Quay crane allocation as a moldable task scheduling problem

1

• Quay crane allocation as a moldable task
  scheduling problem
• J. Blazewicz, M. Machowiak
• Institute of Computing Science, Poznan University
  of Technology,
• C. Oguz,
• Department of Industrial Engineering, Koç
  University, Istanbul

2
Motivation - Crane scheduling problem

• Berth allocation problem is among the most
  important decision problems in a port container
  terminal since a good allocation of berths to the
  incoming ships will enhance ship owners'
  satisfaction and increase terminal productivity,
  leading to higher revenues.
• We model the berth allocation problem as a
  moldable task scheduling problem by considering
  its relation to the crane scheduling problem.
• quay cranes ?? processors
• ships ?? tasks
• turn-around time ?? schedule length

3

• We consider the berth allocation problem as a
  moldable task scheduling problem by incorporating
  the fact that the number of quay cranes allocated
  to a ship will affect its berthing time.
• This approach can simultaneously increase the
  utilization of quay cranes, shorten the
  turn-around time of ships, and decrease the
  waiting time of the containers.

4
Literature

• Despite the strong dependence of the berth
  allocation decisions on the number of quay cranes
  allocated to a ship, the majority of research in
  this area has not taken this relation into
  account when analyzing the berth allocation
  problem (see, for example, Daganzo (1989),
  Peterkofsky and Daganzo (1990), Imai et al.
  (2001, 2003), Li et al. (1998)).
• To the best of our knowledge, only Park and Kim
  (2003) incorporated the crane scheduling
  decisions into the berth allocation problem.

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Moldable Tasks Model

• We consider a set of m identical processors (quay
  cranes) using for executing the set of n
  independent, nonpreemptable moldable tasks
  (ships).
• Each task needs for its execution any number of
  processors (at least one but less or equal to m).
• The total number of processors executing the
  tasks should not exceed m at any time.
• An amount tj gt 0 of work is associated with each
  task Tj.
• fj(r) ? 0 is a non-decreasing processing speed
  function, fj(0) 0.
• fj(r) relates processing speed of task Tj to a
  number of processors allocated.
• The criterion assumed is schedule length.
• The problem is NP-hard (Du Leung, 1989), thus,
  suboptimal algorithms were looked for.

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The solution of continuous problem

• To explain main idea of finding a solution for
  the continuous problem, we introduce set
• of feasible resource allocations and set
• of feasible transformed resource allocations.
  Denote p (p1, , pn)
• Theorem (Weglarz 82)
• Let n ? m, convU be the convex hull of the set
  U, i.e. the set of all convex combinations of the
  elements of U, and u p/C be a straight line in
  the space of transformed resource allocations
  given by the parametric equations uj pj /C, j
  1, , n.
• Then the minimum makespan value for continuous
  problem can be found from

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The solution of continuous problem

• From Theorem it follows that the minimum makespan
  value Ccont for continuous problem is determined
  by the intersection point u0 of the straight line
  u p/C, C gt 0, and the boundary of the set convU
  in the n-dimensional space of transformed
  resource allocations.

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The solution of continuous problem

• In an optimal schedule for continuous problem all
  the tasks are processed in the interval 0,
  Ccont and task Tj uses rj processors, j
  1,...,n.
• The proposed algorithm starts from the continuous
  version of the problem and transforms the
  schedule obtained from the continuous version
  into a feasible schedule for the discrete MT
  model.
• We assume that with each task the concave
  processing speed function is associated.

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Concavity justification
departure time

• Turn around time on 1 processor (crane)
• 2 processors

unloading time
t(1)
berthing time
unloading time
t(2)
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Rounding scheme

• if rj ? 1 then rj 1
• When the number of processors assigned in the
  continuous solution is between 0 and 1 for each
  task, the speed function is linear, and then the
  surface (defined as a product of a task duration
  and a number of processors allocated to it) of a
  task does not change.
• else if rj gt 2 or rj lt 1.5 then rj ?
  rj ?
• In this case, the completion time of a task grows
  but no more than 50 to the continuous solution.
  Moreover, the sum of the processors allocated to
  these tasks is lower than m.
• else (1.5 ? rj ? 2 ) rj 2.
• Here a processing time of a task is smaller than
  in continuous solution but, using the concavity
  assumption, its surface increases. When we have
  many of such tasks, the sum of the discrete
  processors allocated to these tasks is greater
  than m and the Algorithm has to perform many
  steps of rounding off.

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Algorithm

• calculate the Ccont and the optimal continuous
  processor allocations rj.
• round the continuous allocation of processors to
  the integer values.
• calculate the new processing times of the tasks
  and sort them in non-increasing order of
  processing times.
• m ? rj (i1,...,n) (discrete number of
  processors)
• C max(Cj)
• if m m then GoToEnd
• if m lt m then allocate an excess of
  processors to the biggest task GoToEnd
• if m gt m then assign the remaining tasks
  after the tasks already scheduled till the
  max(C, ?tj(1) /m)
• if m m then GoToEnd
• else take the minimum value of the schedule
  length constructed by A or B
• A while m gt m then
• reduce a number of processors assigned to the
  task (group of tasks) with the biggest number of
  processors alloted and assign the tasks on the
  freed processor and on the other processors till
  the finishing time of the longest task (C).
• B the remaining tasks schedule after the tasks
  already scheduled treating them as a new
  instance.
• End Cmax C

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Example

• Let us consider the following data
• n6, m4, T T1, T2, , T6,
• t(1) 32, 34, 14, 92, 18, 55,
• f(r) ? r, the same for each task.
• Ccont 64.46,
• r 0.49 0.53 0.22 1.63 0.28 0.85.
• From this solution we proceed according to the
  rounding algorithm.

Ccont64,46
t3(1)14
t5(1)18
t1(1)32
4
t2(1)34
t4(2)54,7
processors
t6(1)55
time
13

Ccont64,46
t3(1)14
t5(1)18
t1(1)32
4
t2(1)34
t4(2)54,7
processors
t6(1)55
time
14
Ccont64,46
t3(1)14
t1(1)32
4
t2(1)34
t5(1)18
t4(2)54,7
processors
t6(1)55
time
15
Ccont64,46
t1(1)32
t3(1)14
4
t2(1)34
t5(1)18
t4(2)54,7
processors
t6(1)55
time
16
Step A
Ccont64,46
t1(1)32
t3(1)14
4
t2(1)34
t5(1)18
processors
t4(1)92
t6(1)55
time
17
Step A
Ccont64,46
4
t1(1)32
t3(1)14
t2(1)34
t5(1)18
processors
t4(1)92
t6(1)55
time
18
Step B
Ccont64,46
t1(1)32
t3(1)14
4
t2(1)34
t5(1)18
t4(2)54,7
processors
t6(1)55
time
19
Step B
Tasks of new instance
t1(1)32
Ccont64,46
t3(1)14
4
t2(1)34
t5(1)18
t4(2)54,7
processors
t6(1)55
time
20
Step B
Tasks of new instance
Ccont64,46
4
t2(1)34
t5(1)18
t1(3)14,04
t4(2)54,7
processors
t6(1)55
t3(1)14
time
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Computational experiments

• To evaluate the mean behavior of Algorithm we
  use the following measure
• Relative error Cmax / Ccont
• where
• Cmax schedule length obtained by Algorithm
• Ccont optimal schedule length of the
  continuous solution
• Task processing times ti(1) have been generated
  from a uniform distribution in interval 1, 100.
  
• Processing speed functions have been chosen as
• fi(r) ra, 0 lt a ? 1
• Each entry in the table is a mean value for 25
  instances randomly generated.

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Average behavior of the Algorithm for different
shapes of processing speed functions and varying
number of processors
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Conclusion

• The crane allocation as moldable tasks scheduling
  problem has been considered.
• Starting from the continuous version of the
  problem (i.e. where the tasks may require a
  fractional part of the resources),
• we proposed an approximation algorithm with a
  very good average behavior.
• Further investigations in this area will take
  into account a construction of the algorithm with
  a good average and worst case performance
  guarantee for arbitrary non-decreasing processing
  speed function.