Assignment Problem using Hungarian method with solutions, Assignments of Operational Research

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Prepared By: Dr. Solomon Getachew……

Operation Research

Solution of Assignment Problem

using Hungarian method

Lesson-

Learning Objective

• (^) Finding the optimum solution using Hungarian method.

Phase-Two: Optimization Step 1: Row scanning: If there is only one zero in the row, draw square to this zero and mark on the remaining zeros on this column. If not, leave it. Step 2: Column Scanning: If there is only one zero in the column, draw square to this zero and mark on the remaining zeros on this row. If not, leave it. Step 3: Cover all the zeros with minimum number of possible lines. Check whether the number of lines or squares equal to number of rows. Optimality. Step 4: Identify the minimum value of the uncovered values

• (^) Add this minimum value to the intersection point of vertical and horizontal lines.
• (^) Subtract this minimum value from all the uncovered cell values.
• (^) Put other values remain the same. Go to step-1 of phase-2.

Example: Hungarian Method Solution Example : Solve the following assignment problem shown in Table using Hungarian method. The matrix entries are processing time of each man in hours. Job Men 1 2 3 4 5 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

C

min

Step-2: Column Reduction 5 0 8 10 11 0 6 15 10 3 8 5 0 0 0 0 4 2 0 5 3 5 6 0 8 5 0 8 10 11 0 6 15 10 3 8 5 0 0 0 0 4 2 0 5 3 5 6 0 8 Phase 2: Optimization Step-1: Row scanning 5 0 8 10 11 0 6 15 10 3 8 5 0 0 0 0 4 2 0 5 3 5 6 0 8

Phase 3: Cover zeros with minimum number of lines Step-2: Column Scanning 5 0 8 10 11 0 6 15 10 3 8 5 0 0 0 0 4 2 0 5 3 5 6 0 8

Check whether the number of lines or squares equal to number of rows-Optimality

Job Men 1 2 3 4 5 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 Total cost=5+3+2+9+4=

Number of assignment (number of lines)=5 and number of rows=5 equal. Optimality achieved

Problem: To stimulate interest and provide an atmosphere for intellectual discussion, the faculty of mathematical sciences in an institute decides to hold special seminars on four contemporary topics – Statistics, Operations Research, Discrete Mathematics, Matrices. Each such seminar is to be held once a week. However, scheduling these seminars (one for each topic and not more than one seminar per day) has to be done carefully so that the number of students unable to attend is kept to a minimum. A careful study indicates that the number of students who cannot attend a particular seminar on a specific day is as follows: Find an optimal schedule for the seminars. Also find the number of students who will be missing at least one seminar? Monday  Operational Research, Tuesday  Matrices, Wednesday  Statistics, Thursday  Discrete Mathematics

• (^) So the total number of students who will be missing at least one seminar = 33+39+42+33= 147 Statistics Operations Research Discrete Mathematics Matrices Monday 45 33 39 45 Tuesday 51 36 36 39 Wednesday 42 45 30 42 Thursday 48 54 33 51 Exercise