I posted this in the probability and statistics forum, but I didn't get a nibble. Anyone here care to make a suggestion?(adsbygoogle = window.adsbygoogle || []).push({});

The probability that project a will be complete at time t after it begins is given as:

[tex]

\[ p_a(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_1$} \\

\frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2$} \\

1 & \mbox{if t_2 \leq t}

\right. \][/tex]

In other words, the project requires at least [itex]t_1[/itex] to complete and will take no more than [itex]t_2[/itex]. The probability of completion at any time between [itex]t_1[/itex] and [itex]t_2[/itex] is a linear function of t.

The probability that project b will be complete at time t after it begins is given as:

[tex]

\[ p_b(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_3$} \\

\frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4$} \\

1 & \mbox{if t_4 \leq t}

\right. \][/tex]

Project b will begin the moment project a is complete. Given t, what is the probability [itex]p_{ab}(t)[/itex] that both projects will be complete at time t. Obviously,

[tex]

\[ p_{ab}(t) = \left\{ \begin{array}{ll}

0 & \mbox{if t \leq t_1 + t_3$} \\

1 & \mbox{if t_2 + t_4 \leq t}

\right. \][/tex].

What is [itex]p_{ab}(t)[/itex] for [itex]t_1 + t_3 \leq t \leq t_2 + t_4[/itex]?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Scheduling problem

Loading...

Similar Threads - Scheduling problem | Date |
---|---|

A Maximization problem using Euler Lagrange | Feb 2, 2018 |

A Maximization Problem | Jan 31, 2018 |

I A question about Vector Analysis problems | Oct 4, 2017 |

B Square root differential problem | Jul 13, 2017 |

Self-study schedule | May 6, 2010 |

**Physics Forums - The Fusion of Science and Community**