regression test -- data set access (Weighted Geometric Mean)

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Equation / Last modified by MichaelBartmess on 2014/09/19 15:53
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regression test -- data set access (Weighted Geometric Mean)
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MichaelBartmess.regression test -- data set access (Weighted Geometric Mean)

This equation computes the weighted geometric mean of a set of input criteria and their associated weights.  The criteria and weights are input from a Data Set. The user specifies the Data Set to be used by simply making a copy of this equation and adding the Data Set's UUID to the enumerated list of TableIDs that is the input to this equation.  The Data Set used must have the weights in a column labeled "Weights" and the criteria data in a column labeled "Criteria."

The weighted geometric mean is a powerful equation in decision analysis applications, allowing you to weight and combine externally generated values for a set of decision criteria.  This Data set version of the equation is limited only by the number of rows of data provided in the Data Set whose UUID you added to the enumerated set. It is assumed the Criteria data has been normalized. 

Given a set of data,  X = {`x_1, x_2, ... , x_n`}, the corresponding weights are: W = {`w_1, w_2, ... , w_n`}.  This equation computes the resultant weighted geometric mean.

If all the weights are set equal, the weighted geometric mean is equivalent to the common geometric mean.


vCalc has the means to attach an equation, in this case the weighted geometric mean, to columns of data in a data set, so this statistical function can be applied to a much larger set of decision criteria than ten items.    This alternative version of the weighted geometric mean equation which pulls the decision criteria values and decision criteria weights from a Data Set can be easily copied and used for many forms of decision analysis, i.e. making choices amongst complex systems of decision criteria.

This equation's particular implementation of the weighted geometric mean is intended to to support comparisons of combined criteria that are best combined when normalized.  In the example presented here, the criteria are all monetary elements of the problem representing the costs  of various components that may or may not be included in the final combined choices.  Thus the criteria are normalized to the value of a dollar, making the use of these cost components very straightforward.

The weighted geometric mean has the ease-of-use characteristic that any data value whose weight is set to zero will not affect the result.  So, this allows you to do quick what-if analysis where, in addition to trying different combination of data values and weights, you can test the question of "what if I don't consider the n-th criteria at all".

The obvious use in decision support applications is the ability to changes weights representing relative importance of a specific criteria and/or the criteria values themselves and compare the results to another set of weights and criteria values.


In this example a vendor is trying to make a decision whether it is cost effective to sign up to provide their product at an upcoming event.  The event has a fixed fee associated with it, so the fee is the first cost component. If the vendor signs up but then backs out, there is a cost associated with angering the event sponsors and the cost is attributed to the potential revenue generated in future events.

The vendor also has a choice to hire staff to support the event.  The alternative could be to do the event with the vendor themselves the only person supporting the event.  The cost of doing so is a pain function which can be attributed a dollar value equivalent to the number of staff that might support the event.  The vendor estimates they need three staff to fully replace them, so zero pain is equivalent to the cost of three staff for the event.

There are other fixed costs that are associated with choosing to attend the event: hotel costs (for the vendor and any staff, food costs, gas expenses, cost of product.

One additional cost is associated with the time to prepare for the event, which would be taken away from time to produce additional product for sale at later events.  Since the time to prepare for, travel to/from, and support the event can be accurately estimated, it can be associated with the cost of goods produced in that same amount of time.


In this case the vendor chooses to attend the event, and so pays the event fee of $500.  The vendor decides to take hire two staff at eight dollars an hour for the event duration of two 12 hour days.  The drive is approximately 1-3/4 hours and 110 miles.  Travel time is paid time since the staff cannot be using that time themselves.  That makes the staff wages 27.5 total hours at $8 /hour or total of $220.

Because there are two staff, the pain factor is cut by 2/3 for the same duration of the show.  To look at alternatives, the vendor in the final analysis could choose to ignore the pain factor of doing the show (by simply zeroing its weight)  but in this comparison case the vendor chooses to consider their pain factor.

The drive is approximately 1-3/4 hours and 110 miles. Gas at $3.75 / gallon and 15 miles per gallon for the company vehicle make the gas expense $55.

Two rooms are acquired for two nights at $58/night, which totals $232.

The vendor predicts the revenue from this event will be approximately $4500.  The cost of angering the event sponsors would be not being invited back for five years and thus giving up the same revenue for five years.  So, that value is $22,500.