Solving Scheduling Problem - Probability & Statistics Forum

[Jimmy Snyder]

I posted this in the probability and statistics forum, but I didn't get a nibble. Anyone here care to make a suggestion?

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

$$\[ p_a(t) = \left\{ \begin{array}{ll}<br /> 0 & \mbox{if t \leq t_1} \\<br /> \frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2} \\<br /> 1 & \mbox{if t_2 \leq t}<br /> \right. \]$$

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

The probability that project b will be complete at time t after it begins is given as:
$$\[ p_b(t) = \left\{ \begin{array}{ll}<br /> 0 & \mbox{if t \leq t_3} \\<br /> \frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4} \\<br /> 1 & \mbox{if t_4 \leq t}<br /> \right. \]$$


Project b will begin the moment project a is complete. Given t, what is the probability $p_{ab}(t)$ that both projects will be complete at time t. Obviously,
$$\[ p_{ab}(t) = \left\{ \begin{array}{ll}<br /> 0 & \mbox{if t \leq t_1 + t_3} \\<br /> 1 & \mbox{if t_2 + t_4 \leq t}<br /> \right. \]$$.

What is $p_{ab}(t)$ for $t_1 + t_3 \leq t \leq t_2 + t_4$?

 

[mathman]

This is a standard problem of finding the distribution function for the sum of 2 random variables. The solution is obtained by convoluting the distribution functions - in a Stieljes integral form.

In your case, p_ab(t)=(int(t1,t2)p_b(t-s)ds)/(t2-t1).

 

[Jimmy Snyder]

This is a standard problem of finding the distribution function for the sum of 2 random variables. The solution is obtained by convoluting the distribution functions - in a Stieljes integral form.

In your case, p_ab(t)=(int(t1,t2)p_b(t-s)ds)/(t2-t1).

Thanks mathman for taking this up. Do you mean

$$p_{ab}(t) = \int_{t_1}^{t_2}\frac{p_b(t-s)ds}{t_2-t_1}$$

 

[mathman]

Yes. I am sorry that I still haven't gotten the hang of using Latex.

I suggest you take the time to derive it yourself - you'll understand it better.

 

[Jimmy Snyder]

I suggest you take the time to derive it yourself - you'll understand it better.

I really don't know how to derive it. I don't think the equation that I wrote and you agreed to is correct. For one thing, it is not a convolution. Like you, I too expect that the answer might be a convolution, perhaps something like this:

$$p_{ab}(t) = \int p_a(s) \times p_b(t-s)ds$$

But I am not sure if it is correct. Also, I don't know what the limits of integration should be nor what constant factor might be necessary (although if I knew everything else, I could figure that out).

I also have reason to believe that the answer might not be a convolution after all. It seems to me possible that such an integral double counts some of the probability. I came to have that feeling after looking at some finite sums that look like the integral. In short, I still don't really know the answer and I need help.

By the way mathman, I learned the little tex I know by using the quote button on other people's posts. When you do that you can see both the raw tex and the final result.

 

[mathman]

In the integral, the term is not p_a(s)ds. It is dp_a(s), which can be expressed using the derivative,
p'_a(s)ds.

If you had used the density functions (derivative of distributions), then the form you used would be correct to get the density function of the sum.