2016-05-04
May 4 Thread .
Post your answers to Today's Let's Experiment to this thread.
Best, Chris
Post your answers to Today's Let's Experiment to this thread.
Best,
Chris
docs = 50 for n = 8 : k = 15 for n = 4 : k = 25 So by linear interpolation: (((6 - 4)*(25 - 15))/(8 - 4)) + 15 for n = 6 : k = 20
(Edited: 2016-05-04) docs = 50
for n = 8 : k = 15
for n = 4 : k = 25
So by linear interpolation:
(((6 - 4)*(25 - 15))/(8 - 4)) + 15
for n = 6 : k = 20
target size, m= 50 per node retrieval depth (k) should be: (for n = 4)k= 25. (for n = 8), k=15
Averaging (for 6 machines), k = 20
(Edited: 2016-05-04) target size, m= 50
per node retrieval depth (k) should be:
(for n = 4)k= 25.
(for n = 8), k=15
Averaging (for 6 machines),
k = 20
between 15-25. The line should be between the line with n = 4 and n = 8, then find the value when m = 50
between 15-25. The line should be between the line with n = 4 and n = 8, then find the value when m = 50
For size=50 By observing graph, For m=4 we get 25 For m=8 we get 15 By considering this and doing interpolation, For m=6 we will get the answer around 20.
For size=50 By observing graph, For m=4 we get 25 For m=8 we get 15
By considering this and doing interpolation, For m=6 we will get the answer around 20.
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