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)
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)