## Measuring Productivity

I found on the internet that there are 2 main multifactor productivity measures:

1) total output / (labour + capital + intermediate inputs)

2) value added / (labour + capital)

, where value added = total output - intermediate inputs

This two measures make sense but my teacher said that if we want total productivity and use value added we should do:

value added / (labour + capital + intermediate inputs)

This measure makes no sense to me, since intermediate inputs do not produce value added.
What do you think?

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 Recognitions: Homework Help Science Advisor If you need more intermediate inputs to add the same value, something is worse, so you might want some index to go down.

 Quote by mfb If you need more intermediate inputs to add the same value, something is worse, so you might want some index to go down.
Yes, I would include losses of intermediate inputs in the denominator. But the amount I took from numerator shouldn't I take from denominator too? (since I know that amount is represented in both numerator and denominator)

I mean, if the original ratio is: output / input, then if I take the amount "a" from output, (and "a" is also included in input), I should take "a" from input, so it would be:

(output - a) / (input - a)

and not:

(output - a) / input

right?

Recognitions:
Homework Help

## Measuring Productivity

value added = total output - intermediate inputs - (other variables)

You already use (output - a) / (input - a) in your formula, where a are intermediate inputs and (output-a) is "value added".

 Quote by mfb value added = total output - intermediate inputs - (other variables) You already use (output - a) / (input - a) in your formula, where a are intermediate inputs and (output-a) is "value added".
Yes, but my teacher suggested this measure:

value added / (labour + capital + intermediate inputs)

which is:

(output - a) / input

This is my actually question, is he wrong? ^^

 Recognitions: Homework Help Science Advisor If you use "a" as sum of all inputs, it is (output - input) / input = output/input - 1 That is fine, and the value looks interesting.