# 等價製劑

">開始{align} \ max＆\ quad z = X_1 + X_2 + X_3 + X_4 \\\ text {st}＆\ quad y_1 + y_2 + y_3 + y_4 = 2\\＆\ quad X_1 \ leq y_1 \\ && \ quad X_2 \ leq y_1 + y_2 \\ && \ quad X_3 \ leq y_2 + y_3 \\ && \ quad X_4 \ leq y_1 + y_4 \\ && \ quad x，y\ in \ {0,1 \}。\ end {align}

">開始{align} \ max＆\ quad z = y_1 + X_2 + X_3 + X_4 \\\ text {st}＆\ quad y_1 + y_2 + y_3 + y_4 = 2\\＆\ quad X_2 \ leq y_1 + y_2 \\＆\\ quad X_3 \ leq y_2 + y_3 \\＆\\ quad X_4 \ leq y_1 + y_4 \\＆\ quad x，y \ in \ {0,1 \}\ end {align}

1. 優化的方向是最大化，
2. 目標函數係數為$$1$$
3. $$y$$的邊界是$$x$$的和
4. 變量都是二進制的。

This kind of depends on how one defines "equivalent", but in my opinion these formulations are not equivalent. Notice that in the original expression $$X_1$$ can be $$0$$ when $$y_1=1$$. By performing that substitution, you effectively fix that degree of freedom because now your objective will be incremented by $$1$$ if $$y_1=1$$, which was not necessarily the case in the original formulation.

While this might or might not give you the same optimal value, there is no guarantee that your active set will be the same (and although it's hard to tell without solving the problem in this case I suspect they will not be).

In terms of the polytope, you created a different polytope which possibly shares one or more vertices with the original one.