2019-11-12
Nov 13 In-Class Exercise Thread .
Post your solutions to the Nov 13 In-Class Exercise to this thread.
Best,
Chris
Post your solutions to the Nov 13 In-Class Exercise to this thread.
Best,
Chris
2019-11-13
Suppose we have 24 machines and we want to return the top 20 documents with probability at least .999.
For m is the # of docs, n is the # of machines:
Case 1:
For m = 20, n = 16, then we return 7 docs
Case 2:
For m = 20, n = 32, then we return 9 docs
Since n = 24 is half way from 16 to 32, the number of docs should each machine compute their best results till is:
(7 + 9) / 2= 8 docs
(Edited: 2019-11-13) Suppose we have 24 machines and we want to return the top 20 documents with probability at least .999.
For m is the # of docs, n is the # of machines:
Case 1:
For m = 20, n = 16, then we return 7 docs
Case 2:
For m = 20, n = 32, then we return 9 docs
Since n = 24 is half way from 16 to 32, the number of docs should each machine compute their best results till is:
(7 + 9) / 2= 8 docs
To return top 20 (m=20) documents with probability at least .999 we have
k = 9 docs when the number of machines is 16 (n=16)
k = 7 docs when the number of machines is 32 (n=32)
So, for 24 machines we have (7+9)/2 = 8 docs approximately, assuming the distribution shown in the graph.
To return top 20 (m=20) documents with probability at least .999 we have
k = 9 docs when the number of machines is 16 (n=16)
k = 7 docs when the number of machines is 32 (n=32)
So, for 24 machines we have (7+9)/2 = 8 docs approximately, assuming the distribution shown in the graph.
From the graph, to return m=20 top documents with probability at least .999,
when n=16, k is around 8.
when n=32, k is around 5.
Hence when n=24,k should be around (8+5)/2, which is roughly 7 documents should each
machine compute their best results till.
From the graph, to return m=20 top documents with probability at least .999,
when n=16, k is around 8.
when n=32, k is around 5.
Hence when n=24,k should be around (8+5)/2, which is roughly 7 documents should each
machine compute their best results till.
As per the graph,
For n=16,
Target result size, m = 20, k is around 7
For n=32,
Target result size, m = 20, k is around 5
So, when we have machines(n) = 24
Therefore, the number of documents each machine compute their best result =k = (7+5)/2 = 6 approx.
As per the graph,
For n=16,
Target result size, m = 20, k is around 7
For n=32,
Target result size, m = 20, k is around 5
So, when we have machines(n) = 24
Therefore, the number of documents each machine compute their best result =k = (7+5)/2 = 6 approx.
For n = 16, k = 9,
For n = 32, k = 7
So, for n (machines) = 24, each machine will compute their best result until k = (7+9)/2 = 8 documents approx.
For n = 16, k = 9,
For n = 32, k = 7
So, for n (machines) = 24, each machine will compute their best result until k = (7+9)/2 = 8 documents approx.
For top 20 (m=20) documents with probability at least .999,
k = 9 docs, when number of machines is 16 (n=16)
k = 7 docs, when number of machines is 32 (n=32)
Thus,
for 24 machines we have = (7+9)/2 = 8 docs approximately,
assuming the given graph distribution.
(Edited: 2019-11-13) For top 20 (m=20) documents with probability at least .999,
k = 9 docs, when number of machines is 16 (n=16)
k = 7 docs, when number of machines is 32 (n=32)
Thus,
for 24 machines we have = (7+9)/2 = 8 docs approximately,
assuming the given graph distribution.
2019-11-17
For n=16 and m = 20 -----> k = 9
For n=32 and m = 20 -----> k = 7
Hence for n =24 and m =20 ----> it should be approximately in middle i.e. k = 8
For n=16 and m = 20 -----> k = 9
For n=32 and m = 20 -----> k = 7
Hence for n =24 and m =20 ----> it should be approximately in middle i.e. k = 8
From the graph distribution ,
k=6 for n=16 and k=8 for n=32
Hence for n=24, the value of k=(6+8)/2 = 7 approximately.
From the graph distribution ,
k=6 for n=16 and k=8 for n=32
Hence for n=24, the value of k=(6+8)/2 = 7 approximately.
As per the graph, for n =16, k = 9 approximately.
For n= 32, k= 7 approximately.
Thus for n=24, k=(9+7)/2=8 approximately.
As per the graph, for n =16, k = 9 approximately.
For n= 32, k= 7 approximately.
Thus for n=24, k=(9+7)/2=8 approximately.
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