What is the largest n for which one can solve within a day

Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen

ISBN: 9780073383095 37

Solution for problem 16E Chapter 3.3

Discrete Mathematics and Its Applications | 7th Edition

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Discrete Mathematics and Its Applications | 7th Edition | ISBN: 9780073383095 | Authors: Kenneth Rosen

Discrete Mathematics and Its Applications | 7th Edition

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Problem 16E

What is the largest n for which one can solve within a day using an algorithm that requires f(n) bit operations, where each bit operation is carried out in 10−11 seconds, with these functions f(n)?

a) log n

b) 1000n

c) n2

d) l000n2

e) n3

f) 2n

Step-by-Step Solution:

Step 1 of 3

Step-1:

In this problem we need to find the largest n for which can solve with in one second problem using an algorithm requires f(n) bit operations, where each bit operation is carried out $(10{)}^{-11}$ second with these functions f(n).

We know that , in a day there are approximately $(10{)}^{5}$ seconds.So , within a day, the total number of operations is equal to $\frac{(10{)}^{5}}{(10{)}^{-11}}\ =(10{)}^{16}$ , since $\ \frac{{a}^{m}}{{a}^{n}}=\ {a}^{m-n}$ .

So, the operation bit per second is $f(n)\ =(10{)}^{16}$

Step-2:

a)consider,f(n) = log(n).

Calculate the value of n.

$f(n)\ \leq{}(10{)}^{16}$

$log(n)\ \leq{}(10{)}^{16}$

Consider, the base log is 2.

So, $lo{g}_{2}(n)\ \leq{}(10{)}^{16}\ \Rightarrow{}n={2}^{(10{)}^{16}}$ .

Therefore,the largest value of $n={2}^{(10{)}^{16}}$ .

Step-3:

b)consider,f(n) = 1000n.

Calculate the value of n.

$f(n)\ \leq{}(10{)}^{16}$

So, $1000n\ \leq{}(10{)}^{16}$ .

$\Rightarrow{}(10{)}^{3}n\leq{}(10{)}^{16}$

$\Rightarrow{}n=\frac{(10{)}^{16}}{(10){\ }^{3}}\ =(10{)}^{13}$ ,since $\ \frac{{a}^{m}}{{a}^{n}}=\ {a}^{m-n}$ .

Therefore,the largest value of $n=(10{)}^{13}$ .

Step-4:

c)consider, $f(n)={n}^{2}$

Calculate the value of n.

$f(n)\ \leq{}(10{)}^{16}$

${n}^{2}\ \leq{}(10{)}^{16}$

$n\leq{}(10{)}^{16/2}=(10{)}^{8}$

Therefore,the largest value of $n=(10{)}^{8}$

Step-5:

d)consider, $f(n)=1000{n}^{2}$

Calculate the value of n.

$f(n)\ \leq{}(10{)}^{16}$

$1000{n}^{2}\ \leq{}(10{)}^{16}$

$\Rightarrow{}(10{)}^{3}{n}^{2}\leq{}(10{)}^{16}$

$\Rightarrow{}{n}^{2}=\frac{(10{)}^{16}}{(10){\ }^{3}}\ =(10{)}^{13}$ ,since $\ \frac{{a}^{m}}{{a}^{n}}=\ {a}^{m-n}$ .

$\$ $\Rightarrow{}n=(10{)}^{13/2}$ .

$=(10{)}^{6}\sqrt{10}$

$=(3.16)(10{)}^{6}$

Therefore,the...

Step 2 of 3

Chapter 3.3, Problem 16E is Solved

Step 3 of 3

Textbook: Discrete Mathematics and Its Applications

Edition: 7

Author: Kenneth Rosen

ISBN: 9780073383095

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