How does Labview measure dt when combining waveforms? Possible mistake in the NI examples

In the example project NI 9227 Getting Started.lvproj, the example shows how to read a buffer of single point values from all 4 channels. For the host VI, the dt is calculated from parsing the string enumerator and multiplying by 1000 to get the samples per second. The text "2.00 kS/s (NI 9227)" should parse then to 2000 which gives us a dt of 1/2000 or 0.0005.

However the documentation shows that the enumerator is per channel, not all 4.

This leaves me with 2 questions:

1. I want to know if the true maximum sample rate with all 4 channels active, would it be 2.5 kS/s? And the correct dt would be 0.0020 instead?
   
2. In terms of absolute time, would each the 4 decimated arrays have to be shifted to be properly centered? See:
   1. 0.0005*(0.5)
   2. 0.0005*1 + 0.0005*0.5
      
   3. 0.0005*2 + 0.0005*0.5
   4. 0.0005*3 + 0.0005*0.5

The NI Example project can be found under "Hardware Input and Output \ CompactRIO \ Module Specific IO \ Analog Input \NI 9227 Getting Started.lvproj "