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# A Project Management Assignment Focusing On The Project Scheduling (Coursework Sample)

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this sample is about a project management ASSIGNMENT focusing on the project scheduling

source..
Content:

INSTITUTION:
TOPIC:
ASSIGNMENT 1-50%
STUDENT NAME:
STUDENT ID NO:
DATE:
Assignment Outline
This assignment is based on a project case study to be undertaken by Air Control Company. It explore the construction, and use of network diagram in project scheduling. Firstly, this assignment involved constructing the network diagram based on the precedence table given. Secondly, from the constructed diagram different project times are determined namely; earliest start (ES), earliest finish (EF), latest start (LS) and latest finish (LF). Thirdly, the slack/float for each of the activities is determined and then the total project effort and duration is calculated. Fourthly, the critical path in the network diagram is computed and effect of delaying this path on project timeline explored. Fifthly, significance of having a network diagram in a project is illustrated.
Sixthly, a number of ways of reducing project duration are enumerated and explained. Lastly, different ways of projects acceleration when resources are constrained and when resources are not constrained are critically discussed.
SOLUTION
1 Using the precedence table provided a network diagram is constructed as shown. His network diagram is drawn based on critical path method(CPM) CITATION Cha17 \l 1089 (Rayadurg, 2017).
B (15)
17862551079500
23958558699500
9189573241300
C (10)
29959301460500024911054127500
1021513666750050038018071700A (2)
340550515811500 F (15) G (10)
1069340148590004691380698500
39674807493000
898748196290029688194949500507238016192500
D (13)
24149059080500
483425515684500
E (18) H (5)
2367280889000
2948724107240051631855016500
498348014478000
I (0)
Note that I have introduced the activity I in the network diagram, which has a duration of zero to symbol end of the project.
1 Earliest Start (ES) – 5 marks
Let ESi be the earliest start the project will commence for any activity i when all that activities before it have been completed. The values of the earliest start ESi (i=A, B,…,I) are determined by moving forward, from left to right, in the network diagram CITATION Ale17 \l 1089 (Puscasu, 2017) CITATION JEB16 \l 1089 (Beasley, 2016).
Also let Ti be the time it takes to complete any activity i ( e.g. TB ). Also assume the start time of the project is zero days.Then the ESi are calculated as:
ESA=0
ESB= ESA+ TA=0+2=2
ESC= ESA+ TA=0+2=2
ESD= ESA+ TA=0+2=2
ESE= ESA+ TA=0+2=2
ESF=max[ESC+ TC, ESD+ TD =max2+10,2+13=15
ESG=max[ESB+ TB, ESF+ TF =max2+15, 15+15=30
ESH=max[ESG+ TG, ESE+ TE =max30+10,2+18=40
ESI= ESH+ TH=40+5=45
2 Earliest Finish (EF)
Note that the earliest finish time is calculated from the earliest start time of a particular activity. It is given by: EFi= ESi+ Ti i.e. the activity duration in each of network path are added to the early start( ES+ Duration= EF)
EFA=0+2=2
EFB= ESB+ TB=2+15=17
EFC= ESC+ TC=2+10=12
EFD= ESD+ TD=2+13=15
EFE= ESE+ TE=2+18=20
EFF= ESF+ TF=15+15=30
EFG= ESG+ TG=30+10=40
EFH= ESH+ TH=40+5=45
EFI= ESI+ TI=45+0=45
3 Latest Start (LS)
Let LSi be the earliest start the project will commence for activity i such that all its preceding activities have been finished. We calculate the values of the LSi (i=A, B,…,I) by going backward, from right to left, in the network diagram CITATION JEB16 \l 1089 (Beasley, 2016). The earliest start time is therefore given by LS of previous activity – Duration of activity.
LSI=45
LSH=LSI-TH=45-5=40
LSG=LSH-TG=40-10=30
LSF=LSH-TF=40-15=25
LSE=LSH-TE=40-18=22
LSD=LSF-TD=25-13=12
LSC=LSF-TC=25-10=15
LSB=LSG-TB=30-15=15
LSA=min⁡[ LSB-TA,LSC-TA,LSD-TA,LSE-TA]
LSA=min15-2,15-2,12-2,22-2=10
4 Latest Finish (LF)
The latest finish(LF) is taken as the latest start(LS) of the prior activity to the current activity in the network diagram. Therefore, using the latest starts obtained in part c the latest start for he activities are as indicated below:
LFI=45
LFH=LSI=45
LFG=LSH=30
LFF=LSH=30
LFE=LSF=25
LFD=LSE=22
LFC=LSD=12
LFB=LSG=15
LFA=LSB=15
5 Slack/Float on each activity
From a and c we know the earliest start time and latest start time therefore we can determine slack or float time for any activity given by Fi=LSi-ESi . Where float is the time a particular task or activity can be delayed without a resultant on the successive activities connected to it. Therefore, float will show the increase that can be made on the duration of any particular activity without any change in the project completion time. From formula of float time the table below is formulated. As a check all float values should be >= 0.
Activity

Latest start time(LSi)

Earliest Start time(ESi)

Float
Fi=LSi-ESi

A

10

0

10

B

15

2

13

C

15

2

13

D

12

2

10

E

22

2

20

F

25

15

10

G

30

30

0

H

40

40

O

6 Total effort and duration of the project -5marks
The total project duration may be obtained by determining the path with the longest duration, this will give us the shortest possible time the project will take. Determining, the length of each path as:
Path A,B,G,H=2+15+10+5=32
Path A,C,F,G,H=2+10+15+10+5=42
Path A,D,F,G,H=2+13+15+10+5=45
Path A,D,H=2+18+5=25
The total duration= 45 days
Total Effort
Total Effort = Number of work hours or man-hours
Assu...

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