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execute orders with delays, but every day of delay
costs pp = 0.6 SU. Some orders are shifted to the
right on the time axis because of this. The aim is to
be able to schedule trips, as orders come in (the
orders are not known in advance) and calculate
profit. Orders are marked with a number according
to the place in the sequence of entry into the system
and characterized by time of their entry (moment of
entry t), moments of start and finish of order
execution, duration (in days), point of loading and
point of unloading (Table 2).
Table2: Parameters of orders.
Characteristics
Order number
1 2 3 4 5
Time of entry
1 3 5 6 7
Start time of execution
3 4 7 8 9
Finish time of execution
5 5 9 9 10
Where from
4 3 1 4 3
Where to
1 1 4 3 1
Figure 1 shows orders as rectangles, with the
order number and the time of entry. Above each
rectangle ‘where from – where to’ locations are
described. The start and the finish of each rectangle
correspond to the start and the finish of the order
execution.
Figure 1: Diagram of orders entry and scheduling.
Let’s calculate the profit of truck #1 in the Model
#3, where penalties are applied. We will calculate
the profit P at the moments of transition of the truck
from one state to another step by step.
Step 1. Execution of order #1 will require to start at
the moment t=1 from point #1 to point # 4 and will
take 2 days till the moment t=3. At the moment t=3
the profit is P=-q*2=-2.
Step 2. The transportation of cargo from point 4 to
the point 1 will take 2 days, and at t=5 the truck will
arrive at the point 1 with the profit P=-2+(c-q)*2=-
2+2*2=2. Assume that the truck agent assesses
options of further schedule and execution upon
arrival to point 1 at time t=5. Its profit at point 4 is
P=2. By this time order # 3 has been entered at the
moment of time #3. There are two options to execute
it:
Order #2 is to be executed with delay;
Order #2 is rejected, idle time cost is accepted,
order #3 from the same point 1 is to be taken; for
order # 2 can be executed with delay before
execution of order #3, no further options will be
taken into consideration. Let’s take a more precise
look at 2 options.
Step 3. Truck needs to reach point 3, moving from
point 1 (1 day trip), pick up the order and execute it,
going from point 3 to point 1 (1 day). The increase
of profit is dp=-1*q+(c-q)*1=-1+2=1.
Penalty applied because of delay is -pp*2=-2*0.6=-
1.2. As a result the truck will be at the moment t=7
at the point 1 with the profit P=2+1-1.2=1.8.
Execution of the order would seem to be
unprofitable, but one should take into consideration
that in case of cancellation of the order the truck
would stay idle for 2 days, and the profit at the
moment t=7 would be P=2-2*0.3=1.4.
Step 4. That’s why the truck agent is interested in
the execution of order #2 with delay, order #3, t=
7…9 (from point 1 to point 4) - 2 days, profit is
P=1.8+2*(c-q)=1.8+2*2=5.8, and the truck moves to
point 4.
Step 5. At the moment t=9 new order# 5 comes in
at the point 3 with start time of execution t=9; empty
run to its loading point is 1 day, what puts the order
beyond the 10-days scheduling horizon limit. That’s
why the truck agent rejects the order. There is an
outdated order #4 from point 4 to point 3, its
execution start time should be t=8. The truck agent
assesses profit from possible shift of order by a day.
Step 6. Execution of the order #4, empty run is not
required, dp=(3-1)*1=2-penalty 0.6=1.4. If this
order were rejected, the truck would stay idle for 1
day till the end of the scheduling horizon and then
dp=-1*0.3=-0.3. That’s why the truck agent accepts
the order #4.
Outcome: orders #1 and 3 are executed without
delay, order #2 – with allowed delay of 2 days and
order #4 – with allowed delay of 1 day. Order #5 is
rejected. Total profit in 10 days is P=5.8+1.4=7.2.
Final track of the truck is shown on the Figure 2.
The truck starts from the point 1 to the point 4. Then
it executes the order #1 from the point 4 to the point
1 without delay. Then it goes to the point #3 to
execute the order #2. Then it executes the order #2
with delay. After this the truck executes the order #3
from the point 1 to the point 4 without delay. Then it
executes the order #4 with delay. The order #5
remains unfulfilled, because it goes beyond the
scheduling horizon (t=10). The delayed orders on
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