ppt on DMM

7/28/2019 ppt on DMM

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Presented by

Harshdeep Garg

Harshi ta Periwal

Jaiswal Shanky


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An airline that operates seven days a week

has time-table as shown in the next page.The crew must have a minimum layover

time of five hours between flights


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Problem Table

Flight

no.

Delhi-Jaipur

Departure

Delhi-Jaipur

Arrival

Flight

no.

Jaipur -Delhi

Departure

Jaipur -Delhi

Arrival

101 7:00 am 8:00 am 201 8:00 am 9:15 am

102 8:00 am 9:00 am 202 8:30 am 9:45 am

103 1:30 pm 2:30 pm 203 12:00 noon 1:15 pm

104 6:30 pm 7:30 pm 204 5:30 pm 6:45 pm


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Obtain the pairing of flights that

minimizes layover time away from

home. For any giving pairing, the crewwill be based at the city that results in

the smaller layover. For each pair also

mention the town where the crew shouldbe based.


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Solution

Let us presume that each plane makes

exactly one forward trip and one return

trip.

A plane flying from one city to the other

must come back to the starting city atthe earliest possible opportunity to keep

the layover time minimum.


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Case 1

If all crew start from Delhi, the layover time at Jaipurfor various pairing of flights is given below

Flight no 201 202 203 204

101 24 24.5 28 9.5

102 23 23.5 27 8.5

103 17.5 18 21.5 27

104 12.5 13 16.5 22


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Case 2

If all crew start from Jaipur, then layover time at Delhi

for various pairing of flights is given below

Flight no 201 202 203 204

101 21.75 21.25 17.75 12.25

102 22.75 22.25 18.75 13.25

103 28.25 27.75 24.25 18.75

104 9.25 8.75 5.25 23.75


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Assumptions

The crew can be either based at Delhi or Jaipur.

They have to be based at the city that results in

smaller layover time.

We therefore select the smaller layover time for

all pairings of flight.


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I ni tial table

Flight no 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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Hungar ian method is now applied:

Reduced Matrix after subtracting the smallestelement from each row

Flight no 201 202 203 204

101 12.25 11.75 8.25 0

102 14.25 13.75 10.25 0

103 0 0.5 4 1.25

104 4 3.5 0 16.75


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Reduced matrix after deducting the smallest element

from each column

Flight no 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

0

0

0

0

0


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Solution contd ..

Flight no 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

0

0

0

0

0

0


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Optimal basic feasible solution

Flight no 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

0

0

0

00

00

0


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Best pair ing of f l ights

Flight pair Crew based at city Layover time (hrs)

203-101 Jaipur 17.75

102-204 Delhi 8.50

103-201 Delhi 17.50

202-104 Jaipur 8.75

Total Layover time 52.50 hours


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Alternate solution

Flight no 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

0

0

0

0 0

00

0


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Alternate best pair ing of f l ights

Flight pair Crew based at city Layover time (hrs)

202-101 Jaipur 21.25

102-204 Delhi 8.50

103-201 Delhi 17.50

203-104 Jaipur 5.25

Total Layover time 52.50 hours


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Thank you !

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