Computational Intelligence - February 2016 - 62
Without Adaptive Learning
V1
V2
V3
V4
V5
V6
V7
V8
V9
10th Generation
100th Generation
24 0 26 27 30 55 56 59 36 86 87 42 43
25 1 74 78 33 58 84 64 88 20 44
48 3 75 31 80 12 15 17 41 22 94
49 2 4 53 8 32 57 60 61 37 65 67 89 45 47
50 52 7 79 82 38 40 21 46 98
51 5 77 34 10 13 85 19 68 91 70 71 95 96 99
54 83 (s_delete)
72 28 76 11 14 62 63 39 66 90 69 23 97
73 29 6 9 81 35 16 18 92 93
24 0 26 28 27 55 56 59 36 86 87 42 43 97
25 2 1 3 4 5 53 8 57 60 61 63 40 89 45 94 98
48 52 7 79 34 35 33 84 38 39 88 20 96
49 75 80 12 16 66 65 41 22
50 29 77 82 85 19 69 91 70 71 95 47
51 54 78 58 15 17 21 44 99
72 76 10 11 13 62 64 67 90 68 23
73 6 31 30 9 32 81 14 37 18 92 93 46
74 83 (s_delete)
With Adaptive Learning
V1
V2
V3
V4
V5
V6
V7
V8
V9
10th Generation
100th Generation
24 0 26 27 30 55 56 59 36 86 87 42 43
25 1 74 78 33 58 84 64 88 20 44
48 3 75 31 80 12 15 17 41 22 94
49 2 4 53 8 32 57 60 61 37 65 67 89 45 47
50 52 7 79 82 38 40 21 46 98
51 5 77 34 10 13 85 19 68 91 70 71 95 96 99
54 83 (s_delete)
72 28 76 11 14 62 63 39 66 90 69 23 97
73 29 6 9 81 35 16 18 92 93
24 0 73 5 55 58 33 14 62 63 40 90 68 23
25 27 75 8 32 81 15 37 18 92 93 46 99
26 28 4 29 6 56 11 59 36 86 87 42 43 97
48 2 1 3 74 78 12 84 64 88 20 47
49 53 77 10 34 13 85 19 69 91 70 71 44
50 52 7 79 82 38 39 66 65 21 94 95
51 54 30 31 80 60 83 17 41 22 96 98
72 76 9 57 35 61 16 67 89 45 (s_delete)
Figure 5 Without adaptive learning FSEA got trapped in local minima with job 83 in s_delete. With adaptive learning (and remembering to
avoid job 83), FSEA can get out of such local minima and hence reduce the fleet size by one at the 100th generation.
vehicles, regardless of the produced
schedules. However, in the uncertain
case such an evaluation may not be
totally realistic. Different schedules for
the same number of vehicles may not
behave similarly under uncertainty.
Some schedules may show more robustness against changes than others.
To evaluate the robustness of a schedule of vehicles, we propose an MC
approach. The proposed MC evaluates
the schedules of vehicles to answer the
following questions: 1) How robust is the
schedule of vehicles under uncertainty in
travel time? 2) Is it necessary to add
more vehicles to the system to reduce
the possible waiting time of machines
caused by this type of uncertainty? If
those additional vehicles are needed,
how many more should be added?
Before we explain the proposed MC,
we will define some concepts: the frequency of vehicle disruptions and the
time taken to resolve the disruptions. Different types of vehicle disruption can happen in ESTTs, such as breakdowns, collisions and deadlocks. The disruption rate
62
^m (t ) h of a vehicle is defined as the frequency of disruptions until the time t.
Without any loss of generality, we assume
that the disruption rate is a constant value
in the period of evaluation, i.e. m (t ) = m 0.
It is commonly assumed that the disruption of vehicles follows the exponential
distribution with the parameter m 0. Once
disruptions happen to the vehicles, they
will not be available until they get
repaired. Mean time to repair (MTTR) is
a parameter that shows the average of
unavailable time. We use the MTTR in
the MC to estimate the duration for
which vehicles become unavailable.
Algorithm 5 FSEA().
1: initialise population Pt
2: evaluate population Pt
3: while stopping criteria not met
4:
Select elements from Pt to copy into
Pt + 1
5:
Mutate population Pt + 1 by Mutate()
(Alg. 3)
6:
recombine population Pt + 1 by
ReduceJobSequences() (Alg. 2)
7:
evaluate new population Pt + 1
8:
Pt := Pt + 1
9: return the best individual
IEEE ComputatIonal IntEllIgEnCE magazInE | FEbruary 2016
The proposed MC evaluates the
robustness of schedules as follows. First,
for each vehicle the MC estimates the
first moment and the duration of disruptions using a random exponential value
with the parameter m (the given disruption rate and a MTTR value, respectively). Let us assume that the first disruption is at time t 1. This means that the
vehicle can work from time t = 0 until
t = t 1. Then, the vehicle will not be
available for a period equal to MTTR
until the time t 2, t 2 = t 1 + MTTR. This
process is repeated until we reach the
makespan - the time by which the last
job has finished. Then we repeat this
process of simulating disruptions for all
the vehicles until they all reach the
makespan. In order to prevent any delay
in the system, the jobs that are uncovered
due to vehicles being unavailable must
be assigned to other available vehicles.
The MC searches through all the available vehicles in the solution to find suitable substitutions to carry out the
uncovered jobs. If such substitute vehicles are found then MC assigns the
Table of Contents for the Digital Edition of Computational Intelligence - February 2016
Computational Intelligence - February 2016 - Cover1
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Computational Intelligence - February 2016 - Cover3
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