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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

-- Nov 13 In-Class Exercise Thread

 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

-- Nov 13 In-Class Exercise Thread

 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.

-- Nov 13 In-Class Exercise Thread

 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.

-- Nov 13 In-Class Exercise Thread

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.

-- Nov 13 In-Class Exercise Thread

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.

-- Nov 13 In-Class Exercise Thread

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.

-- Nov 13 In-Class Exercise Thread

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

-- Nov 13 In-Class Exercise Thread

 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.

-- Nov 13 In-Class Exercise Thread

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.