1. Slide 1 of 27 Assignment Problem: Hungarian Algorithm and Linear Programming collected from the Internet and edited by Longin Jan Latecki

2. Slide 2 of 27 n Introduction to Assignment Problem n Let C be an nxn matrix representing the costs of each of n workers to perform any of n jobs. The assignment problem is to assign jobs to workers so as to minimize the total cost. Since each worker can perform only one job and each job can be assigned to only one worker the assignments constitute an independent set of the matrix C. n An arbitrary assignment is shown above in which worker a is assigned job q, worker b is assigned job s and so on. The total cost of this assignment is 23. n An arbitrary assignment is shown above in which worker a is assigned job q, worker b is assigned job s and so on. The total cost of this assignment is 23. n Can you find a lower cost assignment? n Can you find the minimal cost assignment? n Remember that each assignment must be unique in its row and column.

3. Slide 3 of 27 Assignment Problem n An assignment problem seeks to minimize the total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is c ij. n It assumes all workers are assigned and each job is performed. n An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs. n The network representation of an assignment problem with three workers and three jobs is shown on the next slide.

4. Slide 4 of 27 Assignment Problem n Network Representation 22 33 11 22 33 11 c 11 c 12 c 13 c 21 c 22 c 23 c 31 c 32 c 33 WORKERSJOBS

5. Slide 5 of 27 Assignment Problem n Linear Programming Formulation Min  c ij x ij Min  c ij x ij i j i j s.t.  x ij = 1 for each resource (row) i s.t.  x ij = 1 for each resource (row) i j  x ij = 1 for each job (column) j  x ij = 1 for each job (column) j i x ij = 0 or 1 for all i and j. x ij = 0 or 1 for all i and j. Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.

6. Slide 6 of 27 n Linear Programming (LP) problems can be solved on the computer using Matlab, and many others. n There are special classes of LP problems such as the assignment problem (AP). n Efficient solutions methods exist to solve AP. n AP can be formulated as an LP and solved by general purpose LP codes. n However, there are many computer packages, which contain separate computer codes for these models which take advantage of the problem network structure.

7. Slide 7 of 27 n A brute-force algorithm for solving the assignment problem involves generating all independent sets of the matrix C, computing the total costs of each assignment and a search of all assignment to find a minimal-sum independent set. The complexity of this method is driven by the number of independent assignments possible in an nxn matrix. There are n choices for the first assignment, n-1 choices for the second assignment and so on, giving n! possible assignment sets. Therefore, this approach has, at least, an exponential runtime complexity. n As each assignment is chosen that row and column are eliminated from consideration. The question is raised as to whether there is a better algorithm. In fact there exists a polynomial runtime complexity algorithm.

8. Slide 8 of 27 n From en.wikipedia.org/wiki/Hungarian_algorithm n The Hungarian method is a combinatorial optimization algorithm which solves the assignment problem in polynomial time. It was developed and published by Harold Kuhn in 1955, who gave the name "Hungarian method" because the algorithm was largely based on the earlier works of two Hungarian mathematicians: Dénes Kőnig and Jenő Egerváry. n James Munkres reviewed the algorithm in 1957 and observed that it is (strongly) polynomial. Since then the algorithm has been known also as Kuhn-Munkres or Munkres assignment algorithm. The time complexity of the original algorithm was O(n 4 ), however later it was noticed that it can be modified to achieve an O(n 3 ) running time. n In 2006, it was discovered that Carl Gustav Jacobi had solved the assignment problem in the 19th century, and published posthumously in 1890 in Latin.

9. Slide 9 of 27 Hungarian Method n The Hungarian method solves minimization assignment problems with m workers and m jobs. n Special considerations can include: number of workers does not equal the number of jobs — add dummy workers/jobs with 0 assignment costs as needednumber of workers does not equal the number of jobs — add dummy workers/jobs with 0 assignment costs as needed worker i cannot do job j — assign c ij = + Mworker i cannot do job j — assign c ij = + M maximization objective — create an opportunity loss matrix subtracting all profits for each job from the maximum profit for that job before beginning the Hungarian methodmaximization objective — create an opportunity loss matrix subtracting all profits for each job from the maximum profit for that job before beginning the Hungarian method

10. Slide 10 of 27 Hungarian Method n Step 1: For each row, subtract the minimum number in that row from all numbers in that row. n Step 2: For each column, subtract the minimum number in that column from all numbers in that column. n Step 3: Draw the minimum number of lines to cover all zeroes. If this number = m, STOP — an assignment can be made. n Step 4: Determine the minimum uncovered number (call it d ). Subtract d from uncovered numbers.Subtract d from uncovered numbers. Add d to numbers covered by two lines.Add d to numbers covered by two lines. Numbers covered by one line remain the same.Numbers covered by one line remain the same. Then, GO TO STEP 3.Then, GO TO STEP 3.

11. Slide 11 of 27 Hungarian Method n Finding the Minimum Number of Lines and Determining the Optimal Solution Step 1: Find a row or column with only one unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.)Step 1: Find a row or column with only one unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.) Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily.Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily. Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment.Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment.

12. Slide 12 of 27 Example 1: AP A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects. Project Project A B C A B C Westside 50 36 16 Westside 50 36 16 Subcontractors Federated 28 30 18 Subcontractors Federated 28 30 18 Goliath 35 32 20 Goliath 35 32 20 Universal 25 25 14 Universal 25 25 14 How should the contractors be assigned to minimize total costs? Note : There are four subcontractors and three projects. We create a dummy project Dum, which will be assigned to one subcontractor (i.e. that subcontractor will remain idle)

13. Slide 13 of 27 Example 1: AP n Network Representation (note the dummy project) 50 36 16 0 28 30 18 0 35 32 20 0 25 25 14 0 West.West. Dum.Dum. CC BB AA Univ.Univ. Gol.Gol. Fed.Fed.

14. Slide 14 of 27 Example 1: AP n Initial Tableau Setup Since the Hungarian algorithm requires that there be the same number of rows as columns, add a Dummy column so that the first tableau is (the smallest elements in each row are marked red): A BC Dummy A BC Dummy Westside 50 3616 0 Westside 50 3616 0 Federated 28 3018 0 Federated 28 3018 0 Goliath35 3220 0 Goliath35 3220 0 Universal252514 0 Universal252514 0

15. Slide 15 of 27 Example 1: AP n Step 1: Subtract minimum number in each row from all numbers in that row. Since each row has a zero, we simply generate the original matrix (the smallest elements in each column are marked red). This yields: A B C Dummy A B C Dummy Westside 50 36 16 0 Westside 50 36 16 0 Federated 28 30 18 0 Federated 28 30 18 0 Goliath 35 32 20 0 Goliath 35 32 20 0 Universal 25 25 14 0 Universal 25 25 14 0

16. Slide 16 of 27 Example 1: AP n Step 2: Subtract the minimum number in each column from all numbers in the column. For A it is 25, for B it is 25, for C it is 14, for Dummy it is 0. This yields: A B C Dummy A B C Dummy Westside 25 11 2 0 Westside 25 11 2 0 Federated 3 5 4 0 Federated 3 5 4 0 Goliath 10 7 6 0 Goliath 10 7 6 0 Universal 0 0 0 0 Universal 0 0 0 0

17. Slide 17 of 27 Example 1: AP n Step 3: Draw the minimum number of lines to cover all zeroes (called minimum cover). Although one can "eyeball" this minimum, use the following algorithm. If a "remaining" row has only one zero, draw a line through the column. If a remaining column has only one zero in it, draw a line through the row. Since the number of lines that cover all zeros is 2 < 4 (# of rows), the current solution is not optimal. A B C Dummy A B C Dummy Westside 25 11 2 0 Westside 25 11 2 0 Federated 3 5 4 0 Federated 3 5 4 0 Goliath 10 7 6 0 Goliath 10 7 6 0 Universal 0 0 0 0 Universal 0 0 0 0 n Step 4: The minimum uncovered number is 2 (circled).

18. Slide 18 of 27 Example 1: AP n Step 5: Subtract 2 from uncovered numbers; add 2 to all numbers at line intersections; leave all other numbers intact. This gives: A B C Dummy A B C Dummy Westside 23 9 0 0 Westside 23 9 0 0 Federated 1 3 2 0 Federated 1 3 2 0 Goliath 8 5 4 0 Goliath 8 5 4 0 Universal 0 0 0 2 Universal 0 0 0 2

19. Slide 19 of 27 Example 1: AP n Step 3: Draw the minimum number of lines to cover all zeroes. Since 3 (# of lines) < 4 (# of rows), the current solution is not optimal. A B C Dummy A B C Dummy Westside 23 9 0 0 Westside 23 9 0 0 Federated 1 3 2 0 Federated 1 3 2 0 Goliath 8 5 4 0 Goliath 8 5 4 0 Universal 0 0 0 2 Universal 0 0 0 2 n Step 4: The minimum uncovered number is 1 (circled).

20. Slide 20 of 27 Example 1: AP n Step 5: Subtract 1 from uncovered numbers. Add 1 to numbers at intersections. Leave other numbers intact. This gives: A B C Dummy A B C Dummy Westside 23 9 0 1 Westside 23 9 0 1 Federated 0 2 1 0 Federated 0 2 1 0 Goliath 7 4 3 0 Goliath 7 4 3 0 Universal 0 0 0 3 Universal 0 0 0 3

21. Slide 21 of 27 Example 1: AP Find the minimum cover: A B C Dummy A B C Dummy Westside 23 9 0 1 Westside 23 9 0 1 Federated 0 2 1 0 Federated 0 2 1 0 Goliath 7 4 3 0 Goliath 7 4 3 0 Universal 0 0 0 3 Universal 0 0 0 3 n Step 4: The minimum number of lines to cover all 0's is four. Thus, the current solution is optimal (minimum cost) assignment.

22. Slide 22 of 27 Example 1: AP The optimal assignment occurs at locations of zeros such that there is exactly one zero in each row and each column: A B C Dummy A B C Dummy Westside 23 9 0 1 Westside 23 9 0 1 Federated 0 2 1 0 Federated 0 2 1 0 Goliath 7 4 3 0 Goliath 7 4 3 0 Universal 0 0 0 3 Universal 0 0 0 3

23. Slide 23 of 27 Example 1: AP The optimal assignment is (go back to the original table for the distances): Subcontractor Project Distance Subcontractor Project Distance Westside C 16 Westside C 16 Federated A 28 Federated A 28 Universal B 25 Universal B 25 Goliath (unassigned) Goliath (unassigned) Total Distance = 69 miles Total Distance = 69 miles

24. Slide 24 of 27 Example 1: AP via LP n In our example the LP formulation is: Min z = 50x 11 + 36x 12 + 16x 13 + 0x 14 + 28x 21 + 30x 22 + 18x 23 + 0x 24 + 35x 31 + 32x 32 + 20x 33 + 0x 34 + + 25x 41 + 25x 42 + 14x 43 + 0x 44 s.t. x 11 + x 12 + x 13 + x 14 = 1(row 1) x 21 + x 22 + x 23 + x 24 = 1(row 2) x 31 + x 32 + x 33 + x 34 = 1(row 3) x 41 + x 42 + x 43 + x 44 = 1(row 4) x 11 + x 21 + x 31 + x 41 = 1(column 1) x 12 + x 22 + x 32 + x 42 = 1(column 2) x 13 + x 23 + x 33 + x 43 = 1 (column 3) x 14 + x 24 + x 34 + x 44 = 1 (column 4) x ij >= 0 for i = 1, 2, 3, 4 and j = 1, 2, 3, 4 (nonnegativity)

25. Slide 25 of 27 Example 1: AP via LP n The solver formulation is:

26. Slide 26 of 27 Example 1: AP via LP n The solver solution is: