# Page History

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title @Risk

Manufactured components = 250 + RiskNegBin(250,1/4). The cost is therefore:

Cost of fulfilling order = 12.50 * (250 + RiskNegBin(250,1/4))

And the cost per unit is:

Cost per unit = 12.50 * (250 + RiskNegBin(250,1/4))/250

Simulating this formula gives the following distribution:

Conclusion: we should quote a per unit price of \$51.80 because there is a 75% chance that the actual outturn cost to us will be less than that figure, and we will therefore make at least some profit. Here's a question for you: If the client changed their mind and said they now want just 100 units, should we recalculate the price?

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title @Risk

The Inverse Hypergeometric distribution has the probability mass function:

LaTeX Math Block
alignment left
f(x)=\frac {\left(
\begin{array}{c}
D \\
s-1
\end{array}
\right)\left(
\begin{array}{c}
M-D \\
x
\end{array}
\right)(D-s+1)}{\left(
\begin{array}{c}
M \\
x+s-1
\end{array}
\right)(M-x-s+1)}

So, once again there is an elegant solution embodied in one distribution.

Trials needed (n) = s + InvHypergeo(s,D,M)

Note that, like the Negative Binomial distribution, the InvHypergeo(s,D,M) distribution models the number of failures. When added to s, we get the total trials needed.

However, the current version of @RISK does not have the InverseHypergeo distribution in its library, so we need to construct the distribution manually and then use the RiskDiscrete distribution of @RISK to sample from it, as shown in Model Image RemovedImage AddedHyperGeoFailures.

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For other processes, there may be elegant solutions to the number of trials needed to achieve a certain number of successes, but it is much more likely that simulation models will need to be built from scratch to determine the distribution. We give three examples here for you to get an idea of the type of techniques that will help you produce such models.

Example 1

You are a government body doing research into the effects of marriage and smoking on peoples' health. You are doing a random telephone survey and you require 50 people from each of the four possible categories. From previous studies you know that 32% of people agree to participate in this type of survey when called. How many calls will you need to make, given that previous studies show the population to be split into the four categories as follows:

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title @Risk

Example 3

This is an extension to this topic. Here we will not only count the failures, but also sum the random variables.

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