered by a line | `0=2-2` cell not covered by a line | cell covered by a line | `3=5-2` cell not covered by a line | |
`E` | cell covered by a line | `3=5-2` cell not covered by a line | `4=6-2` cell not covered by a line | cell covered by a line | `6=8-2` cell not covered by a line | |
`I` | `II` | `III` | `IV` | `V` | ||
`A` | (1) Rowwise cell `(A,II)` is assigned | |||||
`B` | (2) Rowwise cell `(B,I)` is assigned so columnwise cell `(D,I)` crossed off. | |||||
`C` | Rowwise `(C,III)` crossed off because (4) Columnwise cell `(C,V)` is assigned | (4) Columnwise cell `(C,V)` is assigned so rowwise cell `(C,III)` crossed off. | ||||
`D` | Columnwise `(D,I)` crossed off because (2) Rowwise cell `(B,I)` is assigned | (5) Rowwise cell `(D,III)` is assigned | Columnwise `(D,IV)` crossed off because (3) Rowwise cell `(E,IV)` is assigned | |||
`E` | (3) Rowwise cell `(E,IV)` is assigned so columnwise cell `(D,IV)` crossed off. | |||||
`I` | `II` | `III` | `IV` | `V` | ||
`A` | Original cost 10 | Original cost 5 | Original cost 13 | Original cost 15 | Original cost 16 | |
`B` | Original cost 3 | Original cost 9 | Original cost 18 | Original cost 13 | Original cost 6 | |
`C` | Original cost 10 | Original cost 7 | Original cost 2 | Original cost 2 | Original cost 2 | |
`D` | Original cost 7 | Original cost 11 | Original cost 9 | Original cost 7 | Original cost 12 | |
`E` | Original cost 7 | Original cost 9 | Original cost 10 | Original cost 4 | Original cost 12 | |
Work | Job | Cost |
`A` | `II` | |
`B` | `I` | |
`C` | `V` | |
`D` | `III` | |
`E` | `IV` | |
Total | 23 |
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IMAGES
COMMENTS
18 Dec 2019. Whenever the cost matrix of an assignment problem is not a square matrix, that is, whenever the number of sources is not equal to the number of destinations, the assignment problem is called an unbalanced assignment problem. In such problems, dummy rows (or columns) are added in the matrix so as to complete it to form a square matrix.
Here is the video about unbalanced Assignment problem using Hungarian method,In this video we have seen how to solve unbalanced assignment problem using step...
Unit 8: Assignment Problem - Unbalanced. When an assignment problem has more than one solution, then it is Notes (a) Multiple Optimal solution (b) The problem is unbalanced (c) Maximization problem (d) Balanced problem. 8 Unbalanced Assignment Problem. If the given matrix is not a square matrix, the assignment problem is called an unbalanced ...
📒⏩Comment Below If This Video Helped You 💯Like 👍 & Share With Your Classmates - ALL THE BEST 🔥Do Visit My Second Channel - https://bit.ly/3rMGcSAThis vi...
Playlist of all my Operations Research videos-http://goo.gl/zAtbi4Today I'll tell how to solve an Unbalanced Assignment Problem in just 6 Easy Steps! Assignm...
Example: Unbalanced Assignment Problem. Solution. Since the number of persons is less than the number of jobs, we introduce a dummy person (D) with zero values. The revised assignment problem is given below: Table. Now use the Hungarian method to obtain the optimal solution yourself. Ans. = 20 + 17 + 17 + 0 = 54.
The typical textbook solution to the balanced assignment problem is then found using Kuhn's [3] Hungarian method. Problems in which there are more jobs than machines and more than one job can be ...
In Yadaiah and Haragopal [4], they use a different approach to solve the unbalanced assignment problem (see their paper for details). If there are n jobs to be assigned to m machines with n strictly greater than m, then they solve a series of k balanced assignment sub-problems each of size m by m where k is the floor (round down) of n/m. The ...
Abstract. Recently, Yadaiah and Harago pal published in the American Jo urnal of Operations Research a new. approach to solving the u nbalanced assignment problem. They also provide a num erical ...
e minimisation problem.3. The assignment problem wherein the number of rows is not equal to the number of columns is said t. be an unbalanced problem. Such a problem is handled by introducing dummy row(s) if the number of rows is less than the number of columns and dummy column(s) if the number of columns is le.
Tables 2, 3, 4, and 5 present the steps required to determine the appropriate job assignment to the machine. Step 1 By taking the minimum element and subtracting it from all the other elements in each row, the new table will be: Table 2 represents the matrix after completing the 1st step. Table 1 Initial table of a.
The assignment problem consists of finding, in a weighted bipartite graph, a matching of a given size, in which the sum of weights of the edges is minimum. If the numbers of agents and tasks are equal, then the problem is called balanced assignment. Otherwise, it is called unbalanced assignment. [ 1] If the total cost of the assignment for all ...
Unbalanced Assignment problem is an assignment problem where the number of facilities is not equal to the number of jobs. To make unbalanced assignment problem, a balanced one, a dummy facility (s) or a dummy job (s) (as the case may be) is introduced with zero cost or time. Get Quantitative Techniques: Theory and Problems now with the O ...
This is the complete recording of the lecture, Module 3 - Lecture 3 - Unbalanced Assignment Problems in MEE1024 - Operations Research.
solving minimisation problems: Step 1:See whether number. f rows are equal to number of columns. If yes, problem is balanced one; if not, then add a Dummy Row or Column to make the problem a balanced one by allotting zero value to each cell of the D. mmy Row or Column, as the case may be.Step 2: Row Subtraction: Subtract the minimum element of.
Recently, Yadaiah and Haragopal published in the American Journal of Operations Research a new approach to solving the unbalanced assignment problem. They also provide a numerical example which they solve with their approach and get a cost of 1550 which they claim is optimum. This approach might be of interest; however, their approach does not guarantee the optimal solution.
Solve the following assignment problem. Solution: Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is. Here only 3 tasks can be assigned to 3 men.
Unbalanced assignment problem, Hungarian method, Optimal solution. Introduction The assignment problem is a combinatorial optimization problem in the field of operations research. It is a special case and completely degenerate form of a transportation problem, which occurs when each supply is 1 and
Unbalanced Assignment Problems If the number of rows and columns are not equal then such type of problems are called as unbalanced assignment problems. Example A company has 4 machines on which to do 3 jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following table
Unbalanced Assignment Problem Questions - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The document provides information about an unbalanced assignment problem involving road repair work by 5 contractors. It includes the following details: 1) A city corporation needs to repair 4 roads and has received bids from 5 contractors, with each contractor's time and cost ...
How can I balance the following assignment problem (where machines are to be assigned the jobs in optimal way such that the profit is maximized). Cost matrix is given in the problem. First step is to convert it into minimization problem by subtracting all the entries in the matrix from maximum value in the matrix.
Step-18: Stop. 7.. ConclusionThe present paper suggests a modified method for solving the unbalanced assignment problems. The Hungarian method [1] gives us total assignment cost 870 along with the other three jobs assigned to dummy machine, in other words that these three jobs are ignored for further processing, while, when the original problem is divided in the sub-problems, which are ...
Algorithm & Example-1. Algorithm. Hungarian Method Steps (Rule) Step-1: If number of rows is not equal to number of columns, then add dummy rows or columns with cost 0, to make it a square matrix. Step-2: a. Identify the minimum element in each row and subtract it from each element of that row.