# Measuring Productivity

by luis20
Tags: measuring, productivity
 P: 53 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?
 Mentor P: 12,113 If you need more intermediate inputs to add the same value, something is worse, so you might want some index to go down.
P: 53
 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?

 Mentor P: 12,113 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".
P: 53
 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? ^^
 Mentor P: 12,113 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.

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