MBA 675 Week 11 HW
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Bellevue University *
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Course
675
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
xlsx
Pages
8
Uploaded by ElizabethAK
Chapter 7, Problems, #14
1600
1200
Table 7.7
Data for the Billing Process
Activity
A
-
4
1
5,000
8,000
B
-
5
3
8,000
10,000
C
A
1
1
4,000
4,000
D
B
6
3
6,000
12,000
E
B,C
7
6
4,000
7,000
F
D
7
6
4,000
7,000
A. What is the minimum-cost schedule for this project?
Normal Cost Schedule-
Crash Cost Schedule-
Minimum Cost Schedule-
Paths:
Weeks
Costs
Paths:
Weeks
Costs
Paths:
Weeks
Costs
A
4 $
5,000 A
1 $
8,000 A
4 $
5,000
$
5,600
C
1 $
4,000 C
1 $
4,000 C
1 $
4,000
E
7 $
4,000 E
6 $
7,000 E
7 $
4,000
Total:
12 $
13,000
Total:
8 $
19,000
Total:
12 $ 13,000
Critical path
Critical path
Critical path
B
5 $
8,000 B
3 $
10,000 B
3 $ 10,000
D
6 $
6,000 D
3 $
12,000 D
3 $ 12,000
E
7 $
4,000 E
6 $
7,000 E
6 $
7,000
F
7 $
4,000 F
6 $
7,000 F
6 $
7,000
Total:
25 $
22,000
Total:
18 $
36,000
Total:
18 $ 36,000
Longest path:
B-D-E-F
25 Longest path:B-D-E-F
18 Longest path: B-D-E-F
18
Indirect Costs:
1600 $
40,000 Indirect Costs
1600 $
28,800 Indirect Costs:
1600 $ 28,800
Penalty Costs:
13 $
15,600 Penalty Costs
6 $
7,200 Penalty Costs:
6 $
7,200
TOTAL COST:
$
90,600 TOTAL COST:
$
91,000 TOTAL COST:
$ 85,000
You are the manager of a project to improve a billing process at your firm. Table 7.7 contains the data you will need to conduct a cost analysis of the pr
Indirect costs are $1,600 per week, and penalty costs are $1,200 per week after week 12.
Immediate
Predecessor(s)
Normal
Time (wks)
Crash Time
(wks)
Normal
Costs ($)
Crash Costs
($)
B. What is
"normal tim
roject.
the difference in total project costs between the earliest completion time of the project using
me" and the minimum-cost scheduled derived from part (a)?
Chapter 7, Problems, #16
Table 7.9
Project Activity & Cost Data
Normal
Crash
Activity
Costs ($)
Costs ($)
A
12
1,300
11
1,900
1
600
B
13
1,050
9
1,500
4
112.5
C
18
3,000
16
4,500
2
750
D
9
2,000
5
3,000
4
250
E
12
650
10
1,100
2
225
F
8
700
7
1,050
1
350
G
8
1,550
6
1,950
2
200
Paths:
Days
H
2
600
1
800
1
200
ADG*
29
I
4
2,200
2
4,000
2
900
AEH
26
$ 13,050
$19,800
BFH
23
*= Critical Path
CI
22
ADG
27
AEH
26
New Schedule
BFH
23
Activity
Duration
CI
22
A
12
Cost= $
13,450
B
13
C
18
D
7
ADG
25
E
11
AEH
25
F
8
BFH
23
G
6
CI
22
H
2
Cost= $
14,175
I
4
The diagram in Figure 7.12 was developed for the project launch of Kitty Condo, a new product in the luxury cat cage market.
Suppose that you, as project manager, are interested in finding ways to speed up the project at minimal additional cost.
Determine the schedule for completing the project in 25 days at minimum cost. Penalty and project-overhead costs are negligible.
Time and cost data for each activity are shown in Table 7.9.
Time
(days)
Time
(days)
Crash
Days
Crash
Cost/day
Paths & New Durations:
(crashing G by 2 days)
Project Duration of 25 days at a cost of
$14,175
Paths & New Durations:
(crashing D by 2 days and E by 1)
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Time Estimates (days)
Activity
Optimistic
Most Likely
Pessimistic
A
5
8
11
B
4
8
11
C
5
6
7
D
2
4
6
E
4
7
10
What is the expected completion time of the project?
19 days
Activity
Optimistic
Most Likely
Pessimistic
Variance
A
-
5
8
11
8
1.00
B
-
4
8
11
8
1.36
C
A
5
6
7
6
0.11
D
A
2
4
6
4
0.44
E
B,D
4
7
10
7
1.00
Paths:
*=Critical Path
AC
14
ADE*
19
Project Variance=
2.44
BE
15
Chapter 7, Problems, #21 – answer all parts, on part b) provide the probability of completing it in 21 days and also provide the probability of completing it in 21
days or less; on part c) provide the probability of completing it in 17 days and also provide the probability of completing it in 17 days or less;
Recently, you were assigned to manage a project to remodel the seminar room for your company. You have constructed a network diagram
depicting the various activities in the project (Figure 7.13). In addition, you have asked your team to estimate the amount of time that they would
expect each of the activities to take. Their responses are shown in the following table.
Immediate
Predecessor
Expected
Time
z= (given duration - critical path duration) / sqrt (critical path variance)
What is the probability of completing the project in 21 days or les
1.28
0.8996
What is the probability of completing the project in 17 days or les
-1.28
0.1004
Element A:
0.92
Element B:
0.94
Element C:
0.91
A. Draw how these three elements are connected
B. What is the reliability of this system?
R=
0.787
Elements:
A
B
C
=
0.85
0.92
0.94
C
=
0.85
0.8648
C
=
0.85
C
=
0.983
R=
0.982
A piece of electronic equipment used for aviation has three elements connected in series, or sequence. The reliability of each of the three elements is
as follows:
C. Suppose we wanted the reliability to be at least 0.85 by upgrading Element C with a higher reliability. What would the
reliability of Element C need to be?
D. Instead of increasing the reliability of Element C, each of the elements is supported by an identical element in parallel as a back-up.
What is the reliability of this system?
A
B
C
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PM= $
1,200
Weeks between preventive maintenance
Expected # of breakdowns between preventive maintenance
Breakdown:
1
1.1
Week 1=
2200
2
2.52
Week 2=
2400
3
4.63
Week 3=
2600
4
8.27
Week 4=
2800
Total Cost per week:
$
3,620
Weekly preventative maintenance is recommended since it has the lowest total cost per week
Total Costs every 2 weeks:
$
3,624
Total Cost every 3 weeks: $
4,413
Total Cost every 4 weeks: $
6,089
Let’s revisit the preventive maintenance problem discussed in the video. A hospital has seven large identical heat pumps. If one of the
heat pumps malfunctions, it could seriously impact hospital operations. If preventive maintenance is performed weekly, it costs $1200 to
perform preventive maintenance (PM) on all seven of them. If one of the heat pumps breaks down between PM inspections, the
average cost to the company of a breakdown is $2200. The historical data for the heat pumps is as follows:
Now suppose that for each week that preventive maintenance is delayed, the cost to the company of a breakdown increases by $200
(for example, if preventive maintenance is every week, the cost of a breakdown is $2200 and if preventive maintenance is every two
weeks, the cost of a breakdown is $2400, etc.) Based on the above information, what Preventive Maintenance policy do you
recommend? Why?
Discuss the pros and cons of McDonnell’s position as expressed by the VP.
What strategy would you have recommended McDonnell Douglas follow to address the failure? Why?
When a McDonnell Douglas DC-10 crashed over Iowa, a subsequent investigation suggested that the plane’s hydraulic systems
did not provide enough protection. The DC-10 had three separate hydraulic systems, all of which failed when an engine exploded.
The engine threw off shreds of metal that severed two of the lines, and the third line required power from the demolished engine
that was no longer available. The DC-10, unlike other commercial jets, had no shutoff valves that might have stemmed the flow of
hydraulic fluid. Lockheed’s similar L-1011 tri-jet had four hydraulic systems. A McDonnell Douglas VP said at the time. “You can
always be extreme and not have a practical airplane. You can be perfectly safe and never get off the ground.” Discuss the pros
and cons of McDonnell’s position as expressed by the VP.
The viewpoint expressed by the McDonnell Douglas VP highlights the need to strike a balance between practicality and
safety when designing aircraft. The VP argues that extreme safety measures can compromise the functionality and
feasibility of an aircraft. This raises concerns about the practicality and cost-effectiveness of implementing additional
safety features, such as hydraulic shutoff valves. However, the absence of these safety measures in the McDonnell
Douglas DC-10, as revealed during the investigation of a crash in Iowa, exposed a critical design flaw. The incident
highlighted the vulnerability of the DC-10's hydraulic system and prompts questions about whether McDonnell Douglas
could have incorporated alternative designs with greater redundancy, as seen in Lockheed's L-1011 tri-jet. It is worth
noting that the VP's perspective may downplay the importance of safety, potentially eroding public trust in the company
and its products. Ultimately, while considering practicality and operational feasibility in aircraft design is important, it is
crucial to recognize the significant safety implications and maintain public confidence in the aviation industry.
McDonnell Douglas could have addressed the hydraulic system failure in the DC-10 by implementing a
comprehensive strategy. Firstly, a thorough review and redesign of the system should have been conducted to
identify weaknesses, explore alternative designs, and enhance reliability and redundancy. Lessons learned from the
investigation should have guided the implementation of critical safety measures like shutoff valves or alternative
prevention methods. Collaboration with industry experts and regulators would have provided valuable insights and
ensured compliance with safety standards. Transparent communication about the implemented safety measures
would have been crucial for rebuilding trust. Additionally, establishing systems for continuous monitoring, data
analysis, and improvement would have ensured long-term reliability and safety. By following these strategies,
McDonnell Douglas could have demonstrated a proactive commitment to safety and taken the necessary steps to
regain confidence in their aircraft.