Chapter 7 Project Management
Review Questions
1. What are the unique attributes of project management? Frame the answer in terms of other
P/OM work configurations.
Answer: Projects have life-cycle stages that are unlike those of other work configurations.
Specifically, all projects have an endpoint when the goal is achieved. Endpoints can also be
viewed as the starting points of the next set of projects. We will get back to that dynamic
issue after we examine traditional project characteristics.
Projects tend to be unique and aimed at a singular goal, whereas other work configurations
tend to do some jobs, products, or tasks repetitively. The scale of projects frequently dwarfs
that of other work configurations. For example, the project of building a bridge, or a landing
on the moon, is of much longer duration than creating hub caps in a job shop, or assembling
cars in a flow shop. Projects tend to require many participants of numerous disciplines using
varied resources.
The one type of project that now needs to be examined is that of bringing new products to
the marketplace. Formerly, that was done in traditional project mode with a planning phase
at start-up and ending with a product launch which tracks the growth of the new product’s
market share. Within the last two decades, some very successful companies have learned
how to keep projects going. As a project ends—another related project begins. It was seen
that the development of a product platform for continuous updating and upgrading provided
a necessary competitive edge. Ongoing project management keeps the project team always
engaged in the project mode. Product platform planning is discussed in Chapter 11 on NPD.
2. Classify projects by type and describe project life cycles.
Answer: Projects can be simple or complex; they can be one-time-only, first time ever, or it
has-been-done-before. They can be infrequently or frequently done, and they can take days,
months or years to accomplish. The four life-cycle stages are (1) describe the goals; (2) plan
the project; (3) carry out stages; and (4) complete the project. Step 1 requires establishing
the desired outcomes of the project. Step 2 includes full specification of the activities that
make up the project (and the interrelationships among activities). Step 2 also includes the
determination of an appropriate time line for the project (and its critical path). The third step
is the operational aspects of the project. Build the walls, pour the floors after the pipes and
wires have been installed, etc. The last step—for traditional projects—includes shutdown
and team dissolution. However, as previously mentioned, increasingly, the team moves on to
another project which is likely to be the next phase of the prior project.
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3. Why is a project manager considered a leader?
Answer: The project manager in many ways must possess characteristics of a chief
executive. This certainly qualifies as a leadership position. Such a manager is in charge of a
project that has a return only at its conclusion; there may be no revenue stream or profits.
Project managers are guided by strategic management, yet need to work toward cycle-time
reduction in all facets of the project. Project managers must be systems-oriented because
projects combine human skills, technological capabilities, and resources from many parts of
multiple organizations. Projects pose considerable risks, and presage great penalties for
failure. For all of these reasons, projects need leaders, not just managers. Leadership is the
ability to get others (the entire project team) to follow in exactly the right way. The leader
must not only know how to get people to follow; the leader must know “the right way.”
4. What is team work and why is team work repeatedly mentioned when discussing good
project management?
Answer: Projects are large-scale and unique. They utilize large amounts of organizational
resources from many areas of the organization. To be successful—such large events require
the systems philosophy to promote coordination and, hence teamwork. Projects are of long
duration; the workers assigned to a project may stay with it for its lifetime while it evolves
over many stages and phases. Such a semi-permanent combination of workers fulfilling on a
changing set of demands is a team. Projects are so wide-ranging—involving all parts of the
organization and some extra-organizational elements of the supply chain. Scope is huge and
challenging. The systems philosophy is strategic and team work is its tactical response.
Good project management may include concurrent engineering (CE). This subject is not
treated in Chapter 7 but parallel-path project activities are discussed. Wikipedia has a good
explanation of concurrent engineering. For our purposes, we will briefly point out that CE is
inherently a team-oriented method. Concurrent means that several related initiatives are
proceeding together in parallel. Parallel path project management demands coordinated team
efforts to achieve synergy which translates into reduced time for completion with superior
results. (The definition of synergy is: “things working together produce an effect greater
than the sum of their individual effects.) For example, two drugs were tested and each had
some small benefits but when used together, their effectiveness jumped. Parallel-path project
management can have the same effect.
5. What is PERT’s relationship to critical-path methods?
Answer: PERT and CPM (critical path method) were developed at the same time (parallel
needs) but entirely independently. As separate developments in the 1950s PERT and CPM
were commissioned and paid for by different organizations (see section 7.5). PERT uses 3
parameters to estimate activity times and this approach is considered to be better for projects
with stochastic event durations. CPM stressed quick determination of the longest time path
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in the project, called the critical path (CP). CPM was considered to be better for problems
involving “crashing”—spending extra funds to complete selected activities sooner. It is fair
to say that over time differences between PERT and CPM have all but disappeared. Each
has incorporated the uniquely strong points of the other. Both emphasize the value of finding
the critical path and then being able to adjust it by allocating resources in different ways.
6. What does the forward-pass procedure accomplish in critical-path methods?
Answer: The forward pass is the fastest way to score a touchdown in the NFL. However,
since we are talking about project management, the forward pass identifies the earliest start
time of each of the activities in the network. Knowing the durations of the activities, the
earliest finish times are readily calculated. The earliest finish of the last or final activity
determines the target date and time for completion of the project (e.g., of the entire
network). This target date and time determines the length of the critical path. It sets the stage
for the backward pass calculations. Project managers do not call this a lateral pass.
7. What does the backward-pass procedure accomplish in critical-path methods?
Answer: The backward pass calculates the latest finish times for each of the activities in a
project network. By subtracting the durations of the activities, latest start times are
calculated. This is done working backwards from the final activity to the first activity. As a
result of the backward pass, the critical path can be identified. With the forward-pass, only
the length of the critical path was known. After the backward-pass, slack values are known.
The critical path is identified as the set of activities with zero slack.
8. What is a critical path and how does knowing it in detail help project managers?
Answer: The critical path is the set of activities that constitutes the longest time path
through the network. Delay at any one of the activities on the critical path will delay the
entire project. Activities on the critical path have zero slack—as measured by latest start
minus earliest start equals zero. The subtraction of latest finish minus earliest finish will
also equal zero. The managerial importance is in knowing which tasks cannot be delayed
without delaying the entire project. The opposite is also important to the manager, namely,
which tasks can be delayed without disturbing the project’s due date. Such information is
vital to decisions about moving resources from one task to another. This is done to speed up
slow tasks that benefit by being faster at the expense of fast tasks that are unnecessarily so.
9. What is slack and how does knowing exactly what it is and where it resides help project
managers?
Answer: Activities along the critical path have zero slack. Activities that are not on the
critical path have positive slack. Think of slack as allowable slippage which is the time that
an activity can be delayed without impacting the completion time of the project. Knowing
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which activities have very little slack allows management to expedite those activities lest
they become part of the critical path. It is worth considering putting more resources to work
on critical slack activities. Knowing which activities have a lot of slack permits management
to consider swapping resources from large slack activities to minimum slack activities. In
short, it encourages project managers to study if resource swapping can bring better balance
to project activities. The process of moving resources from where they are excessive to
where they are limiting is known as resource leveling.
10. What are the differences between deterministic and probabilistic activity time estimates?
Answer: If an activity time is deterministic, then its duration is a constant. Deterministic
means that there is no variation in that activity time. When activity time is stochastic its
duration is a variable dependent upon a probability distribution. Variation complicates
calculation of critical paths and renders uncertain the length of the critical path as well as the
amount of slack that characterizes each activity. If variation exists in activity times, it is very
likely to be foolhardy to take the averages and act as if they are deterministic activity times.
11. What are the strengths and weaknesses of PERT?
Answer: PERT uses the concept of a critical path—with time estimates adjusted for the
stochastic nature of activity durations. This three parameter adjustment is a significant
strength, when compared to project methods that cannot handle variation in activity times.
PERT offers a reasonable approach for calculating likelihoods of project duration based on
the distributions related to the three parameters. It is critical that that both the Expected time
and the Activity variance be used together to determine the probability that a non-critical
activity can become critical. The major weakness of PERT is that the three parameter
method (which is based on the Beta distribution) does not capture the realities of actual
activity variations. Another weakness is that project managers may use only Expected time
and ignore the Activity variance measure. When probabilities affect which activities will be
part of the actual critical path, it is crucial to plan for both eventualities—even though one
may have a much greater probability than the other.
12. Describe how cost/time trade-off methods can be used given the decision to spend
an additional 10% on the project.
Answer: Cost/Time analysis is the basis for “crashing” activities and projects. In one of its
simple forms, it assumes a linear relationship between extra spending on an activity and the
time reduction that results. If an additional 10% is available, and that money is not sufficient
for a complete crash, managers must choose which critical tasks are shortened and which are
not. Cost/Time trade-off analysis shows where the greatest leverage exists. Less simple is
the non-linear variant of these trade-off functions. These can show expenditures that achieve
large reductions in activity times at first, which then gradually slow up—as more is spent on
resources. The reverse is also possible. Non-linear trade-offs are more complex to calculate.
4
The computers are not the scarce resource; programmers are. It seems reasonable to guess
that the increased accuracy of non-linear trade-off analysis may be well worth the effort.
13. Describe project crashing and contrast it to normal time.
Answer: Crashing is based on assigning extra resources to complete a project in minimum
time. “Crash” time for an activity is its minimum duration, as shortened by added resources.
“Normal” time reflects activities with minimum resource needs. Normal time is associated
with reasonable expenditures. Less than that amount would jeopardize the success of the
project. Smart crashing would have managers use resources where they have the greatest
leverage. Crashing only has leverage with critical (or near-critical) tasks. By calculating the
deltas (change in cost per change in duration) some activities will be preferred to others for
crashing. Managers should start with critical activities having the lowest delta.
14. What advantages can be gained by using crashing?
Answer: The value of crashing is a reduced critical path time. The advantage of a shorter
critical path may be more profit or bonus for the project organization, more prestige for the
early completion, or the avoidance of a penalty for late completion. Most important is the
ability of an early start-up to start generating revenue before Since activities are “crashed”
by adding more resources, managers must be concerned with where these extra resources
can be obtained (let’s talk to finance).
A manager acting without the systems viewpoint could spend more on crashing than is being
saved by the early completion. Also, a manager may crash the wrong activities (a noncritical
activity, or an activity that has become noncritical. Another systems point is that crashing
can shift the critical path. Further, the linearity assumption relating spending on resources to
time reduction may not be valid. In this case, the extra resources may not have the desired
impact. A study to determine the nonlinear shape of cost reduction with increased
expenditure adds to the cost of crashing.
15. What are the dangers of, using crashing?
Answer: The value of crashing is a reduced critical path time. The advantage of a shorter
critical path may be more profit or bonus for the organization, more prestige for the early
completion, or the avoidance of a penalty for late completion. When Boeing delivers a 787
airplane ahead of time a complex sequence of advantages occurs including cash flow to
Boeing, passenger revenues generated by flights flown, on-board and maintenance crew
salaries, etc. Since activities are “crashed” by adding more resources, managers must be
concerned with what these extra resources cost. A manager acting without the systems
viewpoint could spend more on crashing than is being saved by the early completion. Also, a
manager may crash the wrong activities (a noncritical activity, or an activity that has become
noncritical. Another systems point is that crashing can shift the critical path. Further, the
linearity assumption relating spending on resources to time reduction may not be valid. In
5
this case, the extra resources may not have the desired impact. A study to determine the
nonlinear shape of cost reduction with increased expenditure adds to the cost of crashing.
6
Problems
1. Consider the following AON network and the data given in the following table to answer the
next four questions.
Activity
Time
(days)
A
5
B
C
6
7
D
E
F
G
4
9
3
4
B
E
D
A
G
C
a. Identify the critical path.
Answer: A-B-D-E-G
b. Find the earliest completion time of the project.
Answer: 28. This is the length of the critical path.
c. Find ES, EF, LS, and LF of each activity
Answer: See the table below.
d. Find the slack for each activity.
Answer: See the table below.
Solution
List all paths and identify the longest path.
A-B-D-E-G: 28 days (longest path, critical path, gives the project duration)
A-B-D-F-G: 22 days
A-C-F-G: 19 days
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F
Table showing calculations for ES, EF, LS, LF and Slack Time
Activity
Time
ES
EF
LS
LF
A
5
0
5
0
5
Slack
Time
0
B
C
D
E
F
6
7
4
9
3
5
5
11
15
15
11
12
15
24
18
5
14
11
15
21
11
21
15
24
24
0
9
0
0
6
G
4
24
28
24
28
0
8
2. The Delta Company manufactures a full line of cosmetics. A competitor recently developed a
new skin rejuvenating cream that appears to be successful and potentially damaging to Delta’s
skin care position in the marketplace. The sales manager has asked the operations manager what
the shortest possible time would be for Delta to reach the marketplace with a new product
packaged in a competitively redesigned container. The operations manager has drawn up the
following table.
Activity Description
Design product
Design package
Test market package
Distribute to dealers
Order package materials
Fabricate package
Order materials for product
Test-market product
Fabricate product
Package product
Activity
Symbol
A
B
C
D
E
F
G
H
I
J
Immediate
Follower
H and G
E and C
D
None
I and F
D
I and F
J
J
D
Duration
(Days)
30
15
20
20
15
30
3
25
20
4
a. Construct the AON diagram.
Answer: See the diagram below.
b. Find the critical path
Answer: A-G-F-D. This is the longest path (83 days). See the list of paths below.
c. Find the ES, LS, EF, and LF.
Answer: See the table below.
d. What is the project duration?
Answer: 83 days. This is the length of the critical path. All paths are listed below.
e. Determine the slack of each activity.
See the table below.
f.Neither the sales manager nor the P/OM is satisfied with the way the project is designed.
However, the P/OM insists that because of the pressure of time, the company will be
forced to follow this plan. In what ways does this plan violate good practice?
The project has several serious flaws. Materials for both the product and the package are
ordered while they are still being individually test-marketed. The notion of separate test
markets for the product and the package runs counter to good systems practice. They
should be tested together. Further, fabrication of the product and package is scheduled to
begin before either of the test markets has been completed. This will not allow ideas for
improvement gained during test marketing to be incorporated into the product and
package. A further discrepancy arises from the fact that the product is to be packaged
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before the activity “fabricate package” is completed. This difficulty should be cleared up
before the project schedule is approved. Finally, there is too little slack along a number of
paths. That makes the length of the critical path a risky bet which can be dependent upon
relatively minor events in activities A, G, F, and D. Also, activities B and E in adjacent
paths are able to shift the critical path from AGFD to BEFD. This kind of uncertainty
creates a project management nightmare.
Solution
AON Diagram
H
J
A
G
I
D
E
F
B
C
There are following six paths in this network. The numbers in parentheses are the lengths of the
paths.
A-H-J-D (79)
A-G-I-J-D (77)
A-G-F-D (83)
B-E-I-J-D (74)
B-E-F-D (80)
B-C-D (55)
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Table showing calculations for ES, EF, LS, LF and Slack Time
Activity
Description
Design
product
Design
package
Test market
package
Distribute to
dealers
Order
package
materials
Fabricate
package
Order
materials for
product
Test-market
product
Fabricate
product
Package
product
Activity
Symbol
A
Immediate
Follower
H and G
Duration
(Days)
30
ES
EF
LS
LF
0
30
0
30
Slack
Time
0
B
E and C
15
0
15
3
18
3
C
D
20
15
35
43
63
28
D
None
20
63
83
63
83
0
E
I and F
15
15
30
18
33
3
F
D
30
33
63
33
63
0
G
I and F
3
30
33
30
33
0
H
J
25
30
55
34
59
4
I
J
20
33
53
39
59
6
J
D
4
55
59
59
63
4
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3. Consider the data given in the following table. The fixed cost is $ 900 per week. Perform a
time-cost tradeoff analysis. What is the minimum cost to complete the project and what is the
corresponding time?
Immediate
Activity
Predecessor (s)
Normal
time
(Weeks)
Crash
Time
(Weeks)
Normal
Cost
Crash
Cost
A
None
9
6
$13,000
$15,550
B
C
D
E
F
G
H
I
J
None
None
A
B
B
C
D,E
F,G,H
I
5
7
12
8
6
11
5
4
10
4
5
8
5
4
9
4
3
8
$7,000
$15,000
$12,000
$9,000
$5,000
$13,000
$8,000
$2,500
$12,000
$7,900
$15,800
$14,800
$10,500
$6,200
$14,000
$9,000
$3,500
$15,000
Answer
Minimum Cost: $132,050
Time: 33 Days
Solution
The solution of this problem requires crashing of activities and then finding the total cost. The
total cost includes the activity cost and the fixed cost. Various tables showing these calculations
are given below.
Also see the Excel file SMCh07.
12
Table showing calculations of cost of crashing per week and maximum crashing possible
Activity Crashing Data
Activity
Immediate
Predecessor (s)
A
B
C
D
E
F
G
H
I
J
NONE
NONE
NONE
A
B
B
C
D,E
F,G,H
I
Normal
Time
(Weeks)
Crash
Time
(Weeks)
9
5
7
12
8
6
11
5
4
10
Total Cost
6
4
5
8
5
4
9
4
3
8
Activity on Node (AON) Diagram
13
Normal
Cost ($)
Crash
Cost ($)
$13,000
$7,000
$15,000
$12,000
$9,000
$5,000
$13,000
$8,000
$2,500
$12,000
$96,500
$15,550
$7,900
$15,800
$14,800
$10,500
$6,200
$14,000
$9,000
$3,500
$15,000
$112,250
Cost of
Crashing
per Week
($)
Maximum
Crashing
Possible
(Weeks)
$850
$900
$400
$700
$500
$600
$500
$1,000
$1,000
$1,500
3
1
2
4
3
2
2
1
1
2
Crashing Process
Paths
Schedule 1
Schedule 2
Crash D by 4
Weeks
Schedule 3
Crash A by3
Weeks
Schedule 4
Crash H by 1
Week
Schedule 5
Crash I by 1
Week
0
4
3
1
1
2
40
32
25
32
36
32
25
32
33
32
25
32
32
31
24
32
31
30
23
31
29
28
21
29
0
700
850
1000
1000
1000
Increase in activity cost
because of crashing
0
2800
2550
1000
1000
2000
Total Activity Cost
97000
99800
102350
103350
104350
106350
Normal Schedule
Number of weeks
crashed
A-D-H-I-J
B-E-H-I-J
B-F-I-J
C-G-I-J
Cost of Crashing/week
for activities that are
chosen for crashing
Schedule 6*
Crash J by 2 Weeks
The critical path consists of activities A-D-H-I-J with a schedule time of 40 weeks.
Activities are crashed on the critical path in ascending order of their costs of crashing per week.
Path A-D-H-I-J continues to be the critical path until Schedule 3 after D is crashed by 4 weeks and A is crashed by 3 weeks. The project length
is now 33 weeks.
In Schedule 4, one of the three activities, H, I and J can be picked up for crashing. Each one of them has $ 1,000 cost of crashing per week.
Activity H is picked up for crashing by one week using alpha order.
After H is crashed, two paths, A-D-H-I-J and C-G-I-J become critical. The project length is now 32 weeks.
Activity I and J are then crashed by one and two weeks respectively leading to Schedule 6.
At this stage all activities on path A-D-H-I-J have been crashed. Project duration can not be reduced any more.
Fixed Cost per week
Time vs. Cost Analysis
900
Project Time Crashing Cost Activity Cost
29
1000
106850
30
1000
105350
31
1000
104350
32
1000
103350
33
850
102350
34
850
101500
35
850
100650
36
700
99800
37
700
99100
38
700
98400
39
700
97700
40
0
97000
Fixed Cost
26100
27000
27900
28800
29700
30600
31500
32400
33300
34200
35100
36000
Minimum Total Cost
Project Duration
Total Cost
132950
132350
132250
132150
132050
132100
132150
132200
132400
132600
132800
133000
132050
33
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4. Consider the data given in the following table and answer the questions that follow.
Activity
Immediate
Predecessor (s)
Normal Time
(days)
Crash Time
(days)
Normal Cost
$
Crash Cost
$
A
B
C
D
E
None
A
A
B and C
D
4
4
1
3
2
2
3
1
2
1
200
300
200
600
500
400
600
200
650
900
a. What is the normal project time?
Answer: 13 days
b.Identify the critical path?
Answer: A-B-D-E
c. What is the normal cost of the project?
Answer: $ 1,800. This is the total of the normal costs of all activities.
d.If the project time is to be reduced by one day, which activity should be crashed first? What
is the cost of crashing per day of activity E?
Answer: The critical path is A-B-D-E. The cost of crashing activity D is the smallest.
Therefore, crash D. The cost of crashing activity E is $ 400 per day. See the table below.
Solution:
The AON diagram is given below.
B
D
A
E
C
There are 2 paths that are listed below. The numbers in parentheses show the length of the two
paths.
A-B-D-E (13)
A-C-D-E (10)
A-B-D-E is the longer path – the critical path. This gives a project duration of 13 days.
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Table showing calculations for the cost of crashing per day.
Activity
A
B
C
D
E
Immediate
Predecessor
(s)
None
A
A
B and C
D
Normal
Time
(days)
4
4
1
3
2
Crash
Time
(days)
2
3
1
2
1
Normal
Cost $
Crash
Cost $
200
300
200
600
500
400
600
200
650
900
Cost of
Crashing
per Day
100
300
n.a.
50
400
Maximum
Crashing
Possible
2
1
0
1
1
5. There are two paths in a network (see below). The length of Path 1 is 30 days and that of Path
2 is 28 days. You are now told that activity F which currently takes 4 days will actually take 3
days more, that is, 7 days. What will be the project duration with this revised time?
Path 1: A-B-D-E-G
Path 2: A-C-D-F-G
Answer: 31 days.
Solution
The current project duration is 30 days (length of path 1). The length of path2 will become 31
days (28 + 3) if activity F takes 3 days more. The length of path 1 remains unchanged. Therefore,
the project duration becomes 31 days.
6. There are two paths in a network (see below). The length of Path 1 is 30 days and that of Path
2 is 28 days. You are now told that activity D which currently takes 2 days will actually take 3
days more, that is, 5 days. What will be the project duration with this revised time?
Path 1: A-B-D-E-G
Path 2: A-C-D-F-G
Answer: 33 days.
Solution
The length of path 1 will become 33 days and the length of path 2 will become 31 days if activity
D takes 3 more days.
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7. Suppose the length of critical path in a project is 40 days. There are four activities A, B, C
and D on the critical path. The variances of these activities are: A (1.4), B (1.2), C (0.9) and D
(0.4). What is the probability of completing the project within 42 days?
Answer: 0.0844
Solution
Variance of the critical path = 1.4 + 1.2 + 0.9 + 0.4 = 3.9
Standard Deviation of the critical path (σ) = 1.9748.
Due Date (D) = 42 days
Length of critical path (T) = 40 days.
z = (D-T)/ σ = (42-40)/1.9748 = 1.01274.
The probability corresponding to a z value of 1.01274 is 0.8440.
See the z-table for finding the probability.
8. Suppose the length of critical path in a project is 50 days and the standard deviation is 2 days.
What due date should be set for the project so that the probability of completing the project is
90%? Round the answer to the nearest higher integer.
Answer: 52.57
Solution
Length of critical path = 50 days
Standard deviation (σ) = 2
The z-value for a probability of 0.9 is 1.285.
The due date is calculated from the formula z = (D-T)/ σ.
D = z * σ + T = 1.285*2 + 50 = 2.57 + 50 = 52.57.
9. Suppose the length of critical path in a project is 35 days. The probability of completing the
project within 34 days is:
(a) Less than 50%
(b) Greater than 50%
(c) Equal to 50%
(d) Cannot be determined from the above data
Answer: (a) Less than 50%.
Solution: The due date is smaller than the length of the critical path. Therefore, the probability of
completing the project by the due date is less than 50%.
It may be noted that if the due date is equal to the length of the critical path then the probability
of completing the project by the due date will be 50%. In the formula, z = (D-T)/ σ, the value of z
will be 0 (zero) since D =T. For z = 0, the probability is 50%. The value of σ is immaterial.
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