# 8.14. Estimating With Big-O¶

We can use Big-O categories to do an estimation of how long a problem will take to solve based on a smaller version of the problem. We simply need to set up a proportion like the one below and solve it:

\(\frac{\textrm{work for job 1}}{\textrm{work for job 2}} = \frac{\textrm{time for job 1}}{\textrm{time for job 2}}\)

The key is to remember that the work does not necessarily equal the size of the problem. Instead, we have to use the size of the problem and the Big-O of the algorithm we are applying to calculate the approximate amount of work.

For example, say we have a list of 1000 things…

If we want to do a Binary Search, the Big-O is \(O(log_2(n))\). That means the estimated work would be \(log_2(1000)\) or ~9.9657 units of work.

If we want to do a Linear Search, the Big-O is \(O(n)\). That means the estimated work would just be 1,000 units of work.

If we want to do a Selection Sort, the Big-O is \(O(n^2)\). That means the estimated work would be \({1000}^2\) or 1,000,000 units of work.

Note

You can use Wolfram Alpha website to calculate log base 2 by typing something like “log2(1024)”. Try it below.

#### Sample Problem 1

I have timed selection sort on 10,000 items and it takes 0.243 seconds. I want to estimate the time it will take to sort 50,000 items. Because selection sort is an \(O(n^2)\) algorithm, I know I need to square the problem sizes to estimate the amount of work required for each of the two jobs. So I can set up the proportion like this:

\(\frac{ 10000^2 }{ 50000^2 } = \frac{0.243\textrm{ seconds}}{\textrm{time for job 2}}\)

So…

\(\frac{100000000}{2500000000} = \frac{0.243\textrm{ seconds}}{\textrm{time for job 2}}\)

Cross multiplying gives:

\(100000000(\textrm{time for job 2}) = 0.243\textrm{ seconds} \cdot {2500000000}\)

Solving for **time for job 2** gives:

\(\textrm{time for job 2} = 6.075\textrm{ seconds}\)

#### Sample Problem 2

I have timed a linear search on 10,000,000 items and it takes 8.12 seconds (call this job 1). I want to estimate the time it will take to use binary search instead (job 2). The problem sizes are the same for both jobs: 10,000,000 items. However, the algorithms will require different amounts of work. Linear search is a \(O(n)\) algorithm, so the work for job 1 will be 10,000,000. For job 2, we are using a \(O(log_2(n))\) algorithm so the work will be \(log_2(10000000)\)

\(\frac{10000000}{log_2(10000000)} = \frac{8.12\textrm{ seconds}}{\textrm{time for job 2}}\)

So…

\(\frac{10000000}{23.25} = \frac{8.12\textrm{ seconds}}{\textrm{time for job 2}}\)

Cross multiplying gives:

\(10000000(\textrm{time for job 2}) = 8.12\textrm{ seconds} \cdot 23.25\)

Solving for **time for job 2** gives:

\(\textrm{time for job 2} = 0.000019\textrm{ seconds}\)

Significantly faster!