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Factorial design (part 2)
Effects
Now we shall define a concept that is deceptively similar to
runs. We shall again consider the subsets of the letters. But
this time we shall not allow the empty subset, and shall use
capital letters. Thus we shall denote {A,C} by
AC. This is called an effect. There is no effect
called 1. Also, just like runs, we must preserve the
alphabetic order.
Effects with a single letter are called main
effects. Others are called interactions. An
interaction with 2 letters is called a 2-factor
interaction, and so on.
We can multiply two effects just like runs.
The product of two or more effects is called their generalised
interaction. This term is not used for runs, though
the multiplication rule is the same in the both the cases. Note
that a generalised interaction need not be an interaction. For
example, the generalised interaction of AB and A is
just B, which is a main effect. By the way, we cannot
multiply a effect with itself, as that would cancel all the
letters (remember that there is no effect called 1)!
The ``value'' of an effect for a run
This concept is best explained through an example. Take an effect
and a run: ABD and bcd, say. We shall think of the
run bcd as a formula: x2 + x3 + x4 (mod 2). The
effect will
provide values for the variables. Since A,B and D
are
present in ABD we shall take x1 = x2 = x4 = 1 and
x3 = 0. Then the ``value'' of ABD for
bcd is
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1+0+1 (mod 2) = 2 mod 2 = 0.
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Note that the value is always either 0 or 1. It is 0 iff the run
and the effect
have an even number (possibly 0) of letters in common.
| Example:
Find the following values: ABC for abc, A
for bc, BCD for 1.
Solution:

Of course, we could also count number of letters shared by the
effect and the run in each case.
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We shall say that a run clashes with an effect if the
value is 0. For
example, abc clashes with AB, but not with
ABC. The control run 1 clashes with every effect.
The terms ``value'' and ``clash'' are not standard jargon. We shall use them only
for explaining things.
Confounding
A replicate is said to confound an effect if the
effect clashes with all the runs in the control block of
the replicate.
| Example:
Consider the replicate with following control block in a (24 , 22 )
experiment
Does this replicate confound A? CD? ACD?
Solution:
No, it does not confound A because A does not clash
with a (they share an odd number, 1, of
letters).
CD is confounded as it clashes with all the
effects. It shares 0 letter with each run in the block. (0 is an
even number.)
ACD shares exactly one letter with a, and
so is not confounded.
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A replicate can confound more than one effect. It is easy to see
that if two effects are confounded then so is their generalised
interaction. For example, if AB and BC are both
confounded, then so must be AB × BC = AC. We often
call the AB and BC as independent confounded
effects, while AC is a dependent confounded
effect. Of course, we could also call AB and AC the
independent effects, and BC the dependent effect.
It is easy to find all the effects confounded by a given control
block.
Conversely, we can find the control block if we know all the
confounded effects. Just list all the clashing runs. In other
words, the control block consists of all runs sharing an even
number (possibly 0) of letters will all the confounded
effects.
In fact, it is enough to know all the independent confounded
effects.
Creating a replicate from the confounded effects
If we are given a list of all the (independent) effects to be
confounded we can construct the control
block, from which we can construct the entire replicate.
Thus it is meaningful to talk about the replicate
confounding a given list of effects.
Total, partial and balanced confounding
Almost always we shall work with more than one replicate. By an
(2n ,2r ) experiment we shall understand a collection of
such replicates. It is quite possible that an effect that is
counfounded in one replicate is not confounded in another. In
this case we say that the effect is partially confounded
in the experiment. If an effect is confounded in all the
replicates then it is totally confounded. If an effect is not
confounded in any replicate in an experiment then we say that the
experiment does not confound that effect.
We sometimes say that an experiment confounds all the 2-factor
interactions in a balanced way. By this we mean each
2-factor interaction is confounded in the same number of
replicates. For example, in a (23 , 22 ) experiment there
are 3 factors, and so the number of 2-factor interactions is
If we use 3 replicates, and use the
following confounding scheme

then we have balanced confounding of the 2-factor interactions,
as each 2-factor interaction is counfounded in the same number of
replicates (1, in this case).
Sometimes we want to confound both 2-factor and 3-factor
interactions in a balanced way. This means balancing the
2-factor interactions and then balancing the 3-factor
interactions separately: any two 2-factor interactions must be
confounded in an equal number of replicates. Similarly, any
3-factor interactions must be confounded in an equal number of
replicates. It is not required that a 2-factor interaction
is confounded in as many replicates as a 3-factor interaction.
Achieving balance
Suppose that we are required to design a (25 ,22 )
experiment such that
- all 3-factor interactions are balancedly confounded,
- no main effect or 2-factor interaction is confounded,
- the minimum possible number of replicates is used.
We start by listing all the 3-factor interactions:
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ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE.
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In each replicate we have to confound one or more of these. The
more of these we can confound in a single replicate, the better,
because then we shall need fewer replicates. However, we have to
careful about the generalised interactions which will also get
confounded. We must make sure that the generalised interactions
do not turn out to be main effects or 2-factor interactions. For
example, we cannot confound ABC and ABD in the same
replicate, as this would also confound their generalised
interaction CD, which is a 2-factor interaction!
Similarly, no two 3-factor interactions sharing 2 letters can be
confounded in the same replicate.
So we can can pair up the 3-factor interactions that share just a
single letter:

Notice that this requires some trial-and-error. The rest of the
process is now routine.
Construction (2-level)
| Exercise:
A 24 design has been conducted in 22 blocks of
size 4 each. One of the incomplete blocks, with usual
notation, is
Identify the confounded factorial effects and obtain the layout
of the other 3 incomplete blocks.
Give the analysis of the design.
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Show hint.
In a 24 experiment there are 24 = 16 runs:
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1,a,b,c,d,ab,ac,ad,bc,bd,abc,abd,acd,bcd,abcd.
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These are grouped into 4 blocks of size 4. One of the blocks is
given. We have to first find the control block, which is
the block that contains 1. For this pick any run in the given
block and multiply all the runs with it. Suppose we pick d
Then the control block is
To find the confounded effects we have to look for runs with an
even number of letters in common with all the runs in the
control block. These are
We already have 2 blocks, the given block and the control
block. So two more remain to be determined. Pick any run that is
not in these two blocks. Multiply the control block with
it. Suppose that we pick a. Multiplying the control block
with it gives another block:
The remaining block is (by elimination):
| Exercise:
Construct a (24 ,23 ) design with minimum number of
replicates achieving balance over 3-factor and 4-factor
interactions without confounding any main effect and 2-factor
interactions. (Give control blocks only).
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Show hint.
First let us start by listing all the 16 runs:
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1,a,b,c,d,ab,ac,ad,bc,bd,abc,abd,acd,bcd,abcd.
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Let us try to understand what we are supposed to do here. We are
to group these 16 into 2 blocks of size 8 each. Each such
grouping will confound some the effects. We have to do this
grouping a number of times (always 2 blocks of size 8 each, but
the contents of the blocks may differ). The criteria are that we
cannot confound the main effects and 2-factor interactions
(that is, A,B,C,D,AB,AC,AD,BC,BD) and each of
ABC,ABD,ACD,BCD should be confounded an equal number of
times.
Can we confound both ABC and ABD in a single
replicate? No, because then the generalised interaction BD
would also get confounded, which is not allowed. Similarly, no
two 3-factor interactions can be confounded in the same
replicate. So we need at least 4 replicates. Will that suffice?
Can we confound ABCD in the replicate where ABC is
confounded? No, because then the generalised interaction D
would get confounded, which is not allowed. So we need a separate
replicate for confounded ABCD. So we need 5 replicates in
all.
Let us construct the replicate where we confound ABC. We
need to mention only the control block, i.e., the block
with 1 in it. This block is obtained by considering all the runs
with an even number of letters in common with the confounded
effect ABC. These are:
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1, AB, AC, BC, D, ABD, ACD, BCD.
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Similarly, construct the control blocks for the replicates
confounding (separately) ABD, ACD and BCD.
The control block for the replicate confounding ABCD is
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1, AB, AC, AD, BC, BD, CD, ABCD.
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| Exercise:
Construct a balanced (25 ,22 ) design with minimum number
of replications achieving balance over 3-factor and 4-factor
interactions without confounding main effects and 2-factor
interactions. (Give control blocks only.)
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Show hint.
Can you confound in a replicate two 4-factor ineractions? No.
So we need at least 5 replicates. Is that enough? The answer is
yes because there are ten 3-factor interactions, and we can
distribute these 10 over the 5 replicates as follows:
Replicate & 3-factor & 4-factor
\hline
1 & ABE,CDE& ABCD
2 & ACD,BDE& ABCE
3 & ACE,BCD& ABDE
4 & ABD,BCE& ACDE
5 & ABC,ADE& BCDE
We shall construct the control block for replicate 1. It consists
of all runs sharing even numbers of letters with each of ABE,
CDE and ABCD. It is enough to check for ABE and
CDE because ABCD is their generalised interaction.
The control block is
Construct the control blocks for the other replicate similarly.
| Exercise:
In a (25 ,23 ) design only four treatments
(1),bc,de,abc are known. Find the other treatments of the
block. Also find the confounded effects.
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Show hint.
Just form all the products that are not already present.
We have to identify the runs that have an even number of
letters in common with the all the 8 runs in the block. Of
course, it is enough to check with the 4 given runs, as the
others are their products. These are
These are the confounded effects.
| Exercise:
In a (25 ,23 ) design one of the blocks contains
treatments (1), bc, de, abe. Find the other treatments of
the block. Also find the confounded effects.
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Show hint.
The remaining 4 runs are
The confounded effects are
| Exercise:
For a (25 ,23 ) design with the factors A,B,C,D,
and E some treatments in one of the blocks are given by
a,b,bde,ce. Identify the confounded effects and give the
complete allocation for all the blocks.
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Show hint.
The control block contains at least the following runs
The other effects in the control block must be
So the control block is
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1, ab, abde, ace, de, bce, bcd, acd.
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The other 3 blocks are
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a,b,bde,ce,ade,abce,abcd,cd,
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c,abc,adcde,ae,cde,be,bd,ad
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and
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d, abd, abe, acde, e, bcde, bc, ac.
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| Exercise:
In a (25 ,23 ) design only 4 treatments in a block are
known (1),bc,de,abc. Find the other treatments of the
block. Find the confounded effects.
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Show hint.
The given block contains (1), and so is the control block. The
other runs in it are
The confounded effects are
| Exercise:
Discuss total and partial confounding in conection with factorial
experiments. What do you mean by balancing in a confounded design
with more than one replicate? Construct a (24 ,23 ) design
with minimum number of replicates achieving balance over 3-factor
and 4-factor interactions without confounding any main effect and
2-factor interactions.
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Show hint.
All 3-factor interactions and the 4-factor interaction must be
confounded in separate replicates. So we need 5 replicates. The
control block for confounding ABC is
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1, AB, AC, BC, D, ABD, ACD, BCD.
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The control block for confounding ABCD is
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1, AB, AC, BC, AD, BD, CD, ABCD.
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