$$ \begin{array}{cccccccccccc} & & \text{A} \\ & \swarrow & & \searrow \\ \text{B} & & & & \text{C} \\ & \searrow & & \swarrow \\ \downarrow & & \text{D} \\ & & & \searrow \\ \text{F} & & \rightarrow & & \text{E} \\ & & & & \downarrow \\ & & & & \text{G} \end{array} $$
I was told that this is a "workflow diagram" or "workflow graph." When the worker A finishes work on a Thing, B and C begin work. When B and C have both finished their parts of the job, D begins. When B finishes, F starts. When D and F finish, E starts. When E finishes, G starts. (I had an impression that these "workers" were computer programs, but for present purposes that's not what matters.)
I was given the mean and standard deviation of the time each worker needs. I tentatively decided to model the times as gamma-distributed random variables, each with the specified mean and standard deviation. The gamma distribution is $$ \frac1{\Gamma(\alpha)} \left( \lambda t\right)^{\alpha-1} e^{-\lambda t} \left( \lambda\, dt\right) \text{ for } t>0. $$ The parameter $\lambda$ is the "rate" or "intensity."
Since this implies $\text{mean} = \alpha/\lambda$ and $\text{variance} = \alpha/\lambda^2,$ we have \begin{align} \alpha & = \text{mean}^2/\text{variance}, \\[6pt] \lambda & = \text{mean}/\text{variance}. \end{align} If the workflow graph had been a straight line, i.e. corresponded to a linear ordering rather than to a properly partial ordering, then one would add a bunch of independent (I did assume independence here) random variables, and the probability density of the distribution of the total time for the job to get done would just be an ordinary convolution of densities. (And if all the $\lambda\text{s}$ had been the same, then (Exercise:) this would just mean adding up the different $\alpha\text{s}.$ But that doesn't apply here.)
So two questions come up: (1) Is there some other notion of convolution that does something like that for such posets? (2) If there is, might it actually shed any light on the probability distribution of the total time for the job?
The only notion of "convolution" I could recall having heard of for posets like this was the multiplication in the "incidence algebra". The members of the "incidence algebra" are scalar-valued functions whose domain is the set of closed intervals $[a,b] = \{ x : a\le x \le b\}$ and one "multiplies" these by convolving: $$ (f*g)[a,b] = \sum_{x\,:\,a\,\le\,x\,\le\,b} f[a,x]g[x,b]. $$ But this doesn't appear to do the job in this case as far as I know so far, and doesn't say anything about question (2).
