In the following table, there are 12 entries in the form nij, where i = 1, 2, 3 and j 1,2,3,4. Each of these entries denotes the largest integer n such that f(n) milliseconds = does not exceed t, where f(n) is the function corresponding to the row of the entry and t is the time corresponding to the column of the entry. For example, for entry n32, we have f(n) = 2^n and t = 1 minute. Hence n32 should be the largest integer n such that 2^n milliseconds is no more than 1 minute.

- September 04, 2022

Q1 (12 points) In the following table, there are 12 entries in the form nij, where i = 1, 2, 3 and j 1,2,3,4. Each of these entries denotes the largest integer n such that f(n) milliseconds = does not exceed t, where f(n) is the function corresponding to the row of the entry and t is the time corresponding to the column of the entry. For example, for entry n32, we have f(n) = 2^n and t = 1 minute. Hence n32 should be the largest integer n such that 2^n milliseconds is no more than 1 minute.

On the answer sheet, enter the values for nij, i = 1, 2, 3, j = 1, 2, 3, 4.

Q1. Solution:

Aim: Each entry denotes the largest integer n such that f(n) milliseconds do not exceed t.

FIRST ROW...........................................

i) Solving n11 :

Here , f(n) = 100n + 200 And t = 1 second.

so, $100*n + 200 <= t$ where n is an integer

$100*n <= t - 200$

$n <= (1000 - 200)/100$ : [t = 1second = 1000ms]

$n <= 8$

Ans: $n11 = 8$

ii) Solving n12

Here , f(n) = 100n + 200 And t = 1minute = 60sec = 60000ms

$100*n + 200 <= 60000$

$n <= (60000 - 200)/100$

$n <= 59800/100$

$n <= 598$

Ans : $n12 = 598$

iii) Solving n13

Here , f(n) = 100n + 200 And t = 1hr = 60min = 3600sec = 36*10⁵ ms

$100*n + 200 <= 36*10^{5}$

$n <= 35998$

Ans : $n13 = 35998$

iv) Solving n14

Here , f(n) = 100n + 200 And t = 1day = 24hr = 1440min = 86400s = 86400*10³ms

$100*n + 200 <= 86400000$

$n <= 863998$

Ans : $n14 = 863998$

SECOND ROW............................................................

v) Solving n21

Here, $f(n) = 0.1n^{2} + 100n + 8$ and t = 1second = 1000ms

$0.1n^{2} + 100n + 8 <= 1000$

$n^{2} + 1000n - 9920 <= 0$

After solving the above quadratic equation we get :

$n <= 9.8234$

Ans : $n21 = 9$

vi) Solving n22

Here, $f(n) = 0.1n^{2} + 100n + 8$ and t = 1minute = 60sec = 60000ms

$0.1n^{2} + 100n + 8 <= 60000$

$n^{2} + 1000n -599920 <= 0$

After solving the above quadratic equation we get :

$n <= 421.91105$

Ans : $n22 = 421$

vii) Solving n23

Here, $f(n) = 0.1n^{2} + 100n + 8$ and t = 1hr = 60min = 3600sec = 36*10⁵ ms

$0.1n^{2} + 100n + 8 <= 3600000$

$n^{2} + 1000n - 35999920 <= 0$

After solving the above quadratic equation we get :

$n <= 5520.790645$

Ans : $n23 = 5520$

viii) Solving n24

Here, $f(n) = 0.1n^{2} + 100n + 8$ and t = 1day = 24hr = 1440min = 86400s = 86400*10³ms

$0.1n^{2} + 100n + 8 <= 86400000$

$n^{2} + 1000n - 863999920 <= 0$

After solving the above quadratic equation we get :

$n <= 28898.127831547$

Ans : $n24 = 28898$

THIRD ROW...............................................................

ix)Solving n31

Here $f(n) = 2^{n}$ and t = 1second = 1000ms

$2^{n} <= 1000$

$n <= log_{2}1000$

$n <= 9.966$

Ans : $n31 = 9$

x)Solving n32

Here $f(n) = 2^{n}$ t = 1minute = 60sec = 60000ms

$2^{n} <= 60000$

$n <= log_{2}60000$

$n <= 15.873$

Ans : $n32 = 15$

xi)Solving n33

Here $f(n) = 2^{n}$ and t = 1hr = 60min = 3600sec = 36*10⁵ ms

$2^{n} <= 3600000$

$n <= log_{2}3600000$

$n <= 21.78$

Ans : $n33 = 21$

xii) Solving n34

Here $f(n) = 2^{n}$ and t = 1day = 24hr = 1440min = 86400s = 86400*10³ms

$2^{n} <= 86400000$

$n <= log_{2}86400000$

$n <= 26.365$

Ans : $n34 = 26$

The table above is filled as follows:

8	598	35998	863998
9	421	5520	28898
9	15	21	26

IF YOU HAVE ANY DOUBT REGARDING THESE QUESTION YOU CAN ASK IN THE COMMENT BELOW. PLEASE UPVOTE THE ANSWER.

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