IEEE Robotics & Automation Magazine - June 2015 - 71
Definition 3.4
Assume that by starting from initial state S 0, after OD1,
OD2, ..., and OD i are executed, at time x, a system reaches
state S i that is feasible. Then, S i is unsafe if, with S i being
an initial sate, an infeasible state is finally reached no matter
what ODs are performed.
By Definition 3.4, the states of the system are divided into
two mutually exclusive categories: safe and unsafe states. Thus,
if the execution of any OD in a schedule transforms the process from one safe state to another, the schedule must be feasible. In addition, if the state of the process is safe at any time,
there must be a feasible short-term schedule. Thus, to schedule
the crude-oil operations, we need to identify the set of safe
states using the developed hybrid PN model.
Since the operation of any distiller cannot be terminated, a
feasible schedule SCHD 1 = {OD 1, OD 2, f, OD n} that is
found for a time duration 60, a@ does not mean that one can
find a feasible schedule SCHD 2 = {OD 1, OD 2, f, OD n,
OD n +1, f, OD n +k} with SCHD 1 1 SCHD 2 for time duration 60, b@ with b 2 a. For a process with initial state S 0, if a
feasible short-term schedule can be found for a time duration
C = [0, 3], then this process is schedulable.
Definition 3.5
If a process with S being the initial state is schedulable, then S is
safe for the process. By Definition 3.5, safeness is equivalent to
schedulability. Therefore, based on the hybrid PN model, one
can analyze its schedulability. Let DS i denote distiller i with an
oil feeding rate of fdsi, let CTK j denote charging tank j with capacity p j and initial volume g j of oil in it, let Fp (MAX) be the
maximal oil flow rate of the pipeline, and a i = X # fdsi, where
X and K are as given above. It should be pointed out that,
when a refinery is scheduled, the maximal productivity is ideal.
To maximize productivity, we should have enough oil in the
storage tanks or in the coming tankers. Thus, the key to finding
a feasible schedule is to properly decide the ODTs for oil transportation and ODFs for distiller feeding. Then, based on the hybrid PN model developed for the system, we derive the
following schedulability conditions [33].
Distiller
1
For a process of crude-oil operations composed of K distillers DS 1-K with fds 1 ! fds 2 ! f ! fdsK, the system is schedulable if the following conditions are met:
● There are 3K charging tanks CTK 1 -3 K with their capacity
b e i n g p 1 $ 2a 1, p 2 $ 2a 1, p 3 $ 2a 1, f, p 3 i +1 $ 2a i +1,
p 3 i +2 $ 2a i +1, p 3 ( i +1) $ a i +1, 1 # i # K - 1.
● Fp (MAX) = fds 1 + f + fdsK .
● At the initial state, the amount of oil type 1 in CTK1 and
CTK2 is g 1 = a 1 and g 2 = 2a 1, f, the amount of oil in
CTK 3 i +1 and CTK 3 i +2 is g 3 i +1 = a i and g 3 i +2 = 2a i,
1 < i # K - 1, g 3 = g 3 i +3 = 0, 1 < i # K - 1 the oil in
CTK 1, CTK 3 i +1, 1 < i # K - 1, can be discharged for feeding distillers, and CTK3 is ready for charging.
K -1
● fds 1 $ K ` /
f j ( 2 X - K) .
k = 1 dsk
These conditions require that fds 1 ! fds 2 ! f ! fdsK ,
but this does not pose a restriction on the system. In
fact, a process is much easier to schedule if we have
fds1 = fds 2 = f = fdsK . Condition 2 indicates that the
maximum production rate is Fp (MAX) . The initial condition g i = a i means that, when one charges a charging
tank volume, g i = a i should be charged into it, which is
the boundary for schedulabilty. In reality, a tank's capacity is much larger than g i = a i . Thus, to reduce the number of oil-type switches in performing the ODTs and the
number of charging tank switches in performing the
ODFs, when we charge a charging tank it should charge
as much oil as possible. Using this method, we can easily
satisfy constraints.
K -1
For the case of fds 1 $ K ` / k = 1 fdsk j (2X - K), assume
that X = 2K, and there are four distillers, i.e., K = 4. Then, it
requires fds 1 $ ( fds 2 + fds 3 + fds 4) 3. It always holds, as one
can take the largest one as fds1 . Furthermore, if it requires that
a charging tank that is assigned for feeding distiller DS1 is
charged to 5a 1, we have fds 1 $ ( fds 2 + fds 3 + fds 4) 9, which
must hold. In fact, a charging tank's capacity is more than that.
Using these schedulability conditions, we discuss how to find a
short-term schedule by presenting computationally efficient
techniques. For different situations, the schedulability conditions can be found in [29]-[32] and [35].
Oil Number 3 (27,000)
Oil Number 1 (63,000)
2
Oil Number 2 (55,200)
3
Oil Number 5 (55,000)
Oil Number 4 (27,000)
0
20
40
60
80
100
120
Time (h)
140
Oil Number 6 (38,000)
160
180
200
220
240
Figure 6. An example of a refining schedule.
JUNE 2015
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IEEE ROBOTICS & AUTOMATION MAGAZINE
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71
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