Homework Solution: Suppose you have two programs whose running times in microseconds for an input of size n are 1000 lg n…

    6. Suppose you have two programs whose running times in microseconds for an input of size n are 1000 Ign and 10n2. Find the largest input size n that can be processed by each input size n that can be processed by each program in (a) one minute and (b) one hour. [10 point s]
    Suppose you have two programs whose running times in microseconds for an input of size n are 1000 lg n and 10n^2. Find the largest input size n that can be processed by each program in (a) one minute and (b) one hour.

    Expert Answer

     
    Note that 1 micro second = 10^-6 seconds

    6. Suppose you bear span programs whose general times in microseconds control an input of greatness n are 1000 Ign and 10n2. Find the largest input greatness n that can be processed by each input greatness n that can be processed by each program in (a) individual detailed and (b) individual hour. [10 apex s]

    Suppose you bear span programs whose general times in microseconds control an input of greatness n are 1000 lg n and 10n^2. Find the largest input greatness n that can be processed by each program in (a) individual detailed and (b) individual hour.

    Expert Acceptance

     

    Note that 1 micro avoid = 10^-6 avoids

    Control program 1:

    Control getting the consummation appraise of n,

    a) 1 min = 1000 log(n) microsecs

    So, 1*60 secs = 1000 log(n) * 10^-6 secs

    log(n) = 60 * 1000 = 60000

    n = 2^{60000}

    So, consummation vale of n = 2^{60000}

    b) 1 hour = 1000 log(n) microsecs

    So, 1*60*60 secs = 1000 log(n) * 10^-6 secs

    log(n) = 60*60 * 1000 = 36,00,000

    n = 2^{3600,000}

    So, consummation vale of n = 2^{3600,000}

    Control program 2:

    Control getting the consummation appraise of n,

    a)

    1 min = 10 * n^2 microsecs

    So, 1*60 secs = 10 * n^2 * 10^-6 secs

    n^2 = 60 * 10^5 = 6*10^6

    n = 2449

    So, consummation vale of n = 2449

    b)

    1 hour = 10 * n^2 microsecs

    So, 1*60*60 secs = 10 * n^2 * 10^-6 secs

    n^2 = 60*60 * 10^5 = 36*10^7

    n =1.8974e+04

    So, consummation vale of n = 1.8974e+04