Answer:
An input size of N = 128 makes the algorithm run for 14 milliseconds
Explanation:
O(log2N)
This means that the running time for an algorithm of length N is given by:
[tex]F(N) = c\log_{2}{N}[/tex]
In which C is a constant.
Runs for 10 milliseconds when the input size (N) is 32.
This means that [tex]F(32) = 10[/tex]
So
[tex]F(N) = c\log_{2}{N}[/tex]
[tex]10 = c\log_{2}{32}[/tex]
Since [tex]2^5 = 32, \log_{2}{32} = 5[/tex]
Then
[tex]5c = 10[/tex]
[tex]c = \frac{10}{5}[/tex]
[tex]c = 2[/tex]
Thus:
[tex]F(N) = 2\log_{2}{N}[/tex]
What input size makes the algorithm run for 14 milliseconds
N for which [tex]F(N) = 14[/tex]. So
[tex]F(N) = 2\log_{2}{N}[/tex]
[tex]14 = 2\log_{2}{N}[/tex]
[tex]\log_{2}{N} = 7[/tex]
[tex]2^{\log_{2}{N}} = 2^7[/tex]
[tex]N = 128[/tex]
An input size of N = 128 makes the algorithm run for 14 milliseconds
