Evaluating Solver Results

This section explains how to evaluate decision variables, objective functions, and constraints from the results of solver runs returned by the solve() function.

Evaluating decision variables

For the decision variable array used in the formulation, you may want to obtain the values of the variables in the solution contained in the solver’s execution results as a NumPy array with the same shape as the decision variable array. First, as an example, we will perform a solver run using FixstarsClient on a model consisting of an objective function and constraints.

from datetime import timedelta
from amplify import VariableGenerator, equal_to, FixstarsClient, solve

# Create an array of decision variables
gen = VariableGenerator()
q = gen.array("Binary", 5)

# Create an objective function and a constraint
objective = q[0] * q[1] - q[2]
constraint = equal_to(q[0] + q[1] + q[2], 1)

# Define a model
model = objective + constraint

# Create a solver client
client = FixstarsClient()
# client.token = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
client.parameters.timeout = timedelta(milliseconds=1000)

# Obtaining the result of the run
result = solve(model, client)

You can evaluate the values of the best solution in an array of decision variables by passing the values attribute of the Solution object to PolyArray’s evaluate() method as follows.

Tip

The solution in the result can be obtained by retrieving the best result of the run with the best attribute or by accessing the elements in the same way as for a list.

>>> print(result.best.values)
{q_0: 0, q_1: 0, q_2: 1}
>>> q_values = q.evaluate(result.best.values)
>>> print(q_values)
[0. 0. 1. 0. 0.]

If a variable not used in the formulation is included, as in q[3] or q[4] above, it is assigned one of its possible values by default. In the above example, the Amplify SDK assigns 0 as the default value for the binary variable.

The default keyword argument to the evaluate() method can change the value used if no evaluation is performed; if the default keyword argument is given as a number, variables not passed to the solver are assigned that value.

>>> q_values = q.evaluate(result.best.values, default=3)    
Warning: Substituting variable q_3 with 3 is out of bounds.
Warning: Substituting variable q_4 with 3 is out of bounds.
>>> print(q_values)
[0. 0. 1. 3. 3.]

Note

A warning is printed if a value outside the bounds of the variable is given, such as default=3, as shown above.

If the default keyword argument is None, variables not passed to the solver remain as they are. Only in this case the evaluate() method return a PolyArray.

>>> q_values = q.evaluate(result.best.values, default=None)
>>> print(q_values)
[  0,   0,   1, q_3, q_4]

Attention

Some variables may not be passed to the solver even if they are included in the model. This is because model conversions such as penalty function generation or graph embedding can cause terms to cancel each other out.

Polynomial evaluation

The result of evaluating the objective function of the input model with the solution returned by the solve() function can be obtained using the objective attribute of the Solution object.

>>> solution = result.best
>>> solution.objective
-1.0

On the other hand, there are cases where you want to evaluate a polynomial other than the objective function with the solution returned by the solver, for example, when the objective function is expressed as the sum of several polynomials. In this case, we pass the values attribute of the Solution object to Poly’s evaluate() method, just as we would evaluate an array of variables.

>>> objective_1 = q[0] * q[1] # The first term of the objective function
>>> objective_1.evaluate(solution.values)
0.0

Hint

The behavior when the polynomial contains variables not passed to the solver is similar to the evaluate() method of the PolyArray class; the default keyword argument can change this behavior.

Evaluating constraints

If you want to know if the resulting solution satisfies the constraints, you can check with the feasible attribute of the Solution class.

>>> solution.feasible
True

By default, this will always be True because the solve() function retrieves only solutions that satisfy all constraints in the model. To change this behavior and allow the solver to retrieve solutions that do not satisfy the constraints, pass a bool to the filter keyword argument of the solve() function upfront or set filter_solution of the Result class to False afterward.

Let’s test this by adding a constraint to the model that cannot be satisfied inconsistently.

# Create an objective function and constraints
objective = q[0] * q[1] - q[2]
constraint1 = equal_to(q[0] + q[1] + q[2], 1)
constraint2 = equal_to(q[0] + q[1] + q[2], 2)

# Define a model (with conflicting constraints)
model = objective + constraint1 + constraint2

# Get the result of the run
result = solve(model, client)

You cannot retrieve the solution from Result if the solution filter is enabled.

>>> result.best.feasible
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
RuntimeError: result has no feasible solution

Turning off the solution filter allows us to obtain solutions that do not satisfy the constraints.

>>> result.filter_solution = False
>>> result.best.feasible
False

Suppose you obtained a solution that does not satisfy a constraint for some reason, such as model configuration, penalty function weights, or solver settings. In that case, it may be desirable to determine which constraint was not satisfied.

You can check whether the constraint conditions are satisfied by passing the solution returned by the solve() function to the is_satisfied() method of the Constraint class.

>>> constraint1.is_satisfied(result.best.values)
True
>>> constraint2.is_satisfied(result.best.values)
False

In the example above, we see that constraint2 could not be satisfied.

We can also mechanically identify constraints from the list of constraints in the model that are not satisfied, as follows. This method is useful when adjusting and rerunning the penalty function weights.

>>> list(c for c in model.constraints if c.is_satisfied(result.best.values))
[Constraint({conditional: q_0 + q_1 + q_2 == 1, weight: 1, label: ""})]