Solutions - 5160-03

(a)

Compute probability the wait time for a Bundle is at most 1 hr:

$$\begin{array}{c}P[X\le 1]={\int }_0^{1}2e^{-2x}dx\\ \\ \gg \gg {[-e^{-2t}]|}_0^1\gg \gg \approx 0.8647\end{array}$$

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(b)

Erlang distribution:

$$f_Y(y)=\frac{{\lambda }^{\ell }}{(\ell -1)!}{y}^{\ell -1}{e}^{-\lambda y},y\ge 0$$

Desired probability:

$$\begin{array}{c}P[Y\le 1]={\int }_0^1\frac{{10}^5}{4!}{y}^4{e}^{-10y}dy\\ \\ \gg \gg \approx 0.9707\end{array}$$

This integral requires iterated IBP.

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(c)

The results disagree. Method 2 is the correct approach assuming individual calls arrive according to a Poisson process.

It turns out that if you have some Poisson process, bundles of arrivals do not themselves follow a Poisson process.

Think of a Poisson process as some dots scattered on a timeline. Place a heavy dot at the location of each $5^{\text{th}}$ dot in order, and erase all other dots. These heavy dots do not follow a Poisson process.