2. Examples


2.1 Twelve Treatments in Six Blocks of Six

This is the actual problem solved for the Chalco experiments being initiated at the beginning of April 1990.

Treatments:
Twelve, being three sowing dates x four varieties


 V1 V2 V3 V4


DI 1 2 3 4
D2 5 6 7 8
D3 9 10 11 12


The factorial structure is not important for the design of this experiment. All comparisons between
treatment combinations are important. Comparisons between treatments (1,2,5,6,7,8,11 and 12) are
expected to be more important as these combinations are expected to be more successful.

Replicates and Blocks:
Each replicate is to occupy a 4 x 3 grid of plots. The plots are approximately square. It is thought that the
total area of each replicate may be rather too large to be properly homogeneous and that using two blocks
of 2 x 3 within each replicate might provide greater homogeneity within blocks (and correspondingly
differences between blocks within each replicate). If the differences between blocks within replicates turn
out to be negligible the analysis can revert to that for a RCBD since each replicate is considered as a whole
as well as being split into two blocks.

Design:
We wish to divide the twelve treatments into two groups(blocks) of six in each replicate in such a way that
the divisions are as different as possible in the three replicates. The choice is too wide and we may find it
helpful to use the two-factor structure to start the design. For the first replicate let the division be such that
each block includes two combinations for each date and at least one combination for each variety. We try

 Block 1 (1,2,7,8,9,11) Block 2 (3,4,5,6,10,12)

Now the next pair of blocks must each include three combinations from block 1 and three from block 2. It
is still helpful to use the date and variety structure. So we try

 Block 3 (1,3,6,8,11,12) Block 4 (2,4,5,7,9,12)

Quite a number of treatment pairs have not yet occurred together in a block (1 and 4, 2 and 3, 5 and 8, 6
and 7 and so on). We try to remember to include these as well as splitting treatments in the third replicate
so that each block includes three from each of blocks 1, 2, 3 and 4. There are still plenty of good solutions
and we try

 Block 5 (1,4,6,7,9,10) Block 6 (2,3,5,8,11,12)