Inspection
count, random part arrangement, and probability of no inspection
This
post intends to calculate of how many inspections are needed to
meet a false pass percentage when the parts are randomly arranged.
The probability that the part will be the correct side toward the
camera, not blocked by double stacking, or other unknown reason are
considered. It is assumed that the part arrangement is randomized
before each inspection.
Calculating
the minimum number of inspections is based on the average probability
that the part will be arranged in a way suitable for inspection.
Similar to the question, “how many times must a coin be flipped to
get heads?”
The
probability (p) of getting heads is 0.5 (50%).
The
negative result (tails) probability (q) is 1 – p = 1 – 0.5 = 0.5
(50%).
Average
flips (inspections) needed = 1 / p = 1 / 0.5 = 2
The
minimum part inspections must considered probability of wrong side
up, part double stack, and unknown random error.
Like
the coin flip, there is a 50% (0.5) chance of wrong side.
The
double stacking must be mechanically limited to 10% (0.1),
and
the unknown error is guessed to be 5% (0.05) [it's unknown].
These
errors are independent, the probabilities are additive.
The
probability the part can't be inspected is 65% (q = 0.50 +
0.10 + 0.05 = 0.65).
The
probability that the part can be inspected is 1.0 – 0.65 =
0.35 = p = 35%
Only
35% of the randomly arranged parts will be able to be inspected.
The
minimum number of inspections needed is 1/p = 1/0.35 = 2.857 or 3
inspections.
Now
considered the likelihood that the part fails to be inspected 2 times
in a row. The sequential probability is q^ˣ,
where x is the number of repeated inspections.
If
there is a 65% chance the part is not inspected (q = 0.65). Then
randomizing the part arrangement and repeating the test gives a
42.25% chance the part isn't inspected (q^²
= 0.65^² = 0.4225).
Adding more inspections continues to reduce the chance that the part
is not inspected. After the 7th inspection, there is less
than a 5% chance a part has not been inspected. By the 11th
inspection, less than 1% have not been inspected.
How
many inspections are needed depends on the maximum allowable False
Pass Rate and the true Process Defect Rate. If 1 part in 10,000 is
allowed to be bad the False Pass Rate is 0.01% (0.0001). This is
0.01% of all parts. If the Process Defect Rate is 2%, for every
10,000 parts 200 are bad. The “No Inspection” parts are assumed
to have the same evenly distributed 2% defect rate. The maximum
allowable “No Inspect” rate = False Pass Rate / Process Defect
Rate = 0.0001 / 0.02 = 0.005 (0.5%). Based on the chart below, 13
inspections are needed.
To
calculate the number of inspections required for randomly arranged
parts:
Inspection
Count =
log(
False Pass Rate / Process Defect Rate) / log( Probability No
Inspection)
Where:
False
Pass Rate = percentage of maximum allowable bad parts getting through
the process.
Process
Defect Rate = percentage of bad parts made by the process.
Probability
No Inspection = the sum of all error probabilities that would prevent
the part from being inspected. [In this example, Wrong side up +
Double stack + unknown error.]
For
the question “Is it cheaper to add more inspections, or to add
material handling?”
This
calculation is useful when estimating the expense of random
arrangement inspections versus material handling positive part
presentation.
20180228 Lowell Cady
links
minimum-number-of-flips-to-guarantee-heads
logarithms and the calculator
How many coin flips to get heads
logs
