-In sequential the matrix multiplication is performed using three loops. Supposing that $A$, $B$ represent two square matrices, the result of the multiplication of $A \times B$ is
-
-On C2070M Tesla card, this code take 37.68ms to perform the multiplication. On a
-Intel Xeon E31245 at 3.30GHz, it takes 2465ms without any parallelization (using
-only one core). Consequently the speed up between the CPU and GPU version is
-about 65 which is very good regarding the difficulty of parallelizing this code.
+With a sequential programming, the matrix multiplication is performed using
+three loops. Supposing that $A$, $B$ represent two square matrices and that the
+result of the multiplication of $A \times B$ is $C$. The
+element \texttt{C[i*size+j]} is computed as follows:
+\begin{equation}
+C[i*size+j]=\sum_{k=0}^{size-1} A[i*size+k]*B[k*size+j];
+\end{equation}
+
+In Listing~\ref{ch2:lst:ex3}, in the CPU computation, this part of code is
+performed using 3 loops, one for $i$, one for $j$ and one for $k$. In order to
+perform the same computation on a GPU, a naive solution consists in considering
+that the matrix $C$ is split into 2 dimensional blocks. The size of each block
+must be chosen such as the number of threads per block is inferior to $1,024$.
+In Listing~\ref{ch2:lst:ex3}, we consider that a block contains 16 threads in
+each dimension. The variable \texttt{nbTh} represents the number of threads per
+block. So to be able to compute the matrix-matrix product on a GPU, each block
+of threads is assigned to compute the result of the product for the elements of
+this block. So the first step for each thread of a block is to compute the
+corresponding row and column. With a 2 dimensional decomposition, \texttt{int i=
+blockIdx.y*blockDim.y+ threadIdx.y;} allows us to compute the corresponding line
+and \texttt{int j= blockIdx.x*blockDim.x+ threadIdx.x;} the corresponding
+column.
+
+
+On C2070M Tesla card, this code take $37.68$ms to perform the multiplication. On
+a Intel Xeon E31245 at $3.30$GHz, it takes $2465$ms without any parallelization
+(using only one core). Consequently the speed up between the CPU and GPU version
+is about $65$ which is very good regarding the difficulty of parallelizing this
+code.