Description Usage Arguments Value Author(s) Examples
Let us consider a module M of residue class mod(5) having elements 0, 1, 2, 3, 4 and all the elements of M are assigned to each of the n >= 2 classes. This function constructs PBIB designs with the following parameters:
v = 5n, b = 5n, r = 2(n + 1), k = 2(n + 1)
lambda 1 = n + 2, lambda 2 = n + 2, lambda 3 = 3, lambda 4 = 2, lambda 5 = 2n
1 | series3(n)
|
n |
n is the number of classes to which the elements of Module M are assigned |
The function returns the required PBIB design with specified parameters
Parneet Kaur, Davinder Kumar Garg
1 | series3(5)
|
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 1 2 3 4 7 12 17 22 8 13 18 23
[2,] 6 7 8 9 2 12 17 22 3 13 18 23
[3,] 11 12 13 14 2 7 17 22 3 8 18 23
[4,] 16 17 18 19 2 7 12 22 3 8 13 23
[5,] 21 22 23 24 2 7 12 17 3 8 13 18
The Parameters of the design are:
v = 25 b = 25 r = 12 k = 12
lambda[ 1 ] = 7 lambda[ 2 ] = 7 lambda[ 3 ] = 3 lambda[ 4 ] = 2 lambda[ 5 ] = 10
The developed blocks are:
( 1 2 3 4 7 8 12 13 17 18 22 23 )
( 2 3 4 5 8 9 13 14 18 19 23 24 )
( 3 4 5 1 9 10 14 15 19 20 24 25 )
( 4 5 1 2 10 6 15 11 20 16 25 21 )
( 5 1 2 3 6 7 11 12 16 17 21 22 )
( 6 7 8 9 2 3 12 13 17 18 22 23 )
( 7 8 9 10 3 4 13 14 18 19 23 24 )
( 8 9 10 6 4 5 14 15 19 20 24 25 )
( 9 10 6 7 5 1 15 11 20 16 25 21 )
( 10 6 7 8 1 2 11 12 16 17 21 22 )
( 11 12 13 14 2 3 7 8 17 18 22 23 )
( 12 13 14 15 3 4 8 9 18 19 23 24 )
( 13 14 15 11 4 5 9 10 19 20 24 25 )
( 14 15 11 12 5 1 10 6 20 16 25 21 )
( 15 11 12 13 1 2 6 7 16 17 21 22 )
( 16 17 18 19 2 3 7 8 12 13 22 23 )
( 17 18 19 20 3 4 8 9 13 14 23 24 )
( 18 19 20 16 4 5 9 10 14 15 24 25 )
( 19 20 16 17 5 1 10 6 15 11 25 21 )
( 20 16 17 18 1 2 6 7 11 12 21 22 )
( 21 22 23 24 2 3 7 8 12 13 17 18 )
( 22 23 24 25 3 4 8 9 13 14 18 19 )
( 23 24 25 21 4 5 9 10 14 15 19 20 )
( 24 25 21 22 5 1 10 6 15 11 20 16 )
( 25 21 22 23 1 2 6 7 11 12 16 17 )
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