This document summarizes a project to construct a spectrophotometer. It lists the activities required and their durations, predecessors, resource needs, and costs. It constructs a network diagram and identifies critical paths. Time and cost are analyzed through crashing activities on the critical paths. The optimal completion time is 19 days at a cost of 14718 Birr. Resource leveling is also discussed to smooth resource usage before time crashing.

1. PROJECT MANAGEMENT JAN, 2012
BAHIR DAR UNIVERSITY
INSTITUTE OF TECHNOLOGY FOR TEXTILE, GARMENT AND FASHION
DESIGN
TEXTILE ENGINEERING PROGRAMM
OPERATIONS RESEARCH –PROJECT MANAGEMENT
Phone: +251920774757
E_mail:tadeleasmare@yahoo.com
JAN, 2012
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2. PROJECT MANAGEMENT JAN, 2012
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 Activity Preceding Duration Man power Crashing
(days) Normal
(spectrophotometer design) activities required/da
cost(Br) Time Cost (Br)
Normal
y
time (days)
design optical sensor A - 7 1000 8 2 1480
Prepare light source B - 3 800 3 2 860
design signal processor C A 9 1540 3 7 1580
&scanning device
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 H C 5 720 2 4 840
accessories
prepare optical sensor I B,D 6 600 2 4 712
connect prosessor and J F,I 2 300 3 1 330
scanning device
connect optical senor K E,H,G,J 2 400 5 1 440
obtain prisms, slits and L E,H 4 600 3 3 670
softwares vs. accessories
connect prism, slits M L,K 2 200 8 1 240
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3. PROJECT MANAGEMENT JAN, 2012
Time related overhead expense, fixed cost, for this project is Birr 270 per day.
And there is only 8 man power to be allocated.
Normal Day /Man
THE NET WPRK DIAGRAM
5
6/3 E
2
A7/8 5/2 H L4/3
1
D 6/2 9/3 C Dummy 0
4
B3/4 4/2 G 7 2/5K 8 2/8 M 9
F 5/4 2/3 J
3
6
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
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4. PROJECT MANAGEMENT JAN, 2012
I. Project time crashing
 Step 1: establish time- cost relationship
Crashing Cost – time slope = Crashing cost – Normal cost
Normal time - crashing time
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 7 3 9 6 6 5 4 5 6 2 2 4 2
time(days)
Normal poj. 1000 800 1540 800 500 700 750 720 600 300 400 600 200
Cost(Br)
Crashing 4 2 7 3 3 4 2 4 4 1 1 3 1
time(days)
Crashing 1288 800 1580 935 590 790 850 840 712 330 440 670 240
cost(Br)
Max.time an 3 1 2 3 3 1 2 1 2 1 1 1 1
activity to be
crashed(days)
Crash cost – 96 60 20 45 30 90 50 120 56 30 40 70 40
time
slope(Br/day)
 Step 2: Identifying critical paths
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5. PROJECT MANAGEMENT JAN, 2012
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
 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,
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6. PROJECT MANAGEMENT JAN, 2012
TOTAL COST = 16200 +2*20 - 2*270 = 15700 Birr
PCT = 25 days
- Updating the time net work diagram
5
E6
2
A7 5 H L4
1
D 6 C7 Dummy 0
4
B 3 G 4 7 K 2 8 M 2 9
F 5 J2
3
6
I 6
- Now there are three critical paths
Paths Duration (days) State
B-I-J-K-M 15 Non-critical path
A-D-I-J-K-M 25 critical path
A-C-F-J-K-M 25 Critical path
A-C-G- K-M 22 Non-critical path
A-C-H-L-M 25 Critical path
A-E-L-M 19 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
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7. PROJECT MANAGEMENT JAN, 2012
-
- Total cost 15700 + 1*40 – 1*270 = 15470
- PCT = 24 days
6E 5
2
A7 5 H L4
1
D 6 C7 Dummy
4
B3 4G 7 2K 8 1 M 9
F5 J2
3
6
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.
Paths Duration (days) State
B-I-J-K-M 13 Non-critical path
A-D-I-J-K-M 24 critical path
A-C-F-J-K-M 24 Critical path
A-C-G- K-M 21 critical path
A-C-H-L-M 24 Critical path
A-E-L-M 15 Non-critical path
- TOTAL COST = 15470 + 3*96 - 3*270 = 14948
- Now PCT = 21 days
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8. PROJECT MANAGEMENT JAN, 2012
- updating net work diagram
5
6E
2
A4 5 H L4
1
D 6 C7 Dummy
4
B3 4G 7 2K 8 1 M 9
F5 J2
3
6
I 6
Paths Duration (days) State
B-I-J-K-M 13 Non-critical path
A-D-I-J-K-M 21 critical path
A-C-F-J-K-M 21 Critical path
A-C-G- K-M 21 Critical path
A-C-H-L-M 21 Critical path
A-E-L-M 15 Non-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.
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9. PROJECT MANAGEMENT JAN, 2012
critical Activities to MAX. Time Cash cost per day TOTAL Selected
maximum activities
activities be crashed to crash
crashing
days
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
G 2 50 3 G &K
A-C-G-K-M
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.
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10. PROJECT MANAGEMENT JAN, 2012
5
6 E
2
A2 4 H L3
1
D 6 C7 Dummy
4
B3 3G 7 1K 8 1 M 9
F5 J1
3
6
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
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11. PROJECT MANAGEMENT JAN, 2012
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
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12. PROJECT MANAGEMENT JAN, 2012
A 8 C 3 F 4 J 3 5 K 8 M
1 2 4 6 7 8 9
B 3 … . . . . . .. .. .. .. I 2 .. ..
1 3
D 2
2
H 2 L 3
4 5
E 3 .. .. .. .. .. .. .. ..
2
G 2
4
1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7
1 1 1 8 8 8 8 8 8 8 8 8 8 5 5 5 1 1 1 8 6 6 6 8 8 8 8
1 1 1 0 0 0
before levelling time scale graph above and resource histogram below are given
1
011
8
10
6
8
5
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
1 2 4 6 7 8 9
3 B I 2
1 3
D 2 H 2 L 3
2 4 5
O
5
E 3
2
G 2
4
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
8 LEVELLED HISTOGRAM
FG
5
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