Management of Engineering
Projects
Limited Resources
Dr. L. K. Gaafar
3/23/2016
Example
Activity
Duration
PA
Resources
A
B
C
D
3
4
6
7
-A
-C
2
2
3
2
Example
3
A 3 (2)
Activity
Duration
PA
Resources
A
B
C
D
3
4
6
7
-A
-C
2
2
3
2
B4 (2)
0
13
D 7 (2)
C 6 (3)
6
Critical path is CD. Minimum completion time is 13.
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3
B4 (2)
A 3 (2)
0
13
D 7 (2)
C 6 (3)
6
Activity
Time
Resources
Resource Profile 1
5 needed
Time
Schedule 1
Example
3
A 3 (2)
Activity
Duration
PA
Resources
A
B
C
D
3
4
6
7
-A
-C
2
2
3
2
B4 (2)
0
13
Next slide shows 2 possible
schedules to finish the
project on time (13), and the
associated resource profile in
each case.
D 7 (2)
C 6 (3)
6
Critical path is CD. Minimum completion time is 13.
3/23/2016
3
B4 (2)
A 3 (2)
0
13
D 7 (2)
C 6 (3)
6
Activity
Activity
Time
Resources
Time
Resources
Resource Profile 1
5 needed
Resource Profile 2
4 needed
Time
Schedule 1
Time
Schedule 2
Single Resource Allocation
Brooks’ Algorithm
1. Develop the activity network.
2. Determine the activity control time (ACTIM) for all activities. ACTIM is the
longest path to the end of the project starting from the activity.
3. Rank activities in a descending order of their ACTIM with ties broken according
to longest duration followed by largest resource requirements.
4. Set TNOW to 0.
5. Set the early start (TEARL) of all activities that may start at TNOW.
6. Schedule all possible activities that could be started and allocated enough resource
units. Stop when all available activities are scheduled, or when you run out of
resource units for the next available activity. Determine the finishing time for all
scheduled activities (TFIN).
7. Update TEARL of remaining activities based on TFIN of scheduled activities.
8. Advance TNOW to next available TEARL or TFIN and repeat step 5. Continue
until all activities are scheduled.
3/23/2016
Activity
A
B
C
D
Resource allocation
assuming 4 units are
available
ACTIM
7
4
13
7
Order is CDAB
D is ranked before A because it has a longer duration
Activity
ACTIM
Duration
Resources Required
TEARL
TSTART
TFIN
Iteration
TNOW
Resources Available
Activities Allowed
C
13
6
3
0
1
0
4
CA
D
7
7
2
A
7
3
2
0
B
4
4
2
C and A may start at time 0 and their
TEARL is set accordingly.
4 units are available for assignment
Since B depends on A, its TEARL
cannot be determined at this time.
Since D depends on C, its TEARL
cannot be determined at this time.
Algorithm setup at TNOW = 0
3/23/2016
3
B4 (2)
A 3 (2)
0
13
TNOW = 0
D 7 (2)
C 6 (3)
6
Resource allocation assuming 4 units are available
Activity
ACTIM
Duration
Resources Required
TEARL
TSTART
TFIN
Iteration
TNOW
Resources Available
Activities Allowed
3/23/2016
C
13
6
3
0
0
6
D
7
7
2
6
A
7
3
2
0
B
4
4
2
1
0
41
CA
After First Iteration
C and A may be scheduled.
C is allocated 3 units of the resource
and its TSTART and TFIN are set to
0 and 6.
TEARL of D is set to 6 (TFIN of C).
A cannot be scheduled at this time as
only one resource unit is left.
TNOW will be set to 6 for next
iteration
3
B4 (2)
A 3 (2)
0
13
TNOW = 6
D 7 (2)
C 6 (3)
6
Resource allocation assuming 4 units are available
Activity
ACTIM
Duration
Resources Required
TEARL
TSTART
TFIN
C
13
6
3
0
0
6
D
7
7
2
6
6
13
Iteration
TNOW
Resources Available
Activities Allowed
1
0
41
CA
2
6
40
DA
3/23/2016
A
7
3
2
0
6
9
B
4
4
2
9
After Second Iteration
C is done and 3 units of the resource
are recovered. Resource capacity
upgraded to 4.
D and A may be scheduled and
enough resource units are available
TSTART and TFIN of D and A are
updated.
TEARL of B is set to 9 (TFIN of A).
TNOW will be set to 9 for next
iteration
3
B4 (2)
A 3 (2)
0
13
TNOW = 9
D 7 (2)
C 6 (3)
6
Resource allocation assuming 4 units are available
Activity
ACTIM
Duration
Resources Required
TEARL
TSTART
TFIN
C
13
6
3
0
0
6
D
7
7
2
6
6
13
A
7
3
2
0
6
9
Iteration
TNOW
Resources Available
Activities Allowed
1
0
41
CA
2
6
40
DA
3
9
20
B
3/23/2016
After Third Iteration
B
4
4
2
9
9
13
A is done and 2 units of the resource
are recovered. Resource capacity
upgraded to 2.
B may be scheduled and enough
resource units are available
TSTART and TFIN of B are
updated.
All activities have been scheduled.
The project completion time is the
maximum TFIN (13).
Results of applying Brooks’ algorithm
Same as Schedule 2 before
Resources
A (2) B (2)
C (3)
3
A 3 (2)
0
Schedule
D (2)
B4 (2)
13
Resource Profile
D 7 (2)
C 6 (3)
4
6
3/23/2016
3
Time
0
6
13
Time
Example
Bus Shelter Construction
Duration
A
B
C
D
E
F
G
H
I
Activity
Normal
Crash
Resources
PA
Shelter Plate
Shelter Walls
Shelter Roof
Roof Beam
Excavation
Curb and Gutter
Shelter Seat
Paint
Sign work
2
1
2
3
2
2
1
1
1
1
1
1
1
1
1
1
1
1
2
1
2
2
3
3
2
1
2
E
A
B,D
B
-E
D,F
C,G
H
Assume only 4 units of the resource are
available.
3/23/2016
B(1,1)
D(3,2)
C(2,2)
A(2,2)
E(2,3)
H(1,1)
I(1,2)
F(2,3)
G(1,2)
Critical path is EABDCHI, completion time is 12 without resource
constraints
Activity
ACTIM
Duration
Resources Required
TEARL
TSTART
TFIN
E
12
2
3
0
0
2
A
10
2
2
2
2
4
B
8
1
1
4
4
5
D
7
3
2
5
6
9
F
5
2
3
2
4
6
C
4
2
2
9
9
11
G
3
1
2
9
9
10
H
2
1
1
11
11
12
I
1
1
2
12
12
13
Iteration
TNOW
Resources Available
Activities Allowed
1
0
41
E
2
2
42
AF
3
4
40
BF
4
5
1
D
5
6
42
D
6
9
40
CG
7
10
2
8
11
43
H
9
12
42
I
3/23/2016
With resource constraints, completion time is
Example I
I. Given the following Project:
Activity
IPA
Duration
A
B
C
D
E
F
G
--A
--A,C
B,D
A,C
---
5
4
5
7
4
8
16
A5
Resources
Required
1
1
1
1
1
1
1
1. Determine the Critical Path
2. What is the minimum Project completion
time assuming no constraints? 16
3. Use Brooks' algorithm to schedule the
project assuming 3 units of the resource are
available. What is the project completion
time?
4. Use Brooks' algorithm to schedule the
project assuming 2 units of the resource are
available. What is the project completion
time?
B4
E4
C5
D7
F8
G 16
3/23/2016
A5
B4
E4
C5
D7
F8
Assuming 3 resource units
G 16
Activity
ACTIM
Duration
Resources required
TEARL
TSTART
TFIN
Iteration No.
TNOW
Resources available
ACT. ALLOW.
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G
16
16
1
0
0
16
A
16
5
1
0
0
5
1
0
3210
GAC
C
16
5
1
0
0
5
2
5
210
DFB
D
11
7
1
5
5
12
F
8
8
1
5
5
13
3
12
10
B
B
8
4
1
5
12
16
4
13
1
--
E
4
4
1
12
16
20
5
16
32
E
A5
B4
E4
C5
D7
F8
Assuming 2 resource units
G 16
Activity
ACTIM
Duration
Resources required
TEARL
TSTART
TFIN
Iteration No.
TNOW
Resources available
ACT. ALLOW.
3/23/2016
G
16
16
1
0
0
16
A
16
5
1
0
0
5
1
0
210
GAC
C
16
5
1
0
5
10
2
5
10
CB
D
11
7
1
10
10
17
F
8
8
1
10
16
24
3
10
10
D FB
B
8
4
1
5
17
21
4
16
10
FB
E
4
4
1
21
21
25
5
17
10
B
6
21
10
E