Workforce scheduling – Days off scheduling 1. n is the max weekend demand n = max(n 1,n 7 ) Surplus number of employees in day j is u j = W – n j for.

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Presentation transcript:

Workforce scheduling – Days off scheduling 1

n is the max weekend demand n = max(n 1,n 7 ) Surplus number of employees in day j is u j = W – n j for j = 2,…,6 and u j = n-n j for j = 1, 7 Since max weekend demand is n the remaining W-n can take the weekend off See text for the heuristic and an example 2

Shift scheduling m time intervals that are not equal During each time interval i, i = 1,…,m, b i personnel are required n different shift patterns and each employee is assigned to only one pattern j, j = 1,…,n Shift pattern j is denoted as vector (a 1j, a 2j, …, a mj ) where a ij = 1 if period i is a working period in shift j c j is the cost of assigning a person to shift j x j is the number of people assigned to shift j Solve using integer programming Min cx a 11 ……….a in st Ax≥bA= a a 2n x≥0, x = integer a m1 ……….a mn Strongly NP Hard- hence use LP relaxation and solve in polynomial time 3

Cyclic Staffing Problem Minimize the cost of assigning people to am m-period cyclic schedule Sufficient workers are available at time i to meet the demand of b i Each person works a shift of k consecutive periods and is off for the m-k periods Period m is followed by period 1 x j is the number of people assigned to shift j A = for a 7 day cycle with 2 consecutive days off and so on 4

Cyclic Staffing Problem Solve the LP relaxation an obtain x j ’ = x 1 ’,……, x n ’ If x j ’ are integers then it is the optimal solution. STOP Else from two LPs LP’ LP” and add constraint x 1 + x 2 +,…,+ x n = x 1 ’ +,……,+ x n ’ (rounded to the lower side) to LP’ x 1 + x 2 +,…,+ x n = x 1 ’ +,……,+ x n ’ (rounded to the upper side) to LP” LP” has an optimal solution that is integer If LP’ does not have a feasible solution then LP” is optimal If LP’ has a feasible solution, then it has an optimal solution that is integer and the solution to the original problem is the better one of the solutions to LP’ and LP” 5

Crew Scheduling m jobs – flight legs i = I,…,m n - feasible and all possible combinations of flight legs that a crew can handle – these are n feasible and all possible round trips j, j = 1,…,n that can be generated from the flight legs. c j cost of round trip j Each flight leg must be covered by exactly one round trip b i = 1 Minimize cost a ij is 1 if flight leg i is covered by round trip j x j is 0-1 variable and denotes whether a round trip is selected. Min cx a 11 ……….a in st Ax=1A= a a 2n x=0-1, x = integer a m1 ……….a mn NP hard. So use heuristics 6