Initially, some might not see the value of including this second factor, but I found it to be the best way to compare the effort required to bring about savings. To illustrate, imagine there are 100 stores selling ProductX within an area fitting the previously discussed conditions. 99 of them are selling it for $1.00 and one of them for $0.50, resulting in an SF of 0.5. If the distribution is changed, however, and 50 of the retailers price ProductX at $1.00 and 50 at $0.50, the SF would remain 0.5. It's true the potential savings in the area is the same in both instances, but the effort required to bring about those savings is much higher in the former case than in the latter. This fact, intuitive in this example, can be revealed in more complicated data sets by comparing the VFs. For the above cases, the former VF (0.0503) is much less than the latter VF (0.3350), confirming the obvious. Further confirmation of the method is obtained by considering a third case in which 99 prices are $0.50 and one is $1.00. This price distribution results in a VF of 0.0990, correctly indicating an increased degree of difficulty over the more evenly distributed case.
former: 1 @ $0.50, 99 @ $1.00, VF 0.0503.
latter: 50 @ $0.50, 50 @ $1.00, VF 0.3350.
third: 99 @ $0.50, 1 @ $1.00, VF 0.0990.
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