[Recommended]Principles of Project Management

Principles of Project Management Image result for FAB 5 LOGO PMGT 510 Principles of Project Management Group Assignment 9 Part1 Staffing management is much the…

Principles of Project Management
Image result for FAB 5 LOGO
PMGT 510
Principles of Project Management
Group Assignment 9
Part1
Staffing management is much the same as it sounds; it is an approach to deal with the staff on a team or project. The staffing management plan will state when work force will be dropped or added to the project or when they have finished their bit if they will be moved to another level of the project, or if their business is finished. A decent instrument to demonstrate this management plan is to utilize an asset histogram. This will outwardly put the staffing management intend to utilize. An obligation task network would be a decent method to appear and clear up the jobs and duties regarding errands expected to finish the project.
When performing resource leveling, it is helpful for the project manager to consult both the Critical path schedule and resource histogram.
The two techniques for compressing a schedule are:
First schedule compression approach is Fast Tracking. In fast-tracking schedule compression system, basic way exercises are performed in parallel rather than arrangement. This is conceivable just the exercises are not in required reliance. Since, if two exercises are relying upon one another essentially, you can’t do these two exercises in parallel. For example, you can’t begin testing of a screen before finishing advancement. If basic way exercises are relying upon one another due to asset reliance or if there is an optional reliance, you can fast track those exercises to finish remaining exercises faster.
Second schedule compression system is crashing. In crashing schedule compression strategy, there is an exchange off among expense and schedule. On the off chance that the extension is the equivalent and project is behind schedule, another alternative for compacting the schedule is putting additional assets on residual exercises of the project. In such a case that it is conceivable to appoint more than one asset on an action, movement span will diminish individually. This will total the project faster. In any case, since these additional assets were not in the underlying arrangement, there will be an extra expense if crashing is utilized for schedule compression.
When crashing a project schedule, describe the two criteria that are considered when considering which activities will be sped up.
First criteria: Since the critical path indicates the completion time, so choose the right activities for the critical path and is very important.
Second criteria: Some activities can speed up and some cannot, so knowing which activates are worth speed up this can minimize the daily cost and the overall spending.
Part 2: Project network/schedule -scheduling and compressing a schedule (30 points)
You’ve been asked to manage a project that includes conducting a two-week (10 working days) computer training class. Using the information in Table below, complete the following:
Activity
Immediate Predecessor
Normal Time (days)
Normal Cost
Crashed Time (days)
Crashed Cost
A – Obtain Instructor
None
7
$500
4
$800
B – Locate Room
A
3
$200
2
$350
C – Check Coats
None
6
$500
4
$900
D – Room Open?
C
3
$200
1
$500
E – Schedule Class
B and D
2
$300
1
$350
Prepare a network schedule showing:
a. Early Starts (ES) for all activities
a. Late Starts (LS) for all activities
a. Slacks for all activities
a. Critical path and its length
a. Non-critical paths and their length
https://lh3.googleusercontent.com/U0cdt9BO9oJnhXWB2AdlRmZUZn0HGoBF7opYgJFp2R0tWVsZCTtTTsv0a9rEC_H7M9nwuf8ZVvgKixO3hvEJYakEK9_ovwliQ-32HdTFUTSZ1jHxD2ItG_sKnDrNRNklCijkYus
Critical Path is A-B-E, the length is 7+3+2=12 days
Non-Critical Path is C-D-E, the length is 6+3+2=11 days
Schedule Compression:
a. What is the crash cost per day for each activity?
A. (800-500)/(7-4) = 100$
B. (350-200)/(3-2) = 150$
C. (900-500)/6-4 = 200$
D. (500-200)/(3-1) = 150$
E. (350-300)/(2-1) = 50$
b. Which activities should be crashed to meet a project deadline of 10 days with a minimum cost? You may assume that you can partially crash an activity.
Critical path = A-B-E = 7+3+2 = 12 days
E + A = 50 + 100 = 150
Non-critical path = C-D-E = 6+3+2 = 11 days
E = 50
c. What is the new cost after crashing the project?
For critical path
A+B+C+D+E = (500+100) +200+500+200+(300+50) = 1850$
For non-critical path
A+B+C+D+E = (500)+200+500+200+(300+50) = 1750$
d. Fully crash the project to its minimum duration. Prepare a table that shows (for each day the project length is reduced) the project duration, the activities that are crashed, the incremental crash cost, and the total crash cost.
Activity
Immediate Predecessor
Normal Time (days)
Normal Cost
Crashed Time (days)
Crashed Cost
crash cost per day
A – Obtain Instructor
None
7-6-5-4
$500
4
$800
100
B – Locate Room
A
3-2
$200
2
$350
150
C – Check Coats
None
6-5-4
$500
4
$900
200
D – Room Open?
C
3-2-1
$200
1
$500
150
E – Schedule Class
B and D
2-1
$300
1
$350
50
Path
Duration
ABE
12-11-10-9-8-7
CDE
11-10-9-8-7
Project Duration
Activity Crashed
Incremental Crash Cost
Cumulative Crash Cost
12

0
0
11
E
50
50
10
A&D
100+150 (250)
300
9
A&D
100+150(250)
550
8
A&C
100+200(300)
850
7
B&C
150+200(350)
1200
PART 3
The project schedule for given data is shown in the diagram below
https://lh5.googleusercontent.com/L1Sh9-N9bPIYW0pLK2bhqxRWJ4HgZ-Z8ybAHYBbw1E6B1nt8DFOBX5i90VntQL1ubKwrE4sEtA4S2L30mqCqe523_Di3wMuz8rA1Kra01tY2g4y8PgIc7DZ3f8WZ9zYOb4EyIjk
https://lh5.googleusercontent.com/3Lv51SlLm7ReReii86Dsu3c_UM91EmNOWKsQVuEItcxDkwYyGEjT4SySZcDwJc2DoX-jMTWvyB0qCRl4PcdbXFljc7Veodf2EHv3wdPiPz10Ykbd8_JJqDvDzvylP6QBEQphCQ4
Here, the blocks mentioned above follow the legend and its description shown below
https://lh3.googleusercontent.com/k7iI_25kzkijHRPdk0E5tbyF1P9a7fN4DMUz7w22WUnFOxYf-KDx4iNLEw9MSq3pOvd2efwl-zwZLzsZHJBOYfBa-5I3IBCmHUdwSAVLzVOUZU5MoNxDnctoLgMZouKjlBELGo4
§ <Any Activity> can be any activity that is a part of the project
§ <Duration> is the time in the number of days required to finish a task
§ ES stands for ‘Early Start’ and is the earliest possible time on which uncompleted portions of a schedule activity can start
§ LS stands for ‘Late Start’ and is the latest possible time on which uncompleted portions of a schedule activity can start
§ EF stands for ‘Early Finish’ and is the earliest possible time on which uncompleted portions of a schedule activity can start
§ LF stands for ‘Late Finish’ and is known as the latest possible time on which uncompleted portions of a schedule activity can finish
§ Float has the formula: ‘LS-ES’ or ‘Late Start – Early Start’
§ Formula: EF = ES + Duration
§ Formula: LS = LF – Duration
§ Formula: Float = LS – ES
The table mentioned below specifies the slack/float allowed for each activity.
Activity
Float = LS-ES
Float/Slack Allowed
A: Evaluate Freezers
4 minus 0
4
B: Chart Temperatures
0 minus 0
0
C: Review Service Record
4 minus 0
4
D: Consult with HVAC Engineer
6 minus 6
0
E: Develop Construction Plan
9 minus 9
0
F: Complete IC Assignment
22 minus 19
3
G: Complete ROI Analysis
19 minus 19
0
H: Conduct Regulatory review
20 minus 19
1
I: Obtain Construction Approval
24 minus 24
0
Based on the network schedule diagram above, the table below shows duration for each path
Path
Duration Calculation
Final Duration (in days)
A-D-E-F-I
2+3+10+2+2
19
A-D-E-G-I
2+3+10+5+2
22
A-D-E-H-I
2+3+10+4+2
21
B-D-E-F-I
6+3+10+2+2
23
B-D-E-G-I
6+3+10+5+2
26
B-D-E-H-I
6+3+10+4+2
25
C-D-E-F-I
2+3+10+2+2
19
C-D-E-G-I
2+3+10+5+2
22
C-D-E-H-I
2+3+10+4+2
21
As seen in the table above, and as seen in the slack table seen in ‘Answer a’, the critical path will be B-D-E-G-I with a duration of 26 days.

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