基于CP的灵活作业车间OR类型机器同机约束编码实现问询

Hey there! Let's work through fixing that OR machine constraint you're struggling with. First, let's restate the requirement to make sure we're on the same page:

For any job, if Operation 1 is processed on an OR-type machine, then any subsequent operation (Operation 2 or 3) in that job that also chooses an OR-type machine must use the exact same OR machine as Operation 1.

What Was Wrong With the Original Code? #

Your original approach had a few key issues:

• No check for active modes: You were comparing machine IDs regardless of whether the mode (specific machine assignment) was actually selected for the operation. Since modes are optional variables, we only care about constraints when a mode is active.
• Redundant and over-broad logic: The multiple nested forall loops were creating unnecessary constraints and didn't properly target the "if Op1 uses OR, then Op2/Op3 must match if they also use OR" logic.
• Hardcoded operation ID assumptions: While your opID calculation might work, relying on 1+(j-1)*3 is less robust than targeting operations by their position in the job.

Corrected Constraint Implementation #

Replace your problematic constraint block with this code. It focuses on the core logic and uses presenceOf() to only apply constraints when a mode is actually selected:

// Correct implementation of the OR machine constraint
forall(j in Jobs) {
    // Get the three operations for job j using their position in the job
    let o1 = first(o in Ops where o.jobId == j && o.pos == 1);
    let o2 = first(o in Ops where o.jobId == j && o.pos == 2);
    let o3 = first(o in Ops where o.jobId == j && o.pos == 3);

    // If Operation 1 uses an OR machine (mode m1 is active), then any OR mode chosen for Operation 2 must match m1's machine
    forall(m1 in Modes where m1.opId == o1.id && m1.mch in OR) {
        presenceOf(modes[m1]) => 
            forall(m2 in Modes where m2.opId == o2.id && m2.mch in OR) {
                presenceOf(modes[m2]) => (m2.mch == m1.mch);
            }
    }

    // If Operation 1 uses an OR machine (mode m1 is active), then any OR mode chosen for Operation 3 must match m1's machine
    forall(m1 in Modes where m1.opId == o1.id && m1.mch in OR) {
        presenceOf(modes[m1]) => 
            forall(m3 in Modes where m3.opId == o3.id && m3.mch in OR) {
                presenceOf(modes[m3]) => (m3.mch == m1.mch);
            }
    }
}

How This Works #

Let's break down the logic:

• Target operations by position: Using first(o in Ops where o.jobId == j && o.pos == X) ensures we always get the correct operation for each job, regardless of how your opId values are assigned.
• Check active modes: presenceOf(modes[m1]) returns 1 if the mode m1 is selected for Operation 1. We only apply the subsequent constraint when this is true.
• Conditional matching: For each OR mode of Operation 2/3, we say: if this mode is selected (i.e., the operation uses an OR machine), then its machine must match the OR machine used by Operation 1.

Optional: If You Prefer Using OpID Calculations #

If you're confident your operation IDs follow the pattern 1+(j-1)*3 (Op1), 2+(j-1)*3 (Op2), j*3 (Op3), you can replace the let statements with hardcoded opID checks instead:

forall(j in Jobs) {
    int op1Id = 1 + (j-1)*3;
    int op2Id = 2 + (j-1)*3;
    int op3Id = j*3;

    // Rest of the constraint logic remains the same, using op1Id, op2Id, op3Id instead of o1.id, o2.id, o3.id
    forall(m1 in Modes where m1.opId == op1Id && m1.mch in OR) {
        presenceOf(modes[m1]) => 
            forall(m2 in Modes where m2.opId == op2Id && m2.mch in OR) {
                presenceOf(modes[m2]) => (m2.mch == m1.mch);
            }
    }

    forall(m1 in Modes where m1.opId == op1Id && m1.mch in OR) {
        presenceOf(modes[m1]) => 
            forall(m3 in Modes where m3.opId == op3Id && m3.mch in OR) {
                presenceOf(modes[m3]) => (m3.mch == m1.mch);
            }
    }
}

This should correctly enforce your desired constraint without the issues from the original code.

内容的提问来源于stack exchange，提问作者Alex