Blocking and confounding in replicated $2^k$ factorial design Consider the $2^6$ factorial in 8 blocks of eight each runs with ABCD, ACE and ABEF as the independent effects chosen to be confounded with blocks. Generate a design. Find the other effect of confounding.
I can solve the second part of the question. I found the solution of the first part in a solutions manual (question 7-11) for Montgomery's textbook on the Design and Analysis of Experiments (Wiley, 2001), but I could not understand this table. I want some one to explain this table. If we change the number of blocks, how do we construct the design?
 A: 
I want some one to explain this table.

We are studying six factors ($A$, $B$, $C$, $D$, $E$, and $F$) in a full factorial design.  Thus we have $2^6 = 64$ runs.  We want to carry out the experiment in eight blocks of eight runs each and we want to confound $ABCD$, $ACE$, and $ABEF$ with blocks.
The key to understanding the table is to consider the eight blocks as being determined by three more factors, each with two levels, as discussed in Chapter 10 of Montgomery (1991).  The number eight is rather convenient since $2^3=8$.  Sometimes, these extra factors are called "pseudo-factors".
Think about the original design with columns labeled from $A$ to $F$ and with 64 rows.  Now, mentally add three more columns labeled $G$, $H$, and $J$ ($J$ chosen to not conflict with the identity $I$).  Unique combinations of $G$, $H$, and $J$ are going to represent blocks.  
Actually, you should now be able to see that we are essentially designing a $2^{9-3}$ fractional factorial experiment.  Per the exercise, we confound the additional factors as follows:  $G=ABCD$, $H=ACE$, and $J=ABEF$.  This gives us three defining relations:  $I= ABCDG$, $I=ACEH$, and $I=ABEFJ$.  
To obtain the alias pattern, expand the expression
$$(I+ABCDG)(I+ACEH)(I+ABEFJ)$$
to obtain
$$\begin{align} I & =ABCDG = ACEH = BDEGH \\ &= ABEFJ = CDEFGJ = BCFHJ = ADFGHJ.\end{align}$$
Then grind through each of $A$, $B$, $AB$, $C$, $AC$, $BC$, and $ABC$ to find the complete alias pattern for the design.
Now we need the defining contrasts, which are obtained from the defining relations:
$$\begin{align}
L_1 &= x_1 + x_2 + x_3 + x_4 + x_7 \\
L_2 &= x_1 + x_3 + x_5 + x_8 \\
L_3 &= x_1 + x_2 + x_5 + x_6 + x_9 
\end{align}$$
The block containing $(1)$ is the principal block.  You can see that in the table under "Block 6".  Now, the pattern of defining contrasts for $(1)$ is given by $(L_1, L_2, L_3)=(0,0,0)$.  To get the rest of the runs that belong in that block, we need to find all runs that yield the same value for $(L_1, L_2, L_3)$ as $(1)$, namely $(0,0,0)$.  What we could do is just compute $(L_1, L_2, L_3)$ for all 64 runs, and then sort them.
Anyway, notice, for example, that $abcd$ evaluates to $(0,0,0)$, as do the rest of the entries in the table under "Block 6".   Suppose we were given the principal block.  Notice for example that if you take the first run in "Block 3", $a$, and multiply it by each row in the principal block using modulo 2 arithmetic, that you get the rest of the entries.  (E.g., $a \times abcd = bcd$.)
So you could get the rest of the blocks from the principal block by filling in the first row.  Notice in the table that the first row has been filled out as a $2^3$ factorial with factors $A$, $B$, and $C$.
Also notice from the defining relation for blocks, or by inspection of the table, that we do have main effects aliased with second order interactions.  That makes this a resolution III design, not particularly great.  That is, the design in blocks can be implemented as a $2^{9-3}_{III}$ fractional factorial design.

If we change the number of blocks, how do we construct the design?

Well, we would want our number of blocks to fit nicely in a factorial structure like $2^1$, $2^2$, $3^1$, or whatever.  Then there would be some tedious work along the lines above.  Or, more likely,  generate a design using software.
Montgomery, D. (1991) Design and Analysis of Experiments.  Third Edition.  John Wiley & Sons.
