CREW SCHEDULING
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CREW SCHEDULING

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CREW SCHEDULING CREW SCHEDULING Presentation Transcript

  • 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 CUSTOMER CALL OF CUSTOMER CALL OF 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 .
  • 17:00 11 16:00 10 16:00 9 16:00 8 15:00 7 15:00 6 14:30 5 14:00 4 14:00 3 13:00 2 13:00 1 Time of Appointment Customer Number
  • 30 4 25 3 18 2 10 1 Costs($) # of Services in one tour
  • 10 11 11 10 10 10 10 9 9 10 8 8 10 7 7 10 6 6 10 5 5 10 4 4 10 3 3 10 2 2 10 1 1 Cost1 Customer Number Tour Number
  • 18 5,11 32(21) 18 4,11 31(20) 18 4,10 30(19) 18 4,9 29(18) 18 4,8 28(17) 18 3,11 27(16) 18 3,10 26(15) 18 3,9 25(14) 18 3,8 24(13) 18 2,11 23(12) 18 2,10 22(11) 18 2,9 21(10) 18 2,8 20(9) 18 2,7 19(8) 18 2,6 18(7) 18 1,11 17(6) 18 1,10 16(5) 18 1,9 15(4) 18 1,8 14(3) 18 1,7 13(2) 18 1,6 12(1) Cost2 Customer Number Tour Number
  • 25 2,7,11 36(4) 25 2,6,11 35(3) 25 1,7,11 34(2) 25 1,6,11 33(1) Cost3 Customer Number Tour Number
  • 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
    • 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0
    • 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0
    • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0
    • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
    • 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    • 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
    • 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0
    • 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0
    • 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0
    • 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
    • 0 0 1 1
    • 0 0 0 0
    • 0 0 0 0
    • 0 0 0 0
    • 1 0 1 0
    • 0 1 0 1
    • 0 0 0 0
    • 0 0 0 0
    • 0 0 0 0
    • 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(SERVICE2:Y*COST2)+@SUM (SERVICE3:Z*COST3);
    • @FOR(SERVICE(I):X(I)+@SUM(SERVICE2(J):MATRIX1(I,J)*Y(J))+@SUM(SERVICE3(K):MATRIX2(I,K)*Z(K)) = 1);
    • END
  • SOLUTION & ANALYSIS
    • Objective Value = 99 $
    • X5 = 1
    • X10 = 1
    • Y1 = 1
    • Y14 = 1
    • Y17 = 1
    • Z4 = 1
    • X5 CUSTOMER5 at 14:30
    • X10 CUSTOMER10 at 16:00
    • Y1 CUSTOMER1 at 13:00
    • CUSTOMER6 at 15:00
    • Y14 CUSTOMER3 at 14:00
    • CUSTOMER9 at 16:00
    • Y17 CUSTOMER4 at 14:00
    • CUSTOMER8 at 16:00
    • Z4 CUSTOMER2 at 13:00
    • CUSTOMER7 at 15:00
    • CUSTOMER11 at 17:00
    • Objective Value=1*10+1*10+1*18+1*18+1*18+1*25=99
  • 0.0000000E+00 0.0000000E+00 12 0.0000000E+00 0.0000000E+00 11 0.0000000E+00 0.0000000E+00 10 0.0000000E+00 0.0000000E+00 9 0.0000000E+00 0.0000000E+00 8 0.0000000E+00 0.0000000E+00 7 0.0000000E+00 0.0000000E+00 6 0.0000000E+00 0.0000000E+00 5 0.0000000E+00 0.0000000E+00 4 0.0000000E+00 0.0000000E+00 3 0.0000000E+00 0.0000000E+00 2 1.000000 99.00000 1 Dual Price Slack or Surplus Row
  • THANK YOU