In [1] a minimum up-time question is posted:
For \(T=24,n=5\) we have 24 constraints, while enumeration of all feasible solutions would lead to 4316 solutions. NB: I used the Cplex solution pool to enumerate the feasible solutions.
There are a number of extensions one can encounter:
- Assume we have \(T=12\) periods. If we produce in period \(t\) we have a profit of \(c_t\). Some \(c_t\)'s are negative.
- If we turn production on, it should remain on for at least \(n=5\) periods.
- We want to maximize total profit.
- How can we write a constraint that captures this minimum up-time condition?
We can recognize production is turned on in period \(t\) by observing \(x_{t-1}=0\) and \(x_t=1\) where \(x_t\) is a binary variable indicating whether our production process is turned on or off. So our constraint can look like:
\[x_{t-1}=0,x_{t}=1 \Longrightarrow \sum_{k=t}^{t+n-1} x_k = n\>\>\>\forall t \]
In a MIP model we can translate this into:
\[ \sum_{k=t}^{t+n-1} x_k \ge n(x_t-x_{t-1}) \>\>\>\forall t\]
This formulation is not completely obvious. We note that \(x_t-x_{t-1}\) is zero or negative for all cases except \(x_{t-1}=0,x_{t}=1\).
Some notes:
- Behavior near the boundaries is always important. E.g. we could assume \(x_0=0\).
- Let's also assume \(x_{T+1}=0\) so we need to finish our production run in the time horizon.
- The MIP constraint can also be written as: \[ \sum_{k=t+1}^{t+n-1} x_k \ge (n-1)(x_t-x_{t-1}) \>\>\>\forall t\] as we know \(x_t=1\) when this thing kicks in.
- So this gives us 12 constraints for our \(T=12, n=5\) problem.
- We can disaggregate the summation into individual constraints \[x_k \ge x_t-x_{t-1}\text{ for } k=t+1,\dots,t+n-1\] This is tighter and may solve faster. Advanced solvers may use an "implied bound" cut to do this automatically (and smarter).
A complete model can look like:
\[\begin{align} \max \>&\sum_t c_t x_t\\ &\sum_{k=t}^{t+n-1} x_k \ge n(x_t-x_{t-1}) \\ & x_t \in \{0,1\}\end{align}\]
I implemented this in GAMS. GAMS assumes that variables indexed outside their domain are zero. This means I don't have to do anything special for \(t=1\) and for \(t\gt T-n\).
I implemented this in GAMS. GAMS assumes that variables indexed outside their domain are zero. This means I don't have to do anything special for \(t=1\) and for \(t\gt T-n\).
A solution with random \(c_t\) is:
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| Optimal Solution |
For this small case we could have enumerated all possible feasible production schedules and pick the best one. We have 42 possible solutions:
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| All Feasible Solutions ordered by Total Profit |
For \(T=24,n=5\) we have 24 constraints, while enumeration of all feasible solutions would lead to 4316 solutions. NB: I used the Cplex solution pool to enumerate the feasible solutions.
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| Optimal solution for T=24 |
There are a number of extensions one can encounter:
- There may be some initial state. We need more than just the state at \(t=0\). Basically we need to know for how many periods we are in the 'on'-state.
- We may relax the assumption that \(x_{T+1}=0\).
- A maximum up-time is also not too difficult using a similar construct. Say we limit up-time to \(m\) periods. Then we can require:\[ \sum_{k=t}^{t+m} x_k \le m\] I.e. we don't want to see a series of \(m+1\) ones. See the picture below for an example with \(T=24, n=3, m=4\).
- A minimum down-time is sometimes encountered in power generation models. This is close to minimum up-time: make \(y_t=1-x_t\) and apply our tools we described above on \(y_t\).
- Limiting the number of start-ups or shut-downs is quite a standard requirement. E.g. to limit the number of times we start a production run: \[\begin{align} &s_t \ge x_t-x_{t-1}\\& \sum_t s_t \le K \\ & s_t \in \{0,1\} \end{align}\]
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| Run-lengths between 3 and 5. |
References
- Linear programming - optimizing profit with tricky constraints, https://stackoverflow.com/questions/49261669/linear-programming-optimizing-profit-with-tricky-constraints