Post your solutions to the Nov. 28 In-Class Exercise to this Thread.
Best, Chris
These lines look like linear. When n =12, I can find the Diagonal of the n= 8 line and n = 16 line on the graph. use the x= 50. The answer is about 13 or 14.
(Edited: 2018-11-28)Answer - 15
assume a line between n = 8 and n = 16 for n = 12, from that we get k = 15 for m = 50
(Edited: 2018-11-28)The curve for n = 12 will be between the curves for n=8 and n=16 for p = 0.999. This curve meets m = 50 at approximately k = 14. So each machine should compute their best results till 14 documents.
Assuming the case n=12 lies between n=8 and n=16, m=50 on x-axis lies at yaxis '''15''' . So each machine compute their best result is between 13 - 15
(Edited: 2018-12-02)With m=50, p=0.999, and n=12, then k is about 14. This value was obtained by looking at the provided graph screenshot from the textbook and eyeballing the y-axis value at m=50 approximately half way between n=8 and n=16.
(Edited: 2018-11-28)from the figure, for n= 12, take approximation with even split of n=8 and n=16,
per-node retrieval depth k = ( 10 + 15) / 2 = 12.5, round up to 13
Each machine should return about 13 documents. At m = 50, n=8 returns about 15 documents. At m = 50, n=16 returns about 10 documents. Since n=12 is halfway between these two, (10+15)/2 = 12.5 -> 13
(Edited: 2018-11-28)If this scenario is plotted on the graph, the line would lie between the plots for n=8 and n= 16. By estimation, for 50 results to be returned ,each machine should return k=13 or 14 documents
Based off the chart presented, it's reasonable to assume that with 12 machines we would need each to return 12 to 13 results in order to get the top 50 results. The reasoning behind this is simple: it looks like we need 15 results per machine in the 8 machine case and 10 results per machine in the 16 machine case. Twelve is halfway between 8 and 16, so the amount of results should be around halfway between 10 and 15.