View Full Version : Capability and Tool Replacement Please Help!

11-17-2009, 11:24 AM
I think this is the right forum because my course is operations management and and the problems are in the statistics and probability category. If I am wrong please advise where this should have gone.

The chapter I am doing is about quality control, and includes problems on control charts, run tests, and Cp analysis (capability analysis). The problem I am having is that the homework question requires a formula not explained in the book and I don't know where to start.

Here is the problem:
Specs for a metal shaft are much wider than the machine used to make the shaft is capable of. Consequently, the decision has been made to allow the cutting tool to wear a certain amount before replacement. The tool wears at the rate of .004 cm per piece. The process has a natural variation (sigma) of .02 cm and is normally distributed. Specs are 15.0 to 15.2 cm. A three-sigma cushion is set at each end to minimize the risk of output outside of the specs. How many shafts can the process turn out before the tool replacement becomes necessary?

There is a diagram that is something like this:
(Lower spec)|15.0<----3-sigma--->|(0)<--start process mean -- Wear rate = .04/p -- end process mean -->(#Shafts)|<---3-sigma--->|15.2 (Upper spec)

The only formula I have that seems relevant are:
Cp = (Upper spec - Lower Spec) / 6 standard deviations (use if process centered on mean)
or if not centered on mean use smaller of 2 formulas below
Cpk = (Mean - Lower Spec) / 3 standard deviations
Cpk = (Upper Spec - Mean) / 3 standard deviations

I can't see how to use any of those formulas to solve the problem. Sometime the book expects me to come up with algebra out of my head, but I am at a loss and haven't been able to find any help with this. I hope someone will help as I have been trying for several days to get a handle on it.


11-17-2009, 06:03 PM
Thanks all who have viewed my post. I have solved the problem.

To find the answer I did the following:

3-sigma = 3 x .02 cm (sigma) = .06 cm

To get the start process mean I subtract 3-sigma from the Upper spec and add 3-sigma to the lower spec:

15.0 cm + .06 cm = 15.06 cm (start process mean)
15.2 cm - .06 cm = 15.14 cm (end process mean)

#Parts before replacement = (End Mean - Start Mean) / Wear Rate

(15.14 cm - 15.06 cm) / .004 cm = 20 cm

20 parts can be produced by this process before the tool must be replaced.