Question

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

Answer 1

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