updates README

README.md

@@ -232,4 +232,4 @@ function values = solve_(K, H2Otot, NH3tot)
 Which provides a time of ~31 ms, with equpy giving more than a 15x enhancement.
 
 # Considerations on performance
-It is worth noting how a simple MATLAB implementation of equpy (see MATLAB folder) allows to solve the problem in 0.19 ms, for an extra 10x boost in execution time. Running the python implementation here presented from terminal allows for a reduction of execution time both for equpy and chempy of ~30%. It would not be suprising for a simple implementation of the Python algorithm relying on numba or cython to execute in a time comparable or faster than MATLAB.
+While we focused here on benchmarking performance of the approach by Thomas Wayne Wall, it is worth noting how more important than the absolute time is the relative enhancement with respect to more general solvers. A simple MATLAB implementation of equpy (see MATLAB folder) allows to solve the problem in 0.19 ms, for an extra 10x boost in execution time. Running the python implementation here presented from terminal allows for a reduction of execution time both for equpy and chempy of ~30%. It would not be suprising for a simple implementation of the Python algorithm relying on numba or cython to execute in a time comparable or faster than MATLAB.

equpy.py

@@ -132,10 +132,10 @@ def solve(
  """
 
  def plotter(self):
- f, (ax1, ax2) = plt.subplots(1, 2, sharey=False, figsize=(15, 5))
- ax1.plot(np.arange(len(self.residuals)), self.residuals, lw=3, color="black")
+ f, (ax1, ax2) = plt.subplots(1, 2, sharey=False, figsize=(10, 4.5))
+ ax1.plot(np.arange(len(self.residuals)), self.residuals, lw=5, color="black")
  ax1.scatter(
- np.arange(len(self.residuals)), self.residuals, s=120, color="black"
+ np.arange(len(self.residuals)), self.residuals, s=150, color="black"
  )
 
  ax1.set_xlabel("steps", fontsize=14, fontname="Arial")