Mgt 2070 Assignment 4 – Solutions
6.11
Develop a Pareto analysis of the following causes of delay in a production process.
Reason for Delay
Test equipment down
Delay in inspection
Inadequate parts
Lack of personnel available
Awaiting engineering decision
No schematic available
Frequency
22
15
40
3
10
11
To develop the Pareto analysis, we need to rank each reason by frequency and calculate the
cumulative percent defects for each:
Reason
Inadequate parts
Test equipment down
Delay in inspection
No schematic available
Awaiting engineering decision
Lack of personnel available
Frequency
40
22
15
11
10
3
Cum. Frequency
40
62
77
88
98
101
Cum. Percent
40/101 = 39.6%
62/101 = 61.4%
77/101 = 76.2%
88/101 = 87.1%
98/101 = 97.0%
101/101 = 100%
(4 marks)
We can then prepare a chart to aid in our analysis:
Problems
Lack of
personnel
No
schematic
available
Awaiting
engineering
decision
Delay in
inspection
# Defects
Cum. %
Test
equipment
down
100
90
80
70
60
50
40
30
20
10
0
Inadequate
parts
Number of defects
Pareto Analysis
(4 marks)
We can conclude that most of the delay is the result of inadequate parts, and that this is the
problem we should concentrate on solving first. (2 marks)
S6.5 Autopitch devices are made for both major- and minor-league teams to help them
improve their batting averages. When set at the standard position, Autopitch can throw
hardballs toward a batter at an average speed of 60 mph. To monitor these devices and to
maintain the highest quality, Autopitch executives take samples of 10 Autopitch devices at a time.
The average range is 3 mph. Using this information, construct control limits for X-bar and R
charts.
We are given the following in the question: n = 10, desired X = 60, desired R = 3. We then look
up the following values from table S6.1: A2 = 0.308, D4 = 1.777, D3 = 0.233. (2 marks) We then
substitute into the formulae for upper and lower control limits:
UCL X  X  A 2  R  60  0.308  3  60.924
(2 marks)
LCL X  X - A 2  R  60  0.308  3  59.076
(2 marks)
UCL R  D 4  R  1.777  3  5.331
(2 marks)
LCL R  D3  R  0.223  3  0.669
(2 marks)
S6.11 Your supervisor, Lisa Lehrmann, has asked that you report on the output of a machine on
the factory floor. This machine is supposed to be producing optical lenses with a mean weight of
50 grams and a range of 3.5 grams. The following table contains the data for a sample size of n
= 6 taken during the past 3 hours:
Sample Number
1
2
3
4
5
6
7
8
9
10
Sample Average
55
47
49
50
52
57
55
48
51
56
Sample Range
3
1
5
3
2
6
3
2
2
3
What are the X- and R-chart control limits when the machine is working properly? What seems
to be happening? (Hint: Graph the data points. Run charts may be helpful.)
We are given the following in the question: n = 6, desired X = 50, desired R = 3.5. We then look
up the following values from table S6.1: A2 = 0.483, D4 = 0, D3 = 2.004. (2 marks) We then
substitute into the formulae for upper and lower control limits:
UCL X  X  A 2  R  50  0.483  3.5  51.69
LCL X  X - A 2  R  50  0.483  3.5  48.31
UCL R  D 4  R  2.004  3.5  7.014
LCL R  D3  R  0  3.5  0
(2 marks)
We can then prepare control charts, on which we plot the data from our samples:
X-Bar Chart
58
56
54
UCL
52
50
Sample
LCL
48
46
1
2
3
4
5
6
7
8
9
10
(2 marks)
R Chart
6
UCL
4
Sample
LCL
2
0
1
2
3
4
5
6
7
8
9
10
(2 marks)
R values are within control limits. X-bar values are mostly outside the control limits, fluctuating
above and below them. The process is out of control, likely due to an assignmable variation.
Immediate action is required to find and correct the problem with the machine. (2 marks)