Therefore, in order to run the program in the shortest time, you should delay for 3 years
Given that the speed of new computers costing a certain amount of money grows exponentially over time, with the speed doubling every 18 months.
Exponential equation as a function of time can be expressed as
P(t) = [tex]P_{0}[/tex][tex]R^{t/T}[/tex]
Quantity P will be increasing exponentially when R > 1. And it will decrease exponentially when R < 1.
Suppose you need to run a very time consuming program and are allowed to buy a new computer costing a fixed amount on which to run the program. If you start now, the program will take 7 years to finish.
Let the time required to run and complete a program for time t years be P(t).
If the speed doubling every 18 months, which is 3/2 years. This will eventually cause the running time to decrease by half (0.5)
P(t) = 7[tex](1/2)^{3t/2}[/tex] = 7[tex](2)^{-3t/2}[/tex]
The total time T will be sum of the delay time t and the time of running the program. That is
T(t) = t + P(t)
T(t) = t + 7[tex](2)^{-3t/2}[/tex] .......... (2)
We can calculate the minimum time t by differentiating the above equation and make it equal to zero.
dT/dt = 1 + 14/3[tex](2)^{-3t/2}[/tex](ln0.5) = 0
14/3[tex](2)^{-3t/2}[/tex](ln0.5) = -1
[tex](2)^{-3t/2}[/tex](ln0.5) = -3/14
[tex](2)^{-3t/2}[/tex] = -3/14ln0.5
Log (natural logarithm) both sides
-3t/2ln2 = ln(3/14ln2)
make t the subject of formula
t = - 3ln(3/14ln2) ÷ 2ln2
t = 2.5 approximately
To know how long you should delay buying the computer and starting the program so that it will finish in the shortest amount of time from now, substitute t into equation 2
T(2.5) = 2.5 + (7)[tex]2^{-3(2.5)/2}[/tex]
T(2.5) = 2.5 + 0.5202
T(2.5) = 3.02 years
T (2.5) = 3 years approximately
Therefore, in order to run the program in the shortest time, you should delay for 3 years
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