If in the constraint function there is a variable that must be an integer but this variable is not included in the objective function, is the model I created still stated as a mixed integer linear program?
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$\begingroup$ Welcome to OR.SE. I assumed there are other variables, the continuous once, in the model that paper in the objective function. Otherwise, it would be great to elaborate more on the problem details. $\endgroup$A.Omidi– A.Omidi2025-01-28 10:22:31 +00:00Commented Jan 28 at 10:22
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$\begingroup$ You can think of that variable as present in the objective with a cost coefficient of 0.0. $\endgroup$Erwin Kalvelagen– Erwin Kalvelagen2025-01-29 19:17:03 +00:00Commented Jan 29 at 19:17
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2 Answers
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From Science Direct:
Mixed Integer Programming (MIP) is a mathematical technique that involves formulating real-world problems with a linear objective function where at least one variable takes integer or binary values
So yes, it's still a MIP if the integer variable doesn't appear in the objective.
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Besides the mentioned answer, the variables that appear in the objective function, in many cases, are those that form your main goal of using the math model that would need to be maximized or minimized. (In the multi-objective case, both can be used). For example and to name a few, minimizing the total compilation time of tasks on the resources.
One of the best such cases would be the scheduling models in which the objective is usually based on the continuous variable, e.g. flow-time, while integer variables, specifically the binary ones, are used as the intermediate variables in the constraints to deal with some $0/1$ decisions towards the optimization process.
