Project management project scheduling

PROJECT MANAGEMENT JAN, 2012

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PROJECT MANAGEMENT

BY: TADELE ASMARE

Phone: +251920774757

E_mail:[email protected]

JAN, 2012


1

ACTIVITY:

i. identify a specific project title

ii. list out the different tasks to be completed to finish the project

iii. estimate the duration of each of each of the activities

iv. find out the logical relationship

v. construct the network diagram

vi. find a project completion time

vii. carry out time crashing /take own assumptions

viii. estimate resource requirement

ix. apply project leveling

1. MANUFACTURING SPECTROPHOTOMETER

This project is to construct a spectrophotometer: Adjoining is a list and description of

activities for a project constructing (building) a spectrophotometer.

Activity description

(spectrophotometer design)

Activity Preceding

activities

Duration

(days)

Normal

time

Normal

cost(Br)

Man power

required/da

y

Crashing

Time

(days)

Cost (Br)

design optical sensor A - 7 1000 8 2 1480

Prepare light source B - 3 800 3 2 860

design signal processor

&scanning device

C A 9 1540 3 7 1580

obtain optical sensor D A 6 800 2 3 935

design prism and slit E A 6 500 3 3 590

obtain signal processor F C 5 700 4 4 790

obtain scanning device G C 4 750 2 2 850

design softwares

accessories

H C 5 720 2 4 840

prepare optical sensor I B,D 6 600 2 4 712

connect prosessor and

scanning device

J F,I 2 300 3 1 330

connect optical senor K E,H,G,J 2 400 5 1 440

obtain prisms, slits and

softwares vs. accessories

L E,H 4 600 3 3 670

connect prism, slits M L,K 2 200 8 1 240


2

Time related overhead expense, fixed cost, for this project is Birr 270 per day.

And there is only 8 man power to be allocated.

THE NET WPRK DIAGRAM

6/3 E

A7/8 5/2 H L4/3

D 6/2 9/3 C Dummy 0

B3/4 4/2 G 2/5K 2/8 M

F 5/4 2/3 J

6/2 I

- Identify the status of paths

Paths Duration (days) State

B-I-J-K-M 15 Non-critical path

A-D-I-J-K-M 25 Non-critical path

A-C-F-J-K-M 27 Critical path

A-C-G-K-M 24 Non-critical path

A-C-H-L-M 27 Critical path

A-E-L-M 19 Non-critical path

- Project completion time = 27 days

- Critical paths = A-C-F-J-K-M and A-C-H-L-M

1

6

5

3

4

7 8 9

2

Normal Day /Man


3

I. Project time crashing

Step 1: establish time- cost relationship

Max. Time that an activity to be crashed = normal cost –crashing cost

Activities A B C D E F G H I J K L M

Normal

time(days)

7 3 9 6 6 5 4 5 6 2 2 4 2

Normal poj.

Cost(Br)

1000 800 1540 800 500 700 750 720 600 300 400 600 200

Crashing

time(days)

4 2 7 3 3 4 2 4 4 1 1 3 1

Crashing

cost(Br)

1288 800 1580 935 590 790 850 840 712 330 440 670 240

Max.time an

activity to be

crashed(days)

3 1 2 3 3 1 2 1 2 1 1 1 1

Crash cost –

time

slope(Br/day)

96 60 20 45 30 90 50 120 56 30 40 70 40

Step 2: Identifying critical paths

Crashing Cost – time slope = Crashing cost – Normal cost

Normal time - crashing time


4



A-D-I-J-K-M 25 Non-critical path


A-C-G-K-M 24 Non-critical path



Step 3: computing PCT and costs

- Normal Project time = 27 days

- Total normal cost =∑( Normal poject Cost(Br)) = 8910

- Total fixe cost (Br)= 27*270 = 7290

- Crash cost = 0

- Therefore, TOTAL COST = 8910+7290= 16200

Step 4: identify the non critical path with maximum durations next to critical

path. So, A-D-I-J-K-M is the required one with 25 days durations.

- Crash the crtitical paths by 27-25=2 days

- In the critical paths, A-C-F-J-K-M and A-C-H-L-M activity A, C and M are the

common to both activities. But, C has least crashing cost per days and has to be

crashed by 27-25=2 days or its maximum crashing period. And another option is

crashing one activity from each path.

- Computing costs,


5

TOTAL COST = 16200 +2*20 - 2*270 = 15700 Birr

PCT = 25 days

- Updating the time net work diagram

E 6

A7 5 H L4

D 6 C7 Dummy 0

B 3 G 4 K 2 M 2

F 5 J 2

I 6

- Now there are three critical paths



A-D-I-J-K-M 25 critical path


A-C-G- K-M 22 Non-critical path



- A-C-G-K-M is the first non critical path next to critical in duration

- Critical paths, A-D-I-J-K-M, A-C-F-J-K-M and A-C-H-L-M have to be crashed by

25-22 =3 day. Among the activities A and M are the common activities. But, M is

with low crashing cost per day.

- So now crash M

1

6

5

3

4

7 8 9

2


6

-

- Total cost 15700 + 1*40 – 1*270 = 15470

- PCT = 24 days

6 E

A7 5 H L4

D 6 C7 Dummy

B3 4 G 2 K 1 M

F 5 J 2

I 6

# Now A-C-G- K-M is the first non critical path next to critical paths.

- And the critical paths have to be crashed by a maximum number of days = 24 -21 = 3

days. Here, the common activity A has to be crashed since M has been already

crashed.





A-C-G- K-M 21 critical path



- TOTAL COST = 15470 + 3*96 - 3*270 = 14948

- Now PCT = 21 days

1

6 3

4

7 8 9

2

5


7

- updating net work diagram

6 E

A 4 5 H L4

D 6 C7 Dummy

B3 4 G 2 K 1 M

F 5 J 2

I 6





A-C-G- K-M 21 Critical path



- Now A-C-G-K-M is critical path too. And A-E-L-M, takes 15 days, is the first long

non critical path next to citicals and it has to be crashed.

- So crash critical paths by 21-15 =6 days. But it is difficult to crash a critical activity

by once by 6 days. Commonly, all critical activities must be crashed by same number

of days.

1

6 3

4

7 8 9

2

5


8

critical

activities

Activities to

be crashed

MAX. Time

to crash

Cash cost per day

TOTAL

maximum

crashing

days

Selected

activities

A-D-I-J-K-M D 3 45 7

J & K

I 2 56

J 1 30

K 1 40

A-C-F-J-K-M F 1 90 3 J & K

J 1 30

K 1 40

A-C-H-L-M H 1 120 2 H &L

L 1 70

A-C-G-K-M

G

2

50

3

G &K

K 1 40

- Since the maximum crashing days of path A-C-H-L-M is to be 2 days, the project can

be crashed by 2 days now. Common activities are given priority for crashing so as to

minimize crashing cost.

- In A-C-G-K-M, G is only crashed for 1 day.

- PCT = 19 days.


9

6 E

A 2 4 H L 3

D 6 C7 Dummy

B3 3 G 1 K 1 M

F 5 J 1

I 6

- PCT = 19 days.

- To analysis the cost, the crashed activities each for 1 day H, L, G, J and K must be

considered.

- Total cost = 14948+ 1*(120 +70+50+30+40) -2*270

= 14948 + 310 -540

= 14718 Birr

Therefore, Total cost = 14718 Birr

- Since the critical activities in the path A-C-H-L-M are all crashed, the project can’t be

further crashed. So the optimal cost is Birr 14718.

Project cost

16200 .

15700 .

15470 .

14918 .

14718 .

19 21 24 25 27

1

6 3

4

7 8 9

2

5


10

Normal project completion time and its corresponding cost are 27 days and Birr 16200,

respectively.

The project crashing time and optimal time is same, 19 days. As a result, the optimal cost and

maximum crashing cost are equal =14718.

II. Resource levelling – in levelling time – scale graph and resource

histogram are used.

Hence, the resource is tried to be leveled before crashing. To do so,

steps followed are:

draw critical paths on the straight lines on time – scale graph before levelling

draw the non criticals below the criticals

Draw the resource histogram before levelling.

Apply levelling to methods of levelling such as delay non critical activities,

splitting non critical activities in to non sequential.

And draw activities on the time scale graph after levelling

Draw the resource histogram after levelling (the graph below is represented as

follow and the histogram is given in rectangular form below each time scale

graph).

Resource availability

1 2 3 time

8 8 8 resource requirement


11

A 8 C 3 F 4 J 3 5 K 8 M

B 3 … . . . . . .. .. .. .. I 2 .. ..

D 2

H 2 L 3

E 3 .. .. .. .. .. .. .. ..

G 2

1 2 3 4 5 6 7 8 9 1

0

1

1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1

9

2

0

2

1

2

2

2

3

2

4

2

5

2

6

2

7

1

1

1

1

1

1

8 8 8 8 8 8 8 8 8 8 5 5 5 1

0

1

0

1

0

8 6 6 6 8 8 8 8

before levelling time scale graph above and resource histogram below are given

1

0

8

6

5

1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7

A 8 C 3 F 4 3 J 5 K 8 M

3 B I 2

D 2 H 2 L 3

O

E 3

G 2

8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

FG

1 2 4 6 7 8 9

1

3

2

2

4 5

4

11

10

8

6

5

1

1

4 6 7 8 9

3

2

2 4 5

5

2

4

8 LEVELLED HISTOGRAM

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Business

Project management project scheduling