Process Measures — Productivity and Efficiency

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Part of managing processes is to measure their performance.  This article will discuss 2 basic process measures: Productivity and Efficiency.

Worker Productivity

Simply, productivity is measured like this:

Productivity = Outputs / Inputs

For instance,

What is the labor productivity in Hours for each car type?

Car X: (4,000 cars / 20,000 Hours) = 0.2 Cars / Hour
Car Y: (6,000 cars / 30,000 Hours) = 0.2 Cars / Hour

How about in labor Productivity in Dollars?

Car X: [(4,000 * $8,000) / (20,000 * $12)] = $133.33 / Car
Car Y: [(6,000 * $9,500) / (30,000 * $14)] = $135.71 / Car

So, based on the data set above, it appears that Productivity by Hours, both Car X and Car Y are the same; but, Productivity by Dollars, Car X is cheaper to make given the labor hours and hourly rate.

How is this practical?  Below is a direct citation from the Harbour Report, 1998:

Labor hours needed for stamping, power train, and assembly operations:

Harbour Report

(100%	Nissan	27.6 hours
(168%)	GM	46.5 hours
(126%)	Ford	34.7 hours

If GM could operate at Nissan’s level of productivity, they’d save themselves about $4.4 Billion Per Year.  Measured another way, GM has about 55,000 more workers than it needs.

Worker Efficiency

Efficiency is measured with the following equation:

Efficiency = [100% * (actual output / standard output)]

Standard Output in the equation above is a number that is arrived at by looking at historical data for the job and by experience.  One would hope that that number is not an arbitrary one, but a number that is derived by looking at a historical time series.

Here’s an example:

Shmula’s Bodywork does automotive collision work.   An insurance agency, using actuarial data, has determined that the standard time to replace a fender is 2.5 hours (ie, “standard output” = 0.4 fenders per hour), and is willing to pay Shmula $50 per hour for labor (parts and supplies are billed separately).  Shmula pays its workers $35 per hour.

Suppose Shmula’s workers take 4 hours to replace a fender.  What is Shmula’s labor hour efficiency?  Given Shmula’s labor costs, will Shmula make money on the job?

Using the equation and the data above, we get:

(1 fender / 4 hours) / (1 fender / 2.5 hours) = .625 * 100% = 62.5% efficiency

2.5 hours * $50 = $125 paid by insurance
4 hours * $35 = $140 costs

Shmula will lose $15 per fender.

Economic Profits are important to Shmula; Given the answer above, how efficient does Shmula need to be to break even?

[($125) / ($35/hour)] = 3.57 hours to Break Even

We know that, Efficiency = 100% * (actual output / standard output).  So,

(1 fender / 3.57 hours) / (1 fender / 2.5 hours) = .7003 * 100% = 70.03% efficiency.  Shmula will need to improve efficiency to 70% or better to make money.

Often times, I hear people use the phrases “productivity” and “efficiency” carelessly, not completely understanding what they mean.  These terms have technical definitions and are very practical for business.  Still, given the explanation above, one must excercise good judgement in detemining which processes ought to be measured by Productivity and Efficiency and balance these measures with other items that might be important to the individual, firm, and industry.  Some measures might make sense for some activities, but not others.

	This post was written by Pete Abilla