Case Airlines

8/3/2019 Case Airlines

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Case- AirlinesCase- Airlines


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Case DiscussionCase Discussion

We have given a time table of an airline company,We have given a time table of an airline company,

which provides service between Jaipur and Delhiwhich provides service between Jaipur and Delhi

The crew must have a layover time of 5 hoursThe crew must have a layover time of 5 hours The Time Table is shown on the next slideThe Time Table is shown on the next slide


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Time Table of the AirlineTime Table of the Airline


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The cost of service to the Airline company depends on

the Traveling as well as the Waiting time spent by the

crew. We have to connect the airlines in such a way that it

minimizes the waiting time.

This is a typical case of a Standard Assignment

problem which will be solved by the Hungarian

Algorithm.


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For its application it is assumed that:

1) All of the Cs of the starting cost matrix are non

negative integers.2) The assignment problem is of minimization case.

Minimization of layover time is what we have to focuson.

Also we station the crew first at one airport, and thento other to get our initial table.


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Layover Time At Jaipur

201 202 203 204

101 0 .5 4 9.5

102 23 23.5 3 8.5

103 17.5 18 22 3

104 12.5 13 17 22


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Layover Time At DelhiLayover Time At Delhi

201 202 203 204

101 21.75 21.25 17.75 12.25

102 22.75 22.25 18.75 13.25

103 4.25 3.75 .25 18.75

104 9.25 8.75 5.25 25.75


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Combining both the tables and taking the values more

than 5 (waiting time constraint)

201 202 203 204

101 21.75 21.25 17.75 9.5

102 22.75 22.25 18.75 8.5

103 17.5 18 21.5 18.75

104 9.25 8.75 5.25 22


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Taking out the minimum value of each row and

subtracting it from the respective rows

201 202 203 204

101 12.25 11.75 8.25 0

102 14.25 13.75 10.25 0

103 0 .5 4 1.25

104 4 3.5 0 16.75


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Similarly, Taking out the minimum value of each column

and subtracting it from the respective columns

201 202 203 204

101 12.25 11.25 8.25 0

102 14.25 13.25 10.25 0

103 0 0 4 1.25

104 4 3 0 16.75


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201 202 203 204

101 12.25 11.25 8.25 0

102 14.25 13.25 10.25 0

103 0 0 4 1.25

104 4 3 0 16.75


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201 202 203 204

101 4 3 0 0

102 6 5 2 0

103 0 0 4 9.5

104 4 3 0 25


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Case 1Case 1

201 202 203 204

101 1 0 0 0

102 3 2 2 0

103 0 0 7 12.5

104 1 0 0 25


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Case 2Case 2

201 202 203 204

101 1 0 0 0

102 3 2 2 0

103 0 0 7 12.5

104 1 0 0 25


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Flight 101 is connected with flight 202, and stopover time is21 hours 15 minutes at Delhi

Flight 102 is connected with flight 204, and stopover time is

8 hours 30 minutes is at JaipurFlight 103 is connected with flight 201, and stopover time is17 hours 30 minutes at Jaipur

Flight 104 is connected with flight 203, and stopover time is5 hours 15 minutes at Delhi

TOTAL HALT TIME PERTAINING TO MINIMUM COSTCOMESOUT TO BE =

17.75 + 8.5 + 17.5 + 5.25 = 52.5 HRS

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