Chapter 14, Problem 6RE

Explanation of Solution

Given:

The algorithm A to handle a data set of 1000 records it will take around five seconds.

The algorithm A is considered as $\text{O}\left(\text{n}\right)$ algorithm.

To handle data set of 2000 records:

The algorithm is considered as $\text{O}\left(\text{n}\right)$ algorithm.

$\begin{array}{l}\frac{\text{2000}}{1000}\\ =2\\ =2\times 5\\ =10\end{array}$

The algorithm is considered as $\text{O}\left({\text{n}}^2\right)$ algorithm.

$\begin{array}{l}\frac{\text{2000}}{{1000}^2}\\ =4\\ =4\times 5\\ =20\end{array}$

The algorithm is considered as $\text{O}\left({\text{n}}^3\right)$ algorithm.

$\begin{array}{l}\frac{{\text{2000}}^3}{{1000}^3}\\ =8\\ =8\times 5\\ =40\end{array}$

The algorithm A is considered as $\text{O}\left(\text{n log }\left(\text{n}\right)\right)$ algorithm.

$\begin{array}{l}\frac{\text{2000 log}\left(2000\right)}{1000\log \left(1000\right)}\\ =\frac{2\times 3.47}{1\times 3}\\ =2.31\\ =2.31\times 5\\ =11.56\end{array}$

To handle data set of 3000 records:

The algorithm is considered as $\text{O}\left(\text{n}\right)$ algorithm.

$\begin{array}{l}\frac{\text{3000}}{1000}\\ =3\\ =3\times 5\\ =15\end{array}$

The algorithm is considered as $\text{O}\left({\text{n}}^2\right)$ algorithm.

$\begin{array}{l}\frac{\text{3000}}{{1000}^2}\\ =9\\ =9\times 5\\ =45\end{array}$

The algorithm is considered as $\text{O}\left({\text{n}}^3\right)$ algorithm.

$\begin{array}{l}\frac{{\text{3000}}^3}{{1000}^3}\\ =2\\ =27\times 5\\ =135\end{array}$

The algorithm A is considered as $\text{O}\left(\text{n log }\left(\text{n}\right)\right)$ algorithm