1. Assignment Problem How to assign the given jobs to some workers on a one- to-one basis so that the jobs are completed in the least time or at the least cost. How to deal with situations when the number of jobs do not match with the number of job performers. How should the salesmen of a company be assigned to different sales zones so that the total expected sales are maximised? How to schedule the flights or the bus routes between two cities so that the layover times for the crew can be minimised?

2. A) complete enumeration method B) Transportation method C) Simplex method D) Hungarian assignment method 3. Ex. A production supervisor is considering how he should assign the four jobs that are to be performed, to four of the workers. He wants to assign the jobs to the workers such that the aggregate time to perform the jobs is the least. Based on previous experience, he has the information on the time taken by the four workers in performing these jobs, as given in table. Worker Job A B C D 1 45 40 51 67 2 57 42 63 55 3 49 52 48 64 4 41 45 60 55 4. Initial Solution : VAM Worker Job Supply A B C D 1 45 [ ] 40 [ ] 51 67 [1] 1 2 57 42 [1] 63 55 1 3 49 [ ] 52 48 [1] 64 1 4 41 [1] 45 60 55 1 Demand 1 1 1 1 4 Total Time : 1 x 67 + 1 x 42 + 1 x 48 + 1 x 41 = 198 5. Initial Solution : Non-optimal Worker Job Supply ui A B C D 1 45 [ ] 40 [ ] 51 67 [1] 1 0 2 57 42 [1] 63 55 1 2 3 49 [] 52 48 [1] 64 1 4 4 41 [1] 45 60 55 1 -4 Deman d 1 1 1 1 4 vj 45 40 44 67 6. Worker Job Supply ui A B C D 1 45 [ ] 40 [ ] (-7) 51 67 [1] 1 0 2 (-10) 57 42 [1] (-17) 63 (+14) 55 1 2 3 49 [ ] (-8) 52 48 [1] (+7) 64 1 4 4 41 [1] (-9) 45 (-20) 60 (+8) 55 1 -4 Deman d 1 1 1 1 4 vj 45 40 44 67 Initial Solution : Non-optimal 7. Worker Job Supply ui A B C D 1 45 [ ] 40 [1 ] (-7) 51 ( -14) 67 1 0 2 (-10) 57 42 [] (-17) 63 55 [1] 1 2 3 49 [ ] (-8) 52 48 [1] (-7) 64 1 4 4 41 [1] (-9) 45 (-20) 60 (-6) 55 1 -4 Demand 1 1 1 1 4 vj 45 40 44 53 Initial Solution : optimal Total Time = 1 x 40 + 1 x 55 + 1 x 48 + 1 x 41 = 184 8. Hungarian Assignment Method (HAM) Step 1 : Locate the smallest cost element in each row of the cost table. Now subtract this smallest element from each element in that row. As a result, there shall be at least one zero in each row of the new table, called the Reduced cost table. Step 2: In this reduced cost table obtained, consider each column and locate the smallest element in it. Subtract the smallest value from every other entry in the column. As a result, there would be at least one zero in each of the rows and columns of the second reduced cost table. Step 3: Draw the minimum number of horizontal and vertical lines that are required to cover all the Zero. If the number of lines drawn is equal to n (the number of rows/columns) the solution is optimal, and proceed to step 6. If the number of lines drawn is smaller than n, go to step 4. Step 4: Select the smallest uncovered (by the lines) cost element. Subtract this element from all uncovered elements including itself and add this element to each value located at the intersection of any two lines. Step 5: Repeat steps 3 and 4 until an optimal solution is obtained. 9. Step 6: Make the job assignments to Zero elements. a) Locate a row which contains only one zero element. Assign the job corresponding to this element. Cross out the zeros, if any, in the column corresponding to the element b) Repeat (a) for each of such rows which contain only one zero. Similarly, perform the same operation in respect of each column containing only one zero, crossing out the zero, if any, in the row c) If there is no row or column with only a single Zero left, then select a row/column arbitrarily and choose one of the jobs and make the assignment. Now cross the remaining zeros in the column and row in respect of which the assignment is made. d) Repeat steps (a) through (c) until all assignments are made. e) Determine the total cost with reference to the original cost table 10. Ex. Worker Job A B C D 1 45 40 51 67 2 57 42 63 55 3 49 52 48 64 4 41 45 60 55 Step 1 : Locate the smallest cost element in each row of the cost table. Now subtract this smallest element from each element in that row. As a result, there shall be at least one zero in each row of the new table, called the Reduced cost table. 11. Worker Job A B C D 1 5 0 11 27 2 15 0 21 13 3 1 4 0 16 4 0 4 19 14 Step 1 Step 1 : Locate the smallest cost element in each row of the cost table. Now subtract this smallest element from each element in that row. As a result, there shall be at least one zero in each row of the new table, called the Reduced cost table. 12. Worker Job A B C D 1 5 0 11 27 2 15 0 21 13 3 1 4 0 16 4 0 4 19 14 Step 2: In this reduced cost table obtained, consider each column and locate the smallest element in it. Subtract the smallest value from every other entry in the column. As a result, there would be at least one zero in each of the rows and columns of the second reduced cost table. 13. Step 2 Worker Job A B C D 1 5 0 11 14 2 15 0 21 0 3 1 4 0 3 4 0 4 19 1 Step 2: In this reduced cost table obtained, consider each column and locate the smallest element in it. Subtract the smallest value from every other entry in the column. As a result, there would be at least one zero in each of the rows and columns of the second reduced cost table. 14. Step 3 & 4 Worker Job A B C D 1 5 0 11 14 2 15 0 21 0 3 1 4 0 3 4 0 4 19 1 Step 3: Draw the minimum number of lines covering all zeros. As a general , we should first cover those rows/columns which contain larger number of zeros. Step 3: Count the number of lines. Which is 4, equal to number of rows/column, so optimal solution is obtained. 15. Step 4 Worker Job A B C D 1 5 0 11 14 2 15 0 21 0 3 1 4 0 3 4 0 4 19 1 Step 4: Make the assignment after scanning the rows and columns for unit zeros. Total Time = 40 + 55 + 48 + 41 = 184 minutes 16. Ex. Using the following cost matrix, determine (a) optimal job assignment and (b) the cost of assignments. Machini st Job 1 2 3 4 5 A 10 3 3 2 8 B 9 7 8 2 7 C 7 5 6 2 4 D 3 5 8 2 4 E 9 10 9 6 10 17. Step 1: Obtain row reduction Machini st Job 1 2 3 4 5 A 8 1 1 0 6 B 7 5 6 0 5 C 5 3 4 0 2 D 1 3 6 0 2 E 3 4 3 0 4 18. Step 2: Obtain column reduction Machini st Job 1 2 3 4 5 A 7 0 0 0 4 B 6 4 5 0 3 C 4 2 3 0 0 D 0 2 5 0 0 E 2 3 2 0 2 Since the number of lines covering all zeros is less than the number of columns/rows, we have to modify the table The lest of the uncovered cell values is 2. this would be subtracted from each of the uncovered values and added to each value lying at the intersection of lines. 19. Step 3:Reduced cost table Machini st Job 1 2 3 4 5 A 7 0 0 2 6 B 4 2 3 0 3 C 2 0 1 0 0 D 0 2 5 2 2 E 0 1 0 0 2 Number of lines is 5 and number of rows/columns are 5, so solution is optimal 20. Step 4:Assignment of Job Machini st Job 1 2 3 4 5 A 7 0 0 2 6 B 4 2 3 0 3 C 2 0 1 0 0 D 0 2 5 2 2 E 0 1 0 0 2 Total cost associated = 3 + 2 + 4 + 3 + 9 = 21 21. Solve the following assignment Problem by Hungarian assignment method Worker Job 1 2 3 A 4 2 7 B 8 5 3 C 4 5 6 22. Step 1 : Obtain Row reduction Worker Job 1 2 3 A 2 0 5 B 5 2 0 C 0 1 2 Ans. A-2, B-3, C-1, Total Time = 2 + 3 + 4 = 9 Minutes. 23. ABC company is engaged in manufacturing 5 brands of packed snacks. It has five manufacturing setups, each capable of manufacturing any of its brands one at a time. The cost to make a brand on these setups vary according to the following table. Machini st Setup S1 S2 S3 S4 S5 B1 4 6 7 5 11 B2 7 3 6 9 5 B3 8 5 4 6 9 B4 9 12 7 11 10 B5 7 5 9 8 11 Assuming five setups and five brands, find the optimum assignment of products on these setups resulting in the minimum cost 24. Step 1: Obtain row reduction Machini st Setup S1 S2 S3 S4 S5 B1 0 2 3 1 7 B2 4 0 3 6 2 B3 4 1 0 2 5 B4 2 5 0 4 3 B5 2 0 4 3 6 25. Step 2: Obtain column reduction Machini st Setup S1 S2 S3 S4 S5 B1 0 2 3 0 5 B2 4 0 3 5 0 B3 4 1 0 1 3 B4 2 5 0 3 1 B5 2 0 4 2 4 Since the number of lines covering all zeros is less than the number of columns/rows, we have to modify the table The lest of the uncovered cell values is 1. This would be subtracted from each of the uncovered values and added to each value lying at the intersection of lines. 26. Machini st Setup S1 S2 S3 S4 S5 B1 0 3 4 0 5 B2 4 1 4 5 0 B3 3 1 0 0 2 B4 1 5 0 2 0 B5 2 0 4 1 3 Step 3: Reduce cost table Since the number of lines covering all zeros is equal to columns/rows, so solution is optimal 27. Machini st Setup S1 S2 S3 S4 S5 B1 0 3 4 0 5 B2 4 1 4 5 0 B3 3 1 0 0 2 B4 1 5 0 2 0 B5 2 0 4 1 3 Step 4: Assignment Ans: B1-S1, B2-S5, B3-S4, B4-S3, B5-S2, Total cost = 4+5+6+7+5 = 27 28. Unbalanced assignment Problems When number of columns and number of rows are equal, the problem is balanced problem, and when not, it is called an unbalanced problem. In such situations, dummy column/row, whichever is smaller in number, are inserted with zeros as the cost elements. 29. Constrained Assignment problems Sometimes a worker cannot perform a certain job or is not to be assigned a particular job. To cope with this situation, the cost of performing that job by such person is taken to be extremely large (M). 30. Person Job J1 J2 J3 J4 J5 P1 27 18 X 20 21 P2 31 24 21 12 17 P3 20 17 20 X 16 P4 22 28 20 16 27 31. Person Job J1 J2 J3 J4 J5 P1 27 18 M 20 21 P2 31 24 21 12 17 P3 20 17 20 M 16 P4 22 28 20 16 27 P5 (Dummy) 0 0 0 0 0 Step 1: Reduce row table 32. Person Job J1 J2 J3 J4 J5 P1 9 0 M 2 3 P2 19 12 9 0 5 P3 4 1 4 M 0 P4 6 12 4 0 11 P5 (Dummy) 0 0 0 0 0 Step 1: Reduce row/column table Since the number of lines covering all zeros is not equal to columns/rows, so solution is not optimal 33. Person Job J1 J2 J3 J4 J5 P1 9 0 M 6 3 P2 15 8 5 0 1 P3 4 1 4 M 0 P4 2 8 0 0 7 P5 (Dummy) 0 0 0 4 0 Step 2: Reduce cost table Since the number of lines covering all zeros is equal to columns/rows, so solution is optimal 34. Person Job J1 J2 J3 J4 J5 P1 9 0 M 6 3 P2 15 8 5 0 1 P3 4 1 4 M 0 P4 2 8 0 0 7 P5 (Dummy) 0 0 0 4 0 Step 3: Assignment Ans: Assignment is P1-J2, P2-J4, P3-J5, P4-J3, J1 would remain unassigned. Total Cost = 18 + 12 + 16 + 20 = 66 35. In the modification of a plant layout of a factory four new machines M1, M2, M3, and M4 are to be installed in the machine shop. There are five vacant places A, B, C, D and E available. Because of limited space, machine M2 cannot be placed at C and M3 cannot be placed at A. The cost of locating a machine at a place in hundreds of rupees is as under A B C D E M1 9 11 15 10 11 M2 12 9 -- 10 9 M3 -- 11 14 11 7 M4 14 8 12 7 8 Find the optimal assignment schedule. 36. Step 1: Reduce row/column table A B C D E M1 0 2 6 1 2 M2 3 0 M 1 0 M3 M 4 7 4 0 M4 7 1 5 0 1 M5 (dummy) 0 0 0 0 0 Since the number of lines covering all zeros is equal to columns/rows, so solution is optimal 37. Step 2: Assignment A B C D E M1 0 2 6 1 2 M2 3 0 M 1 0 M3 M 4 7 4 0 M4 7 1 5 0 1 M5 (dummy) 0 0 0 0 0 Ans: Assignment is M1-A, M2-B, M3-E, M4-D, Place C will be unassigned. Total cost in hundreds of rupees = 9+9+7+7=32. 38. Unique vs Multiple optimal solutions In the process of making assignments, we select a row/column with only a single zero to make an assignment. However, a situation may arise wherein the various rows and columns have multiple zeros. In such cases, we get multiple optimal solutions to the given problem. 39. Worker Job (Cost in Rs.) 1 2 3 4 5 A 25 18 32 20 21 B 34 25 21 12 17 C 20 17 20 32 16 D 20 28 20 16 27 40. Worker Job (Cost in Rs.) 1 2 3 4 5 A 25 18 32 20 21 B 34 25 21 12 17 C 20 17 20 32 16 D 20 28 20 16 27 E (dummy) 0 0 0 0 0 Step 1: Reduce row table 41. Worker Job (Cost in Rs.) 1 2 3 4 5 A 7 0 14 2 3 B 22 13 9 0 5 C 4 1 4 16 0 D 4 12 4 0 11 E (dummy) 0 0 0 0 0 Since the number of lines covering all zeros is not equal to columns/rows, so solution is not optimal 42. Worker Job (Cost in Rs.) 1 2 3 4 5 A 7 0 14 6 3 B 18 9 5 0 1 C 4 1 4 20 0 D 0 8 0 0 7 E (dummy) 0 0 0 0 0 43. Worker Job (Cost in Rs.) 1 2 3 4 5 A 7 0 14 6 3 B 18 9 5 0 1 C 4 1 4 20 0 D 0 8 0 0 7 E (dummy) 0 0 0 0 0 Solution 1 44. Worker Job (Cost in Rs.) 1 2 3 4 5 A 7 0 14 6 3 B 18 9 5 0 1 C 4 1 4 20 0 D 0 8 0 0 7 E (dummy) 0 0 0 0 0 Solution 2 45. Solution 1 Worker Job Cost A 2 18 B 4 12 C 5 16 D 1 20 Job left 3 Total cost 66 Solution 2 Worker Job Cost A 2 18 B 4 12 C 5 16 D 3 20 Job left 1 Total cost 66 46. To stimulate interest and provide an atmosphere for intellectual discussion, the finance faculty in a management school decides to hold special seminars on four contemporary topics. Such seminars would be held once per week in the afternoons. However, scheduling these seminars (one for each topic, and not more than one seminar per afternoon) has to be done carefully so that the number of students unable to attend is kept to a minimum. A careful study indicated that the number of students who cannot attend a particular seminar on a specific day is as follows: Day Leasing Portfolio Managem ent Private Mutual Funds Swaps and Options Monday 50 40 60 20 Tuesday 40 30 40 30 Wednesday 60 20 30 20 Thursday 30 30 20 30 Friday 10 20 10 30 Find an optimal schedule of the seminars. Also find out the total number of students who will be missing at least one seminar. 47. Day Leasing Portfolio Management Private Mutual Funds Swaps and Options Monday 50 40 60 20 Tuesday 40 30 40 30 Wednesday 60 20 30 20 Thursday 30 30 20 30 Friday 10 20 10 30