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: ""})]