Transcript Ch5-1 SHW Solutions

Chapter 5 – Part 1
Solutions to SHW
1. What do we mean by a process that is
consistently on target?
USL
LSL
Target
Amount of Toner
Process is consistently on target if the distribution is
tightly clustered around the target, or, equivalently,
if the variance around the target is small.
2. What do we mean by a process that
is consistently on target?
USL
LSL
Target
Mean
Amount of Toner
Process is consistently off target if the distribution is
off target but tightly clustered—or has a small
variance—around its mean.
3. What do we mean by a process that is
haphazardly on target?
USL
LSL
Target
Process is haphazardly on target if the distribution is
on target but exhibits a great deal of variation around
The target.
4. Refer to Jane and Sam in the notes to
Chapter 5, Part 1. Draw the distribution
of two machines, one behaving like Jane
and the other like Sam. Which machine
is easier to fix? How would you fix it?
Jane
(Machine B)
Sam
(Machine A)
Target
Problem 4 - Continued
• Easier to adjust mean to target than to
reduce variance.
• Machine A: reduce variance
• Machine B: adjust mean to target
5. How does better quality increase
productivity?




As quality improves, rework and scrap
decrease
This results in fewer inputs being used to
produce a units of output.
Also, more good units are produced the first
time.
Since better quality means high output of good
units and fewer inputs, productivity—output
divided by inputs—increases.
Problem 6
a) Find the loss function
LSL = 0.6
USL = 1.6
Target = (USL+ LSL)/2 =(1.6+.6)/2 = 1.1
USL = Target + a
1.6 = 1.1 + a
a=0.5
Problem 6
L  k(X  T )
2
R
$12
k 2 
 $48
2
a
(0.5)
L  $48( X  1.1)
2
Problem 6
b) If X = 1.3,
L  $48(1.3  1.1)  1.92
2
c) If the company ships a brake pad with a
thickness of 1.3 inches, the company will
impose a loss of $1.92 on the customer.
The loss is due to the thickness being
0.2 inches off target.
Problem 6
d) Expected loss
E ( Loss )  k
 ($48)(0.75)
 $36

The company imposes, on average, a loss of
$36 on its customers from shipping off target units.
(Note that some units will have a loss Greater than
$36 but other units will a loss of less than $36.
The average of the losses of all units shipped
will be $36.)
Problem 7
•
•
•
•
Ashi Newspapers on April 17, 1979 reported
that: Identical sets were assembled by Sony in
a plant in Japan and in a plant in the U.S. with
same design, and the same parts.
U.S. customers preferred TV’s assembled in
Japan to those assembled in U.S., because of
better color.
The U. S. plant performed 100% inspection.
As a result, none of the sets assembled in U.S.
were out of specs.
However, Japan shipped all set “as is” without
inspection.
Problem 7
• The result was that 0.0027% of sets assembled
in Japan were out-of-spec, and thus defective.
• The color density (the quality characteristic of
interest, X) of sets produced in Japan was
normally distributed, while the color density of
the sets produced in the U.S. had a uniform
distribution. The specification limits are 10 and
20.
• The distributions at each plant are shown below.
Problem 7
Color Density
Distribution
Japan-built sets:
.3% out of limits
US-built sets:
100% within limit
y
10
15
20
Problem 7
The cost of repairing a TV set that is at the specification limits to the target value of 15
was $6 at both plants.
a. Find the expected loss at each plant and explain the meaning of your answer.
(Ans. U.S. plant, $2.00; Tokyo plant, $0.69)
b. Why is the expected loss less at the plant that has a higher percentage of out-ofspec TV sets?
c. Is your result consistent with the response of consumers?
Hint: If the random variable X has a uniform distribution over the range a to b, variance
of the distribution is
(b  a) 2
 
12

For the Tokyo plant, you need to use the appropriate table to find the z value (standard
normal random variable) that corresponds to the desired area under the normal
distribution. Once you find the z value, you should be able to obtain the standard
deviation.
Solution to 7a –Tokyo
 Defect rate = .003. Split between two tails
beyond spec limits = .0015 in each tail.
LSL
USL
.0015
.0015
10
15
20
Use Appendix B, p. 652 to find z. Since the tail area
above USL is .0015 and Appendix B gives the area
between 0 and z, we Look up the area between 0
and z, which is .5000 -.0015 =.4985
Table area =.4985
.0015
.0015
LSL
USL
0
z = 2.965
From Appendix B, the z value is z = 2.96 or 2.97,
so split z to get 2.965. See next slide.
z Table (Text, p. 652)
z
.00
.01
.02
.
.
.
.06
.07
0.0
0.1
0.2
.
.
.
2.9
.4985 .4985
Solution to 7a -Tokyo
Use formula for z to solve for standard
deviation:
z
USL  mean
2.965 

20  15

   1.69
Solution to 5a - Tokyo
 Expected loss
R $6
k  2  2  $0.24
a
5
E ( Loss )  k
2
 ($0.24)(1.69)
 $0.69

Solution to 5a-Tokyo
• On average, each TV set shipped from the
Tokyo plant imposes a loss $0.69 on the
customer.
Solution to 5a – U.S. Plant
(b  a)
(20  10)
 

 8.33
12
12
2
2
2
E ( Loss )  k
 ($0.24)(8.33)
 $2.00

Solution to 5b
• Why is the expected loss less at the plant
that is producing a higher percentage of
out of spec sets—the Tokyo plant?
Solution to 5b
• E(Loss -Tokyo) =$.69
• E(Loss - U.S.) =$2.00
• The reason why the expected loss is less
at the Tokyo plant than at the U.S. plant is
because the variance is smaller at the
Tokyo plant, so there is less variability
around the target.
• The sets are therefore more consistently
on target, even if 3 out of 1000 sets are
out-of-spec.
Solution to 5b
• At the U.S. plant, all sets are in spec. but
they are haphazardly on target.
• Since the U.S. distribution is uniform, the
percentage of sets on target is the same
as the percentage of set near either on of
the spec. limits.
• This means that the customer is just as
likely to get a set that is on target as one
that it on the lower or upper spec. limit.
Solution to 5b
• Note that, although the U.S. plant is
performing 100% inspection and is not
producing any out-of-spec sets, it still has
a higher expected loss. Why?
• The reason is that mass inspection will not
reduce the expected loss because it does
not reduce the variance of color density.
• To reduce variance, we must improve the
process by, for example, using better parts
and/or materials, better maintenance of
tools and equipment, etc.