assignment problem

ASSIGNMENT CASE STUDY

ASSIGNMENT CASE STUDY

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.

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.

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)

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.

Case

Flight No.

Delhi Departure

Kolkata Arrival

Flight No.

KolkataDeparture

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

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.

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 1100202 0900 1100 102 1000 1200203 1330 1530 103 1530 1730204 1930 2130 104 2000 2200

24


Dep /ArrFlight No.

101 102 103 104

201 24202

203

204


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 1100202 0900 1100 102 1000 1200203 1330 1530 103 1530 1730204 1930 2130 104 2000 2200

25


Dep /ArrFlight

101 102 103 104

201 24 25 6.5202

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.


Dep /ArrFlight

101 102 103 104

201 24.0 25.0 6.5 11.0202 22.0 23.0 28.5 9.0203 17.5 18.5 24.0 28.5204 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

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 1100202 0900 1100 102 1000 1200203 1330 1530 103 1530 1730204 1930 2130 104 2000 2200

20 hrs


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 1100202 0900 1100 102 1000 1200203 1330 1530 103 1530 1730204 1930 2130 104 2000 2200

22 hrs


Arr /DepFlight

101 102 103 104

201 20.0 19.0 13.5 9.0202 22.0 21.0 15.5 11.0203 26.5 25.5 20.0 15.5204 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

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.

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

Solution Step 4-A

Row Reduction Method

From each row we identify the smallest element and subtract it from each element of the row

Solution Step 4-A (Row reduction)


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 1Likewise the same method is followed for each row

Solution Step 4-A (Row Reduction)


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 after Row Reduction

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

Solution Step-B (Column Reduction)


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)

For example the minimum value in column 1 is 1 and this is subtracted from each element of column 1. We follow the same procedure in the remaining columns also

Solution Step 4-C (Opportunity Cost Matrix)


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 after column reduction

Solution Step 5 (Optimality Criterion)


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.

Solution Step 6


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

Solution Step 7


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.

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 NumberWaiting Time

(in hours)

1 Delhi 201 – 103 6.52 Delhi 202 – 104 9.03 Delhi 203 – 101 17.54 Kolkata 204 - 102 7.5

Total 40.5

assignment problem

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