# DPO without multiplying by number of units?

In a book, Six Sigma for Dummies, the authors define DPO as number of defects per number of opportunities but they do not multiply that by number of units.

So basically they say that if you have 14550 opportunities to defect and you find 153 defects, the DPO is 0,011.

I think it does not make sense. What if I have a paperclip with one opportunity and the one I have is faulty, thus DPO=1 and DPMO=1000000?

This reference defines the DPO as

``````DPO = N_defects / (Number of units * Number of opportunities per unit).
``````

I'll base my answer on this definition.

The number of opportunities by itself already contains information about the number of units, since

``````N_opportunities = Number of units * Number of opportunities per unit.
``````

In your example, one paperclip (one unit) has a certain probability of containing a defect. There may be, say, 3 kinds of defects (bent, broken, brittle), so

``````N_opportunities = 1 unit * 3 defects per unit = 3 opportunities
``````

If you found your one paperclip was bent then the DPO would be

``````DPO = 1 defect / 3 opportunities = 0.33 (2 decimal places)
``````

Of course, this metric is only useful for considering samples which are sufficiently large that the probabilities of the defects manifest in a meaningful way e.g. so that expectation values start to become useful quantities.

For example, consider 2 samples (A and B), each of 10000 paperclips. If sample A had 20 bent paperclips, 3 broken paperclips, and 7 brittle paperclips from a sample of 10000 then

``````DPO_A = 30 defects / (10000 units * 3 opportunities per unit) = 0.001
``````

whereas, if sample B had 10 bent paperclips, 2 broken paperclips, and 3 brittle paperclips

``````DPO_B = 15 defects / (10000 units * 3 opportunities per unit) = 0.0005
``````

The DPO now tells you something interesting about the samples and, perhaps, the way in which the samples were created.