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### assignment problem

1. 1. ASSIGNMENT CASE STUDY ASSIGNMENT CASE STUDY
2. 2. Introduction An Assignment Problem is a particular case of transportation problem where 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 of allocation. The problem of assignment arises because available resources have varying degrees of efficiency for performing different activities.
3. 3. Applications Some of the problems where assignment technique may be useful are :  Assignment of workers to machines,  Salesmen to different sales area,  Clerks to various checkout counters,  Vehicles to routes,  Contract to bidders,  Pairing of crew with flights schedule, etc.
4. 4. Solving Method Hungarian Method (Developed by D. Konig) The Hungarian method of assignment works on the principle of reducing the given the cost matrix to a matrix of opportunity costs. Opportunity Costs show the relative penalties associated with assigning a resource an activity as opposed to making the best or least cost assignment. If we can reduce the cost matrix to the extent of having at least one zero in each row and column, it will be possible to make optimal assignments (opportunity costs are all zero)
5. 5. Case : Assigning of crew to flights An airline that operates seven days a week has a timetable as shown in the next slide. Crews must have a minimum rest of six hours between flights. Obtain the pairing of flights that minimize waiting time away from the city. For any given pairing the crew will be based at the city that results the smaller waiting time. For each pair also mention the city where the crew should be based.
6. 6. Case Fligh t No. Delhi Departure Kolkata Arrival Flight No. Kolkata Departure Delhi Arrival 201 07:00 A.M. 09:00 A.M. 101 09:00 A.M. 11:00 A.M. 202 09:00 A.M. 11:00 A.M. 102 10:00 A.M. 12:00 Noon 203 01:30 P.M. 03:30 P.M. 103 03:30 P.M. 05:30 P.M. 204 07:30 P.M. 09:30 P.M. 104 08:30 P.M. 10:30 P.M. Table – 1: Time table of the airline
7. 7. Problem Solving Approach  As the service time is constant, it does not affect the decision of stationing the crew.  Assume that the crew of all the flights are based at Delhi.  For each combination if the onward and return flight, find the time the crew has to wait in Kolkata to catch the next available flight to return keeping in mind that the minimum time the crew has to rest at Kolkata is 6 hours.  Perform the above step assuming that the crew of all the flights are based at Kolkata.  For each combination record the minimum of the two durations.  Apply the Assignment model technique to minimize the waiting time away from the city.
8. 8. Solution Step 1 (Crew is based at Delhi) The crew for flight 201 reaches Kolkata at 0900 hrs and thus can start from Kolkata to Delhi earliest at 0900 + 6 hrs = 1500 Hrs. So if the crew has to return by Flight 101, it has to wait for the next day till 0900 Hrs for exactly 24 Hrs. Flight no. Delhi Departure Kolkata Arrival Flight no. Kolkata Departure Delhi Arrival 201 0700 0900 101 0900 1100 202 0900 1100 102 1000 1200 203 1330 1530 103 1530 1730 204 1930 2130 104 2000 2200 24
9. 9. Solution Step 1 (Crew is based at Delhi) Dep /Arr Flight No. 101 102 103 104 201 24 202 203 204
10. 10. Solution Step 1 (Crew is based at Delhi) The crew for flight 201 reaches Kolkata at 0900 hrs and thus can start from Kolkata to Delhi earliest at 0900 + 6 hrs = 1500 Hrs. So if the crew has to return by Flight 102, it has to wait for the next day till 1000 Hrs for exactly 25 Hrs. Flight no. Delhi Departure Kolkata Arrival Flight no. Kolkata Departure Delhi Arrival 201 0700 0900 101 0900 1100 202 0900 1100 102 1000 1200 203 1330 1530 103 1530 1730 204 1930 2130 104 2000 2200 25
11. 11. Solution Step 1 (Crew is based at Delhi) Dep /Arr Flight 101 102 103 104 201 24 25 6.5 202 203 204 Similarly, if the crew has to return on flight 103 which departs from Kolkata at 1530 Hrs, There will be a gap of 6.5 Hrs which is acceptable as per the rules and the crew can return on the same day.
12. 12. Solution Step 1 (Crew is based at Delhi) Dep /Arr Flight 101 102 103 104 201 24.0 25.0 6.5 11.0 202 22.0 23.0 28.5 9.0 203 17.5 18.5 24.0 28.5 204 11.5 12.5 18.0 22.5 Similarly, the waiting duration for all the combinations of arrival and departure flights can be calculated. Table – 2: Waiting time when the full crew is based at Delhi
13. 13. Solution Step 2 (Crew is based at Kolkata) In the second step we assume that the crew for all the flights are based in Kolkata and have to return from Kolkata after a minimum stay of 6 Hrs in Delhi. If the crew of flight 101 has to return by the flight 201 it has to stay in Delhi for at least 20 Hrs. Flight no. Delhi Departure Kolkata Arrival Flight no. Kolkata Departure Delhi Arrival 201 0700 0900 101 0900 1100 202 0900 1100 102 1000 1200 203 1330 1530 103 1530 1730 204 1930 2130 104 2000 2200 20 hrs
14. 14. Solution Step 2 (Crew is based at Kolkata) If Instead the crew which flew by flight 101 has to return by flight 202 then the minimum duration of stay at Delhi would be 22 Hrs. Flight no. Delhi Departure Kolkata Arrival Flight no. Kolkata Departure Delhi Arrival 201 0700 0900 101 0900 1100 202 0900 1100 102 1000 1200 203 1330 1530 103 1530 1730 204 1930 2130 104 2000 2200 22 hrs
15. 15. Solution Step 2 (Crew is based at Kolkata) Arr/Dep Flight 101 102 103 104 201 20.0 19.0 13.5 9.0 202 22.0 21.0 15.5 11.0 203 26.5 25.5 20.0 15.5 204 8.5 7.5 26.0 21.5 Similarly, the waiting duration for all the combinations of arrival and departure flights can be calculated, assuming that the crew is based at Kolkata. Table – 3: When the full crew is based at Kolkata
16. 16. Solution Step3(Developing Opportunity Cost Matrix) We have calculated the waiting time when all crew members are asked to reside at Delhi and in the other case at Kolkata. As the crew can be asked to reside at Delhi or Kolkata, the minimum waiting time from Tables 1 & 2 can be obtained for the different route connections by choosing minimum value out of the waiting times. These values of waiting times are shown in Table 3.
17. 17. Solution Step 3 (Opportunity Cost Table) Arrival/ Departure Flight No. 101 102 103 104 201 20(K) 19(K) 6.5(D) 9(K) 202 22(D/K) 22(K) 15.5(K) 9(D) 203 17.5(D) 18.5(D) 20(K) 15.5(K) 204 8.5(K) 7.5(K) 18(D) 21.5(K) Where : K stands for Kolkata and D stands for Delhi Table – 4: Opportunity Cost Table
18. 18. Solution Step 4-A Row Reduction Method From each row we identify the smallest element and subtract it from each element of the row
19. 19. Solution Step 4-A (Row reduction) Arrival/ Departure Flight No. 101 102 103 104 201 20(K) 19(K) 6.5(D) 9(K) 202 22(D/K) 22(K) 15.5(K) 9(D) 203 17.5(D) 18.5(D) 20(K) 15.5(K) 204 8.5(K) 7.5(K) 18(D) 21.5(K) In row 1 the smallest element is 6.5 and this value is subtracted from each element of row 1 Likewise the same method is followed for each row
20. 20. Solution Step 4-A (Row Reduction) Arrival/ Departure Flight No. 101 102 103 104 201 13.5(K) 12.5(K) 0.0(D) 2.5(K) 202 13.0(D/K) 12(K) 6.5(K) 0.0(D) 203 2.0(D) 3.0(D) 4.5(K) 0.0(K) 204 1.0(K) 0.0(K) 10.5(D) 14.0(K) Table -5 : Opportunity Cost Matrix afterRow Reduction
21. 21. Solution Step 4-B Column Reduction Method From each column of the row reduced matrix we identify the smallest element in that column and subtract it from each element of the column
22. 22. Solution Step-B (Column Reduction) Arrival/ Departure Flight No. 101 102 103 104 201 13.5(K) 12.5(K) 0.0(D) 2.5(K) 202 13.0(D/K) 12(K) 6.5(K) 0.0(D) 203 2.0(D) 3.0(D) 4.5(K) 0.0(K) 204 1.0(K) 0.0(K) 10.5(D) 14.0(K) Forexample the minimum value in column 1 is 1 and this is subtracted fromeach element of column 1. We follow the same procedure in the remaining columns also
23. 23. Solution Step 4-C (Opportunity Cost Matrix) Arrival/ Departure Flight No. 101 102 103 104 201 12.5(K) 12.5(K) 0.0(D) 2.5(K) 202 12.0(D/K) 12(K) 6.5(K) 0.0(D) 203 1.0(D) 3.0(D) 4.5(K) 0.0(K) 204 0.0(K) 0.0(K) 10.5(D) 14.0(K) Table – 6 : Opportunity Cost Matrix aftercolumn reduction
24. 24. Solution Step 5 (Optimality Criterion) Arrival/ Departure Flight No. 101 102 103 104 201 12.5(K) 12.5(K) 0.0(D) 2.5(K) 202 12.0(D/K) 12(K) 6.5(K) 0.0(D) 203 1.0(D) 3.0(D) 4.5(K) 0.0(K) 204 0.0(K) 0.0(K) 10.5(D) 14.0(K) Making Assignments in the Opportunity Cost Matrix The solution is not optimal because only three assignments are made which is not equal to the number of rows (or columns) which is equal to four. Hence, the solution is not optimal.
25. 25. Solution Step 6 Arrival/ Departure Flight No. 101 102 103 104 201 12.5(K) 12.5(K) 0.0(D) 2.5(K) 202 12.0(D/K) 12(K) 6.5(K) 0.0(D) 203 1.0(D) 3.0(D) 4.5(K) 0.0(K) 204 0.0(K) 0.0(K) 10.5(D) 14.0(K) Revising the Opportunity Cost Matrix
26. 26. Solution Step 7 Arrival/ Departure Flight No. 101 102 103 104 201 12.5(K) 12.5(K) 0.0(D) 3.5(K) 202 11.0(D/K) 11.0(K) 5.5(K) 0.0(D) 203 0.0(D) 2.0(D) 3.5(K) 0.0(K) 204 0.0(K) 0.0(K) 10.5(D) 15.0(K) Revised Opportunity Matrix The solution is optimal because the number of assignments and the number of rows each are equal to four.
27. 27. Final Solution The pattern of optimal assignments among routes with respect to the waiting time is given in the following table:- Crew Residence at Route Number Waiting Time (in hours) 1 Delhi 201 – 103 6.5 2 Delhi 202 – 104 9.0 3 Delhi 203 – 101 17.5 4 Kolkata 204 - 102 7.5 Total 40.5