3
$\begingroup$

If I have a linear system of equations $Ax=b$ where $A \in \mathbb{R} ^{n\times n}, x \in \mathbb{R} ^{n}, b \in \mathbb{R} ^{n} $ this system can be solved for $x$ via an LU decomposition: $$A = LU$$ where $U \in \mathbb{R} ^{n\times n}$ is upper triangular and $L \in \mathbb{R} ^{n\times n}$ is lower triangular.

I understand a forward substitution is then required where one first solves:

$$Ly=b$$ for $y$.

And then we solve: $$Ux=y$$ for $x$.

I am currently trying to determine the operation count or the FLOPS for each of the forward substitution and backward substitution. I have seen that the correct value is approximately given by $\mathcal{O}(n^{2})$ flops but I am unsure how one can arrive at this value.

I can see that for the backward substitution, for example, the system is represented as:

$$\begin{bmatrix} u_{11} & u_{12} & \cdots & u_{1n} \\ 0 & u_{22} &\cdots &u_{2n} \\ \cdots& \cdots & \ddots &\vdots \\ 0 & 0 & \cdots & u_{nn} \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\\ \vdots \\ x_{n} \end{bmatrix} = \begin{bmatrix} y_{1}\\ y_{2}\\ \vdots \\ y_{n} \end{bmatrix}$$

From which:

$$x_{i} = \frac{1}{u_{ii}} \left ( y_{i} - \sum_{j=i+1}^{n}u_{ij}x_{j} \right ); i = n, ..., 1$$

From an equation like this, how can one identify the approximate operation count?

$\endgroup$
3
$\begingroup$

If your matrix is $n\times n$ you have the following operations

  1. $n$ divisions,
  2. $\frac{n^2-n}{2}$ sums,
  3. $\frac{n^2-n}{2}$ multiplications.

The number of divisions is clear because you have $n$ rows. The number of sums and multiplications are $\sum_{i=1}^{n-1}n^2=\frac{n^2-n}{2}$. It comes from $\sum_{j=1+1}^{n}u_{ij}x_j$.

The total number of operations is $2n^2-n=\mathcal{O}(n^2)$

I hope it solves your doubts

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.