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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def calc_actualremoterelieve_v1(self):
 """Calculate the actual amount of water released to a remote location to relieve the dam during high flow conditions. Required control parameter: |RemoteRelieveTolerance| Required flux sequences: |AllowedRemoteRelieve| |PossibleRemoteRelieve| Calculated flux sequence: |ActualRemoteRelieve| Basic equation - discontinous: :math:`ActualRemoteRelease = min(PossibleRemoteRelease, AllowedRemoteRelease)` Basic equation - continous: :math:`ActualRemoteRelease = smooth_min1(PossibleRemoteRelease, AllowedRemoteRelease, RemoteRelieveTolerance)` Used auxiliary methods: |smooth_min1| |smooth_max1| Note that the given continous basic equation is a simplification of the complete algorithm to calculate |ActualRemoteRelieve|, which also makes use of |smooth_max1| to prevent from gaining negative values in a smooth manner. Examples: Prepare a dam model: Prepare a test function object that performs seven examples with |PossibleRemoteRelieve| ranging from -1 to 5 m³/s: We begin with a |AllowedRemoteRelieve| value of 3 m³/s: Through setting the value of |RemoteRelieveTolerance| to the lowest possible value, there is no smoothing. Instead, the relationship between |ActualRemoteRelieve| and |PossibleRemoteRelieve| follows the simple discontinous minimum function: | ex. | possibleremoterelieve | actualremoterelieve | | 1 | -1.0 | 0.0 | | 2 | 0.0 | 0.0 | | 3 | 1.0 | 1.0 | | 4 | 2.0 | 2.0 | | 5 | 3.0 | 3.0 | | 6 | 4.0 | 3.0 | | 7 | 5.0 | 3.0 | Increasing the value of parameter |RemoteRelieveTolerance| to a sensible value results in a moderate smoothing: | ex. | possibleremoterelieve | actualremoterelieve | | 1 | -1.0 | 0.0 | | 2 | 0.0 | 0.0 | | 3 | 1.0 | 0.970639 | | 4 | 2.0 | 1.89588 | | 5 | 3.0 | 2.584112 | | 6 | 4.0 | 2.896195 | | 7 | 5.0 | 2.978969 | Even when setting a very large smoothing parameter value, the actual remote relieve does not fall below 0 m³/s: | ex. | possibleremoterelieve | actualremoterelieve | | 1 | -1.0 | 0.0 | | 2 | 0.0 | 0.0 | | 3 | 1.0 | 0.306192 | | 4 | 2.0 | 0.634882 | | 5 | 3.0 | 1.037708 | | 6 | 4.0 | 1.436494 | | 7 | 5.0 | 1.788158 | Now we repeat the last example with a allowed remote relieve of only 0.03 m³/s instead of 3 m³/s: | ex. | possibleremoterelieve | actualremoterelieve | | 1 | -1.0 | 0.0 | | 2 | 0.0 | 0.0 | | 3 | 1.0 | 0.03 | | 4 | 2.0 | 0.03 | | 5 | 3.0 | 0.03 | | 6 | 4.0 | 0.03 | | 7 | 5.0 | 0.03 | The result above is as expected, but the smooth part of the relationship is not resolved. By increasing the resolution we see a relationship that corresponds to the one shown above for an allowed relieve of 3 m³/s. This points out, that the degree of smoothing is releative to the allowed relieve: | ex. | possibleremoterelieve | actualremoterelieve | | 1 | -0.01 | 0.0 | | 2 | 0.0 | 0.0 | | 3 | 0.01 | 0.003062 | | 4 | 0.02 | 0.006349 | | 5 | 0.03 | 0.010377 | | 6 | 0.04 | 0.014365 | | 7 | 0.05 | 0.017882 | One can reperform the shown experiments with an even higher resolution to see that the relationship between |ActualRemoteRelieve| and |PossibleRemoteRelieve| is (at least in most cases) in fact very smooth. But a more analytical approach would possibly be favourable regarding the smoothness in some edge cases and computational efficiency. """

    con = self.parameters.control.fastaccess
    flu = self.sequences.fluxes.fastaccess
    d_smoothpar = con.remoterelievetolerance*flu.allowedremoterelieve
    flu.actualremoterelieve = smoothutils.smooth_min1(
        flu.possibleremoterelieve, flu.allowedremoterelieve, d_smoothpar)
    for dummy in range(5):
        d_smoothpar /= 5.
        flu.actualremoterelieve = smoothutils.smooth_max1(
            flu.actualremoterelieve, 0., d_smoothpar)
        d_smoothpar /= 5.
        flu.actualremoterelieve = smoothutils.smooth_min1(
            flu.actualremoterelieve, flu.possibleremoterelieve, d_smoothpar)
    flu.actualremoterelieve = min(flu.actualremoterelieve,
                                  flu.possibleremoterelieve)
    flu.actualremoterelieve = min(flu.actualremoterelieve,
                                  flu.allowedremoterelieve)
    flu.actualremoterelieve = max(flu.actualremoterelieve, 0.)