ParSA Results

Interferograms
The set of images in Fig. 1 and Fig. 2 depict the test interferogram analyzed and the best interferogram solution found by simulated annealing. When compared visually, the two interferograms are nearly indistinguishable. This comparison shows that the solution found by simulated annealing is a good one.

Surfaces
Fig. 3 and Fig. 5 show the front and back surfaces of the diamond wafer used to create the test interferogram depicted in Fig. 1. Fig. 4 and Fig. 6 show the solutions found by the ParSA algorithm which generated the interferogram shown in Fig. 2.  The incident light used to create each of the interferograms is incident in the z-direction. The test diamond surfaces are the same shape and are apparently planar, with the exception that they have a small contribution from a second order Legendre polynomial. This slight curvature can be seen in the complex dark and light band structure depicted in Fig. 1. If the surfaces were actually planar as they appear, the resulting interferogram would not have these rich features and would simply be a series of straight, equally-spaced, light and dark bands.

The solutions found by simulated annealing feature two ambiguities. The first ambiguity results from the fact that a mirror transformation of the surfaces about a vertical plane do not change the interferogram. The second ambiguity is a result of the periodic nature of light and can be seen in differences present in the distances between surfaces between the test and solution surfaces.

MIR Performance
The ParSA library documentation gives the following equation which can be used to estimate performance of an MIR run

where P is the probability of non-convergence, $$\chi_n$$ is a solution of a run of length n, $$Cost_{min}$$ is the minimum acceptable solution, and K and $$\alpha$$ are problem specific parameters. K and $$\alpha$$ can be determined by plotting the Bayesian estimator for P versus n on a log scale and determining the slope and y-intercept. The expression for the Bayesian estimator $$ \hat{p} $$ is given by

where $$n_f$$ is the number of runs whos cost function value is greater than $$Cost_{min}$$ and N is the total number of runs.

 alpha = 0.5 

Fig. 7 shows the log plot of probability of non-convergence versus run length for a MIR run with the alpha cooling parameter set to 0.5. The slope and y-intercept of the linear least-squares fit are $$-0.0273 \pm 0.0103$$ and $$0.297 \pm 0.127$$ with a reduced $$\chi^2$$ value of 1.0. K and $$\alpha$$ were then determined to be $$(5.31 \pm 2.4) \times 10^4$$ and $$0.0273 \pm 0.0103$$.

 alpha = 0.9 

Fig. 8 shows the log plot for a MIR run with the cooling parameter alpha set to 0.9. The slope and y-intercept of the linear least-squares fit are $$0.0047 \pm 0.0060$$ and $$-0.070 \pm 0.074$$ with a reduced $$\chi^2$$ value of 0.55. K and $$\alpha$$ were then determined to be $$(0.03 \pm 1.2) \times 10^8$$ and $$-0.0047 \pm 0.0060$$. Runs of greater lengths will be needed to determine these parameters for this value of alpha.

Sequential Performance
Using the K and $$\alpha$$ parameters determined for the alpha = 0.5 run, the number of steps that would be needed to achieve $$50\%$$ probability of convergence for a sequential run can be determined to be on the order of $$10^{19}$$. It would take a single processor 500 million years to take this number of steps!

Optimizing Run Length
The optimal run length can be determined now that the K and $$\alpha$$ parameters are known. Given that the the total number of steps T is given by the following expression

where R is the number of runs and n is the run length. The probability of obtaining one successful run out of R runs is given by

Using the method of Lagrange multipliers, the optimal step number can be found to be

$$ n = K e $$

and given the K parameter determined from the alpha = 0.5 run, the optimal step number can be determined to be $$ (1.44 \pm 0.65) \times 10^5 $$.