In [1], a problem is stated that can be interpreted as a Set Partitioning Problem [2].
Consider the following data:
Note: zeros are not printed.
The problem is:
This is slightly different than the data in [1], in order to make the problem feasible.
Before we can work on the model itself, we need to develop some derived data. We introduce the boolean parameter \(\mathit{Cover}_{i,t}\) defined by: \[\mathit{Cover}_{i,t}=\begin{cases} 1 & \text{if job $i$ covers hour $t$}\\ 0 & \text{otherwise}\end{cases}\] This parameter looks like:
We further introduce the binary decision variables: \[x_i = \begin{cases} 1 & \text{if job $i$ is selected} \\ 0 & \text{otherwise}\end{cases}\] With this we can formulate:
When we solve this problem using the above data set we get the solution:
It is noted that the model can also be stated using a set representation of the coverage data. Such a model can look like:
Although this looks somewhat different, it is really exactly the same. This is actually the form I use most of the time.
It is quite easy to see data sets that cause the problem to be infeasible. Indeed, the data set shown in [1] is actually yielding a model that is not feasible.
There are two ways in which the model can be infeasible. The first is the easy one: we have no jobs at all that cover a time period \(t\). Consider the data:
We see that hours 15, 16, 17, and 18 are not covered by any job. This can be easily checked in advance. Even GAMS is complaining when generating the model:
There is another form of infeasibility that is more difficult to diagnose. When we use the data set:
we get:
There is not much in diagnostics for most solvers. Cplex at least reports:
This is better than most solvers, but may be difficult to digest for an end-user. In applications, it may be better to make sure the problem is never infeasible. Basically, allow the constraint to become infeasible but at a price. This is sometimes called an elastic formulation. There are two ways to do this for this model:
We can add expensive filler-jobs to our data set as follows:
When we solve this, we see the following solution:
So we have trouble covering 2 hours with jobs. If we allow then to remain like that, the best solution is to use jobs 12, 15 and 19 for the other hours.
You can see we added a slack variable \({\mathit{Uncov}}_t\), and included this in the objective with a penalty (999). When we run this model, we see:
I use this type of elastic formulations a lot. This approach prevents a user from seeing "sorry model was infeasible". Instead, it gives back a meaningful result.
Consider the following data:
---- 16 PARAMETER data input data
start length cost
job1 2547
job2 19520
job3 11238
job4 5214
job5 5840
job6 31026
job7 6368
job8 19561
job9 1080
job10 10437
job11 22270
job12 12778
job13 22267
job14 17735
job15 1417
job16 13419
job17 1868
job18 4959
job19 14812
job20 8682
Note: zeros are not printed.
The problem is:
Find a subset of jobs that don't overlap and cover each hour 0 through 23, and that minimize cost.
This is slightly different than the data in [1], in order to make the problem feasible.
Set Partitioning Model
Before we can work on the model itself, we need to develop some derived data. We introduce the boolean parameter \(\mathit{Cover}_{i,t}\) defined by: \[\mathit{Cover}_{i,t}=\begin{cases} 1 & \text{if job $i$ covers hour $t$}\\ 0 & \text{otherwise}\end{cases}\] This parameter looks like:
---- 16 PARAMETER cover coverage of hours by jobs
h00 h01 h02 h03 h04 h05 h06 h07 h08
job1 11111
job4 11
job5 1111
job6 111111
job7 111
job9 111111111
job15 1111
job17 11111111
job18 11111
job20 1
+ h09 h10 h11 h12 h13 h14 h15 h16 h17
job3 11
job5 1111
job6 1111
job9 1
job10 1111
job12 111111
job14 1
job16 1111
job18 1111
job19 1111
job20 11111
+ h18 h19 h20 h21 h22 h23
job2 11111
job8 11111
job11 11
job12 1
job13 11
job14 111111
job19 1111
We further introduce the binary decision variables: \[x_i = \begin{cases} 1 & \text{if job $i$ is selected} \\ 0 & \text{otherwise}\end{cases}\] With this we can formulate:
| Set Partitioning Model |
|---|
| \[\begin{align}\min & \sum_i \color{darkblue}{\mathit{Cost}}_i \cdot \color{darkred}x_i \\ & \sum_i \color{darkblue}{\mathit{Cover}}_{i,t} \cdot \color{darkred}x_i = 1 &&\forall t \\ & \color{darkred}x_i \in \{0,1\} \end{align}\] |
When we solve this problem using the above data set we get the solution:
---- 31 VARIABLE x.L selected jobs
job9 1, job10 1, job13 1, job19 1
---- 31 VARIABLE tcost.L = 196 total cost
It is noted that the model can also be stated using a set representation of the coverage data. Such a model can look like:
| Set Partitioning Model 2 |
|---|
| \[\begin{align}\min & \sum_i \color{darkblue}{\mathit{Cost}}_i \cdot \color{darkred}x_i \\ & \sum_{i|\color{darkblue}{\mathit{Cover}}(i,t)} \color{darkred}x_i = 1 &&\forall t \\ & \color{darkred}x_i \in \{0,1\} \end{align}\] |
Although this looks somewhat different, it is really exactly the same. This is actually the form I use most of the time.
Infeasibilities
It is quite easy to see data sets that cause the problem to be infeasible. Indeed, the data set shown in [1] is actually yielding a model that is not feasible.
There are two ways in which the model can be infeasible. The first is the easy one: we have no jobs at all that cover a time period \(t\). Consider the data:
---- 16 PARAMETER data input data
start length cost
job1 21263
job2 19585
job3 11231
job4 5870
job5 5280
job6 3437
job7 6920
job8 19555
job9 524
job10 10589
job11 22234
job12 12236
---- 16 PARAMETER cover coverage of hours by jobs
h00 h01 h02 h03 h04 h05 h06 h07 h08
job1 1111111
job4 1111
job5 11
job6 1111
job7 111
job9 11111
+ h09 h10 h11 h12 h13 h14 h19 h20 h21
job1 11111
job2 111
job3 11
job4 1111
job7 111111
job8 111
job10 11111
job12 11
+ h22 h23
job2 11
job8 11
job11 11
We see that hours 15, 16, 17, and 18 are not covered by any job. This can be easily checked in advance. Even GAMS is complaining when generating the model:
**** Exec Error at line 24: Equation infeasible due to rhs value
**** INFEASIBLE EQUATIONS ...
---- ecover =E=
ecover(h15).. 0 =E= 1 ; (LHS = 0, INFES = 1 ****)
ecover(h16).. 0 =E= 1 ; (LHS = 0, INFES = 1 ****)
ecover(h17).. 0 =E= 1 ; (LHS = 0, INFES = 1 ****)
ecover(h18).. 0 =E= 1 ; (LHS = 0, INFES = 1 ****)
There is another form of infeasibility that is more difficult to diagnose. When we use the data set:
---- 16 PARAMETER data input data
start length cost
job1 2667
job2 19552
job3 11547
job4 5220
job5 5238
job6 3814
job7 61040
job8 19326
job9 868
job10 101061
job11 22280
job12 12237
job13 22270
job14 17278
job15 11167
job16 13435
job17 1417
job18 4819
job19 14968
we get:
S O L V E S U M M A R Y
MODEL m OBJECTIVE tcost
TYPE MIP DIRECTION MINIMIZE
SOLVER CPLEX FROM LINE 28
**** SOLVER STATUS 1 Normal Completion
**** MODEL STATUS 10 Integer Infeasible
**** OBJECTIVE VALUE NA
There is not much in diagnostics for most solvers. Cplex at least reports:
Infeasibility row 'ecover(h08)': 0 = 1.
This is better than most solvers, but may be difficult to digest for an end-user. In applications, it may be better to make sure the problem is never infeasible. Basically, allow the constraint to become infeasible but at a price. This is sometimes called an elastic formulation. There are two ways to do this for this model:
- use a data-driven method: introduce expensive emergency jobs used to make the constraint feasible
- change the constraint directly
Data-driven Elastic model
We can add expensive filler-jobs to our data set as follows:
---- 21 PARAMETER data input data
start length cost
job1 2667
job2 19552
job3 11547
job4 5220
job5 5238
job6 3814
job7 61040
job8 19326
job9 868
job10 101061
job11 22280
job12 12237
job13 22270
job14 17278
job15 11167
job16 13435
job17 1417
job18 4819
job19 14968
notcov00 1999
notcov01 11999
notcov02 21999
notcov03 31999
notcov04 41999
notcov05 51999
notcov06 61999
notcov07 71999
notcov08 81999
notcov09 91999
notcov10 101999
notcov11 111999
notcov12 121999
notcov13 131999
notcov14 141999
notcov15 151999
notcov16 161999
notcov17 171999
notcov18 181999
notcov19 191999
notcov20 201999
notcov21 211999
notcov22 221999
notcov23 231999
When we solve this, we see the following solution:
---- 36 VARIABLE x.L selected jobs
job12 1, job15 1, job19 1, notcov00 1, notcov23 1
---- 36 VARIABLE tcost.L = 2170 total cost
So we have trouble covering 2 hours with jobs. If we allow then to remain like that, the best solution is to use jobs 12, 15 and 19 for the other hours.
Elastic model: change constraint
Instead of changing the data, we can also do the same by changing the covering constraint:
| Set Partitioning Model, elastic version |
|---|
| \[\begin{align}\min & \sum_i \color{darkblue}{\mathit{Cost}}_i \cdot \color{darkred}x_i + \color{darkblue}{\mathit{Penalty}} \cdot \sum_t\color{darkred}{\mathit{Uncov}}_t \\ & \sum_i \color{darkblue}{\mathit{Cover}}_{i,t} \cdot \color{darkred}x_i + \color{darkred}{\mathit{Uncov}}_t = 1 &&\forall t \\ & \color{darkred}x_i \in \{0,1\} \\ & \color{darkred}{\mathit{Uncov}}_t \in [0,1] \end{align}\] |
You can see we added a slack variable \({\mathit{Uncov}}_t\), and included this in the objective with a penalty (999). When we run this model, we see:
---- 44 VARIABLE x.L selected jobs
job12 1, job15 1, job19 1
---- 44 VARIABLE uncov.L uncovered hours
h00 1, h23 1
---- 44 VARIABLE tcost.L = 2170 total cost (objective)
PARAMETER jobcost = 172 cost related to running jobs
PARAMETER penalties = 1998 for uncovered hours
References
- Profit maximising job scheduling using Python, https://stackoverflow.com/questions/62297792/profit-maximising-job-scheduling-using-python
- Garfinkel, R. S., and G. L. Nemhauser. “The Set-Partitioning Problem: Set Covering with Equality Constraints.” Operations Research, vol. 17, no. 5, 1969, pp. 848–856.
