Standing consumption 200 kWh/week Product A 500 kWh/tonne Product B 300 kWh/tonne Product C 1000 kWh/tonneIf the output volumes in a given week were, say, 100 tonnes of A, 20 tonnes of B and 5 tonnes of C, the expected gas consumption for that week would be:

200 + (100 x 500) + (20 x 300) + (5 x 1000) = 61,200 kWh

`
However, this multiple-driver procedure can be avoided. Instead of processing the production figure for each product,
we can add them together on a weighted-sum basis. We can convert the production figures into a
single number representing the `*equivalent output volume of one product*. For example, we could choose
Product A as the 'index product' (it accounts for most of the consumption).

So as we made 20 tonnes of Product B and its gas requirement is 300/500 that of Product A, we can say that this was equivalent to making 20 x 300/500 = 12 tonnes of A. The 5 tonnes of Product C, on the same basis, was the equivalent (in terms of gas requirement) of 5 x 1000/500 = 10 tonnes of A. Hence the total production could be expressed as

`
100 + 12 + 10 = 122 tonnes of A equivalent
`

Product A requires 500 kWh/tonne and the standing consumption is 200 kWh per week, so the target for 122 tonnes-of-A-equivalent is

`
200 + (122 x 500) = 61,200 kWh
`

Which agrees with the result obtained on the multiple-driver basis. However, it is considerably more convenient to have a single figure for the driving factor, which the weighted-sum method achieves.

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