get value from linear expressions

mip/entities.py

@@ -60,6 +60,16 @@ class LinExpr:
     .. code:: python
 
      m += x1 + x2 + x3 == 1
+
+    If used in intermediate calculations, the solved value of the linear
+    expression can be obtained with the ``x`` parameter, just as with
+    a ``Var``.
+
+    .. code:: python
+
+     a = 10*x1 + 7*x4
+     print(a.x)
+
     """
 
     def __init__(
@@ -348,6 +358,18 @@ def violation(self: "LinExpr"):
 
         return viol
 
+    @property
+    def x(self) -> Optional[numbers.Real]:
+        """Value of this linear expression in the solution. None
+        is returned if no solution is available."""
+        x = self.__const
+        for var, coef in self.__expr.items():
+            var_x = var.x
+            if var_x is None:
+                return None
+            x += var_x * coef
+        return x
+
 
 class Constr:
     """ A row (constraint) in the constraint matrix.

test/mip_test.py

@@ -591,3 +591,31 @@ def test_variable_bounds(solver: str, val: int):
     assert round(m.objective_value) == val
     assert round(x.x) == 2 * val
     assert round(y.x) == val
+
+
+@pytest.mark.parametrize("val", range(1, 4))
+@pytest.mark.parametrize("solver", SOLVERS)
+def test_linexpr_x(solver: str, val: int):
+    m = Model("bounds", solver_name=solver)
+
+    x = m.add_var(lb=0, ub=2 * val)
+    y = m.add_var(lb=val, ub=2 * val)
+    obj = x - y
+
+    assert obj.x is None  # No solution yet.
+
+    m.objective = maximize(obj)
+    m.optimize()
+
+    assert m.status == OptimizationStatus.OPTIMAL
+    assert round(m.objective_value) == val
+    assert round(x.x) == 2 * val
+    assert round(y.x) == val
+
+    # Check that the linear expression value is equal to the same expression
+    # calculated from the values of the variables.
+    assert abs((x + y).x - (x.x + y.x)) < TOL
+    assert abs((x + 2*y).x - (x.x + 2*y.x)) < TOL
+    assert abs((x + 2*y + x).x - (x.x + 2*y.x + x.x)) < TOL
+    assert abs((x + 2*y + x + 1).x - (x.x + 2*y.x + x.x + 1)) < TOL
+    assert abs((x + 2*y + x + 1 + x/2).x - (x.x + 2*y.x + x.x + 1 + x.x/2)) < TOL