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# 数学代写|数值分析代写Numerical analysis代考|Operation count for the elimination step of Gaussian elimination

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## 数学代写数值分析代写Numerical analysis代考|Operation count for the elimination step of Gaussian elimination

The elimination step for a system of $n$ equations in $n$ variables can be completed in $\frac{2}{3} n^3$ $+\frac{1}{2} n^2-\frac{7}{6} n$ operations.

Normally, the exact operation count is less important than order-of-magnitude estimates, since the details of implementation on various computer processors differ. The main point is that the number of operations is approximately proportional to the execution time of the algorithm. We will commonly make the approximation of $\frac{2}{3} n^3$ operations for elimination, which is a reasonably accurate approximation when $n$ is large.
After the elimination is completed, the tableau is upper triangular:
$$\left[\begin{array}{cccc:c} a_{11} & a_{12} & \ldots & a_{1 n} & b_1 \ 0 & a_{22} & \ldots & a_{2 n} & b_2 \ \vdots & \vdots & \ddots & \vdots & \vdots \ 0 & 0 & \ldots & a_{n n} & b_n \end{array}\right]$$
In equation form,
\begin{aligned} a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n & =b_1 \ a_{22} x_2+\cdots+a_{2 n} x_n & =b_2 \ & \vdots \ a_{n n} x_n & =b_n, \end{aligned}
where, again, the $a_{i j}$ refer to the revised, not original, entries. To complete the computation of the solution $x$, we must carry out the back-substitution step, which is simply a rewriting of $(2.8)$ :
\begin{aligned} x_1 & =\frac{b_1-a_{12} x_2-\cdots-a_{1 n} x_n}{a_{11}} \ x_2 & =\frac{b_2-a_{23} x_3-\cdots-a_{2 n} x_n}{a_{22}} \ & \vdots \ x_n & =\frac{b_n}{a_{n n}} . \end{aligned}

## 数学代写|数值分析代写Numerical analysis代考|Operation count for the back-substitution step of Gaussian elimination

The back-substitution step for a triangular system of $n$ equations in $n$ variables can be completed in $n^2$ operations.

The two operation counts, taken together, show that Gaussian elimination is made up of two unequal parts: the relatively expensive elimination step and the relatively cheap back-substitution step. If we ignore the lower order terms in the expressions for the number of multiplication/divisions, we find that elimination takes on the order of $2 n^3 / 3$ operations and that back substitution takes on the order of $n^2$.

We will often use the shorthand terminology of “big-O” to mean “on the order of,” saying that elimination is an $O\left(n^3\right)$ algorithm and that back substitution is $O\left(n^2\right)$. This usage implies that the emphasis is on large $n$, where lower powers of $n$ become negligible by comparison. For example, if $n=100$, only about 1 percent or so of the calculation time of Gaussian elimination goes into the back-substitution step. Overall, Gaussian elimination takes $2 n^3 / 3+n^2 \approx 2 n^3 / 3$ operations. In other words, for large $n$, the lower order terms in the complexity count will not have a large effect on the estimate for running time of the algorithm and can be ignored if only an estimated time is required.

Estimate the time required to carry out back substitution on a system of 500 equations in 500 unknowns, on a computer where elimination takes 1 second.

Since we have just established that elimination is far more time consuming than back substitution, the answer will be a fraction of a second. Using the approximate number $2(500)^3 / 3$ for the number of multiply/divide operations for the elimination step, and $(500)^2$ for the back-substitution step, we estimate the time for back substitution to be
$$\frac{(500)^2}{2(500)^3 / 3}=\frac{3}{2(500)}=0.003 \mathrm{sec} .$$

## 数学代写数值分析代写Numerical analysis代考|Operation count for the elimination step of Gaussian elimination

$$a_{11} x_1+a_{12} x_2+\cdots+a_{1 n} x_n=b_1 a_{22} x_2+\cdots+a_{2 n} x_n \quad=b_2 \vdots a_{n n} x_n \quad=b_n,$$

$$x_1=\frac{b_1-a_{12} x_2-\cdots-a_{1 n} x_n}{a_{11}} x_2=\frac{b_2-a_{23} x_3-\cdots-a_{2 n} x_n}{a_{22}} \vdots x_n \quad=\frac{b_n}{a_{n n}} .$$

## 数学代写|数值分析代写Numerical analysis代考|Operation count for the back-substitution step of Gaussian elimination

$$\frac{(500)^2}{2(500)^3 / 3}=\frac{3}{2(500)}=0.003 \mathrm{sec}$$

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