GLPK模型无原始可行解求助：疑Stope_i约束表述有误

Hey there, let's dig into why your GLPK model is throwing that frustrating "No Primal Feasible Solution" error—you’re right to suspect those commented-out Stope_* constraints are the root cause. Let’s break this down step by step.

First, Let’s Confirm the Culprit #

Your base model (with only the StopeB constraint active) should run fine—StopeB just limits each 2x2 block (since k=2, l=2) to have at most one active x[i,j] variable, which is totally feasible for binary 0/1 variables. The error hits as soon as you uncomment any of the Stope_i, Stope_j, Stope_ij, or Stope_im constraints, so those are definitely where we need to focus.

Why Those Constraints Break Feasibility #

Let’s take the Stope_i constraint as an example (the same logic applies to the others):

subject to Stope_i{g in P, h in Q}:
 x[g,h] - sum{j in h+1..h+q} x[g,j] = 1;

Since q = l-1 = 1, this simplifies to:
x[g,h] - x[g,h+1] = 1

Rearranged, that’s x[g,h] = 1 + x[g,h+1]. But x variables are binary (0 or 1)! If x[g,h+1] is 1, x[g,h] would need to be 2—which violates the <=1 bound on x. Even if x[g,h+1] is 0, this forces x[g,h] to be 1 for every g in P and h in Q—which would immediately conflict with the StopeB constraint (since multiple 1s would end up in the same 2x2 block).

The other constraints have identical issues:

• Stope_j simplifies to x[g,h] - x[g+1,h] = 1, requiring x[g,h] = 1 + x[g+1,h]—again impossible for binary variables.
• Stope_ij becomes x[g,h] - x[g+1,h+1] = 1, same feasibility conflict.
• Stope_im uses an invalid index range: with l=2 and m=8, when h=7 (part of Q which is 1..7), h+l=9 is way outside the J set (1..8)—this creates invalid variable references that break the model entirely.

Fixes to Get Your Model Feasible #

1. Rewrite Constraints to Match Your Actual Business Logic #

First, clarify what you’re trying to enforce with these Stope_* rules. For example:

• If you want to ensure the top-left cell of a block is only active if no other cells in that row/column are active, use an inequality instead of equality:

  # Example: x[g,h] can only be 1 if x[g,h+1] is 0
  subject to Stope_i{g in P, h in Q}:
   x[g,h] + x[g,h+1] <= 1;
  

• If you need exactly one cell active per block, replace StopeB with an equality constraint:

  subject to StopeB{g in P, h in Q}:
   sum{i in g .. g+p, j in h..h+q} x[i,j] = 1;
  

2. Fix Invalid Index Ranges #

For Stope_im, adjust the j range to stay within valid indices. For example, if you meant to reference cells to the right of the block, use:

subject to Stope_im{g in P, h in Q}:
 x[g,h] - sum{j in h+q+1..m} x[i,j] <= 0;

(Adjust the logic to match your intended rule—this example just ensures x[g,h] can’t be 1 if any cells to the right in the same row are active.)

3. Test Incrementally #

Start with the working base model (only StopeB), then add one modified Stope_* constraint at a time. Run the model after each addition to catch feasibility issues early.

内容的提问来源于stack exchange，提问作者ayePete