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CE413 Assignment #4
CE413: CONSTRUCTION METHODS
Due 11/9/2023
Use Excel spreadsheets for each question and submit the hardcopies of the spreadsheets as your
answers. Also, make certain that you clearly state all of the days considered for resource allocation.
1. Solve the following by the SERIES method of resource allocation. The resource limits are as follows:
5M and 3H.
6
7
B
4
2
3
1
1
A
1
0 0
3M,2H
2
2
0 1
2M,1H
C
6
2
2
0 0
2M,2H
D
5
0 1
1M
I
3
11
14
12 0 1 15
2M,1H
11
12
6
7
E
2
6
13 6 7
3M
8
8
F
2
8
10
10 4 2 12
3M,1H
K
2
14
16
15 1 1 17
4M,3H
8
15
G
3
8
11
11 3 3 14
1M,2H
8
8
M
1
17
18
17 0 0 18
1M,2H
J
10
13
3
12 1 2 15
4M,1H
L
3
14
17
14 0 0 17
3M,1H
H
6
14
0 0 14
1M,1H
2. Solve this resource scheduling problem with the PARALLEL method. The resource limits are as
follows: 4C and 3L.
I
2
6
8
13 11 11 19
1L
4
6
1
1
A
1
0
0
3L
2
2
2
2
B
2
0 0
3C,2L
4
4
4
4
D
3
0 2
1C,3L
C
4
0
0
7
9
8
8
3C
4
9
5
2C
8
8
F
3
0
0
6
11
H
6
3
11 0 5
3C
J
10
17
2 2 19
1C,1L
11
11
G
11
13
2
11 0 0 13
2C,2L
9
14
L
9
14
5
14 5 5 19
2C,1L
2L
E
2
0
7
9
K
13
19
6
13 0 0 19
1C,2L
M
19
1
19 0 0
1C
20
20
CE413 Assignment #4
3. Solve the following network by the Brooks method. The resource limit is 6.
2
5
B
3
0
3
5
8
E
5
5
13 5 8
4H
4
4
8
8
8
4H
1
1
A
1
0
0
2
2
2
6
C
2
4
2H
2
2
F
7
0
0
3H
1H
D
6
G
3
0
8
8
0
8
12 4
1H
4
H
3
15
18 3 3
6H
10
18
18
21
J
21
1
21 0 0
1H
15
15
11
15
I
6
15
15 0
0
22
22
21
21
1H
5H
4. Level the resources assigned to the activities on the network below. Please make sure that the
improvement factors for all free floats (and back floats) available for activities should be calculated.
(Note: Do not reschedule any activity with an IF of “zero” for resource leveling.)
1
1
A
4
0
2R
5
5
5
5
C
6
0
7R
11
11
F
4
11
11 0
4R
PS
1
3
B
2
0
3R
3
5
5
7
3
5
D
4
0
5R
E
2
0
4R
5
7
14
17
I
15
2
15 0
3R
17
17
15
15
9
11
G
3
9
12 3
3R
H
3
11
14 3
4R
12
15
5
7
J
10
2
3R
15
17
K
17
2
17 0
3R
19
19
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CE 413: Construction Methods
General
• While activities may be scheduled initially based upon physical
constraints and with the general assumption of an early start, the
necessity to manage limited resources imposes further constraints on
the scheduling of activities.
3
CE 413: Construction Methods
General
Typical Project Labor Utilization
4
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CE 413: Construction Methods
Definitions
• Resource Allocation: the distribution of limited resources to the
activities of a project by shifting activity timing
• Resource Leveling: efficient use of the required resources within the
constraint of a fixed project duration by shifting activity timing
5
CE 413: Construction Methods
Definitions
Resource Allocation: the distribution of limited resources to the activities of a project by
shifting activity timing
Resource Leveling: efficient use of the required resources within the constraint of a fixed
project duration by shifting activity timing
6
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CE 413: Construction Methods
Resource Allocation Methods
• Series Method
• Parallel Method
• Brooks Method
7
CE 413: Construction Methods
Series Method
• Activity is scheduled to start immediately following it’s predecessors as resource
capacity allows
• Activities CANNOT be interrupted (suspended)
• Selection criteria sequence for competing activities:
• Activity already begun has priority
• If activities not begun, the one with earliest late start is given priority
• If same late start, priority goes to activity with least total float
• If tie persists, priority goes to activity with largest number of resources
• If tie still persists, begin with the activity that was first in the input order
8
4
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CE 413: Construction Methods
Resource Allocation
2
2
1
1
6
0 6
6
7
7
1 8
6
6
10
0 10
7
8
9
1 10
10
10
16
0 16
2
0 2
16
16
2
8
4
6 10
4
11
7
7 14
4
10
8
6 14
8
14
17
0 17
10
6 16
EF
4
12
6
8 14
LS
9
Resource Allocation Example – Initial Schedule
10
5
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Resource Allocation Example – Initial Schedule
Resource limits:
5 masons
2 helpers
11
Resource Allocation Example – Series Method
∕ 2/1
Resource limits:
5 masons
2 helpers
12
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Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
Resource limits:
5 masons
2 helpers
Activity A followed by both
activities B and C since
their combined resource
requirements (4 masons, 2
helpers) do not exceed
what is available.
13
Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
Resource limits:
5 masons
2 helpers
After activity C is
complete, then activities
F, G, and H are ready to
start. According to
earliest late start priority,
G should be initiated
first, but it’s resource
requirements (when
combined with activity B)
exceed limits.
14
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Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
∕
3/03/03/0
Resource limits:
5 masons
2 helpers
Therefore, activity F
starts next.
15
Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
∕
3/03/03/0
Resource limits:
5 masons
2 helpers
After activity B is
completed, activities D
and E join G and H on the
list of ready activities.
According to the earliest
late start criteria, E
should go first, but it
requires too many
masons.
16
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Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
1/0
∕
∕
∕
Resource limits:
5 masons
2 helpers
3/03/03/0
1/21/21/21/2
17
Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
Resource limits:
5 masons
2 helpers
After activity D is
completed
(making activity J
ready to start),
activity E is
started because
it has an earlier
late start than
activities H and J.
18
9
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Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
∕
Resource limits:
5 masons
2 helpers
2/2 2/2
19
Resource Allocation Example – Series Method
∕ 2/1
∕ 2/1 2/1 2/1 2/1
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
∕
Resource limits:
5 masons
2 helpers
2/2 2/2
20
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Resource Allocation Example – Series Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/1 2/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/12/1 2/12/1 2/1
∕
21
Resource Allocation Example – Series Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/1 2/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/12/1 2/12/1 2/1
∕
22
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Resource Allocation Example – Series Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/1 2/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/12/1 2/12/1 2/1
∕
3/2 3/2
∕
23
Resource Allocation Example – Series Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/1 2/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/03/0
∕
1/21/21/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/12/1 2/12/1 2/1
∕
3/2 3/2
∕
2/1
∕
24
12
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CE 413: Construction Methods
Parallel Method
• Activity is scheduled to start immediately following it’s predecessors as resource
capacity allows
• Activities CAN be interrupted (suspended)
• Selection criteria sequence for competing activities:
• Activity with the earliest late start is given priority
• If same late start, priority goes to activity with least total float
• If tie persists, priority goes to activity with largest number of resources
• If number of resources is the same, priority is given to the activity already begun
• If tie still persists, begin with the activity that was first in the input order
25
CE 413: Construction Methods
Resource Allocation
2
2
1
1
6
0 6
6
7
7
1 8
6
6
10
0 10
7
8
9
1 10
10
10
16
0 16
2
0 2
16
16
2
8
4
6 10
4
11
7
7 14
4
10
8
6 14
8
14
17
0 17
10
6 16
EF
4
12
6
8 14
LS
26
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Resource Allocation Example – Parallel Method
∕ 2/1
Resource limits:
5 masons
2 helpers
Only Activity A is eligible for
scheduling
27
Resource Allocation Example – Parallel Method
∕ 2/1
2/1 2/1
∕
2/1 2/1
Resource limits:
5 masons
2 helpers
Both Activities are scheduled
on days 2 and 3
28
14
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Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
Resource limits:
5 masons
2 helpers
Activity B has not been
completed and Activities F, G,
and H are included for
scheduling.
3/03/0
29
Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
3/0
Resource limits:
5 masons
2 helpers
On day 6, Activities D and E
are eligible for scheduling with
Activities F, G, and H.
3/03/0
1/2
30
15
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Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
Resource limits:
5 masons
2 helpers
3/0 3/0
On day 7, Activity J is eligible
for scheduling because Activity
D is completed.
3/03/0
1/2
2/2
31
Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
Resource limits:
5 masons
2 helpers
3/0 3/0 3/0
On day 8, Activity L still is not
eligible for scheduling.
3/03/0
1/2
∕
2/2 2/2
32
16
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Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
Resource limits:
5 masons
2 helpers
3/03/0
1/2
∕
1/2
2/2 2/2
33
Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
∕
1/2
Resource limits:
5 masons
2 helpers
On day 10, Activity K is eligible
for scheduling. Activities G and
K have same late start so
check on the activity with least
TF
2/2 2/2
2/1
34
17
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Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
∕
∕
Resource limits:
5 masons
2 helpers
1/2
1/1 1/1
2/2 2/2
2/1 2/1 2/1
35
Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
∕
∕
Resource limits:
5 masons
2 helpers
1/2
1/1 1/1
2/2 2/2
2/1 2/1 2/1 2/1 2/1
36
18
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Resource Allocation Example – Parallel Method
∕ 2/1
∕ 2/1 2/1 2/12/1
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
∕
∕
∕
Resource limits:
5 masons
2 helpers
1/2
1/1 1/1
2/2 2/2
2/1 2/1 2/1 2/1 2/12/1
37
Resource Allocation Example – Parallel Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/12/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
1/2
1/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/1 2/1 2/1 2/12/1
∕
38
19
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Resource Allocation Example – Parallel Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/12/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
∕
3/0 3/0 3/0 3/0
3/0
3/03/0
∕
1/2
1/2
1/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/1 2/1 2/1 2/12/1
∕
3/2 3/2
∕
39
Resource Allocation Example – Parallel Method
∕ 2/1
Resource limits:
∕ 2/1 2/1 2/12/1
5 masons
2 helpers
∕ 2/1 2/1
1/0
∕
3/03/03/0 3/0
∕
3/03/0
3/0
∕
1/2
1/2
1/21/2
∕
1/1 1/1
∕
2/2 2/2
∕
2/1 2/12/1 2/12/1 2/1
∕
3/2 3/2
∕
2/1
∕
40
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CE 413: Construction Methods
Brooks Method
• The Brooks method, a tabular procedure, produces the same result as
the series method due to having the same standard priorities and
assumptions. Constraints can be imposed, however, to yield different
results.
41
CE 413: Construction Methods
Key Terms in Brooks Method
• ACTIM value: difference between the late start date and the project
completion
• TNow: current time under consideration; the particular day being
scheduled
• Act.ready: those activities eligible for scheduling because their
immediate predecessors have already been scheduled
• Act.sched: activities selected for scheduling from the list of eligible
activities
42
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CE 413: Construction Methods
Brooks Method
• Activities are arranged (prioritized) across the top of the table
according to the largest ACTIM value. When activities have the same
ACTIM value, priority is given to the activity with the least total float.
• Activities are scheduled following completion of their predecessors
according to the highest ACTIM value and as allowed by the resource
constraints.
• Typically, activities are not suspended.
43
CE 413: Construction Methods
Resource Allocation
2
2
1
1
6
0 6
6
7
7
1 8
6
6
10
0 10
7
8
9
1 10
10
10
16
0 16
2
0 2
16
16
2
8
4
6 10
4
11
7
7 14
4
10
8
6 14
8
14
17
0 17
10
6 16
EF
4
12
6
8 14
LS
44
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Resource Allocation Example – Brooks Method
Lesser total float.
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
4
A
2
1
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
6
E,J
2
1
4
C
1
2
3
C
3
0
2
2
1
C F,G,H K,L
1
3
2
1
2
1
5
2
45
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
4
A
2
1
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
6
E,J
2
1
4
C
1
2
3
C
3
0
2
C
1
1
2
1
1
5
2
A
A
2
1
First TNow is day 1
F,G,H K,L
3
2
2
1
Resource limits allow the ready
activities to start. Not an issue
here with only one activity.
46
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Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
6
E,J
2
1
4
C
1
2
3
C
3
0
2
C
1
1
2
1
1
5
2
A
A
2
1
2
F,G,H K,L
3
2
2
1
Activity duration is added to
TStart to obtain TFin
Next smallest TFin
becomes new TNow
47
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
6
E,J
2
1
4
C
1
2
3
C
3
0
2
C
1
1
2
1
1
5
2
A
A
2
1
2
F,G,H K,L
3
2
2
1
B,C
Resource limits allow both ready
B,C
activities B and C to start
4
Sums of resources in use for
2
scheduled activities
48
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Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
3
C
3
0
2
C
1
1
2
1
1
5
2
A
A
2
1
2
F,G,H K,L
3
2
2
1
TStart taken from TNow for both
newly scheduled activities and
durations added for TFin
B,C
B,C
4
2
49
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
3
C
3
0
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
B,C
B,C
4
2
BGFH
B,F
5
1
F,G,H K,L
3
2
2
1
Next smallest TFin
becomes new TNow
B continues while G, F. and H become ready after end
of C. G has highest ACTIM but requires too many
resources. F is next and can be started. No room for H
at this time.
50
25
10/16/2023
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
3
C
3
0
4
7
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
B,C
B,C
4
2
BFGH
F,G,H K,L
3
2
2
1
B,F
5
1
51
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
3
C
3
0
4
7
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
6
B,C
B,C
4
2
F,G,H K,L
3
2
2
1
BFGH FEDGH
B,F FDG
5
5
1
2
52
26
10/16/2023
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
6
7
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
6
10
3
C
3
0
4
7
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
6
B,C
B,C
4
2
F,G,H K,L
3
2
2
1
BFGH FGHDE
B,F FDG
5
5
1
2
53
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
1
B
1
0
6
7
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
6
10
3
C
3
0
4
7
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
6
7
B,C
B,C
4
2
F,G,H K,L
3
2
2
1
BFGH FGHDE
B,F FDG
5
5
1
2
54
27
10/16/2023
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
7
11
1
B
1
0
6
7
2
D
2
2
2
A
2
1
2
4
6
E,J
2
1
4
C
1
2
6
10
3
C
3
0
4
7
2
C
1
1
2
1
1
5
2
A
A
2
1
2
4
6
7
B,C
B,C
4
2
F,G,H K,L
3
2
2
1
BFGH FGHDE GEJH
B,F FDG G,E
5
5
4
1
2
2
55
Resource Allocation Example – Brooks Method
A
16
B
15
E
11
D
10
J
9
C
9
K
7
G
7
F
6
H
5
L
3
M
1
A
B
E
D
J
C
K
G
F
H
L
M
1
2
1
1
2
4
A
2
1
2
6
4
B
3
0
7
11
1
B
1
0
6
7
2
D
2
2
10
12
2
A
2
1
2
4
6
4
E,J C
2
1
1
2
12 6
18 10
3
C
3
0
4
7
2
2
1
C F,G,H K,L
1
3
2
1
2
1
12 18 20
14 20 21
1
5
2
A
A
2
1
2
4
6
7
10
11
14
18
20
B,C BFGH FGHDE GHEJ EJH J,H K,H K,L
B,C B,F FDG G,E E,J J
K,H K
4
5
5
4
5
2
3
2
2
1
2
2
2
2
2
1
L
L
3
2
M
M
2
1
12
56
28
10/16/2023
CE 413: Construction Methods
Resource Leveling
Concept
• Minimum Moment Algorithm
• Resource requirements smoothed by making use of available free float
• Systematic evaluation of impact of using any free float associated with any activity
• Assumptions
• Activities cannot be interrupted
• Resource consumption rate is constant for each activity
• Network logic is not questioned unless final solution is unacceptable
57
CE 413: Construction Methods
Resource Leveling
Procedure
• Minimum‐moment algorithm procedure
• Begins with early start schedule
• Performs a backward pass, calculating the improvement factor (IF) for each
activity that could occur on the day in question
58
29
10/16/2023
CE 413: Construction Methods
Resource Leveling
Procedure
59
CE 413: Construction Methods
Resource Leveling
Procedure
• Only one resource leveled at a time
• Largest non‐negative IF determines the number of FF days to use
• If more than one activity, priority to largest IF activity
• If IF tied, priority to activity with largest resource/day requirement
• If resource/day the same, priority to activity using greatest number of FF days
• If FF days same, priority to activity with latest start date
• If latest start date same, priority by input order
• After a tie is broken and the priority activity is rescheduled, remaining activities
are re‐evaluated.
60
30
10/16/2023
CE 413: Construction Methods
Resource Leveling
Procedure
• Backward pass rescheduling produces back float—free float that
precedes the start of the activity. During the iterations of the
backward pass, an accounting of the back float must be maintained
so that a forward pass can be performed.
61
CE 413: Construction Methods
Resource Leveling
1
1 0
2
2
2
6 4
4
8
2
4 0
4
4
2
7
3
8
5
4
4 0
Example
8
8
8
14 6
9
15
8
8 0
11
11
8
15 7
10
17
11
11 0
13
13
13
13 0
15
15
15
15 0
17
17
17
17 0
18
18
62
31
10/16/2023
CE 413: Construction Methods
Resource Leveling
Example
63
CE 413: Construction Methods
Resource Leveling
Example
Important to note that activities F and H
cannot start any sooner, and activities C
and D cannot end any later (see network).
Begin backward pass, checking each
day until you encounter free float. At
that point, begin calculating
improvement factors (IFs) for the
corresponding activities. Activity H is
the first one encountered on Day 16.
64
32
10/16/2023
CE 413: Construction Methods
Resource Leveling
Example
IF(H,1) = 3 [(4+5+3) – 5-3(1)]=12
IF(H,2) = 3 [(4+5+3+5+3) – (5+1)-3(2)]=24
IF(H,3) = 3 [(4+5+3+5+3) – (1+1)-3(2)]=36
IF(H,4) = 3 [20 – (1+3) -3(2)]=30
IF(H,5) = 3 [20 – (3+3) – 6]=24
IF(H,6) = 3 [20 – (3+4) – 6] =21
IF(H,7) = 3 [20- (4+4)-6]=18
Highest non‐negative IF is associated with using
3 days of free float.
65
CE 413: Construction Methods
Resource Leveling
Example
Activity H shifted to use 3 days of free float and
total resource requirements are updated.
IF(F,1) = 4 [(4+5) – 5-4(1)]=0
IF(F,2) = 4 [(4+5) – 5-4(1)]=0
IF(F,3) = 4 [9 – (1+3)-4]=4
IF(F,4) = 4 [9 – (1+3) -4]=4
IF(F,5) = 4 [9 – 3 – 4]=8
IF(F,6) = 4 [9 – 3 – 4] =8
Continue backward pass. Activity
F free float encountered next. Use
highest non‐negative IF that
corresponds to the most free float.
66
33
10/16/2023
CE 413: Construction Methods
Resource Leveling
Example
Activity F shifted to use 6 days of free float and
total resource requirements are updated.
IF(C,1) = 24
IF(C,2) = 36
IF(C,3) = 36
IF(C,4) = 36
IF(D,1) = 0
IF(D,2) = 21
IF(D,3) = 21
IF(D,4) = 21
IF(D,5) = 21
Continue backward pass. Free float for
activities C and D encountered next.
Calculate IF for each considering other
in original position. Then compare for
highest non‐negative IF.
67
CE 413: Construction Methods
Resource Leveling
Example
Activity C shifted to use 4 days of free float and
total resource requirements are updated.
IF(D,1) = 0
IF(D,2) = 9
IF(D,3) = 9
IF(D,4) = -3
IF(D,5) = -3
Recalculate IF for activity D with
activity C in its new position.
68
34
10/16/2023
CE 413: Construction Methods
Resource Leveling
Example
Activity D shifted to use 3 days of free float and
total resource requirements are updated.
69
CE 413: Construction Methods
Resource Leveling
Example
Begin forward pass,
calculating IFs only for the
back float. Activities C and
D are the first to consider.
IF(C,1) = -12
IF(C,2) = -12
IF(C,3) = -12
IF(C,4) = -24
IF(D,1) = 0*
IF(D,2) = -9
IF(D,3) = -9
70
35
10/16/2023
CE 413: Construction Methods
Resource Leveling
Recalculate IF for
activity C with activity D
in it’s new position.
Example
Activity D shifted to use 1 days of back float and
total resource requirements are updated.
IF(C,1) = 0
IF(C,2) = -12
IF(C,3) = -24
IF(C,4) = -24
Can move C one day if
you like although it does
not help any.
71
CE 413: Construction Methods
Resource Leveling
IF(F,1) = 0
IF(F,2) = -4
IF(F,3) = -4
IF(F,4) = -8
IF(F,5) = -8
IF(H,1) = -12
IF(H,2) = -24
IF(H,3) = -24
Example
Can move F back one day
if you like although it
does not help any.
72
36
10/16/2023
CE 413: Construction Methods
Resource Leveling
Concept
73
37
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