Python algorithm to approximate closest parallel equivalence of resistors from a list

The formula for series equivalence of resistors:

series equivalence = sum(resistors)

For parallel it is 1/(sum(1/resistors[i]))

I wrote code to return a list of resistors that is closest to a specified target value from a list, within a specified tolerance.

percentage_difference = lambda xi,xf: 100*(xf-xi)/xi

def series_equivalance(R,target,tolerance):
"""
R = list of resistors present
target = target value
tolerance = range += % of target that is acceptable 
This function returns a list of resistors
"""

tol = tolerance/100 #converting tolerance to decimal

if target < min(R):
    return "Your target is too small for series equivalence, Try parallel equivalence"

else:

    r = R #dummy/copy R
    toriginal = target #dummy values for arguments made to not change arguments 
    approximate = 0 #this is for exit condition, target in and of itself could be used but that would make algo unstable
    resistors_list = [] #list to return at the end
    
    while True: #Infinite loop because multiple exit conditions
    
        if (approximate >= (1-tol)*target and approximate <= (1+tol)*target)  :#exit condition
            break
        
        if len(R) == 0: #If all values are used up
            return "All values used up, list: {}, approximate: {}".format(resistors_list,series_sum(resistors_list))
 
        difference_from_target = [abs(toriginal-i) for i in R] #finding absolute difference of target from list of R values
        
        for i,v in enumerate(difference_from_target):
            if v == min(difference_from_target): #adding lowest differences to list
                approximate += r[i] #increment approximate by value from resistors with least difference
                toriginal -= r[i] #remove that from target dummy target
                resistors_list.append(r[i]) #adding to list to be returned
                r.remove(r[i]) 
                break
        
    return "Resistors to use are {}, Approximated value: {}, %Δ of {}%".format(resistors_list,sum(resistors_list),percentage_difference(target,int(sum(resistors_list))))
    

So for example series_equivalance([1,2,3,4,5],7,0)will return [5,2].

I want to a function that can do the same for parallel equivalence. How would I go about it?

Answer

Edit: I made a blog post which expands on the mip solution and can solve for minimum resistors satisfying a tolerance.

I’ve solved this two ways

1. Using your function and feeding it inverse values
2. Using a mixed integer linear program

Using your function and feeding it inverse values

This is a half way solution that just feeds in 1/R for the resistor values and gives it a target resistance of 1/target. Then, take reciprocal of the result and you have your resistor values. Doesn’t work with the tolerance value properly. Need to comment out the “Your target is too small” check for it to work.

def parallel_equivalance(R, target, tolerance):

    R_recip = [1/x for x in R]
    target_recip = 1/target
    tolerance_recip = tolerance  # TODO: have a think about how to handle this.

    result_recip = series_equivalance(R_recip, target_recip, tolerance_recip)
    # resistors_to_use = [1/x for x in result_recip]

print(parallel_equivalance([1, 2, 3, 4, 5, 6, 7], 1.5555, 1e-2)) gives Resistors to use are [5, 2], Approximated value: 7, %Δ of 0.0%

Using a mixed integer linear program

For this method I use a mixed integer linear program to pick out which resistors to use, such that (1/r1 + 1/r2 + …) is as close as possible to 1/target. Solves very quickly (<1s) even when given ten thousand resistors to choose from.

You could modify this to pick the least number of resistors (minimise sum(R_in_use)) with the constraint that the error value must be within some margin (say, error >= -eps and error <= +eps)

You’ll need to pip install mip

import mip


def parallel_equivalance(R, target):

    R_recip = [1/x for x in R]
    target_recip = 1/target

    m = mip.Model()  # Create new mixed integer/linear model.

    # Will take value of 1 when corresponding resistor is in use, otherwise 0.
    R_in_use = [m.add_var(var_type=mip.BINARY) for _ in R_recip]
    opt_r = sum([b * r for b, r in zip(R_in_use, R_recip)])  # This will be the optimal resistance
    error = opt_r - target_recip  # Want to minimise the absolute value of this error.

    # create a variable which is greater than than the absolute value of the error.
    # Because we will be minimizing, this will be forced down to equal the
    # absolute value. Common trick, google "linear programming absolute value".
    abs_eror = m.add_var(lb=0)
    m += abs_eror >= error
    m += abs_eror >= -1 * error

    # Objective of the optimisation is to minimise the absolute error.
    m.objective = mip.minimize(abs_eror)
    m.verbose = False  # Turn off verbose logging output.
    sol_status = m.optimize()
    print(sol_status)  # This should be `optimal`.

    # Get the solution values telling us which resistors are in use.
    R_in_use_sol = [float(v) for v in R_in_use]

    # Pick out the values of the resistors corresponding to the resistors
    # that the optimiser decided to use.
    R_to_use = [r for r, i in zip(R, R_in_use_sol) if i > 0]

    solved_resistance = 1/sum(1/x for x in R_to_use)
    solved_error = 100 * (solved_resistance - target) / target
    print(f'Resistors {R_to_use} in parallel will produce '
          f'R={solved_resistance:.3f}. '
          f'Aiming for R={target:.3f}, '
          f'error of {solved_error:.2f}%')
    return R_to_use


def main():
    print(f'mip version {mip.version}')
    sol = parallel_equivalance([1, 2, 3, 4, 5, 6, 7], 1.5555)
    sol = parallel_equivalance([1, 2, 3, 4, 5, 6, 7], 1.9)
    sol = parallel_equivalance(list(range(1, 100)), 123)
    sol = parallel_equivalance(list(range(1, 1000)), 5.954520294)
    sol = parallel_equivalance(list(range(1, 10_000)), 5.954520294)


if __name__ == '__main__':
    main()

mip version 1.13.0
OptimizationStatus.OPTIMAL
Resistors [2, 7] in parallel will produce R=1.556. Aiming for R=1.556, error of 0.00%
OptimizationStatus.OPTIMAL
Resistors [3, 5] in parallel will produce R=1.875. Aiming for R=1.900, error of -1.32%
OptimizationStatus.OPTIMAL
Resistors [99] in parallel will produce R=99.000. Aiming for R=123.000, error of -19.51%
OptimizationStatus.OPTIMAL
Resistors [27, 40, 41, 68, 69, 83, 123, 166, 172, 219, 277, 384, 391, 435, 453, 782, 837] in parallel will produce R=5.954. Aiming for R=5.955, error of -0.01%
OptimizationStatus.OPTIMAL
Resistors [7, 2001, 2021, 2065, 2130, 2152, 2160, 2176, 2191, 2202, 2216, 2245, 2270, 2279, 2282, 2283, 2313, 2342, 2351, 2381, 2414, 2417, 2497, 2728, 2789, 3449, 3514, 3566, 3575, 3621, 3701, 3789, 3812, 3868, 3879, 3882, 3903, 3936, 3952, 3959, 4128, 4145, 4152, 4158, 4183, 4373, 4382, 4430, 4441, 4498, 4525, 4678, 4722, 4887, 4953, 5138, 5178, 5253, 5345, 5358, 5543, 5593, 5620, 5774, 6002, 6247, 6364, 6580, 6715, 6740, 6819, 6904, 7187, 7293, 7380, 7468, 7533, 7782, 7809, 7846, 7895, 7914, 8018, 8067, 8242, 8309, 8414, 8507, 8515, 8590, 8627, 8872, 8893, 8910, 8952, 9171, 9282, 9311, 9376, 9477, 9550, 9657, 9736, 9792, 9822, 9876, 9982, 9988] in parallel will produce R=5.957. Aiming for R=5.955, error of 0.04%