Post on 03-Apr-2018
transcript
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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 !