Measuring Productivity


by luis20
Tags: measuring, productivity
luis20
luis20 is offline
#1
Nov11-12, 08:07 AM
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?
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mfb
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#2
Nov11-12, 10:45 AM
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P: 10,840
If you need more intermediate inputs to add the same value, something is worse, so you might want some index to go down.
luis20
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#3
Nov11-12, 10:57 AM
P: 53
Quote Quote by mfb View Post
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?

mfb
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#4
Nov11-12, 11:17 AM
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P: 10,840

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".
luis20
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#5
Nov11-12, 11:32 AM
P: 53
Quote Quote by mfb View Post
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? ^^
mfb
mfb is offline
#6
Nov11-12, 03:29 PM
Mentor
P: 10,840
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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