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From Algorithms to Z-Scores - matloff - University of California, Davis

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164 CHAPTER 8. MULTIVARIATE PMFS AND DENSITIES<br />

Scale time so that for hosts A and B above, the mean time <strong>to</strong> infection is 1.0. Since in state i there<br />

are i(g-i) such pairs, the time <strong>to</strong> the next state transition is the minimum <strong>of</strong> i(g-i) exponentiallydistributed<br />

random variables with mean 1. Thus the mean time <strong>to</strong> go from state i <strong>to</strong> state i+1 is<br />

1/[i(g-i)].<br />

Then the mean time <strong>to</strong> go from state 1 <strong>to</strong> state g-1 is<br />

Using a calculus approximation, we have<br />

g−1<br />

1<br />

1<br />

x(g − x)<br />

dx = 1<br />

g<br />

g−1<br />

<br />

i=1<br />

g−1<br />

1<br />

1<br />

i(g − i)<br />

( 1<br />

x<br />

(8.84)<br />

1 2<br />

+ ) dx = ln(g − 1) (8.85)<br />

g − x g<br />

The latter quantity goes <strong>to</strong> zero as g → ∞. This confirms that the behavior seen by the student<br />

in simulations holds in general. In other words, (8.84) remains bounded as g → ∞. This is a very<br />

interesting result, since it says that the mean time <strong>to</strong> alert is bounded no matter how big our group<br />

size is.<br />

So, even though our model here was quite simple, probably overly so, it did explain why the student<br />

was seeing the surprising behavior in his simulations.<br />

8.3.9 Example: Electronic Components<br />

Suppose we have three electronic parts, with independent lifetimes that are exponentially distributed<br />

with mean 2.5. They are installed simultaneously. Let’s find the mean time until the last<br />

failure occurs.<br />

Actually, we can use the same reasoning as for the computer worm example in Section 8.3.8: The<br />

mean time is simply<br />

8.3.10 Example: Ethernet Again<br />

1/(3 · 0.4) + 1/(2 · 0.4) + 1/(1 · 0.4) (8.86)<br />

In the Ethernet example in Section 8.3.3, we assumed that transmission time was a constant, 0.1.<br />

Now let’s account for messages <strong>of</strong> varying sizes, by assuming that transmission time T for a message<br />

is random, exponentially distributed with mean 0.1. Let’s find P (X < Y and there is no collision).

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