CPLEX中定义为dvar flaot+的变量返回负值及OPL约束报错问题求助

Let's break down what's going wrong here and fix the issue step by step.

Root Cause of the Problem #

Looking at your constraint and variable setup, two key issues are causing the unexpected negative values and model infeasibility:

1. Constraint Logic Conflict
   Your current constraint for savings is:

   forall(h in truck) sum(n in order) (x_[h,n])*fixed_cost- fixed_cost == savings[h];
   

   When a truck h isn't assigned any orders (i.e., sum(n in order) x[h,n] = 0), the left-hand side calculates to 0*75 -75 = -75. But you defined savings as dvar int+ (a non-negative integer variable). This creates a contradiction: the constraint demands savings[h] = -75, which violates the variable's non-negative requirement. CPLEX returns the negative value because it prioritizes satisfying the equality constraint over the variable's type definition.

2. Infeasibility When Adding Non-Negative Constraint
   When you added forall(h in truck) savings[h] >= 0;, the model becomes impossible to solve. For unused trucks, you’re now asking CPLEX to satisfy both savings[h] = -75 (from your original constraint) and savings[h] >= 0—a logical impossibility.

Also, note the possible typo: you defined x[truck, order] but referenced x_[h,n] in the constraint. Double-check that this matches in your actual code.

Solution: Fix the savings Constraint Logic #

You want savings[h] to be 0 for unused trucks, and fixed_cost*(sum(x[h,n]) -1) for used trucks. Here are two clean ways to implement this:

Option 1: Use OPL's Conditional Expressions (Simplest)

OPL supports ternary operators to directly encode your desired logic:

int fixed_cost = 75;
dvar int+ x[truck, order] in 0..1;
dvar int+ savings[truck];

forall(h in truck) {
  // If the truck is used (has at least one order), calculate savings; else set to 0
  savings[h] == (sum(n in order) x[h,n] >= 1) ? (sum(n in order) x[h,n]*fixed_cost - fixed_cost) : 0;
}

Option 2: Use Linear Constraints (For Strict Linear Models)

If you need fully linear constraints (e.g., for compatibility with solvers that don’t support logical expressions), add a binary variable to track whether a truck is used:

int fixed_cost = 75;
int num_orders = ...; // Replace with the total number of your orders
dvar int+ x[truck, order] in 0..1;
dvar int+ savings[truck];
dvar boolean used[truck]; // 1 = truck is used, 0 = unused

// Link the `used` variable to order assignments
forall(h in truck) {
  used[h] <= sum(n in order) x[h,n]; // If the truck has orders, it must be marked as used
  sum(n in order) x[h,n] <= used[h] * num_orders; // If unused, no orders can be assigned
}

// Calculate savings only for used trucks
forall(h in truck) {
  savings[h] == (sum(n in order) x[h,n] * fixed_cost - fixed_cost) * used[h];
}

Why This Works #

• For unused trucks, used[h] (or the conditional check) ensures savings[h] = 0, matching your expected result.
• For used trucks, savings[h] uses your original calculation, which will be non-negative (since sum(x[h,n]) >= 1 implies sum*75 -75 >= 0).

After implementing either fix, your model should return the expected savings = [0, 75] without infeasibility issues.

内容的提问来源于stack exchange，提问作者FrT