Long multiplication school method: number of primitive operations to calculate first partial product? Having got some basics down in regard to addition and explaining it in terms of primitive operations (addition and multiplication), I am now again stuck on understanding the more complicated long multiplication.

I have read in my introductory book (Mehlhörn's 'Algorithms and Data Structures') that each partial product requires 2n + 1 primitive operations.

Now I understand, I think, where the +1 comes from for the first partial product at least, as in there being one extra digit in the result for the first partial product, but I don't understand why it is "2n" + 1, as it seems to me it would be more like n + 1, so I am defintely missing something important here. By 'partial product', I mean the result arising from the first intermediate calculation of the multiplication problem.

I am of course also totally confused about quantifying the rest of the primitive operations required to get the final result/product! But I would just like to understand where the 2n is coming from for now for the first partial product. Just to add my guess from what I have seen in addition: does the 2 in 2n refer to a maximum guiding number, rather than a definitive one? 

Here is a link to the chapter in the book I am learning from: http://people.mpi-inf.mpg.de/~mehlhorn/ftp/Toolbox/Appetizer.pdf

I understood primitive operations to mean: addition and multiplication. 
The author does write at bottom of pg.1 

"...we have two primitive operations at our disposal: the addition of three digits with a two-digit result (this is sometimes called a full adder), and the multiplication of two digits with a two-digit result"

The product of two n-digit numbers when they multiplied together needs then: 3n^2 + 2n primitive operations.

It is the algorithm as per the method taught at school for long multiplication problems.

(All of this is on pgs 1 and 2 of chapter)
Perhaps the following example will be useful. Suppose that $n = 3$, and we want to multiply $a_2 a_1 a_0$ by $b$. We first compute
$$
\begin{align*}
a_0 \times b &amp;= c_0 d_0 \\
a_1 \times b &amp;= c_1 d_1 \\
a_2 \times b &amp;= c_2 d_2 
\end{align*}
$$
(Here $c_0 d_0$ is a two-digit integer.)
So far we have done 3 operations (for general $n$, this will be $n$ operations).

Now we wish to perform the addition
$$
\begin{array}{cccc}
0&amp;d_2&amp;d_1&amp;d_0\\
c_2&amp;c_1&amp;c_0&amp;0 \\\hline
r_3&amp;r_2&amp;r_1&amp;r_0
\end{array}
$$
The book states that you need 4 operations for this (and in general, $n+1$), though 2 are enough (in general, $n-1$):
$$
\begin{align*}
r_0 &amp;= d_0 \\
s_1 r_1 &amp;= d_1 + c_0 \\
s_2 r_2 &amp;= d_2 + c_1 + s_1 \\
r_3 = s_2
\end{align*}
$$
Here $s_1,s_2$ are carry bits, and the first and last operations are for free (though the author charges one operation for each).

Summarizing, if you only charge addition and multiplication, only $2n-1$ operations are needed. If you also charge moves, $2n+1$ operations are needed. (The moves can be eliminated, though. Exercise.)