PROC PLAN: Specifying Factor Structures :: SAS/STAT(R) 9.22 User's Guide

The PLAN Procedure

Specifying Factor Structures

By appropriately combining features of the PLAN procedure, you can construct an extensive set of designs. The basic tools are the factor-selections, which are used in the FACTORS and TREATMENTS statements. Table 65.1 summarizes how the procedure interprets various factor-selections (assuming that the ORDERED option is not specified in the PROC PLAN statement).

Table 65.1 *Factor-Selection* Interpretation
Form of
Request	Interpretation	Example	Results
name= $\text{[math]}$	produce a random permutation of the integers $\text{[math]}$	`t=15`	lists a random ordering of the numbers $\text{[math]}$
name= $\text{[math]}$ cyclic	cyclically permute the integers $\text{[math]}$	`t=5 cyclic`	selects the integers 1 to 5. On the next iteration, selects 2,3,4,5,1; then 3,4,5,1,2; and so on.
name= $\text{[math]}$ of $\text{[math]}$	choose a random sample of $\text{[math]}$ integers (without replacement) from the set of integers $\text{[math]}$	`t=5 of 15`	lists a random selection of 5 numbers from 1 to 15. First, the procedure selects 5 numbers and then arranges them in random order.
name= $\text{[math]}$ of $\text{[math]}$ ordered	has the same effect as name= $\text{[math]}$ ordered	`t=5 of 15` `ordered`	lists the integers 1 to 5 in increasing order (same as t=5 ordered)
name= $\text{[math]}$ of $\text{[math]}$ cyclic	permute $\text{[math]}$ of the $\text{[math]}$ integers	`t=5 of 30` `cyclic`	selects the integers 1 to 5. On the next iteration, selects 2,3,4,5,6; then 3,4,5,6,7; and so on. The 30th iteration produces 30,1,2,3,4; the 31st iteration produces 1,2,3,4,5; and so on.
name= $\text{[math]}$ perm	produce a list of all permutations of $\text{[math]}$ integers	`t=5 perm`	lists the integers 1,2,3,4,5 on the first iteration; on the second lists 1,2,3,5,4; on the 119th iteration lists 5,4,3,1,2; and on the last (120th) lists 5,4,3,2,1.
name= $\text{[math]}$ of $\text{[math]}$ comb	choose combinations of $\text{[math]}$ integers from $\text{[math]}$ integers	`t=3 of 5` `comb`	lists all combinations of 5 choose 3 integers. The first iteration is 1,2,3; the second is 1,2,4; the third is 1,2,5; and so on until the last iteration 3,4,5.
name= $\text{[math]}$ of $\text{[math]}$ cyclic (initial-block)	permute $\text{[math]}$ of the $\text{[math]}$ integers, starting with the values specified in the initial-block	`t=4 of 30` `cyclic` `(2 10 15 18)`	selects the integers 2,10,15,18. On the next iteration, selects 3,11,16,19; then 4,12,17,20; and so on. The thirteenth iteration is 14,22,27,30; the fourteenth iteration is 15,23,28,1; and so on.
name= $\text{[math]}$ of $\text{[math]}$ cyclic (initial-block) increment	permute $\text{[math]}$ of the $\text{[math]}$ integers. Start with the values specified in the initial-block, then add the increment to each value.	`t=4 of 30` `cyclic` `(2 10 15 18)` `2`	selects the integers 2,10,15,18. On the next iteration, selects 4,12,17,20; then 6,14,19,22; and so on. The wrap occurs at the eighth iteration. The eighth iteration is 16,24,29,2; and so on.

In Table 65.1, in order for more than one iteration to appear in the plan, another name=j factor selection (with $\text{[math]}$ ) must precede the example factor selection. For example, the following statements produce six of the iterations described in the last entry of Table 65.1.

proc plan;
   factors c=6 ordered t=4 of 30 cyclic (2 10 15 18) 2;
run;

The following statements create a randomized complete block design and output the design to a data set.

proc plan ordered;
   factors blocks=3 cell=5;
   treatments t=5 random;
   output out=rcdb;
run;

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