In [1] two formulations were discussed for a scheduling problem with multiple machines. Here we add a few more. Some of them are a bit strange. None of them really works better than the ones in [1].
So this is a collection of formulations that may sound reasonable at first, but are really not suited for this problem. If you want to read about good models, skip this post.
Summary of the input data
---- 29 PARAMETER jobdata
proctime release duedate weights
job1 461701
job2 959781
job3 7671
job4 521351
job5 56171
job6 4191
job7 557681
job8 941591
job9 3121
job10 767821
job11 10161
job12 79251
job13 1040571
job14 958781
job15 4261
job16 8131
job17 4631
job18 5661
job19 8451
job20 625421
job21 5321
job22 5321
job23 45211
job24 4941
job25 722441
job26 960811
job27 4371
job28 8211
job29 961781
job30 54161
job31 3281
job32 7101
job33 424341
job34 939551
job35 5231
job36 5251
job37 724401
job38 8381
job39 8391
job40 6971
job41 61001
job42 333431
job43 528431
job44 364801
job45 531461
job46 4931
job47 82201
job48 759761
job49 9151
job50 551621
---- 33 SET precedenceprecedence constraints: i must execute before j
job4 job8 job9 job10 job11 job13 job16 job20 job33
job1 YES
job4 YES
job6 YES
job7 YES
job8 YES YES YES
job11 YES
job15 YES
job18 YES
job28 YES
job30 YES
+ job36 job37 job44
job33 YES YES
job40 YES
We implement a number of objectives, so we combine them and get something that is similar to what is used in practical situations.
In [1] we discussed two models:
- A time-indexed model that used the start of a job as central variable: \[\color{darkred}x_{j,m,t} = \begin{cases} 1 & \text{if job $j$ starts at time $t$ on machine $m$}\\ 0 & \text{otherwise}\end{cases}\] This model did very well.
- A continuous-time model, where we model no-overlap using standard big-M constraints based on the Alan Manne formulation to sequence jobs such that they don't overlap [2]. This model did not do as well. It found good solutions fairly quickly but finding and proving the global optimum was challenging.
 |
| An optimal solution |
In this post I show some results with the following models:
- M3: A version of a time-indexed model that can be used when the run length of a job is not known in advance. That more general approach leads to a slower model.
- M4: A version of a continuous-time model from [3]. It is more complicated but not better than the model I used in [1].
- M5: A positional formulation. It works, but it is not competitive.
M3: A slow time-indexed model
Here we try to use as central variable: \[\color{darkred}x_{j,m,t} = \begin{cases} 1 & \text{if job $j$ occupies machine $m$ at time $t$} \\ 0 &\text{otherwise}\end{cases}\] This approach makes the no-overlap constraint very straightforward: \[\sum_j \color{darkred}x_{j,m,t} \le 1 \>\>\forall m,t\] To find the start and end time we can do: \[\begin{align} & \color{darkred}{\mathit{xstart}}_{j,m,t}\ge\color{darkred}x_{j,m,t} - \color{darkred}x_{j,m,t-1} \\ & \color{darkred}{\mathit{xend}}_{j,m,t}\ge\color{darkred}x_{j,m,t} - \color{darkred}x_{j,m,t+1} \\ & \sum_{m,t} \color{darkred}{\mathit{xstart}}_{j,m,t} \le 1&&\forall j \\ &\sum_{m,t}\color{darkred}x_{j,m,t} = \color{darkblue}{\mathit{proctime}}_j && \forall j \\ & \color{darkred}x_{j,m,t} \in \{0,1\} \\ & \color{darkred}{\mathit{xstart}}_{j,m,t}, \color{darkred}{\mathit{xend}}_{j,m,t} \in [0,1] \end{align}\] The first two inequalities implement the implications \[\begin{align} & \color{darkred}x_{j,m,t-1}= 0 \> \mathbf{and}\> \color{darkred}x_{j,m,t}=1 \Rightarrow \color{darkred}{\mathit{xstart}}_{j,m,t} = 1 \\ & \color{darkred}x_{j,m,t}= 1 \>\mathbf{and}\>\color{darkred}x_{j,m,t+1}= 0 \Rightarrow \color{darkred}{\mathit{xend}}_{j,m,t} = 1\end{align}\] The condition that we have maximum one start has two implications: first we have one start and thus one consecutive production run (without holes) and second that the production run is executed on one machine.
 |
| Relation between x, xstart and xend. |
It is noted that \(\color{darkred}x_{j,m,t}\) is a real binary variable. The variables \(\color{darkred}{\mathit{xstart}}_{j,m,t}\) and \(\color{darkred}{\mathit{xend}}_{j,m,t}\) can be continuous between 0 and 1.
I also added the constraint \[\sum_{m,t} \color{darkred}{\mathit{xend}}_{j,m,t} \le 1\] to enforce that otherwise unrestricted \(\color{darkred}{\mathit{xend}}\) variables are set to zero. This makes it easier to calculate the \(\color{darkred}{\mathit{obj}}_{\mathit{numtardy}}\) objective.
The complete model can look like:
| Model M3: Slow time indexed model |
|---|
| \[\begin{align} \min&\sum_{\mathit{objs}} \color{darkblue}{\mathit{objw}}_{\mathit{objs}} \cdot \color{darkred}{\mathit{obj}}_{\mathit{objs}} && \\ \hline & \color{darkred}{\mathit{obj}}_{\mathit{completion}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{completion}}_j && \\ & \color{darkred}{\mathit{obj}}_{\mathit{sumtardy}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{tardy}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{numtardy}} = \sum_{j,m,t|t\ge\color{darkblue}{\mathit{due}}(j)}\color{darkred}{\mathit{xend}}_{j,m,t} \\ & \color{darkred}{\mathit{obj}}_{\mathit{maxtardy}} \ge \color{darkred}{\mathit{tardy}}_j && \forall j \\ & \color{darkred}{\mathit{obj}}_{\mathit{makespan}} \ge \color{darkred}{\mathit{completion}}_j && \forall j\\ \hline& \color{darkred}{\mathit{xstart}}_{j,m,t}\ge\color{darkred}x_{j,m,t} - \color{darkred}x_{j,m,t-1} \\ & \color{darkred}{\mathit{xend}}_{j,m,t}\ge\color{darkred}x_{j,m,t} - \color{darkred}x_{j,m,t+1} \\ & \sum_{m,t} \color{darkred}{\mathit{xstart}}_{j,m,t} \le 1&&\forall j \\ & \sum_{m,t} \color{darkred}{\mathit{xend}}_{j,m,t} \le 1&&\forall j \\ &\sum_{m,t}\color{darkred}x_{j,m,t} = \color{darkblue}{\mathit{proctime}}_j && \forall j \\ &\sum_j \color{darkred}x_{j,m,t} \le 1 && \forall m,t\\ & \color{darkred}{\mathit{start}}_j = \sum_{m,t} t \cdot \color{darkred}{\mathit{xstart}}_{j,m,t} && \forall j \\ & \color{darkred}{\mathit{completion}}_j = \color{darkred}{\mathit{start}}_j +\color{darkblue}{\mathit{proctime}}_j && \forall j\\ &\color{darkred}{\mathit{tardy}}_j \ge\color{darkred}{\mathit{completion}}_j -\color{darkblue}{\mathit{due}}_j&& \forall j\\ &\color{darkred}{\mathit{completion}}_i \le\color{darkred}{\mathit{start}}_j && \forall i,j|\color{darkblue}{\mathit{precedence}}_{i,j} && \\ \hline & \color{darkred}x_{j,m,t} \in \{0,1\} && \\ &\color{darkred}{\mathit{xstart}}_{j,m,t}, \color{darkred}{\mathit{xend}}_{j,m,t} \in [0,1]\\ & \color{darkred}{\mathit{obj}}_{\mathit{objs}} \ge 0 \\ & \color{darkred}{\mathit{tardy}}_j \ge 0 \\ &\color{darkred}{\mathit{start}}_j \ge \color{darkblue}{\mathit{release}}_j \end{align}\] |
This model works, but it is slow. This formulation is more often used in situations where the run lengths are not fixed. An example is: a power generator can only be turned on at most \(n\) times during a time period to prevent excessive wear and tear. Here the situation is slightly different: we know there is exactly one production run for a given job and its length is fixed.
| model m1 time-indexed | model m2 continuous-time | model m3 time-indexed | |
|---|---|---|---|
| Equations | 626 | 10,026 | 40,726 |
| Variables | 20,158 | 1,633 | 60,158 |
| binary | 20,000 | 1,475 | 20,000 |
| Nonzero elements | 191,024 | 49,896 | 250,752 |
| Objective | 324.096 | 324.102 | 437.303 |
| completion | 2,096 | 2,102 | 2,303 |
| sumtardy | 322 | 322 | 435 |
| maxtardy | 84 | 84 | 87 |
| numtardy | 7 | 7 | 14 |
| makespan | 97 | 97 | 100 |
| Time (seconds) | 20 | 1800 (time limit) | 1800 (time limit) |
| Gap | optimal | 0.13% | 30% |
| Nodes | 529 | 1,262,868 | 5,317 |
| Iterations | 28,553 | 5,463,876 | 13,311,418 |
The conclusion is that relatively minor variations in a model can lead to very different performance. Also, a formulation that exploits as much as possible of the problem at hand (in this case: fixed run lengths), is likely the best.
M4: an alternative, complex continuous-time model
In [1] a very simple continuous-time model was implemented, based on the implication:
if jobs \(i\) and \(j\) execute on the same machine then job \(j\) ends before job \(i\) starts or job \(j\) starts after job \(i\) ends.
This was implemented as: \[\begin{align} &\color{darkred}{\mathit{completion}}_i \le \color{darkred}{\mathit{start}}_j + \color{darkblue} M \cdot (1-\color{darkred}x_{i,m}) +\color{darkblue} M \cdot (1-\color{darkred}x_{j,m}) + \color{darkblue} M \cdot \color{darkred} \delta_{i,j} && \forall i\lt j, \forall m \\ & \color{darkred}{\mathit{completion}}_j \le \color{darkred}{\mathit{start}}_i + \color{darkblue} M \cdot (1-\color{darkred}x_{i,m}) +\color{darkblue} M \cdot (1-\color{darkred}x_{j,m}) + \color{darkblue} M \cdot (1- \color{darkred} \delta_{i,j}) && \forall i\lt j, \forall m \\ & \color{darkred} \delta_{i,j} \in \{0,1\}\end{align}\] Note that all comparisons are only for \(i\lt j\). Here \(\color{darkred}x\) indicates the assignment of jobs to machines and \(\color{darkred} \delta\) implements the or condition. The constant \(\color{darkblue}M\) should be equal to the length of the planning period.
In [3], a more convoluted version of this approach is proposed. Let's see how it behaves.
We introduce the following variables: \[\begin{aligned}& \color{darkred}x_{j,m}=\begin{cases} 1 & \text{if job $j$ is assigned to machine $m$} \\ 0 & \text{otherwise}\end{cases}\\ & \color{darkred}\delta_{i,j} = \begin{cases} 1 & \text{if job $i$ executes before job $j$ on the same machine}\\ 0 & \text{otherwise}\end{cases} \\ &\color{darkred}d_{i,j} = \begin{cases} 1 & \text{if jobs $i$ and $j$ execute on different machines}\\ 0 & \text{otherwise}\end{cases}\end{aligned}\]
The first constraint is \[\color{darkred}\delta_{i,j} + \color{darkred}\delta_{j,i}+\color{darkred}d_{i,j} = 1\>\>\forall i \lt j\] This means: jobs \(i\) and \(j\) are on different machines or they don't overlap. The second constraint, the transitivity constraint, is a bit of mystery to me: \[\color{darkred}\delta_{i,j} +\color{darkred}\delta_{j,k} + \color{darkred}\delta_{k,i}\le 2 \>\>\forall i \lt j \lt k\] I don't think this constraint is needed. It may have been added to strengthen the formulation. The next constraint \[\color{darkred}x_{i,m}+\color{darkred}x_{j,m}+\color{darkred}d_{i,j}\le 2 \>\>\forall i\lt j\] can be considered as an implication \[\color{darkred}x_{i,m}=\color{darkred}x_{j,m}=1 \Rightarrow \color{darkred}d_{i,j}=0\] I.e. if jobs \(i\) and \(j\) are on same the machine, we need to have \(\color{darkred}d_{i,j}=0\). This is used in the no-overlap (first) constraint. Obviously we have the assignment constraint: \[\sum_m \color{darkred}x_{i,m}=1\>\>\forall i\] Finally, we have the completion times to consider: \[\color{darkred}{\mathit{completion}}_j \ge \color{darkred}{\mathit{completion}}_i + \color{darkblue}{\mathit{proctime}}_j(\color{darkred}\delta_{i,j}+\color{darkred}x_{i,m}+\color{darkred}x_{j,m}-2) - \color{darkblue}M(1-\color{darkred}\delta_{i,j})\>\>\forall i,j,m\] Instead, we can use the simpler: \[\color{darkred}{\mathit{start}}_j \ge \color{darkred}{\mathit{completion}}_i - \color{darkblue}M(1-\color{darkred}\delta_{i,j})\>\>\forall i\ne j, m\] Note that I explicitly excluded the case \(i=j\) which is a bit more precise.
The complete model can be summarized as follows:
| Model M4: Alternative continuous-time model |
|---|
| \[\begin{align}\min&\sum_{\mathit{objs}} \color{darkblue}{\mathit{objw}}_{\mathit{objs}} \cdot \color{darkred}{\mathit{obj}}_{\mathit{objs}} && \\ \hline & \color{darkred}{\mathit{obj}}_{\mathit{completion}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{completion}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{sumtardy}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{tardy}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{numtardy}} = \sum_j \color{darkred}{\mathit{isTardy}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{maxtardy}} \ge \color{darkred}{\mathit{tardy}}_j && \forall j\\ & \color{darkred}{\mathit{obj}}_{\mathit{makespan}} \ge \color{darkred}{\mathit{completion}}_j && \forall j\\ \hline &\color{darkred}\delta_{i,j} + \color{darkred}\delta_{j,i}+\color{darkred}d_{i,j} = 1&& \forall i \lt j \\ &\color{darkred}\delta_{i,j} +\color{darkred}\delta_{j,k} + \color{darkred}\delta_{k,i}\le 2 &&\forall i \lt j \lt k\\ & \color{darkred}x_{i,m}+\color{darkred}x_{j,m}+\color{darkred}d_{i,j}\le 2 && \forall i\lt j \\ & \sum_m \color{darkred}x_{j,m}=1 && \forall j \\ & \color{darkred}{\mathit{completion}}_ j = \color{darkred}{\mathit{start}}_ j + \color{darkblue}{\mathit{proctime}}_ j && \forall j \\ & \color{darkred}{\mathit{start}}_j \ge \color{darkred}{\mathit{completion}}_i - \color{darkblue}M(1-\color{darkred}\delta_{i,j})&&\forall i\ne j, m \\ & \color{darkred}{\mathit{tardy}}_j \ge\color{darkred}{\mathit{completion}}_j -\color{darkblue}{\mathit{due}}_j && \forall j\\ &\color{darkred}{\mathit{tardy}}_j \le \color{darkblue}M \cdot\color{darkred}{\mathit{isTardy}}_j&& \forall j \\ &\color{darkred}{\mathit{completion}}_i \le\color{darkred}{\mathit{start}}_j && \forall i,j|\color{darkblue}{\mathit{precedence}}_{i,j} \\ \hline & \color{darkred}x_{j,m}\in \{0,1\} \\ & \color{darkred}\delta_{i,j} \in \{0,1\} \\ & \color{darkred}d_{i,j} \in [0,1]\\ &\color{darkred}{\mathit{isTardy}}_j\in \{0,1\} \\ & \color{darkred}{\mathit{obj}}_{\mathit{objs}} \ge 0 \\ & \color{darkred}{\mathit{tardy}}_j \ge 0 \\ & \color{darkred}{\mathit{start}}_j \ge \max(1,\color{darkblue}{\mathit{release}}_j) \end{align}\] |
The performance is as follows:
| model m1 time-indexed | model m2 continuous-time | model m4 continuous-time | |
|---|---|---|---|
| Equations | 626 | 10,026 | 28,451 |
| Variables | 20,158 | 1,633 | 4,133 |
| binary | 20,000 | 1,475 | 2,750 |
| Nonzero elements | 191,024 | 49,896 | 85,571 |
| Objective | 324.096 | 324.102 | 324.139 |
| completion | 2,096 | 2,102 | 2,139 |
| sumtardy | 322 | 322 | 322 |
| maxtardy | 84 | 84 | 84 |
| numtardy | 7 | 7 | 7 |
| makespan | 97 | 97 | 97 |
| Time (seconds) | 20 | 1800 (time limit) | 1800 (time limit) |
| Gap | optimal | 0.13% | 0.15% |
| Nodes | 529 | 1,262,868 | 114,732 |
| Iterations | 28,553 | 5,463,876 | 3,455,261 |
Notes:
- The performance is very close to model M2. This formulation does not add much value above the more intuitive and simpler model M2. The M2 model is easier to remember. More complex models are not necessarily better.
- I don't know what the purpose is of the transitivity constraint.
- Some of the constraints should have \(\forall i\ne j\) instead of \(\forall i,j\).
- Of course, we can add ordering constraints to reduce symmetry. I used: machine \(m\) has more jobs (or the same number) than machine \(m+1\). As the machines in my model are identical, we can renumber machines.
M5: positional formulation.
This is also from [3]. Here we use as central variable: \[\color{darkred}x_{j,m,p}=\begin{cases} 1 & \text{if job $j$ has position $p$ on machine $m$}\\ 0 & \text{otherwise}\end{cases}\] The assignment constraints related to this variable are: \[\begin{align}&\sum_{m,p}\color{darkred}x_{j,m,p}=1 &&\forall j \\ &\sum_j\color{darkred}x_{j,m,p}\le 1 &&\forall m,p \end{align}\] The corresponding completion time is denoted by \(\color{darkred}f_{m,p}\). We have: \[\color{darkred}f_{m,p} \ge \color{darkred}f_{m,p-1} + \sum_j \color{darkblue}{\mathit{proctime}}_j\cdot \color{darkred}x_{j,m,p}\] We assume here the first position is 1 and \(\color{darkred}f_{m,0}=1\) (this is because my first time period is 1 and completion time is at the beginning of a time period). Finally we link the job completion time through: \[\color{darkred}{\mathit{completion}}_j \ge \color{darkred}f_{m,p} - \color{darkblue}M(1-\color{darkred}x_{j,m,p}) \>\>\forall j,m,p \]
To allow for release dates and precedence constraints, we need to add: \[\begin{align}& \color{darkred}f_{m,p} \ge \sum_j (\color{darkblue}{\mathit{release}}_j+\color{darkblue}{\mathit{proctime}}_j)\cdot \color{darkred}x_{j,m,p}\\ & \color{darkred}f_{m,p} \ge \color{darkred}{\mathit{completion}}_i + \color{darkblue}{\mathit{proctime}}_j\cdot \color{darkred}x_{j,m,p} - \color{darkblue}M\cdot (1-\color{darkred}x_{j,m,p})&& \forall i,j|\color{darkblue}{\mathit{precedence}}_{i,j} \end{align}\] The last constraint is not in [3]: they only consider release dates but no precedence constraints.
| Model M5: Positional model |
|---|
| \[\begin{align} \min&\sum_{\mathit{objs}} \color{darkblue}{\mathit{objw}}_{\mathit{objs}} \cdot \color{darkred}{\mathit{obj}}_{\mathit{objs}} && \\ \hline & \color{darkred}{\mathit{obj}}_{\mathit{completion}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{completion}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{sumtardy}} = \sum_j \color{darkblue}w_j \cdot \color{darkred}{\mathit{tardy}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{numtardy}} = \sum_j \color{darkred}{\mathit{isTardy}}_j \\ & \color{darkred}{\mathit{obj}}_{\mathit{maxtardy}} \ge \color{darkred}{\mathit{tardy}}_j && \forall j\\ & \color{darkred}{\mathit{obj}}_{\mathit{makespan}} \ge \color{darkred}{\mathit{completion}}_j && \forall j\\ \hline &\sum_{m,p}\color{darkred}x_{j,m,p}=1 &&\forall j \\ & \sum_j\color{darkred}x_{j,m,p}\le 1 &&\forall m,p \\ & \color{darkred}f_{m,p} \ge \color{darkred}f_{m,p-1} + \sum_j \color{darkblue}{\mathit{proctime}}_j\cdot \color{darkred}x_{j,m,p} &&\forall m,p\\ & \color{darkred}f_{m,0}=1 \\ & \color{darkred}f_{m,p} \ge \sum_j (\color{darkblue}{\mathit{release}}_j+\color{darkblue}{\mathit{proctime}}_j)\cdot \color{darkred}x_{j,m,p} && \forall m,p \\ & \color{darkred}f_{m,p} \ge \color{darkred}{\mathit{completion}}_i + \color{darkblue}{\mathit{proctime}}_j\cdot \color{darkred}x_{j,m,p} - \color{darkblue}M\cdot (1-\color{darkred}x_{j,m,p})&& \forall i,j|\color{darkblue}{\mathit{precedence}}_{i,j},m,p \\ & \color{darkred}{\mathit{completion}}_j \ge \color{darkred}f_{m,p} - \color{darkblue}M(1-\color{darkred}x_{j,m,p}) && \forall j,m,p \\ & \color{darkred}{\mathit{completion}}_ j = \color{darkred}{\mathit{start}}_ j + \color{darkblue}{\mathit{proctime}}_ j && \forall j \\ & \color{darkred}{\mathit{tardy}}_j \ge\color{darkred}{\mathit{completion}}_j -\color{darkblue}{\mathit{due}}_j && \forall j\\ &\color{darkred}{\mathit{tardy}}_j \le \color{darkblue}M \cdot\color{darkred}{\mathit{isTardy}}_j&& \forall j \\ &\color{darkred}{\mathit{completion}}_i \le\color{darkred}{\mathit{start}}_j && \forall i,j|\color{darkblue}{\mathit{precedence}}_{i,j} \\ \hline & \color{darkred}x_{j,m,p}\in \{0,1\} \\ & \color{darkred}f_{m,p} \ge 1 \\ &\color{darkred}{\mathit{isTardy}}_j\in \{0,1\} \\ & \color{darkred}{\mathit{obj}}_{\mathit{objs}} \ge 0 \\ & \color{darkred}{\mathit{tardy}}_j \ge 0 \\ & \color{darkred}{\mathit{start}}_j \ge \max(1,\color{darkblue}{\mathit{release}}_j) \end{align}\] |
We can add a few refinements. First, we can renumber the machines as they are identical. I reduce symmetry by ordering the machines by how many jobs they execute. The second source of symmetry is that we can have "empty" positions in this model. I added a constraint to make sure the empty, unused positions are at the end and not in the middle. This also adds a symmetry-breaking effect to the model. After this we see:
| model m1 time-indexed | model m2 continuous-time | model m5 positional formulation | |
|---|---|---|---|
| Equations | 626 | 10,026 | 5,742 |
| Variables | 20,158 | 1,633 | 4,368 |
| binary | 20,000 | 1,475 | 4,050 |
| Nonzero elements | 191,024 | 49,896 | 36,564 |
| Objective | 324.096 | 324.102 | 324.135 |
| completion | 2,096 | 2,102 | 2,135 |
| sumtardy | 322 | 322 | 322 |
| maxtardy | 84 | 84 | 84 |
| numtardy | 7 | 7 | 7 |
| makespan | 97 | 97 | 97 |
| Time (seconds) | 20 | 1800 (time limit) | 1800 (time limit) |
| Gap | optimal | 0.13% | 0.14% |
| Nodes | 529 | 1,262,868 | 49,186 |
| Iterations | 28,553 | 5,463,876 | 2,368,537 |
Notes:
- This model is not as effective as the models in [1].
- There are some interesting possibilities to reduce symmetry in this model.
Conclusion
There are many formulations for this problem, choosing the right one can pay off.
Having access to different formulations can be an excellent debugging tool: they all should deliver the same optimal objective. We can plug in a solution from one model into another (using mipstart or by fixing): it should be accepted without hesitancy. Complex models are not so easy to get right immediately, so an extra tool to verify solutions is valuable.
References
- Parallel machine scheduling I, two formulations, https://yetanothermathprogrammingconsultant.blogspot.com/2021/03/parallel-machine-scheduling-i-two.html
- Manne, Alan S. “On the Job-Shop Scheduling Problem.” Operations Research, vol. 8, no. 2, 1960, pp. 219–223.
- Yasin Unlu, Scott J. Mason, Evaluation of mixed integer programming formulations for non-preemptive parallel machine scheduling problems, Computers & Industrial Engineering 58 (2010) 785–800.