Listing 3

1000。045 1000。691 4999。001 5000。999

5000。000 5000。000

1050。440 1055。666 5049。001 5055。999

5050。000 5055。000

1700。089 1720。742 5699。001 5720。999

5700。000 5720。000

2010。130 2012。919 6009。001 6012。999

6010。000 6012。000

2055。714 2056。395 6054。001 6056。999

6055。000 6056。000

2560。952 2567。627 6559。001 6567。999

6560。000 6567。000

3000。324 3001。379 6999。001 7001。999

7000。000 7001。000

3005。255 3007。646 7004。001 7007。999

7005。000 7007。000

The relationship of the CVR event to the adjusted and raw FDR event is shown in Figure 4。 The rFDR points show the 1-second samples of microphone keyings by the FDR; in Figure 4 this is shown in points 2 and 3。 The aFDR points show the maximum possible extent of the microphone keyings given the FDR sample rate。 The aFDR points are infinitesimally close to the adjacent sampling frame。  The  CVRs    point  must  lie  somewhere    between

Figure 4。 Relationship of CVR, aFDR, and  rFDR。

aFDRs and rFDRs; and the CVRe point must lie somewhere between rFDRe  and aFDRe。

In the previous section, Step 1–Matching, the term perturbed test sets was introduced, Figure 4 permits an expanded discussion of this term。 A perfectly aligned CVR and FDR event is defined by CVRs 5 FDRs and CVRe 5 FDRe。 This type of perfect alignment has no slack—shifting the points left or right causes infeasibility。 A random perturbation transforms the perfectly aligned CVR event by CVR(s,e) ¡ e, where e is a random number greater than 0 and less than 1。 This simulates a near real world scenario considering the qualitative nature of CVR transmission identification。 The perfectly perturbed set is CVRs – e and CVRe + e。 Perfectly perturbed events represent the ideal CVR- FDR alignment problem, given the constraints defined in this paper。 While a perfectly perturbed and perfectly aligned set of events do have feasible solutions, a random perturbation will likely not have a feasible solution。

The point relationship shown in Figure 4 is the basis for a linear optimization model。 The objective function  is

Min

Min S ð12Þ

and the second time, Equation 5 is modified  to

Max S ð13Þ

The additional two runs push the solution towards a feasible, constrained solution either towards the left or right, as shown in Figure 5。 The terms left and right are used metaphorically for explanation purposes consistent with Figure 5。 The two additional solutions produce two additional displacements, S, for a total of three displace- ments。 The two additional displacements represent the extreme limits of all feasible solutions, with the solution containing Equation 59s objective function providing a balanced offset considering both start and end  times。