Crew Scheduling

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CREW SCHEDULING

İ.HAKAN KARAÇİZMELİ

GENERAL VIEW

• • • • • •

CREW SCHEDULING TYPES FLEXIBLE MANAGEMENT STRATEGIES DESCRIPTION OF PROBLEM FORMULATION OF PROBLEM MODEL IN LINGO SOLUTION & ANALYSIS

CREW SCHEDULING

• Airline Crew Scheduling
1.The most appropriate pairings.
2.Equal workloads. 3.Minimum crew COSTS.

• Mass Transit Crew Scheduling
1.Railway track maintenance problems.
2.Mathematical program. 3.Tabu search.

• Generic Crew Scheduling
1.Manpower scheduling problems.
2.Mixed integer program. 3.Mimimum manpower. 4.Package programs(CPLEX..).

FLEXIBLE MANAGEMENT STRATEGIES
• Functional Flexibility


• •

-Deployment on different tasks. Numerical Flexibility -Variable working hours. Temporal Flexibility -Career breaks,job sharing,term-time works.. Wage Flexibility -Performance related pays.

DESCRIPTION OF PROBLEM
-Algorithm of Problem:
SOFTWARE COMPANY SOFTWARE COMPANY

CALL OF CUSTOMER CALL OF CUSTOMER

CUSTOMER

ASSIGN SERVICE ENGINEER

Informations about problem

• Service engineering is not different job . All of


• •

Software engineers may go services . Service time includes times which pass on the way too . We see that service times did not pass over 2 hours according to old datas . This problem include assignments only for an afternoon .

Customer Number 1 2 3 4 5 6 7 8 9 10 11

Time of Appointment 13:00 13:00 14:00 14:00 14:30 15:00 15:00 16:00 16:00 16:00 17:00

# of Services in one tour 1

Costs($) 10

2
3 4

18
25 30

Tour Number 1 2 3 4 5 6 7 8 9 10 11

Customer Number

1 2 3 4 5 6 7 8 9 10 11

Cost1 10 10 10 10 10 10 10 10 10 10 10

Tour Number 12(1) 13(2) 14(3) 15(4) 16(5)

Customer Number 1,6 1,7 1,8 1,9 1,10

Cost2 18 18 18 18 18

17(6)
18(7) 19(8) 20(9)

1,11
2,6 2,7 2,8

18
18 18 18

21(10)
22(11) 23(12) 24(13) 25(14) 26(15) 27(16) 28(17) 29(18) 30(19) 31(20) 32(21)

2,9
2,10 2,11 3,8 3,9 3,10 3,11 4,8 4,9 4,10 4,11 5,11

18
18 18 18 18 18 18 18 18 18 18 18

Tour Number

33(1)
34(2) 35(3)

Customer Number 1,6,11 1,7,11 2,6,11

Cost3

25
25 25

36(4)

2,7,11

25

After these informations we describe our mathematical model:

• Decison Variables :
-X : Number of 1 Customer Service in One Tour ( X=1..11 ) -Y : Number of 2 Customer Services in One Tour ( Y=1..21 ) -Z : Number of 3 Customer Services in One Tour ( Z=1..4 )

• Objective Function:
-Zmin=∑(X*Cost1) + ∑(Y*Cost2) + ∑(Z*Cost3)

• Constraints:
– For customer 1 : X1 + Y1 + Y2 + Y3 + Y4 +Y5 + Y6 + Z1 + Z2 =1 – For customer 2 : X2 + Y7 + Y8 + Y9 + Y10 + Y11 + Y12 + Z3 + Z4 = 1 – For customer 3 : X3 + Y13 + Y14 + Y15 + Y16 = 1 – For customer 4 : X4 + Y17 + Y18 + Y19 + Y20 = 1 – For customer 5 : X5 + Y21 = 1 – For customer 6 : X6 + Y1 + Y7 + Z1 + Z3 = 1 – For customer 7 : X7 + Y2 + Y8 + Z2 + Z4 = 1 – For customer 8 : X8 + Y3 + Y9 + Y13 + Y17 = 1 – For customer 9 : X9 + Y4 + Y10 + Y14 + Y18 = 1 – For customer10: X10 + Y5 + Y11 + Y15 + Y19 = 1 – For customer11: X11 + Y6 + Y12 + Y16 + Y20 + Y21 + Z1 + Z2 + Z3 + Z4=1

MODEL IN LINGO
SETS: SERVICE/1..11/:COST1,X; SERVICE2/1..21/:COST2,Y; LOOK(SERVICE,SERVICE2):MATRIX1; SERVICE3/1..4/:COST3,Z; LOOK2(SERVICE,SERVICE3):MATRIX2; ENDSETS

DATA: COST1=10 10 10 10 10 10 10 10 10 10 10; MATRIX1=1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 000000111111000000000 000000000000111100000 000000000000000011110 000000000000000000001 100000100000000000000 010000010000000000000 001000001000100010000 000100000100010001000 000010000010001000100 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1;

COST2=18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18; MATRIX2=1 1 0 0 0011 0000 0000 0000 1010 0101 0000 0000 0000 1 1 1 1; COST3=25 25 25 25; ENDDATA

@FOR(SERVICE:@BIN(X)); @FOR(SERVICE2:@BIN(Y)); @FOR(SERVICE3:@BIN(Z)); MIN=@SUM(SERVICE:X*COST1)+@SUM(SERVIC E2:Y*COST2)+@SUM (SERVICE3:Z*COST3); @FOR(SERVICE(I):X(I)+@SUM(SERVICE2(J):MAT...
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