Six Sigma Quality at Flyrock Tires Susan Douglas, vice president of quality at F
ID: 1732805 • Letter: S
Question
Six Sigma Quality at Flyrock Tires
Susan Douglas, vice president of quality at Flyrock Tires, was wondering how to explain the
value of Six Sigma quality to the people within her organization. Flyrock, a major manufacturer
of tires in the United States, had five manufacturing facilities where tires were made and another
20 facilities for various components and material used in tires. Quality had always been
emphasized at Flyrock, but lately quality was a bigger issue because of recent fatal accidents
involving tires made by other manufacturers due to tread separation. These tire problems had
received significant coverage in the popular press and all manufacturers were under pressure to
show they were working to ensure that problems did not arise in the future. Several lawsuits had
been filed on behalf of the families of victims in these accidents and much of the discussion i
these lawsuits was likely to involve the way these tires were built. Douglas wanted to push
through a Six Sigma program and significantly improve the level of quality at Flyrock, and she
needed to build a solid case for it first.
Manufacturing Tires
While tire technology had made many advances over the decades from old-fashioned,
inexpensive, bias-ply tires to today’s steel-belted radials, the basic process of making tires had
essentially remained unchanged and was shared by all manufacturers.
Rubber, the major component of tires, arrived in bales at the tire factory where it was mixed
with chemicals and other ingredients. Each manufacturer had its own recipe for this process.
Large blades in a machine broke down the raw rubber and mixed the chemicals, much like a
kitchen mixer would. The rubber compound was then rolled into a milling machine where it was
forced back and forth through a series of rollers to more thoroughly mix the chemicals added
earlier.
Tread referred to the part of the tire that touched the road, while sidewall referred to the sides
of the tire. Tread and sidewall rubber went through the same process but with a different chemical
mix. After further processing, each type of rubber was extruded into a single continuous strip.
These strips were put onto a conveyor system with the strips of tread rubber in the middle and the
sidewall rubber on either side.
The long strips of tread and sidewall rubber were then sent through a coating apparatus that
applied adhesive. After the strips had cooled, they were shrunk and cut to the exact size needed.
While the rubber was cooling, strips of fabric, usually polyester, were coated with another rubber
blend. These strips were used to make the body of the tire after the layering process began.
Another key component was the steel hoop for steel-belted radial tires. The hoop formed the
backbone, or the metal skeleton, of the tire. The steel strands were aligned into a ribbon coated
with rubber for adhesion and then formed into loops that were wrapped together.
With the ingredients prepared, the tire was assembled on a building drum. Radial tires started
with a layer of synthetic gum rubber, called an inner liner, which sealed in air. This is why the tire
did not need a tube. Next came the rubber-coated ply fabric, followed by two more layers to help
further stiffen the tire. The material was then put into a tire-building machine, which shaped the
tire to its near-final dimensions and made sure all the components were in proper position for the
final molding. The tire was then inspected, whereupon it was ready for curing.
During the curing process, the tire received its final shape and tread pattern. In a process
called vulcanizing, the tire was put into a hot mold that engraved the tread pattern and sidewall
markers. The tire was heated, or cured, at more than 300 degrees Fahrenheit for 12 to 25 minutes,
depending on its size. When the curing was done, it was sent to a final finish area for another
inspection. If there were any blemishes or outside flaws, the tire was rejected.
Adhesion flaws and other internal problems were not visible, so manufacturers randomly
pulled tires from the production line and cut them apart to look for defects.
Quality Issues at Flyrock
Given the large number of steps in building a tire, errors tended to accumulate. As a result, even
if each step produced only 1 percent defects, at the end of 20 steps only 81.2 percent of the product would be of good quality. Since each factory produced about 10,000 tires per hour, such a process would result in about 1,880 defective tires per hour. Thus, it was very important that each stage had a high level of quality.
Another quality issue was related to the settings on various machines. Over time, these settings
tended to vary because of wear and tear on the machines. In such a situation, a machine would
produce defective product even if the machine had the correct setting. To detect such situations,
Douglas implemented a statistical process control (SPC) program. At the extruder, the rubber for the
AX-527 tires had thickness specifications of 400 ± 10 thousandth of an inch (thou). Douglas and her
staff had analyzed many samples of output from the extruder and determined that if the extruder
settings were accurate, the output produced by the extruder had a thickness that was normally
distributed with a mean of 400 thou and a standard deviation of 4 thou.
Conditions:
When the extruder setting is accurate, the specification limits are:
LSL=390
USL=410
Mean=400
STD Deviation=4
The proportion of rubber extruded will be .9876
When Douglas is asked to take a sample of 10 sheets of rubber each hour from the extruder and measure the thickness of each sheet, and uses 3 sigma control limits, the control limits used will be:
UCL=.0124+3*.035=.117
LCL=-0.093
Questions
The questions are designed to help Susan Douglas understand the current capability of the
extrusion process and quantify the benefit that may be achieved if the extruder could be
transformed to a Six Sigma process.
3. If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting
is 400 thou. Under this condition, what proportion of defective sheets will the extruder
produce? Assuming the control limits in question 2, what is the probability that a sample
taken from the extruder with the worn bearings will be out of control? On average, how many
hours are likely to go by before the worn bearing is detected?
4. Now consider the case where extrusion is a Six Sigma process. In this case, the extruder
output should have a standard deviation of 1.667 thou. What proportion of the rubber
extruded will be within specifications in this case?
5. Assuming that operators will continue to take samples of 10 sheets each hour to check if the
process is in control, what control limits should Douglas set for the case when extrusion is a
Six Sigma process?
6. Return to the case of the worn bearing in question 3 where extrusion produces a mean
thickness of 403 thou when the setting is 400 thou. Under this condition, what proportion of
defective sheets will the extruder produce (for the Six Sigma process)? Assuming the control
limits in question 5, what is the probability that a sample taken from the extruder with the
worn bearings will be out of control? On average, how many hours are likely to go by before
the worn bearing is detected?
Explanation / Answer
Answer:
Question 1
= 400 - 3* 4/SQRT(10) - 403 / 4/SQRT (10) = - 5.37
likewise upper limit = 0.6282
Hnce Z value = 0.7324
Hence Probability of example leaving out of control = 1-0.7324 = 26.76 %
Now the standard runtime before this is ddetected = 1/0.2676 = 3.73 hours
Question 2
Upper manage limit = Mean + 3 Sigma/ Sqrt (10) = 400 + 3*4/Sqrt(10) = 403.79
inferior control limit = Mean - 3 sigma = 400- 3*4/Sqrt(10) = 396.20
Qeustion 3
We will use same procedure as in question 1
assuming so as to standard deviation of procedure remain same
We will calcualte Z = USL - Mean / Standard devitaion = 410-403 /4=7/4 = 1.75
We will seem for the value in Z bench we get Z value as 0.9599 hence
% defectives =( 1-0.9599)* 100 = 4.01
Now to calculate the likelihood that the value will fall beyound control limits
we calclulate as below
= Mean - 3 Sigma/sqtr(10) -( Mean+ Sigma ) /Sigma/Sqrt(10)
= 400 - 3* 4/SQRT(10) - 403 / 4/SQRT (10) = - 5.37
likewise upper limit = 0.6282
Hnce Z value = 0.7324
Hence Probability of example leaving out of control = 1-0.7324 = 26.76 %
Now the standard runtime before this is ddetected = 1/0.2676 = 3.73 hours
Question 4
Upper requirement Limit in this container will be 400+ 3 * 1.667/Sqrt(10) = 400+ 1.581 = 401.581
Z = (401.581 - 400) /4 = 1.581/4 =0.3953 look for Z score in this = 0.6517
sample falling exterior control limit = 1-0.6517=0.34 that is 34