Assume that the speed of new computers costing a certain amount of money grows e

Question

Assume that the speed of new computers costing a certain amount of money grows exponentially over time, with the speed doubling every 18 months. 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 9 years to finish. How long should you delay buying the computer and starting the program so that it will finish in the shortest amount of time from now? Also, what will the shortest finishing time be?

Explanation / Answer

Solution:

At 0 months the time to completion is

Total_Time = Delay + 72 months = 72 Months

At X months delay, you know that the time exponentially decreases by a factor of 2 every 18 months.

Total_Time = Delay + (72 Months / 2(Delay/18))

So you get this simplified equation.

T = D + 72/ 2(D/18)

All you need to do now is calculate the derivative. Simple right? Let's make a rough approximation first.

@ 0 months --- 72 months

@ 18 months --- 36 months + 18 months = 54 months

@ 36 months --- 18 months + 36 months = 54 months

@ 54 months --- 9 months + 54 months = 63 months

You can roughly guess that the ideal month lies between 18 and 36 months or somewhere around 27 months but

let's do the math and verify that. Here's where the fun starts and I hope you were able to get this far on your own.

Using substitution on 2(D/18), let's set x = D/18 to make calculating the derivative easier.

T = D + 72 / 2x, x = D/18

dx = 1/18 dD

dT = dD + dx( 72/2x)

That 72/2x is still kind of ugly so let's solve that first.

y = 72/2x = 72 * 2-x

y/72 = 2- x

ln( y/72 ) = - x ln(2)

dy/y = - ln(2) dx

dy = - y ln(2) dx

dy = -(72/2x) ln(2) dx            <-- We substitute y back into the equaton.

dy = - (72/2(D/18)) ln(2) (1/18)dD        <-- We substitute x and dx back into the equation.

We can simplify that to

dy = - 4/2(D/18) ln(2) dD                <-- The 72 and 1/18 reduce.

4 is the same as 22 so we can simplify even further.

dy = - 4/2(D/18)   ln(2) dD

= - 22 / 2(D/18) ln(2) dD

= - 2(2-D/18) ln(2) dD

dy = -2( 2-D/18 ) * ln(2) dD

Now going back to the original equation.

dT = dD + dx( 72/2x)

dT = dD - 2( 2-D/18) * ln( 2 ) dD

dT/dD = 1 - 2( 2-D/18) * ln( 2 )

Solving for dT/dD = 0 you get D = 26.48 months which is really close to the 27 months that we predicted.

Substitute 26.46 months back into the original equation.

T = D + 72/2(D/18)

T = 26.46 + 72/2(26.46/18) = 52.45 months

So there is a delay of 26.48 months with a completion time of 52.45 months.

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