Name:_______________
PMAN 639: PROJECT QUALITY MANAGEMENT
FINAL EXAM-
- Please answer the following True or False questions (each worth 2 points)
A process in control does not produce unacceptable output.
True ______ False___FALSE___
Random variation falls within the control limits.
True ___TRUE___ False______
A process is capable if the variations fall within the control limits.
True ___TRUE___ False______
If X bar chart is in controlthen R chart is also in control.
True __TRUE___ False________
- Control charts are used to determine if there are unexpected changes in the process.
True ___TRUE___ False______
For defective units situations, aP chart is used when the sample size varies.
True ___TRUE___ False______
A standard Six Sigma Process has a 4.5 Sigma performance
True __TRUE___ False________
A process is not capable if the variations exceeds the mean
True ___TRUE___ False______
A histogram shows the trendline of a process
True _____ False____FALSE____
A Fishbone diagram used to organize data in categories
True ___TRUE__ False________
Please answer the following multiple choice questions (each worth 4 points)):
Which of the following tools is used to identify process frequency distribution?
A) Matrix diagram
B) Histogram
C)Pareto Chart
D) Check sheets
E) Process map
Which of the following tools is used to organize ideas.
A) Matrix diagram
B) Histogram
C) Affinity Diagram
D) Check sheets
E) Process map
- Which of the following tools is used to analyze cycle time
Control Chart
B) Fishbone diagram
C) Pareto Chart
D) Process map
E) Root Cause Analysis
Which of the following is true of common cause variation?
A) It is uncontrollable.
B) It is centeredaround the mean.
C) It is inherent in the process.
D) All of the above.
E) None of the above
- A true 3 Sigma Process should not havemore than_______defects per million opportunities
3.4
270
2700
45500
- Please respond to the following questions. Each question is worth 10 points:
97.
- Please create a process map for withdrawing cash from an ATM machine as described below:
Janet needed some cash and walks up to an ATM machine. She inserts her debit card. The machine returns the card and the monitor reads: “could not read the card try again”. Janet recognizes that she has entered the card in the wrong direction. She takes the card and turns it around and insert the card again. The machine keeps the card and the monitor reads: “enter pass code”. Janet enters a 4 digit code. The monitor reads: “ error, try again”.
Janet enters the code again. The monitor displays: Withdrawal? Yes No
Janet selects Yes. Displays shows:
“ $100, $200, $300, $400, $______”.
Janet selects $200. The machine pushes out 10 $20 bills. Monitor then displays “Receipt? Yes No”. Janet selects No. Monitor displays “Take your cash” .
The machine returns the card. Display changes to “Thank You”.
Janet takes the cash and her card and walks away.
Type of Problem
Frequency
(number of times)
Call went to voice mail
18
Items damaged when received
15
Literature not in the box
7
Parts missing
13
Billing error
6
Unit not working
2
Delivery was late
3
Construct a Pareto chart including the cumulative % frequency.What is the cumulative percentage of the two highest priority issues?
1
6.10
6.15
5.35
5.98
5.95
2
6.35
6.02
6.20
6.10
5.70
3
5.85
5.90
6.20
6.12
6.08
4
6.00
5.80
5.70
6.04
6.17
5
6.15
5.60
6.25
5.99
6.01
6
6.13
6.05
5.89
6.00
5.87
7
5.88
6.03
5.76
6.1 3
6.00
8
6.12
5.45
6.15
6.04
5.91
Sample
Task1
Task2
Task3
Task4
Task5
total
x bar
R
1
6.1
6.15
5.35
5.98
5.95
29.53
5.906
0.8
2
6.35
6.02
6.2
6.1
5.7
30.37
6.074
0.65
3
5.85
5.9
6.2
6.12
6.08
30.15
6.03
0.27
4
6
5.8
5.7
6.04
6.17
29.71
5.942
0.47
5
6.15
5.6
6.25
5.99
6.01
30
6
0.65
6
6.13
6.05
5.89
6
5.87
29.94
5.988
0.26
7
5.88
6.03
5.76
6.13
6
29.8
5.96
0.37
8
6.12
5.45
6.15
6.04
5.91
29.67
5.934
0.7
Sum=
47.834
4.17
average=
5.97925
0.52125
We have from calculation table
= 5.97925
= 0.52125
R CHART
Centreline == 0.52125
Upper Class Limit =UCL = D4 * =2.114*(0.52125) = 1.1019225
Lower Class Limit =LCL = D3 * = 0*(0.52125) = 0
X – bar CHART
Centreline = = 5.97925
Upper Class Limit =UCL = + A2( )
= 5.97925 + 0.577* (0.52125)
= 5.97925 + 0.030076125
= 6.28001125
Lower Class Limit =LCL = – A2( )
= 5.97925 – 0.030076125
= 5.67848875
- For the above problem if the true mean and standard deviation of the process are 6.05 and 0.04,
- Is the process capable of meeting its specification? Calculate its CP and CPk.
The process is capable of meeting the specification laid out.
Cpk = min(USL – μ, μ – LSL) / (3σ)
Cpk=min(6.28001125-0.1, 0.1-5.67848875)/3×0.04
Cpk=
1.068
- What % of the cylinders if any are out of specs?
- 4.6%
- A project manager at Global Design Corporation has collected the following data on the size of software programs and the length of programming time. The company is bidding on a new system that is estimated to consist of 300,000 lines of code. Use the data to find the correlation function of coding time to the program size and estimate the number of days it would take to code the system.
Module
Size in 1000 lines of code (KLOC)
X
Number of days to code
Y
SQUARE OF X
SQUARE OF Y
1
160
70
2
158
66
3
150
70
4
135
59
5
178
72
6
170
63
7
158
69
8
138
65
9
200
72
10
195
68
11
189
65
12
173
68
13
159
70
14
163
55
15
150
66
16
140
65
17
206
73
18
144
64
19
157
70
20
183
74
21
195
75
22
190
77
23
182
69
24
152
65
25
174
69
We should apply Pearson’s correlation coefficient in the following question
Module
Size code (KLOC), x
Number of codes, y
X^2
Y^2
XY
1
150
72
22500
5184
(150*72) =10800
2
158
66
24964
4356
(158*66) =10428
3
148
65
21904
4225
(148*65) =9620
4
120
60
14400
3600
(120*60) =7200
5
178
75
31684
5625
(178*75) =13350
6
170
68
28900
4624
(170*68) =11560
7
152
68
23104
4624
(152*68) =10336
8
138
65
19044
4225
(138*65) =8970
9
200
80
40000
6400
(200*80) =16000
10
195
75
38025
5625
(195*75) =14625
11
189
65
35721
4225
(189*65) =12285
12
173
68
29929
4624
(173*68) =11764
13
159
62
25281
3844
(159*62) =9858
14
163
71
26569
5041
(163*71) =11573
15
150
65
22500
4225
(150*65) =9750
16
140
65
19600
4225
(140*65) =9100
17
206
74
42436
5476
(206*74) =15244
18
144
64
20736
4096
(144*64) =9216
19
157
70
24649
4900
(157*70) =10990
20
183
76
33489
5776
(183*76) =13908
21
195
75
38025
5625
(195*95) =14625
22
190
77
36100
5929
(190*77) =14630
23
185
73
34225
5329
(185*73) =13505
24
152
58
23104
3364
(152*58) =8816
25
178
70
31684
4900
(178*70) =12460
Sum
4173
1727
708573
120067
290613
m = 0.1949227
- A project manager has developed the following data on a
project tasks and durations by weeks.
- Please create an AON diagram, identify the critical path and the expected project duration.
- The customer has asked for a 99% confidence on the project completion date. Calculate the number of additional weeks the project manager needs to make that commitment.
Time estimation in weeks
Task
Predecessor
optimistic
Most likely
Pessimistic
A
3
4
5
B
A
1
8
15
C
A
2
4
6
D
A
4
8
12
E
B
7
9
11
F
C,B
2
5
8
G
E,D
6
8
10
H
F,G
4
7
10
Activity
expected time
variance
A
4
0.11
B
8
5.44
C
4
0.44
D
8
1.77
E
9
0.44
F
5
1
G
8
0.44
H
7
1
D G
A B E H
C
Path duration
ADGH 4+8+8+7=27
ABEGH 4+8+9+8+7=36
ABFH 4+8+5+7=24
ACFH 4+4+5+7=20
Critical Path take the longest path= 36
Expected completion of the project = 36 weeks
- It takes one week to have less time =36-1. To be 35 weeks. The variance of activities =0.11+5.44+0.44+0.44+1= 7.43
Standard deviation would be = 2.723
Z score would be 35-36/2.723 = -0.37
From table probability would be= 0.35569
Which is the same as 35.569
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