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ITS Sys Admin Pat is dealing with a computer virus outbreak. Pat esti- mates that every second at least 1 uninfected computer becomes infected for every 5 computers already infected on campus. To be exact there will be (C/5 newly infected computers in the next second if there are currently C infected computers. Pat is able to disinfect at most 10 computers per second. If Pat knows that there are at least 60 infected computers now, prove by induction that in n seconds there will be at least 12 6 10 + 50 infected computers. 5

1 Answer

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Solution :

Let the base case, n = 0

Given at
$t=0,$ at least
60 computers are infected

So,
$12* \left((6)/(10)\right)^0+50=60$

Therefore, n = 0 is not true

Let suppose, it is true for n = k

Then,
$C(k) \geq 50+12(0.6)^k$

We have to show that it is also true for
$n=(k+1)$

That is at the end of (k+1) second,


$C(k+1) \geq 50+12 * (0.6)^(k+1)$

Now, the number of the infected computer that is added between
n=k and
$n=(k+1)$ second =
$(1)/(5)\left[50+12(0.6)^k\right]$

Now the number of the infected computer decreased between
n=k and
$n=(k+1)$ second is = 10.

Hence at the end of the
$(k+1)$ second.

Number of the computers
$\geq 50+12(0.6)^k + 1/5[50+12(0.6)^k]-10$


$\geq 50+12(0.6)^k +10+ 0.5[12(0.6)^k]-10$


$\geq 50+12(0.6)^k+0.5[12(0.6)^k]$


$\geq 50+12(0.6)^k+2(0.6)^k$


$ \geq 50+12 * (0.6)^(k+1)$

Hence Proved

User Draconar
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