Task Assignment Problem for Security
Screening Officers at Airports
A mathematical approach to minimize immediate changeovers and optimize the
positioning of idle time if demand has a sequencing pattern.
Masters Thesis Econometrics, Operations Research and Actuarial Studies Supervisor: Prof. Dr. K.J. Roodbergen
Task Assignment Problem for Security
Screening Officers at Airports
A mathematical approach to minimize immediate changeovers and optimize the
positioning of idle time if demand has a sequencing pattern.
Jurian Onderdijk
Abstract
In this thesis, an assignment problem is studied in which tasks are assigned to teams. The teams work a predefined shift of which the start, end and break times are known. In this thesis, the teams have to staff the lanes of a security screening filter at an airport. For each moment in time a team is at work, they have to be assigned to one of the lanes which have to be open to search the departing passengers. Since the demand for open lanes does not perfectly match with the available teams, idle time is incorporated in the shifts. Since there are always enough teams available, all required open lanes can be staffed.
To manage passenger flow, the lanes are opened according to an opening sequence. Additionally, the last opened lane is the first one to be closed. When a lane closes, the team either changes to another lane, takes a break, ends its shift, or becomes idle. It can happen that a team has to switch from one lane to another in consecutive time periods, which is called an instant switch.
Those instant switches disrupt the screening process, because closing a lane and opening another lane takes a few minutes. This is highly undesirable. In this study, a solution method is shown which primarily minimizes the number of instant switches and, as a secondary objective optimizes the positioning of idle time. Idle time is preferably bundled to longer periods and/or placed at the beginning or end of a shift.
We present a network formulation of the problem which can then be solved as a MIP. We present a formulation in which the breaks are fixed in advance. Furthermore, an alternative formulation is given in which small changes in the positioning of the breaks are allowed. We prove that if the objective of the problem only depends on the number of instant switches and the positioning of idle time, then instant switches are only required at time periods in which the demand decreases. This drastically reduces the number of arcs of the network formulation. This reduction of the size of the network formulation impacts the computation times significantly.
Contents
1 Introduction 5
2 Literature review 7
3 Problem description 9
4 Base model formulations 10
4.1 Fixed break model . . . 10
4.2 Variable break model . . . 12
4.3 Constraints matrix is not totally unimodular . . . 14
5 Graph reductions 15 5.1 Incorporating predefined shifts into the model . . . 15
5.2 Omitting nodes when there is no demand . . . 16
5.3 Instant switches only from closing lanes . . . 16
6 Problem extensions 18 6.1 Favouring consecutive periods of idle time . . . 18
6.2 Favouring idle time at beginning or end of shifts . . . 18
7 Computational results 19 7.1 Instances . . . 19
7.2 Results with fixed breaks . . . 21
7.3 Effect of variable breaks . . . 22
7.4 Effect of arc reductions on computation time . . . 25
8 Case study: Planning of security screening officers at airports 26 8.1 Case description . . . 26
8.2 Fixed break model compared to current practice . . . 28
8.3 Effect of variable breaks . . . 30
9 Conclusions 32
10 Further research 33
11 References 34
12 Tables and Figures 35
Appendix A: Shift specifications case study 39
1
Introduction
Passengers travelling by airplane have to undergo security screening at the departing airport. Both the passenger and the hand luggage are checked for safety reasons in this process. Typically, an airport has multiple areas where security screening can take place, these areas are known as filters. There may, for example, be a filter for domestic flights and a filter for international flights. Or a filter may be present for each terminal. Each filter is subdivided into a number of lanes. A lane is composed by a scanner for passengers, a scanner for hand luggage, equipment for conveying luggage, and is staffed by a team of security officers.
The volume of passengers to be screened fluctuates during the day. Therefore not every lane needs to be staffed by a team of security officers at all times. A working day is usually divided into a number of time periods of for example 5 minutes each. For each time period, the number of open lanes is decided by the airport for each filter. Security companies provide the teams that staff the lanes, and must consequently determine a staff planning based on the number of open lanes per time period as dictated by the airport. High salary costs and tight labour market conditions (Kleinjan and Baas, 2018) make it critical to create an efficient staff planning. The continuous growth of the airports passengers volume (Sondermeijer, 2019) will presumably increase the necessity for this, underlining the relevance of an efficient planning.
At Schiphol airport, a team of security officers, required to staff a single lane, is composed of four security agents and one team leader. The team works together for the entire day. They start and end their shift at the same time and take their breaks together. When they start their shift, they are assigned to a specific lane. Each team has to take a break at least every two hours due to regulations (see Appendix A). After the break, the team is assigned to the same or another lane, at the same or another filter. If, based on the airports schedule, a lane has to be closed, the team assigned to that lane either changes to another lane, takes a break, ends its shift, or becomes idle. An example of a shift is given in Figure 1.
To manage the passenger flow, the airport requires open lanes to be connected. Or stated differently, the airport prescribes a strict opening sequence, in which the last opened lane is the first one to be closed. In case the lane numbering coincides with the opening sequence, this implies that lane i can only be open if lanes 1, . . . , i are all open. As a direct consequence, lanes cannot temporarily close when teams take a break; another team must continue the operation at that lane.
The staff planning made by security companies at an airport must (1) meet the staffing requirements set by the airport, which includes the opening sequence of the lanes, (2) meet regulations on shifts lengths and break times. Due to the variable demand requirements, minimum shift length requirements, and difficult planning process, it is practically impossible to cover the demand exactly with the shifts. This results in idle time. Preferences exist for positioning idle time. Ideally, some idle time for a team is planned when switching from one lane to another. This reduces the number of so-called instant switches, in which a team is supposed to switch from one lane to another lane in zero time. Furthermore, we favour idle time to be at the beginning or end of a shift such that shifts might be shortened in practice. When idle time is placed in the middle of a shift, we favour a consecutive hour of idle time for a single team rather than distributing the idle time over multiple teams, since this hour can then be used for a training course.
Currently, at Schiphol airport, the planning of security staff is performed in a two-step approach. The two steps are depicted in Figure 2. The first step is the filter planning. In this step the shift for each team is determined, which consists of a start and an end time, the positioning of the break(s) and to which filter the team is assigned for each working period. In this step, the best possible schedule is calculated to minimize the idle time of all teams, with the requirement that all demand is fulfilled. In the second step, given the shifts and filter assignments from the first step, all teams are assigned to a specific lane for each time period.
Filter planning Airport Security company Lane planning Demand Sequencing pattern Staff limits
Figure 2: Setup of the planning process.
integer due to the structure of the problem.
The security filters are closed during the night or only have a single lane open between 23:00 and 1:00 Hence, a single team is assigned to cover the day changeover. Consequently, this means that the planning for a single day has no dependencies on the planning of the day before or the day after. Therefore, we will perform the planning on a daily basis.
Deciding which team works when at which lane can be seen as an assignment problem in which tasks are assigned to workers. The assignment problem is extensively studied (Sinha, 2009; Bard and Wan, 2006; Campbell and Diaby, 2002). However, this research is a clear contribution to the existing literature. In the airport setting assignments need not only meet required capacity, but must also adhere to a prescribed sequencing pattern. To the best of the authors knowledge, no other research exists with this characteristic. Though, the sequencing pattern is important in this research.
This thesis is organized as follows. We explain the background literature of the problem in Section 2. The literature is followed by a complete description of the problem in Section 3. We present two base model formulations for the TAP-IBS in Section 4, one with fixed breaks and another with variable breaks. These base models can be reduced in size as explained in Section 5. Next to that, we provide problem extensions related to the positioning of idle time in Section 6. The model formulations and extensions will be tested on several instances of which the results are given in Section 7. Afterwards, a case study with practical data from Schiphol airport is provided in Section 8. Finally, in Section 9 we draw our conclusions and in Section 10 we elaborate on possible extensions of the model which have not been included in this thesis.
2
Literature review
In this study we assign lanes to teams. This can be related to a general assignment problem. Assignment problems arise both in staff scheduling and machine scheduling. These types of problems have many similarities. However, there are some differences as machines, in general, do not take breaks. Both, staff and machine scheduling, have been studied extensively.
A review of the staff scheduling and rostering problems is given by Ernst, Jiang, Krishnamoorthy, and Sier (2004). They present several methods and models where especially the so-called task assignment problem (TAP) is applicable to this study. An extension to this TAP for the staff planning and scheduling in the service industry is given by Bard and Wan (2006). The research is applied to the US Postal Service mail processing and distribution centers. A two-phase approach is presented, in which in the first phase the weekly schedule is calculated for the employees. In this phase, the employees are assigned their workdays, starting and ending times and break windows. A deeper analysis on this phase is given in Bard, Binici, and deSilva (2003). The second phase of the research of Bard and Wan (2006) focuses on assigning specific tasks to these employees. Due to variations in demand from one time period to the next, it is necessary to re-position the workforce to maximise the use of equipment and manpower.
of workers which have predefined shifts. Workers can be assigned to any of the workstations. This can be directly translated to the security staff planning, since teams with a predefined shift have to be assigned to lanes and there are no restrictions that a team cannot be assigned to a certain lane. The sequencing pattern is an important aspect for the TAP-IBS, this pattern is not included in Bard and Wan (2006).
Different solution methods to the task assignment problem have been presented by Bard and Wan (2006) to solve the problem. They present a delayed idle period assignment, with and without daily decomposition, and a tabu-search. For these solution methods, the problem is formulated in a network design. This network design formulation is taken as a basis in this research. Our extension of the sequencing pattern of the demand does not impact the solution methods. However, the sequencing pattern can be used to simplify the network design formulation. Next to this, the idle period assignment can be directly incorporated into the network formulation.
A similar scheduling problem as in Bard and Wan (2006) is studied by Wan and Bard (2007). They incorporate restrictions defining which employee can be assigned to which type of work. They present an alternative solution method based on a large-scale integer program which is solved by using column-generation. However, the presented method focuses mainly on the restrictions of the employees. Whereas in the TAP-IBS, each team can work at any lane. This makes the solution method less applicable to incorporate for the TAP-IBS.
We found that the study of Bard and Wan (2006) is the most applicable to our study. Other studies related to the problem often differ from our characteristic that breaks or idle time are incorporated. Sinha (2009) researches assignment problems with changeover costs in machine scheduling. Two alternatives are presented, one when changeover costs are dependent on the operator, but indepen-dent of the jobs and another when changeover costs are depenindepen-dent on the jobs. The latter might be interesting for the TAP-IBS, since a job can be seen as working at a lane. The instant switches can be seen as a changeover which is undesirable and will have high costs. When this would be the only factor in the objective, this is a very applicable method. However, the TAP-IBS is a multi-objective optimization problem in which the positioning of idle time is also incorporated. For example, we favour a consecutive hour of idle time over distributing the idle time among multiple teams. Sinha (2009) focuses solely on the changeovers. This makes the methods of Sinha (2009) less applicable. Another problem where changeovers are incorporated in the task assignment is in home healthcare, since multiple patients have to be visited. Mutingi and Mbohwa (2014) present an optimization approach for that problem, which relates to the basic variant of the TAP as defined in Ernst et al. (2004). In this task assignment problem, the travel times are incorporated. Incorporating travel times in the task assignment for the security staff planning might be interesting since there are multiple filters which require walking times. By incorporating this, we might have a solution approach which solves multiple filters at once. However, Mutingi and Mbohwa (2014) treat the traveling time as part of the task, this is not a problem in their research, since their main focus is that the team is unavailable for other tasks during traveling. However, at the security this is a problem since this would mean that we do assign the team to a lane when the team is still traveling. Consequently, in practice the lane is not open in time. Mutingi and Mbohwa (2014) define time intervals in which the tasks must be performed. In our approach, the tasks must be done at exact times which is an important aspect of the problem.
multi-criteria with exact times. In their approach there are three type of workers which have to do two types of jobs. The first two types of workers can do only one of the job types, whereas the last type of workers can do both. This is not an aspect which is important in our research, but we can see all workers as one of the type which can perform all tasks. However, staff is unable to switch between tasks, which has to be possible in our problem. The same problem arises in Campbell and Diaby (2002). They present an assignment heuristic for allocating cross-trained workers to multiple departments at the beginning of a shift. The assignment procedure at the beginning of the shifts is similar as in the lane planning. However, both Cai and Li (2000) and Campbell and Diaby (2002) assign workers to the same task for the whole shift, whereas it is critical for the TAP-IBS that teams switch tasks.
3
Problem description
The TAP-IBS problem considers n teams which work a shift from time period psk until ptk for all k ∈ {1, . . . , n} = K. The teams have to be assigned to lanes, in total there are m lanes. We define variable xpki= 1 when team k is assigned to lane i at time period p and 0 otherwise. Of course, the assignment can only be made when team k is available, that is xpki≤ ykp, in which ypkis a parameter with ykp= 1 if team k is at work at time period p and 0 otherwise. We define the last time period by t giving P = {1, . . . , t} as the set of all time periods.
The main goal is to minimize the number of instant switches, that is switching from one lane to another in consecutive periods. An instant switch for team k takes place when xpki= xp+1kj = 1 with i 6= j and i, j ∈ ¯D. In which ¯D is the set of all lanes.
A secondary goal is optimizing the positioning of idle time. The shifts are planned such that at each moment in time there are enough teams available to cover the demand. That is P
k∈Ky p k ≥ Dp
for all p ∈ P , in which Dp defines the number of open lanes at time period p. It does not occur
that there are multiple teams required at the same lane at the same time. There are time periods in which there are more teams available than required, that isP
k∈Ky p k > Dp. At these periods P k∈Ky p
k− Dpteams are idle.
When there is idle time, we define an artificial lane Q to which one or several teams can be assigned. A team can be assigned to multiple periods of idle time. Furthermore, multiple teams can be idle at the same time. For a team, a long consecutive period of idle time is favoured over several short periods such that a team can do something else during the ‘idle’ time. Next to that, idle time is favoured to be connected to the start or end of a shift such that the shift might be shortened when possible.
There are three factors in the optimization, minimizing the number of instant switches, maximizing the periods of idle time of one hour and maximizing the number of idle time periods connected to the start and end of a shift. By assigning costs to these factors, one can favour one over the other. In this thesis the costs structure is such that minimizing the number of instant switches outweighs all other costs.
Within the demand there is a sequencing pattern. We sequence all lanes according to this sequencing pattern. That is when the demand at time period p is Dp, then lanes 1 to Dp are open. Due to
only lane b is open. We define the demand for each individual lane i at time p by dpi which is 1 if i ≤ Dp and 0 otherwise.
The set of all lanes including the idle lane is defined by D = ¯D ∪ {Q}. Every team has to be assigned to a lane or idle time at each period that he is at work, that is P
i∈Dx p ki = y
p k for all
k ∈ K and p ∈ P . Furthermore, the demand for each lane in each time period has to be covered givingP
k∈Kx p ki= d
p
i for all i ∈ ¯D and p ∈ P .
In practice, most shifts contain a break. During this break a team is unavailable to work at a lane. When team k has a break in period ¯p, then we have ypk¯= 0.
4
Base model formulations
4.1
Fixed break model
The problem described in Section 3 can be seen as an extension of the basic model of Bard and Wan (2006). They present a task assignment problem with unrestricted movement between workstation groups as a variant of the minimum cost, multi commodity network flow problem of which the main difference is that there are additional demand constraints. We present a base model which includes idle time directly into the formulation. In this base model only the main objective of the TAP-IBS, minimizing the number of instant switches, is taken into consideration. The incorporation of the secondary goal, optimizing the positioning of idle time, is explained in Section 6.
For each team k ∈ K a directed subgraph Gk = {Nk, Ak} is defined with node set Nk and arc
set Ak. The nodes v ∈ Nk represent a lane-time period pair for that team, that is the triplet
v = (i, p, k) represents the node for team k at lane i at time period p. Besides the m lanes for which there is demand sometimes a team has to be idle, an additional artificial lane Q is introduced for that. For completeness, a source node (sk) and sink node (tk) for each team are included giving
|Nk| = 2 + t(m + 1). Each arc (v, w) ∈ Ak represents a permissible transition from one lane to
another which has costs cvw. These transition costs from one node to another are the only costs
in the model. The transition costs can be different for each team, but in this thesis it is chosen to have identical costs for each team. Other costs can be incorporated as well, but in this research it is chosen to neglect these. Consequently, the sum of all transition costs gives the objective function. The graph for a single team is given for a day with a 5-minute interval in Figure 6. The number in square brackets next to each node is the demand dpi for lane i at time period p. No demand is prescribed for the idle nodes. For simplicity, in this model we initially include all time periods for all teams, with ypk set to appropriate values to account for actual working times. In Section 5.1 we explain how the graph can be reduced by tuning the graph to the actual start and end times of a shift.
If we see the network representation of Figure 6 as a two-dimensional grid (since all nodes in this grid are for the same team k), then the third dimension represent copies of this grid with a single copy for each team. This defines the full graph G = (N, A) = S
k∈K
Gk= ({N1∪· · ·∪Nn}, {A1∪· · ·∪An}).
If the supply at sk is 1 and the demand tk is 1 for each k ∈ K, then the problem is a minimum cost
i ∈ ¯D and time p ∈ P . From the first phase of the problem when defining the start and end times of shifts and the positions of the breaks, we know that there are always enough teams to cover the demand. Hence, it is guaranteed that there exists a feasible solution.
We assume that each day can be solved as a separate problem. This can be done when there are no operations during the night changeover or the demand and available teams is exactly the same at the night changeover. In case of the latter the assignment during the night can be fixed. By fixing this, the continuation across days is not an issue.
Given the network formulation above, a complete mathematical representation of the problem can be specified.
Indices
• v, w = indices for nodes • i = index for lanes
• p = index for time periods
Sets
• N = set of all nodes • S = set of all source nodes • T = set of all sink nodes
• ¯D = set of all lanes excluding the artificial idle lane • P = set of all time periods
• Nip = set of nodes corresponding to lane i at time period p
• F (v) = set of nodes that are immediate successors of node v • P (v) = set of nodes that are immediate predecessors of node v
Parameters
• cvw = cost of a transition from node v to node w
• dpi = demand for lane i at time period p Decision variables
Model 1 Minimize z =X v∈N X w∈F (v) cvwxvw (1a) subject to X w∈F (v) xvw− X w∈P (w) xwv = 0, for all v ∈ N \ {S ∪ T } (1b) X w∈F (v) xvw= 1, for all v ∈ S (1c) X v∈P (w) xvw= 1, for all w ∈ T (1d) X v∈Nip X w∈F (v)
xvw= dpi, for all i ∈ ¯D, p ∈ P (1e)
xvw∈ {0, 1}, for all v ∈ N, w ∈ F (v) (1f)
The goal is to get the best schedule as possible. In the objective function (1a) we minimize all transition costs from one node to another. As an example, by this we can apply high costs for switching from one lane to another in successive time periods and low costs when staying at the same lane. Furthermore, different costs can be applied when moving to/from an idle node. Considering the formulation of the problem in a network flow setting, constraint (1b) takes care of the fact that no shift will stop or end at an intermediate time period. For every team, the whole day is specified since each team starts, constraint (1c), and stops, constraint (1d) at the source and sink nodes. Of course, the most important constraint is to satisfy all demand which is preserved by constraint (1e). Finally, the binary constraint (1f) must be included, a team ‘crosses’ an arc or not, we cannot put two half teams at one lane.
Most shifts in practice incorporate at least one break in which the team cannot work at a lane. When a team takes a break, it cannot be assigned to work at a lane. This is incorporated in the arc set Ak. When team k takes a break from time period p0 until p00 then all arcs for these periods
are removed. Consequently, the nodes in period p0− 1 are connected to those in period p00+ 1.
4.2
Variable break model
In the formulation in Section 4.1 we assume that the breaks are fixed in advance. In practice, this is most often not the case. Seldom it is impossible to take a break five minutes earlier or later. Incorporating this variability in the breaks can reduce the number of instant switches. However, incorporating this into the model results into more variables and more constraints.
node v0∈ N0\ {S ∪ T } is a quadruplet with an additional index for each layer of the specific team
giving v0= (i, p, k, l) with l ∈ Lk which is the set of all different layers for team k.
Each of the layers defines a shift with a different break position. Of these multiple layers l ∈ Lk
for any team k ∈ K, only a single one can be chosen. We can link the source node for team k, sk,
to all nodes in N0 defined for team k in the first time period. Since there are no arcs across layers and only one arc connected to sk can be crossed, we are ensured that arcs are only crossed in one
of the layers for each team.
The mathematical formulation for the problem in which the breaks are variable is similar to the formulation of the fixed break model. We defined N0 as the set of all nodes for the variable break model. Compared to the fixed break formulation, the immediate successors and predecessors of a node v ∈ N0 are defined by F0(v) and P0(v), respectively. The set of nodes corresponding to lane i ∈ ¯D in time period p ∈ P is defined by Nip0 . The variable break model formulation is defined in Model 2. Model 2 Minimize z =X v∈N0 X w∈F0(v) cvwxvw (2a) subject to X w∈F0(v) xvw− X w∈P0(w) xwv = 0, for all v ∈ N0\ {S ∪ T } (2b) X w∈F0(v) xvw= 1, for all v ∈ S (2c) X v∈P0(w) xvw= 1, for all w ∈ T (2d) X v∈N0 ip X w∈F0(v)
xvw= dpi, for all i ∈ ¯D, p ∈ P (2e)
xvw∈ {0, 1}, for all v ∈ N0, w ∈ F0(v) (2f)
The main difference in the formulation can be seen in constraints (2c) and (2d). Due to these constraints, only one of the layers in Lk for each team k ∈ K can be used. If breaks are only
allowed to move one time period later or earlier and a team has a single break, the set Lk contains
three layers. One needs to mention that when there are more breaks within a shift and breaks are allowed to start even earlier/later, the size of this set grows exponentially.
The transition costs cvw can depend on the team k and/or the layer l. In this thesis, these costs
are identical for each team and each layer and solely depend on the type of transition for the team.
4.2.1 Variable break model with single layer is identical to fixed break model
difference is that all nodes v ∈ N0\ {S ∪ T } have an additional layer index compared to the nodes defined in v ∈ N \ {S ∪ T }. However, since each team k ∈ K has only a single layer index in Lk,
there are no differences due to the additional index. Hence, Model 1 can be seen as a special variant of Model 2.
4.3
Constraints matrix is not totally unimodular
Solving a MIP can take a lot of computation time due to integrality constraints. If the variables are continuous, the computation times can be drastically reduced since the problem can be solved as an LP by for example the simplex method. Many network design problems have a special structure which allow that the integrality constraints can be ignored. If the constraints matrix of a problem is (1) totally unimodular (TUM), i.e. every square sub matrix of it has determinant +1, -1, or 0, and (2) the right-hand sides of the constraints are integers, then the optimal solution to the LP relaxation will be an integer solution when solved by a simplex approach (Schrijver, 1998). Clearly, the right-hand sides of all constraints in Model 1 are integer. Hence, if we can show that the constraints matrix of Model 1 is TUM the problem can be solved with a linear relaxation. However, despite the network structure, the constraints matrix of Model 1 is not TUM. Ignoring constraint (1e) gives a constraints matrix of which it is known that it is TUM. However, constraint (1e) must be included to satisfy all demand.
Though, we have solved 10, 000 random instances (with between 10 and 25 teams) by using the simplex method and a linear relaxation to check whether the optimal solutions are integer. For most of these (9, 704) this is the case. However, for 296 we obtained a non-integer solution. These 296 instances have also been solved with integrality constraints. 215 of these instances have the same objective value as in the linear relaxation and 81 have a higher objective value. Given that the linear relaxation has a better objective value than the integer problem for some instances, it is clear that the constraints matrix is not TUM.
In Section 4.2.1 we showed that the fixed break model is a special case of the variable break formulation. That is, the constraints matrix of Model 1 is a subset of the constraints matrix of Model 2. Since the constraints matrix of the fixed break model formulation is not TUM, the constraints matrix of the variable break model cannot be either.
Over 97% of the random instances have an optimal solution which is integer when the simplex method is applied on a linear relaxation. This is due to the TUM structure which is part of the constraint matrix. Due to this, the computation time advantages of applying a linear relaxation can still be used. Trying the simplex method first and applying a branch-and-bound method only on those instances which have no integer optimal solution can reduce the computation times. This is due to the fact that the linear relaxation is faster than solving the model with taking the integer constraints directly into account.
been to include additional nodes for the break options within the layer for each team which requires fewer additional arcs and nodes. However, this does require additional constraints to ensure that breaks are positioned correctly and have the correct duration. These additional constraints will most likely change the constraint matrix such that most of the TUM structure disappears, losing the computation time advantages of that.
5
Graph reductions
In Section 4 we present two base model formulations. One when the breaks of the predefined shifts are fixed in advance and a second when breaks are variable. The number of arcs and nodes in these formulations can be reduced. We present two examples why several nodes, and connecting arcs, can be omitted in Sections 5.1 and 5.2. Furthermore, the number of arcs can be reduced by disallowing certain changeovers which is explained in Section 5.3. In this section we focus on the fixed break model, all results can be applied to the variable break model as well.
5.1
Incorporating predefined shifts into the model
The fixed break formulation in Model 1 has many arcs and nodes. In Section 4.1 we assumed that each team has a shift from time period 1 to t. Of course, in practice this is impossible. We can delete several nodes and corresponding arcs from the network given the predefined shifts of the teams.
We define the node set Nipk⊂ N as the set of all nodes corresponding to lane i ∈ D at time period
p ∈ P for team k ∈ K. For the fixed break model this is a single node, for the variable break model there can be multiple nodes in the set.
Suppose team k starts their shift at time period ps
kand finishes the shift after time period p t k. Then
we define the set Pk0 as the set with all time periods ˆp with ˆp < ps
k or ˆp > p t
k. At these periods, the
team cannot be assigned to any lane. Therefore, we can delete all nodes v ∈ S
p∈P0 k
S
i∈D
Nipk from the
node set N . All arcs connected to these nodes can be ignored as well. Of course, these nodes can be deleted for all teams k ∈ K. In order that a path will exist from the source node sk to the sink
node tk for team k ∈ K, arcs from sk to all nodes v ∈ S i∈D
Ni,ps
k,k will be included such that a team
can ‘start’ a shift. Next to that, arcs from all nodes v ∈ S
i∈D
Ni,pt
k,k to the sink node tk have to be
included such that the shift can be ‘ended’. These additional arcs are incorporated for all teams k ∈ K.
5.2
Omitting nodes when there is no demand
Suppose at time period p there is a demand for Dp lanes. Due to the sequencing pattern, it is
known which Dp lanes are required to be open. In this thesis these are lanes 1 to Dp. We define
the set of these lanes 1 to Dpby D0p.
Clearly, a team can never be assigned to a lane i ∈ ¯D \ D0p in time period p. This means that all nodes v ∈ S k∈K S p∈P S i∈ ¯D\D0 p
Ni,p,kcan be deleted from N . Ignoring all these nodes and corresponding
arcs reduces the size of the network significantly since it rarely occurs that all available lanes are open.
5.3
Instant switches only from closing lanes
The prescribed model consists of many arcs with instant switches in which a team can switch from one lane to another in consecutive time periods. These are the arcs from node (i, p, k) to (j, p + 1, k) with i 6= j, 1 ≤ i ≤ Dp, 1 ≤ j ≤ Dp+1 and k ∈ K. Proposition 1 shows that the number of instant
switches does not increase when instant switches are prohibited at moments when a lane does not close.
Proposition 1. Any feasible solution to the TAP-IBS can be rewritten into a solution in which in-stant switches occur only when demand decreases, without increasing the number of inin-stant switches. Proof. Suppose we have a feasible solution for the TAP-IBS in which a team does have an instant switch when the demand does not decrease. In this solution team A switches from lane a in time period p − 1 to lane b in time period p with Dp−1≤ Dp. Since the demand does not decrease, this
means that another team B works from period p at lane a.
Let the first time period after p in which one of the lanes (a or b) is closed be ˆp. Suppose we switch all assignments from time period p until ˆp of lane a to lane b and vice versa. This means that the instant switch for team A is prevented at time period p. In the time periods between p and ˆp the number of instant switches remains the same. We need to find out what happens between time period ˆp and ˆp + 1 due to the permutation.
There are three different situations. Lane a closes earlier than lane b, both lanes close at the same time, or lane b closes earlier than lane a. These are given below.
1. Lane a and b close at the same time ˆp.
If both lanes end at the same time, then no instant switch can occur from time period ˆp to ˆ
p+1 from these lanes due to the permutation. This means that we have rewritten the solution with fewer instant switches.
2. Lane b closes at time ˆp, lane a stays open.
time period ˆp + 1. However, the number of instant switches is equal and the instant switch is now at the time when demand decreases.
3. Lane a closes at time ˆp, lane b stays open.
Suppose at time ˆp team C works at lane b. When team C does not work at lane b in the following time period ˆp + 1, we have found a solution with fewer instant switches. When team C does work at lane b in time period ˆp + 1, this means that after the permutation there will be an instant switch for team C since it will then go from lane b in time period ˆp to lane a in time period ˆp + 1. However, the number of instant switches is the same and the instant switch is now at the time when demand decreases.
This permutation can be done for all instant switches which occur at time periods in which the demand does not decrease. Iteratively, each of these instant switches can then be prevented or rewritten into an instant switch at the moment the demand decreases. This proves the proposition.
Given Proposition 1, it seems intuitive to omit all arcs with an instant switch when the lane stays open in these time periods. This decreases the size of Model 1 significantly.
Though, the number of instant switches is not the only factor in the objective. We also apply costs for the positioning of idle time as explained in the problem extensions in Section 6. Theorem 1 shows that if the costs only depend on the number of instant switches and the positioning of idle time, then all arcs for switching to another lane when the lane remains open can be ignored without having an impact on the objective value of an optimal solution.
Theorem 1. If the costs of the TAP-IBS only depend on the number of instant switches and the positioning of idle time, in which a solution with i instant switches is better than any solution with j > i instant switches. Then an optimal solution for the TAP-IBS with instant switches only allowed when demand decreases is also an optimal solution when all instant switches are allowed. Proof. Suppose the costs of a solution to the TAP-IBS only depend on the number of instant switches and the positioning of idle time. For this proof, the model without instant switches at periods when demand does not decrease is called TAP-Without and the model with these instant switches is called TAP-With.
Let Xw and Xwo be the optimal solutions to the TAP-With and TAP-Without model with
cor-responding costs c(Xw) and c(Xwo). The solution space for the TAP-Without is a subset of the
solution space for the TAP-With. Hence, Xwo is a solution for the TAP-With. Since Xw is an
optimal solution to that problem, we must have c(Xw) ≤ c(Xwo).
Then it remains to show that c(Xw) ≮ c(Xwo). Due to Proposition 1 and the fact that a solution
is always worse if it has more instant switches than another, the number of instant switches in Xw
and Xwo is equal. Hence, we only have to compare the costs for idle time between these solutions.
Suppose c(Xw) < c(Xwo). Then Xwmust have an instant switch from a lane a to b when the demand
does not decrease. As described in the proof of Proposition 1, this solution can be rewritten into a solution Xw0 with the same number of instant switches which is also a solution to the TAP-Without (Xw0 = Xwo0 ). As can be seen in the proof of Proposition 1, in rewriting the solution Xwinto Xw0, the
is also a solution to the TAP-Without this means c(Xw) = c(Xw0) = c(Xwo0 ) ≥ c(Xwo). This
contradicts c(Xw) < c(Xwo). Hence, we must have c(Xw) = c(Xwo) which proves the theorem.
6
Problem extensions
Both model formulations from Section 4 focus on the main objective, which is minimizing the number of instant switches. As explained in Section 3, there is a secondary objective in optimizing the positioning of idle time. In this section we present two extensions which can be used to improve this positioning of idle time. We explain how this can be applied to the fixed break model. The same analysis can be applied to the variable break model.
6.1
Favouring consecutive periods of idle time
As the term suggest, idle time is undesirable for a company. However, idle time in the planning might be used effectively in practice. When a team is idle for a longer time, they can do other tasks which are not defined to be done at a certain time point. For example, one can think of doing an online training course.
For this thesis, we favour a complete hour of idle time over shorter blocks of idle time. In this research we use a five-minute time interval, meaning we require 12 consecutive time periods of idle time for a team.
For a team k, we can include an additional arc in the arc set Ak from the idle node v = (Q, p, k) to
the idle node w = (Q, p + 11, k). Defining the costs for this arc lower than the costs of the arcs of individual time periods, this option will be chosen when possible.
This arc only has to be included if the team is available for work in all these time periods, P11
i=0y p+i
k = 12. Furthermore, at all the intermediate time periods there must be more teams
available than there is demand, giving P
k∈Ky p+i
k > Dp+i for all i ∈ {0, . . . , 11}. If there is an
intermediate time period in which the number of available teams equals the demand, the arc does not have to be included since it is not possible to schedule a consecutive hour of idle time at that moment in any feasible solution.
6.2
Favouring idle time at beginning or end of shifts
In the positioning of idle time, another consideration can be made. The idle time can be scheduled between work at lanes, connected to a break or at the beginning or end of a shift. For all these possibilities of connecting arcs different costs can be incorporated to favour one over the other. For this research, we prefer having idle time at the start or end of a shift. By this the planning department can still decide to, if possible, shorten shifts. This does lead to more arcs as in Section 6.1.
For the fixed break model, team k starts its shift at time period ps
k. Then in the network formulation
include an additional arc in the arc set Ak from the source node sk to the idle node (Q, psk+ 1), we
know that if this arc is included in the final solution there are two periods of idle time at the start of a shift. When the costs for this arc are c1 with c1< c0, we favour starting with two periods of
idle time over a single period.
Of course, this can be extended by including arcs from sk to the idle nodes (Q, psk+ i) with costs
ci < ci+1 for i = 1, . . . , x in which time period ˆp = psk+ x + 1 is the first time period after p s k
that the number of available teams equals demandP
k∈Ky ˆ p
k= Dpˆor the team is unavailable to be
assigned to a lane ykpˆ= 0. By applying this into the cost structure, we favour having longer periods of idle time at the beginning or end of the shift. This can be applied to all teams k ∈ K.
Similar reasoning applies to connecting idle nodes at the end of the shift to the sink node. Further-more, this result can be applied to every layer in the model with variable breaks.
7
Computational results
We have presented different formulations for the problem. In the following sections we discuss the results of these formulations. In Section 7.1 we present instances which will be solved in the remainder of this section. In Section 7.2 we present the results to these instances for the fixed break formulation including the graph reductions and problem extensions. Section 7.3 illustrates the effects of making the breaks variable. Finally, we show the effect of Theorem 1 on computation times in Section 7.4.
The Gurobi solver has been used to solve the instances in these sections. All instances have been solved with the same time limit of ten minutes. The specifications of the soft- and hardware which is used can be found in Appendix B.
7.1
Instances
7.1.1 CharacteristicsWe have created twenty instances to test the model formulations. In these instances we create n shifts of 8 hours including a single break of thirty minutes. The start times of the shifts and breaks are randomly assigned following a uniform distribution. We do not allow the shifts to cover two days. This leads to the fact that the number of available teams is low at the start of the day, increases until noon and decreases at the end of the day. The shifts and breaks can start at any five minute interval giving t = 288.
available teams and the corresponding demand of the fifth instance is given. Clearly, the demand is constant for every hour and there are always more teams available than there is demand.
0 24 48 72 96 120 144 168 192 216 240 264 288 0 4 8 12 16
Figure 3: Demand and number of available teams with n = 20 and demand constant per hour.
We first generate the shifts and afterwards calculate the demand. Due to this, the longer the constant demand interval, the higher the percentage of idle time. In reality, the opposite would be more realistic. However, it does not have an effect for testing on these instances. A case study with practical data is given in Section 8.
7.1.2 Costs structure
In this research the only costs are the transition costs for crossing the arcs. A team can have multiple transitions. There are three options, from the source node to an intermediate node, from one intermediate node to another and from an intermediate node to the sink node. In this thesis, each team k has the same costs structure. Suppose team k goes from intermediate node v = (i, p, k) to w = (j, p0, k). Then the following costs structure applies.
• p0= p + 1 – i = j: cvw= 0 – i 6= j and i 6= Q 6= j: cvw= 100 – i 6= j and i = Q or j = Q: cvw= 1 • p0> p + 1 – p0 = p + 11 and i = j = Q : cvw = −1 – p0 6= p + 11 and i 6= Q 6= j: cvw = 1 – p0 6= p + 11 and i = Q or j = Q: cvw= 0.5
lane to another, the so-called instant switch, the costs are 100. When the team switches from a lane to idle time or vice versa, the costs are 1.
When p0 > p + 1, this means that either the assigned lane is the ’artificial’ idle lane i = j = Q, and we have 12 consecutive periods of idle time (see Section 6.1) or that the team takes a break. We favour having some idle time around the break over going from one lane to the break room and directly returning to another lane.
Arcs from the source node v = sk to a node w = (i, p, k) have the following costs structure.
• i 6= Q: cvw= 1
• i = Q: cvw= 0.5 − 0.05 ∗ (p − psk)
Next to that, an arc from a node v = (i, p, k) to the sink node w = tk has a similar costs structure.
• i 6= Q: cvw= 1
• i = Q: cvw= 0.5 − 0.05 ∗ (ptk− p)
Starting and ending a shift from a lane has a costs of 1. We favour starting or ending a shift with idle time of which the costs are at most 0.5. As explained in Section 6.2, we favour positioning longer periods of idle time at the beginning or end of shifts. The costs decrease when more consecutive periods of idle time are incorporated.
The costs of 100 for an instant switch are high enough such that we will always minimize the number of instant switches before optimizing the positioning of idle time.
7.2
Results with fixed breaks
First we present the results of solving the model with fixed breaks. As shown in Section 4.3, it can be interesting to solve the instances by using a linear relaxation to reduce computation times. All instances up to fifty teams can be solved optimally within ten minutes. The instances with 75 teams cannot be solved within ten minutes except for instance 20. However, the latter can only be solved when applying a linear relaxation. In Table 1 the results are shown. All these optimal solutions of the linear relaxation were integer, and thus have the same objective value as the IP formulation.
Table 1: Result of the fixed break model.
Ins. n Min. Idle % Obj. Switches Time LR Time IP
1 10 60 21.0 46.9 0 0 0 2 10 30 11.1 57.45 0 0 0 3 10 15 3.1 165.25 1 0 0 4 10 10 2.0 263.8 2 0 0 5 20 60 17.2 96.9 0 3 4 6 20 30 9.9 115.25 0 1 2 7 20 15 4.0 132.45 0 1 1 8 20 10 1.9 423.75 3 1 1 9 30 60 15.4 143.1 0 32 34 10 30 30 7.7 278.2 1 12 14 11 30 15 3.7 281.05 1 10 11 12 30 10 2.0 592.9 4 9 11 13 50 60 13.0 233.75 0 399 + 14 50 30 6.7 263.65 0 120 222 15 50 15 2.7 402.35 1 67 118 16 50 10 1.6 782.3 5 55 72 17 75 60 10.3 - - + + 18 75 30 6.7 - - + + 19 75 15 2.6 - - + + 20 75 10 1.6 723.15 3 591 +
- : No solution was found.
+ : Time limit of ten minutes reached without a feasible solution.
7.3
Effect of variable breaks
Table 2: Results when breaks can move by at most 5 minutes. Fixed breaks Variable breaks Warm start
Ins. n Min. Obj. Time Obj. Time Obj. Time
1 10 60 46.9 0 45.9 6 45.9 3 2 10 30 57.45 0 56.65 8 56.65 3 3 10 15 165.25 0 161.3 26 161.3 5 4 10 10 263.8 0 262.85 81 262.85 8 5 20 60 96.9 3 94.85 166 94.85 18 6 20 30 115.25 1 112.65 107 112.65 16 7 20 15 132.45 1 130.4 231 130.4 39 8 20 10 423.75 1 421.3 410 421.3 18 9 30 60 143.1 32 - + 136.6 124 10 30 30 278.2 12 - + 172.55 88 11 30 15 281.05 10 - + 276.2 + 12 30 10 592.9 9 - + 592.9 + 13 50 60 233.75 399 - + 233.75 + 14 50 30 263.65 120 - + 263.65 + 15 50 15 402.35 67 - + 402.35 + 16 50 10 782.3 55 - + 782.3 +
- : No solution was found.
+ : Time limit of ten minutes reached.
If the breaks are fixed, a solution can be found relatively fast. Therefore, we introduce a two-phase approach. Initially we solve the model with fixed breaks. The solution from this model can be used as an initial solution to start solving the model with variable breaks. The results of using the initial solution, a warm start, in the variable break model are in the column ‘Warm start’ of Table 21. Due to this warm start, for all instances in which the fixed break model has a solution, we are guaranteed to have a feasible solution. Clearly, the computation times are smaller than without applying the warm start. Therefore, there is no reason not to do the two-phase approach. Though, as can be seen in Table 2, not all instances can be solved optimally within the time limit. Some instances have improved from the result of the fixed break model. However, after applying the warm start, not all instances improve within the time limit.
Comparing the variable break models with and without a warm start, applying the warm start does lead to two more instances (9 and 10) being solved optimally and one more instance (11) which has a better solution. For instance 10 we even see that the instant switch is removed due to the variable breaks. Given the two-phase approach, it can be chosen to only do the second phase when there are instant switches in the result of the first phase to try to decrease the number of instant switches.
Table 3: Results with variable breaks when using a warm start.
5 min. 10 min. 15 min.
Ins. n Min. Obj. Time Obj. Time Obj. Time
1 10 60 45.9 3 45.7 6 43.9 75 2 10 30 56.65 3 56.4 7 55.25 71 3 10 15 161.3 5 159.2 11 159.2 21 4 10 10 262.85 8 261.35 11 260.4 21 5 20 60 94.85 18 89.35 31 89.55 + 6 20 30 112.65 16 107.3 60 106.95 + 7 20 15 130.4 39 127.15 37 126.55 163 8 20 10 421.3 18 420.8 34 420.8 54 9 30 60 136.6 124 143.1 + 141.5 + 10 30 30 172.55 88 278.2 + 278.2 + 11 30 15 276.2 + 281.05 + 281.05 + 12 30 10 592.9 + 592.9 + 592.9 + 13 50 60 233.75 + 233.75 + 233.75 + 14 50 30 263.65 + 263.65 + 263.65 + 15 50 15 402.35 + 402.35 + 402.35 + 16 50 10 782.3 + 782.3 + 782.3 +
- : No solution was found.
+ : Time limit of ten minutes reached.
In the results of Table 2 the breaks are only allowed to be repositioned by 5 minutes. Of course it is questionable what happens if the breaks are more variable. These results are given in Table 3. When optimally solved, the objective value decreases when breaks are more variable. However, the more variability, the more difficulties arise in finding better (or optimal) solutions.
Given the warm start, there is a solution for all these instances. However, several instances have not improved within the time limit. For example instances 9 and 10 do not benefit from having more freedom within the allowed computation time. Allowing the breaks to move only 5 minutes can be optimally solved, but allowing the breaks in these instances to change by 10 or 15 minutes gives worse solutions. This is due to the fact that the size of the network formulation increases exponentially and can therefore not be solved optimally within the time limit.
7.3.1 Linear relaxation for variable break model
Applying a linear relaxation to the fixed break model does lead to a decrease in computation times. We have also applied a linear relaxation to the variable break model when breaks are variable by 5 minutes. The results are given in Table 8.
times are larger, fewer instances can be solved within ten minutes and several instances have a non-integer solution.
7.4
Effect of arc reductions on computation time
In Section 5.3 we have shown that the number of instant switches does not increase when we allow the instant switches to occur only when a lane closes. Given the costs structure defined in Section 7.1.2, the solution will even have the same optimum as a result of Theorem 1. Hence, comparing the solutions with and without other instant switches makes no sense. However, ignoring the arcs does have a significant effect on the computation times. The computation times for solving without removing the unnecessary arcs for instant switches can be found in Table 4.
Table 4: Computation times in seconds with and without omitting arcs for the IP formulation and the linear relaxation.
Reduced model All switches
Inst. LR IP LR IP 1 0 0 0 0 2 0 0 0 0 3 0 0 1 1 4 0 0 1 1 5 3 4 7 11 6 1 2 4 6 7 1 1 5 6 8 1 1 5 7 9 32 34 32 48 10 12 14 23 41 11 10 11 26 62 12 9 11 38 73 13 399 + + + 14 120 222 288 + 15 67 118 358 + 16 55 72 491 + 17 + + + + 18 + + + + 19 + + + + 20 591 + + +
+ : Time limit of ten minutes reached.
(LR) and without (IP) a linear relaxation. All instances which have been solved within ten minutes with the linear relaxation have an integer optimal solution.
The computation times are significantly larger in the model without ignoring the unnecessary arcs. This is the case for both the integer programming formulation and the linear relaxation. Furthermore, more instances can be solved within the time limit due to the model reductions. Ignoring these arcs is definitely beneficial.
8
Case study: Planning of security screening officers at
air-ports
As mentioned in Section 1, the problem studied is relevant for the lane planning of the security screening at airports. In this section we test the solution method on practical instances from Schiphol airport, the largest airport in the Netherlands. For this, we require some background on the setup of the planning which is given in Section 8.1. We compare the new method with fixed breaks to the method currently used in practice in Section 8.2 and check the effects of making the breaks variable in Section 8.3. In the case study, we apply the same costs structure as applied to the instances in Section 7, which can be found in Section 7.1.2.
8.1
Case description
The planning of the security screening officers at airports is performed in a two-phase approach. In the first phase, the filter planning, we create shifts for teams. Within that shift the start time, break times, walking times and filter assignments are given. The goal of the filter planning is to find a feasible set of shifts which ensures enough teams at the filters at all times by using the least amount of working hours.
The airport decides the required number of open lanes, based on the expected departures and arrivals of the flights for departing and transferring passengers, respectively. The demand patterns for each of the filters is different. Furthermore, the patterns differ for the several days of the week and across weeks as well. There are no two days in a year in which the demand is exactly the same. Therefore, the planning has to be performed for every day.
The demand of the airport is given per half an hour. In general, the demand can be described by a steep increase in demand in the early hours of the day which starts to decrease at the end of the morning. During the afternoon the demand goes slightly up and down and spikes up around dinner time. After 20:00 the demand start to decrease until a single or no lanes are open, since flights during the night occur seldom.
Given the laws and regulations (see Appendix A) on the shifts and the walking times in Table 9, the filter planning is scheduled on a five-minute basis. There is no reason to change this when assigning the exact lanes to the teams.
having a long break of forty minutes. The continuation of the day is visible in Figure 4 and the team finishes at filter 1 at 16:30, completing their 8.5 hour workday.
7:00 8:00 9:00 10:00 11:00 12:00 13:00 14:00 15:00 16:00 17:00 Monday Filter 1 Short break Filter 0 Long break Filter 2 Short break Filter 1
Figure 4: Timetable for a team on a single day from the filter planner.
earlier. By comparing the results of the filter planner with the demand, we know that there is idle time in these shifts. The total idle hours have been listed in Table 11.
Figure 5: Demand pattern for a Saturday in different weeks.
8.2
Fixed break model compared to current practice
In practice, another approach is used to solve the lane planning. For each team multiple possible assignments are created for each working block, the uninterrupted time in which a team is assigned to a filter. Multiple kind of assignments exist. These are being assigned to a lane for the whole working block, being assigned to a lane and then move to idle time (and vice versa), being assigned to one lane and switch to another within the working block (instant switch) and being assigned to one lane, then move to idle time and afterwards move to another lane. From all of these options for each working block, the variables, exactly one has to be chosen in the final solution. Each filter is solved as a separate instance.
For each working block, the amount of options is large. This is due to the fact that one can switch to multiple different lanes within each block and that these switches can occur at any point in time within the block. This IP problem has many variables, and a limited amount of constraints. The constraints define that for each point in time the demand for each lane must be satisfied and for each working block of each shift only one of the created options is allowed.
The number of instant switches is shown for the instances in Table 5, when solved by the method from current practice. For two instances no solution was found within thirty minutes2. This method
solely focuses on minimizing instant switches. Comparing results on the positioning of idle time within this section is therefore irrelevant.
Clearly, for the busy weeks there are hardly any instance switches. However, in the non vacation week there are multiple. This can be explained by the fact that there are fewer shifts scheduled throughout the day, hence the possible options to prevent an instant switch are limited. In Table 5 the total computation time is shown as well. Especially for Filter 1 the computation times are large, for most instances more than fifteen minutes.
Table 5: Number of instant switches with computation time in seconds for the old method.
Small vacation Summer vacation Non vacation
Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2
Mon. 0 0 0 0 0 0 0 - 1 (35) (1111) (228) (32) (1094) (201) (6) (-) (97) Tue. 0 0 0 0 1 0 1 1 0 (35) (1368) (389) (31) (1057) (172) (5) (863) (53) Wed. 0 0 1 0 0 0 3 2 0 (36) (952) (1393) (41) (966) (189) (8) (1745) (50) Thu. 0 - 0 0 0 0 1 0 0 (35) (-) (212) (32) (1048) (329) (8) (314) (59) Fri. 0 0 0 0 0 0 0 0 1 (35) (1104) (234) (35) (1116) (190) (10) (345) (67) Sat. 0 0 0 0 0 0 0 0 0 (39) (975) (193) (26) (916) (185) (5) (161) (73) Sun. 0 0 0 0 0 0 0 0 0 (40) (1186) (534) (37) (999) (198) (7) (948) (87)
- : No solution was found within thirty minutes.
We apply the fixed break model including all model reductions and problem extensions to the same instances. We first apply the linear relaxation to these instances. When this does not result into an integer optimal solution, the problem is solved as a MIP. The results can be found in Table 6. All instances can be solved optimally within two minutes.
For all instances, besides those which were not solved by the old method, the number of instant switches is the same in both methods. However, the differences in computation time are large. There is not a single instance which was solved faster by the old method than the new method. We see that for 9 instances the linear relaxation has a non-integer solution. Hence, these have also been solved with integrality constraints.
Table 6: Number of instant switches with computation time in seconds for the fixed break model.
Small vacation Summer vacation Non vacation
Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2
Mon. 0 0 0 0 0 0 0 2 1* (1) (48) (3) (1) (51) (4) (0) (8) (3) Tue. 0 0 0 0 1 0 1 1 0 (1) (39) (8) (1) (26) (4) (0) (4) (1) Wed. 0 0 1 0 0 0 3 2 0 (1) (32) (3) (1) (30) (3) (0) (6) (0) Thu. 0 0* 0 0 0 0 1 0 0 (1) (60) (12) (0) (39) (6) (0) (15) (1) Fri. 0* 0 0 0* 0 0* 0 0* 1 (2) (64) (9) (1) (37) (3) (0) (12) (1) Sat. 0 0* 0* 0 0 0 0 0 0 (1) (74) (18) (0) (19) (3) (0) (2) (1) Sun. 0 0 0 0 0 0 0 0* 0 (1) (67) (9) (1) (24) (3) (0) (12) (2)
* : No integer solution with linear relaxation.
It is clear that computation times are not an issue for these instances. Not all instant switches can be prevented if the positions of the breaks defined in the filter planning are fixed. In total there are fourteen instant switches. Especially for the non vacation week with low demand there are several. Sometimes even multiple at the same filter at the same day.
8.3
Effect of variable breaks
There are 14 instant switches in the results when the instances are solved with the fixed break model. However, computation times are small so it is interesting to see if the number of instant switches can be decreased by allowing the breaks to be repositioned. From Section 7 we know that the best results are obtained when a warm start is applied, hence this is used in this case study. We present the results when the breaks are variable by five minutes in Section 8.3.1. In Section 8.3.2 the results are given when breaks are more variable.
8.3.1 Breaks variable by five minutes
Table 7: Number of instant switches with comp. time in seconds with breaks 5 minutes variable.
Small vacation Summer vacation Non vacation
Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2 Fil. 0 Fil. 1 Fil. 2
Mon. 0 0 0 0 0 0 0 1 0 (49) (+) (22) (5) (*) (31) (1) (37) (10) Tue. 0 0 0 0 1 0 1 1 0 (13) (+) (108) (5) (+) (61) (0) (145) (12) Wed. 0 0 1 0 0 0 3 0 0 (8) (+) (18) (4) (+) (35) (1) (55) (5) Thu. 0 0 0 0 0 0 1 0 0 (4) (+) (29) (10) (+) (27) (1) (48) (7) Fri. 0 0 0 0 0 0 0 0 0 (5) (+) (79) (4) (*) (22) (1) (433) (7) Sat. 0 0 0 0 0 0 0 0 0 (8) (+) (+) (4) (+) (15) (0) (15) (43) Sun. 0 0 0 0 0 0 0 0 0 (10) (*) (26) (23) (+) (19) (1) (*) (10)
+ : Time limit reached without finding a better solution than the warm start. * : Second phase not optimal within time limit.
Clearly, there are no problems in solving for the instances of filters 0 and 2. However, we see that applying the variable break model to most of the instances for filter 1 in the vacation weeks does not have an effect due to the size of the problem.
Of the fourteen instant switches which occurred in the solution of the fixed break approach, five are prevented when breaks are allowed to be variable by five minutes. Of course this gives rise to the question of what happens when the breaks are more variable.
8.3.2 Breaks variable by up to ten or fifteen minutes
The same analysis has been applied to these instances when breaks are allowed to be repositioned by up to 10 or 15 minutes. These results are given in Table 12 and Table 13, respectively. The more variable the breaks are, the more difficult it is to solve the model due to the increases in the number of variables and constraints. We see that more instances cannot be optimally solved within ten minutes, multiple instances have not even improved from the fixed break model which was used as a warm start.
to be ‘wasted’ on it. This saves computation time on models which have a solution without instant switches with fixed breaks, which can then be spent on the larger models with instant switches.
9
Conclusions
We have studied a task assignment problem with idle time, breaks and a sequencing pattern in the demand. In this problem, lanes (tasks) have to be assigned to teams (or employees) which have a predefined shift. Breaks can be positioned at fixed moments within these shifts. Next to that, an alternative formulation is given in which breaks are variable. The exact demand has to be fulfilled resulting in idle time for overcapacity. Due to the fact that demand has to be exactly fulfilled and one cannot deviate from the opening sequence, instant switches arise. This means that one team has to close one lane and start a different lane at the exact same time. Of course, in practice this gives complications. In this thesis, the goal is to develop a solution method which focuses on minimizing instant switches and optimizing the positioning of idle time. This solution method is found and works for small to medium size instances.
A network approach is given to solve the problem, which yields promising results. For each team a layer in the network is designed in which nodes represent a lane at a certain time. Within the network, a team starts at a source node and has to go to a sink node passing the lane-time period nodes. The predefined shifts limit the node and arc sets for each team, such that the team starts and ends at the correct time and takes a break when required. The arcs in the model represent changeovers, the goal is to travel the least amount of arcs in which one has an instant switch, i.e. going from one lane to another in successive time periods. To check whether the demand is satisfied for each lane at each time, the ‘flow’ passing through each lane-time period node is summed for all teams. By using idle nodes for each time period, it is possible to exactly satisfy the demand. Furthermore, one can assign certain costs to arcs going in and out of these idle nodes to optimally position the idle time in the shifts. Within this method it is possible to favour a long period of successive idle time for a team and to position the idle time at the beginning or end of a shift. Due to the network structure, solving with a linear relaxation gives an integer solution for most instances with a fixed break formulation. This reduces computation times significantly. A two-phase approach is presented in which the result of a model with fixed breaks is used as a start solution for the model with variable breaks. Unfortunately, the computation time benefits of applying a linear relaxation do not apply to the variable break formulation.
In this research we show that if costs only depend on the number of instant switches and the positioning of idle time, we only require instant switches when a lane closes. This result drastically reduces the number of arcs and thus the size of the problem. This is relevant since we found that the size of the problem is crucial for the solvability of the problem. Hence, this allows us to solve larger instances. This result is an extension to the existing literature on assignment problems. Testing on several instances shows that relatively large instances can be solved in a model with fixed breaks. Variablizing the breaks gives better results with fewer instant switches and a better positioning of idle time for relatively small instances, but makes large instances unsolvable within the time limits.
of the solutions is equally good or better than the method currently used in practice. In addi-tion, the computation times are significantly smaller. The results indicate that incorporating the method described in this thesis could improve the security operation by decreasing the amount of disturbances caused by instant switches combined with a more effective usage of idle time.
10
Further research
Most instances can be solved relatively fast with high quality solutions. However, as shown in Section 7, there are still instances which cannot be solved without instant switches from one lane to another. Ideally, these are all prevented. In this section we outline further research possibilities which could help to achieve this objective.
In our research the start and end times of the shifts have been fixed. By making these variable, one might reduce the number of instant switches. This can be done while fixing the lengths of the shifts, but these can also be variable depending on the situation.
One could also think of shortening/lengthening the shifts. In this research the idle time is favoured to be placed at the beginning or end of shifts such that shifts can possibly be shortened. When the lengths of the shifts are not restricted by regulations, this can be directly incorporated in the model. Then we can also lengthen shifts to reduce the number of instant switches when applicable. Of course, one would need to minimize these extended shifts.
Besides solution quality, computation speed can possibly be improved upon as well. For the fixed break model this is a small issue. However, when the breaks are variable the computation times increases drastically as shown in Section 7.3. In our proposed model formulation, as presented in Section 4.2, a new layer is defined in the network for each possible break position, increasing the size of the problem significantly, but keeping the TUM structure intact (see Section 4.3). The variable breaks can also be included within a single layer. In this approach additional constraints are required to ascertain that the breaks comply with the required characteristics. Modelling the variable breaks in this way could reduce the computation times due to fewer nodes and arcs. Next to that, in Section 4.3 we showed that the constraints matrix is not totally unimodular. However, most of the random instances can be solved when ignoring the integrality constraints. Applying a linear relaxation to the fixed break model does reduce computation times as shown in Section 7.2. This does not apply to the variable break model. It has to be investigated whether the proportion of integer solutions when solving with a linear relaxation can be increased by changing the parameters of the solver. One could think for example of defining the demand constraints as lazy constraints.
11
References
Bard, Jonathan F, Canan Binici, and Anura H deSilva (2003). Staff scheduling at the united states postal service. Computers & Operations Research 30 (5), 745–771.
Bard, Jonathan F and Lin Wan (2006). The task assignment problem for unrestricted movement between workstation groups. Journal of Scheduling 9 (4), 315–341.
Cai, X and KN Li (2000). A genetic algorithm for scheduling staff of mixed skills under multi-criteria. European Journal of Operational Research 125 (2), 359–369.
Campbell, Gerard M and Moustapha Diaby (2002). Development and evaluation of an assignment heuristic for allocating cross-trained workers. European Journal of Operational Research 138 (1), 9–20.
Ernst, Andreas T, Houyuan Jiang, Mohan Krishnamoorthy, and David Sier (2004). Staff scheduling and rostering: A review of applications, methods and models. European journal of operational research 153 (1), 3–27.
Kleinjan, Bastiaan and Thijs Baas (2018). Beveiligingsbedrijven worstelen met personeelstekort. BNR.
Mutingi, Michael and Charles Mbohwa (2014). A fuzzy-based particle swarm optimisation approach for task assignment in home healthcare. South African Journal of Industrial Engineering 25 (3), 84–95.
Schrijver, Alexander (1998). Theory of linear and integer programming. John Wiley & Sons. Sinha, Pritibhushan (2009). Assignment problems with changeover cost. Annals of Operations
Research 172 (1), 447.
Sondermeijer, Vincent (2019). Aantal reizigers schiphol vorig jaar gestegen naar 71 miljoen. NRC . Wan, L and Jonathan F Bard (2007). Weekly staff scheduling with workstation group restrictions.
12
Tables and Figures
Table 8: Comparison of computation times with and without a linear relaxation for the variable break model with and without a warm start.
Variable break Ins. IP-WS LR-WS LR 1 3 5 6 2 3 9 6 3 5 27 33 4 8 - -5 18 - -6 16 - 370 7 39 - -8 18 341 300 9 124 + + 10 88 + + 11 + + + 12 + + + 13 + + + 14 + + + 15 + + + 16 + + + 17 + + + 18 + + + 19 + + + 20 + + +
- : Optimal solution is not integer. + : Time limit of ten minutes reached.
Table 9: Walking times around breaks in the case study. Filter 0 Filter 1 Filter 2
Filter 0 5 5 10
Filter 1 5 5 5
