assignment problem explained

1. ASSIGNMENT CASE
STUDY
ASSIGNMENT CASE
STUDY

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. 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. 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. 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. 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. 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. 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. Solution Step 1 (Crew is based at Delhi)
Dep /Arr
Flight No.
101 102 103 104
201 24
202
203
204

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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