How does this solution relate to the actual real-life stats problem?

Question

4.62 in Newbold (8 ed):

A new warehouse is being designed and a decision concerning the number of loading docks is required. There are two models based on truckarrival assumptions for the use of this warehouse, given that loading a truck requires 1 hour. Using the first model, we assume that the warehouse could be serviced by one of the many thousands of independent truckers who arrive randomly to obtain a load for delivery. It is known that, on average, 1 of these trucks would arrive each hour. For the second model, assume that the company hires a fleet of 10 trucks that are assigned full time to shipments from this warehouse. Under that assumption the trucks would arrive randomly, but the probability of any truck arriving during a given hour is 0.1. Obtain the appropriate probability distribution for each of these assumptions and compare the results.

The solution to this is that in the first case it's the Poisson cumulative distribution function with mean =1, while in the second model it's binomial with n=10 and p=0.1, similar but not quite identical. Great. What I don't understand is 1. What do the x/n numbers mean in this case; and 2. how does this help us make a decision about the number of loading docks?

What I mean is that e.g. with the Poisson it's
0 - 0.367
1 - 0.735
2 - 0.919
.
8 - 1.000

So... does this mean that there is a 0.367 chance that zero trucks turn up, and 1.0 chance that 8 trucks turn up?! Even though on average it's 1 an hour?! And how would this help us decide how many loading docks we need? So, it's a classic problem: I can (kind of) do the maths, I just have no idea how any of it relates to real life...

Answer 1

Here is one possible way to look at your question.

The main question is whether one loading dock will suffice. You begin to run into inefficiencies if more than one truck arrives in an hour. Then trucks will have to queue up for unloading.

Poisson model. According to your first model with $X \sim \mathsf{Pois}(\lambda = 1),$ you have $E(X) = 1,\,$ $SD(X) = 1,\,$ and (as computed in R, where ppois is a Poisson CDF), $P(X > 1) = 1- P(X \le 1) = 0.2642.$ Similarly, $P(X > 2) = 0.0803.$

1 - ppois(1:2, 1)
[1] 0.2642411 0.0803014

Binomial model. According to your second model with $Y \sim \mathsf{Binom}(n=10, p=.1),$ you have $E(Y) = 1,\,$ $SD(X) = 1.3077,\,$ and (as computed in R, where pbinom is a binomial CDF), $P(Y > 1) = 1- P(Y \le 1) = 0.2639.$ Similarly, $P(Y > 2) = 0.0702.$

1 - pbinom(1:2, 10, .1)
[1] 0.26390107 0.07019083

Comments. With either model, it seems there will be queueing at a single loading dock during slightly more than a quarter of the hours. With two loading docks, more of the trucks can be unloaded immediately. Under either model, the 2-dock scenario seems tidier to me. But I have no idea how realistic either model is, what it costs if a truck has to wait to unload, or what the future prospects for increasing or declining truck traffic may be.

Addendum: Another way to look at the problem is to notice that the number of trucks arriving per day under the Poisson model is $D \sim \mathsf{Pois}(\lambda = 8).$ Then $P(D > 8) = 0.4075.$ So with only one loading dock, on more than 40% of days, there will be more than 8 truck arrivals a day, so that there will still be trucks waiting at the end of an 8 hour day. However, with two loading docks the probability that not all trucks could be accommodated during an 8-hour day is negligible.

1 - ppois(c(8, 16), 8)
[1] 0.407452659 0.003718021

Because the binomial model assumes there are only ten trucks, the congestion would be less. If a truck can appear for unloading only once a day, there would never be a day when the total required time at a loading dock exceeds 10 hours.