SCM 302
SPC Practice Solutions
Problem 1
a. Calculate the 3σ upper and lower control limits and the center lines for both the mean ( ̅ ) and range (R) charts. Use
the “factors” from the class slides or textbook assuming a normal distribution.
Sample Number
1
2
3
4
5
6
7
1
469.92
457.34
473.96
480.06
467.46
473.06
456.27
Rotation Time (msec)
2
3
4
5
468.67 479.76 454.38 469.58
454.37 475.28 453.46 480.03
459.26 460.42 462.04 450.60
469.86 456.42 460.63 465.66
476.56 474.01 465.34 475.27
475.86 472.97 454.93 470.73
476.37 479.50 459.86 470.73
6
454.46
480.40
451.52
466.99
462.97
466.24
452.35
Mean
(X-bar)
Range
(R)
466.128
466.813
459.633
466.603
470.268
468.965
465.847
25.380
26.940
23.360
23.640
13.590
20.930
27.150
Using ̅ and R values, we get ̿=466.32 and ̅ =23.00. And the factors can be read from the slides given in class for
sample size=6:
Factors
A2
D3
D4
0.48
0.00
2.00
Then the following line and limits can be calculated:
̅ bar chart
̿=466.32
Center Line (CL)
Upper Control Limit (UCL)
Lower Control Limit (LCL)
̿
̿
̅
̅
1
477.36
455.28
R chart
̅ =23.00
̅
̅
46.00
0.00
SCM 302
SPC Practice Solutions
b.
1. Using the upper and lower control limits and center lines you computed in part a), plot the sample mean and
sample range for each of the 10 samples on the appropriate control chart.
Mean Chart (X-Bar)
485.000
480.000
475.000
470.000
465.000
460.000
455.000
450.000
445.000
440.000
1
2
3
4
5
Sample Mean
6
CL
7
8
UCL
9
10
LCL
Range Chart (R)
50.000
45.000
40.000
35.000
30.000
25.000
20.000
15.000
10.000
5.000
0.000
1
2
3
4
Sample Range
5
6
CL
7
UCL
8
9
10
LCL
2. Using your results from both charts, is the production process still in control? Why or why not?
Although all observations are between the UCL and LCL of the chart, two points are above the UCL of ̅ chart.
Therefore the process is no longer in control. The R chart shows that the dispersion (range) has not changed,
but the X-bar chart shows that the mean has shifted. While the procurement costs for the key component may
be lower with the new supplier it appears that the change to the new supplier’s component is also leading to
slower machine (in terms of higher msec per rotation).
2
SCM 302
SPC Practice Solutions
Problem 2
a. Calculate the 3σ upper and lower control limits and the center lines for the proportion of defective batteries.
1000
15
Sample size n
# of samples
Week
1
2
3
4
5
6
7
8
# of defects per lot
17
14
9
3
20
6
6
4
Proportion
1.70% 1.40% 0.90% 0.30% 2.00% 0.60% 0.60% 0.40%
Week
9
10
11
12
13
14
15
Total Average
# of defects per lot
17
2
10
2
5
18
1
134
8.93
Proportion
1.70% 0.20% 1.00% 0.20% 0.50% 1.80% 0.10% 13.4%
0.89%
From the above table, ̅ = 0.0089
√
̅
̅
p chart :
Center Line (CL)
Upper Control Limit (UCL)
= ̅ = 0.89%
=̅
= 0.0089+ 3 *
= 1.79%
=max(0, ̅
)
= max(0, 0.0089- 3 *
)=0
Lower Control Limit (LCL)
b. Process Control
p-Chart
Fraction Defective (p)
2.50%
2.00%
1.50%
1.00%
0.50%
0.00%
16
17
18
Proportion (p)
19
20
CL
21
22
UCL
23
24
LCL
The process is no longer in control because 3 data points have exceeded the upper control limit.
3
25
SCM 302
SPC Practice Solutions
Problem 3
a. Calculate Cp. Is the process capable of 3σ performance based on Cp? Explain your answer.
The appropriate parameters are USL=96, LSL=36 and =8. Using formula on page 203,
Cp =
(96-36)/(6*8)=1.25.
Since the Cp measure is greater than 1, the process variability is less than the specifications and the process is capable of
3 performance.
b. Calculate Cpk. Does your answer from part (a) regarding process capability change when you also use C pk? Explain
your answer.
Cpk =
(
)
(
)
Since Cpk is less than 1, the process is not capable of 3 performance. The Cp measure alone leads to an incorrect
conclusion because the process mean is not centered but nearer to the LSL.
4