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UNIT II TOPIC ASSIGNMENT PROBLEM gaurav sonkar Gaurav Sonkar ASSIGNMENT PROBLEM The assignment problem is a particular case of transportation problem in which the objective is to assign a number of resources to an equal number of activities so as to minimize total cost or maximize total profit. Typical examples of decision making situations where assignment model can be used are: workers to machines sales personnel to different sales areas products to factories jobs to machines classes to rooms 2 A GENERAL ASSIGNMENT PROBLEM LOOKS Gaurav Sonkar LIKE: 1 2 ………. n 1 C11 C12 ………….. C1n 1 2 C21 C22 …………… C2n 1 : . . . . . . . . …………… Cn1 Cn2 Cnn n . 1 1 1 ……….. 1 3 Gaurav Sonkar HOW TO SOLVE AN A.P. Step I: Row Operation: Subtracting smallest element in each row from all other elements of that row. Column Operation: Subtracting smallest element in each column from all other elements of that column. 4 Step II: MAKING ASSIGNMENTS: Gaurav Sonkar Search for an optimal assignment in the finally modified cost matrix as follows: Examine the first row. If there is only one zero in it, then enclose this zero in a box ( ) cross (X) all the zero of that column passing through the enclosed zero. Similarly done with other rows. If any row has more than one zero, then do not touch that row & passes on the next row. Repeat the same procedure with columns. 5 Step III: Gaurav Sonkar If each row and each column of the reduced matrix has exactly one enclosed zero, then the enclosed zero yield an optimal assignment. If not, then go to next step. Step IV: Draw minimum number of horizontal and/or vertical lines to cover all the zeros of the reduced matrix. Since the assignment is not optimal, the number of lines will be less than n (i.e. order of matrix). In order to move towards optimality, generates more zeros as follows: Find the smallest of the elements of the reduced matrix not covered by any of the lines. Let this element be α. Subtract α from each of the element not covered by the lines and add α to the element at the intersection of these lines. Do not change the remaining elements 6 Gaurav Sonkar Step V: Go to step II and repeat the procedure till an optimal assignment is achieved. 7 ILLUSTRATION 1 Gaurav Sonkar A department head has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimate of the times each man would take to perform each task is given in the effectiveness matrix below. How should the task be allocated, one to a man, so as to minimize the total man hours? I II III IV A 8 26 17 11 B 13 28 4 26 C 38 19 18 15 D 19 26 24 10 8 SOLUTION Gaurav Sonkar Step I: Row Operation: Subtracting smallest element in each row from all other elements of that row. I II III IV I II III IV A A B B C C D D Column Operation: Subtracting smallest element in each column from all other elements of that column. 9 Step II: MAKING ASSIGNMENTS: Gaurav Sonkar I II III IV A 0 14 9 3 B 9 20 0 22 C 23 0 3 0 D 9 12 14 0 Since every row and every column have one assignment, hence the current solution is optimal. 10 , THE OPTIMAL SOLUTION OBTAINED IS Gaurav Sonkar AS UNDER Task Subordinate Man Hour TOTAL MAN HOURS 11 ILLUSTRATION 2 Gaurav Sonkar A department has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulties. The estimates of time (in man-hours) each man would take to perform each task is given by: SUBORDINATES JOBS I II III IV A 18 26 17 11 B 13 28 14 26 C 38 19 18 15 D 29 26 24 10 How should the tasks to be allocated to men so as to optimize the total man-hours? 12 SOLUTION Gaurav Sonkar STEP I: Subtracting the smallest element of each row from all the elements of that row, we obtain the reduced matrix (a) given below: Further subtracting the smallest element of each column from all the elements of that column, we obtain the reduced matrix (b) given below: I II III IV I II III IV A 7 15 6 10 A 7 11 5 0 B 0 15 1 13 B 0 11 0 13 C 23 4 3 0 C 23 0 2 0 D 19 16 14 0 D 19 12 13 0 13 STEP II: Gaurav Sonkar Examine the rows of matrix (b) successively until a row with exactly one zero is found, enclose that zero in a box and cross all the other zero in its column. Thus we obtain matrix (c) I II III IV A 7 11 5 0 B 0 11 0 13 C 23 0 2 0 D 19 12 13 0 Further examine the columns of matrix (c) successively until a column with exactly one unmarked zero is found, enclose that zero in a box and cross all the other zero in its row 14 STEP III: Gaurav Sonkar Matrix (c) does not provide an optimal solution, since fourth row as well as third column does not have an enclosed zero. STEP IV: Now we draw minimum number of horizontal and / or vertical lines to cover all the zeros of the reduced matrix (c) I II III IV A 7 11 5 0 B 0 11 0 13 C 23 0 2 0 D 19 12 13 0 15 The smallest element in matrix (d) not covered by any of the lines above is 5. Gaurav Sonkar Subtracting this element from all uncovered elements and adding the same to the elements lying at the intersection of these lines, we obtain matrix (e) given below: I II III IV A 2 6 0 0 B 0 11 0 18 C 23 0 2 5 D 14 7 8 0 STEP V: Following the procedure of enclosing and crossing the zeros as in (step II) in matrix (e). 16 STEP VI: Gaurav Sonkar Since each row and each column in matrix (g) has exactly one enclosed zero, we have attained the following optimal assignment schedule: A III, B I, C II, D IV The minimum total time for this assignment schedule is: Z = 17+13+19+10= 59 man-hours. 17

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posted: | 4/4/2011 |

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