Is it possible to add an "explicit output" to an implicit component without extra computational effort compared to an explicit component? - openmdao

Question

While trying to figure out if the code can be simplified to avoid some duplication, I was wondering if it is possible to add an explicit output to an implicit component without adding extra computation effort compared to an explicit component. Explicit output may not be a fully correct term here though, since it depends on another output that is determined implicitly. Taking the node implicit component example from the docs:
class Node(om.ImplicitComponent):
"""Computes voltage residual across a node based on incoming and outgoing current."""
def initialize(self):
self.options.declare('n_in', default=1, types=int, desc='number of connections with + assumed in')
self.options.declare('n_out', default=1, types=int, desc='number of current connections + assumed out')
def setup(self):
self.add_output('V', val=5., units='V')
for i in range(self.options['n_in']):
i_name = 'I_in:{}'.format(i)
self.add_input(i_name, units='A')
for i in range(self.options['n_out']):
i_name = 'I_out:{}'.format(i)
self.add_input(i_name, units='A')
def setup_partials(self):
#note: we don't declare any partials wrt `V` here,
# because the residual doesn't directly depend on it
self.declare_partials('V', 'I*', method='fd')
def apply_nonlinear(self, inputs, outputs, residuals):
residuals['V'] = 0.
for i_conn in range(self.options['n_in']):
residuals['V'] += inputs['I_in:{}'.format(i_conn)]
for i_conn in range(self.options['n_out']):
residuals['V'] -= inputs['I_out:{}'.format(i_conn)]
When we would like to calculate the power going through the node, one option would be to create an explicit component that takes the node voltage and each of the node current in and outs as inputs to calculate the power, and group it with the implicit component. However, since all of the parameters are already available inside the implicit component, and this approach duplicates some current in/out loops between the components, I was wondering if this can be done directly within the implicit component. Since the docs example mentions "The solve_nonlinear method provides a way to explicitly define an output within an implicit component":
def solve_nonlinear(self, inputs, outputs):
total_abs_current = 0
for i_conn in range(self.options['n_in']):
total_abs_current += np.abs(inputs['I_in:{}'.format(i_conn)])
for i_conn in range(self.options['n_out']):
total_abs_current += np.abs(inputs['I_out:{}'.format(i_conn)])
outputs['P_total'] = total_abs_current * outputs['V'] / 2
Reading on further, the docs say it is still necesarry to also add a power residual under the apply_nonlinear() method. Hence, something like:
def apply_nonlinear(self, inputs, outputs, residuals):
residuals['V'] = 0
total_abs_current = 0
for i_conn in range(self.options['n_in']):
residuals['V'] += inputs['I_in:{}'.format(i_conn)]
total_abs_current += np.abs(inputs['I_in:{}'.format(i_conn)])
for i_conn in range(self.options['n_out']):
residuals['V'] -= inputs['I_out:{}'.format(i_conn)]
total_abs_current += np.abs(inputs['I_out:{}'.format(i_conn)])
residuals['P_total'] = outputs['P_total'] - total_abs_current * outputs['V'] / 2
But will the component actually use this function to "solve" for the power, even when solve_linear() specifies/calculates the power already explicitely? Will this implementation then therefore require more computational resources compared to the explicit component approach? And when specifying the partials through the linearize() method, should they follow the apply_nonlinear() or solve_nonlinear() calculation?

Answer 1

I typically call this kind of situation a pseudo-implicit output. You have an analytic expression so you don't really need it to be implicit, but you want to stick the calculation in with a bunch of other implicit stuff. You have the basic layout right. You write a solve_nonlinear method that does the calculation for you, and you add the residual form in the apply_nonlinear.
But will the component actually use this function to "solve" for the
power, even when solve_linear() specifies/calculates the power already
explicitly?
Yes .. and no :) The simple answer is that (in most cases) the solve_nonlinear method will ultimately provide the value for the pseudo-implicit output as part of the whole global nonlinear solve. The residual form will effectively always return 0 for that particular variable. This holds true if you are using a block gauss-seidel solver, or a newton solver with solve_subsystems turned on.
The more subtle situation happens if you use a pure newton method (without solve_subsystem). In that case the residual form is actually driving the entire calculation, and the solve_nonlinear method of any implicit component is not ever getting called. This is not a super common mode of running the newton solver, but it does come up often enough.
I would say that the pseudo-implicit output gives you the flexibility to work either way with no real loss of performance. As I'll discuss below, there isn't any practical difference between this and just breaking it out into an explicit component anyway.
Will this implementation then therefore require more computational
resources compared to the explicit component approach?
The short answer is no, at least not by any meaningful amount. The long answer requires diving into the math of newton solvers and understanding how OpenMDAO really does ExplicitComponents. For all the details, you should check out section 5.1 of the OpenMDAO paper along with the implicit transformation that OpenMDAO does internally for all ExplicitComponents.
In summary, explicit components in OpenMDAO do the exact same thing that you did in the apply_linear is what OpenMDAO does internally anyway when it needs to compute a residual. So your implementation doesn't really add anything more or less than OpenMDAO already does in the background.
residuals['P_total'] = outputs['P_total'] - total_abs_current * outputs['V'] / 2
There is one caveat here though. I'll exaggerate to make the situation clear. Lets say you had a single scaler implicit relationship in a component, that you then add 1e6 pseudo-implicit outputs to it as well. In that case, you are better of splitting them up because you're making the newton system a lot larger and more expensive. But generally, adding a few extra pseudo-explicit outputs won't make much of a difference at all.
When specifying the partials through the linearize() method,
should they follow the apply_nonlinear() or solve_nonlinear()
calculation?
Differentiate the apply_nonlinear. Don't worry about what you did in the solve_nonlinear at all in the context of derivatives for implicit components!