Taguchi Loss Function example

Marc

Hunkered Down for the Duration with a Mask on...
Staff member
Admin
#1
I got an email about a page here on the site that the guy from the old Red Road (who sorta disappeared some years ago) did. I'm not knowledgeable in this so I'm asking you folks for some help.

The page in question is: Taguchi Loss Function example

Here is the email:

There seems to be an error in the calculations.

Here's my rationale below:

The average of the parts is 0.501 and the standard deviation is about 0.022.(I AGREE)
find k,

using Lx = k(x-t)^2
$0.45 = k (0.550 - 0.500)^2
k = 180 (NOT 18000)
next,
using the Average loss equation: L=k * (s^2 + (pm - t)^2)
L = 180 (.022^2 + (.501 - .500)^2) = .0873 (NOT 8.73)
So the average loss per part in this set is about $0.09. (NOT $8.73)

For the loss of the total 30 parts produced,
= L * number of samples
=.0873 X 30 = $2.62 (NOT $261.90- which makes sense even if you scrapped all 30 parts at $0.45, your total loss would be less than $15)

Other than the loss being zero at 0.500", assuming that s also equals zero, I'm totally at a loss at how you obtained the numbers in the last 3 paragraphs.

I triple and quadruple checked my math and unless I'm not seeing something. I'd really appreciate an explanation as to how you got those numbers.


Can anyone help me with this?
 
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Marc

Hunkered Down for the Duration with a Mask on...
Staff member
Admin
#3
I think you're right, Kevin.

Has anyone else taken a minute to look this over?
 

Tim Folkerts

Super Moderator
#4
I agree that there seems to be a mistake in the original page.

The simplest solution would seem to be to change the failure cost to $45 in the description of the situation and then change the one line

$0.45 = k (0.550 - 0.500)^2

to
$45 = k (0.550 - 0.500)^2

Then I think the rest of the page would end up correct.

Tim F
 
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