2019-10-29
Project Management
Class – 8
Resource
Allocation
Learning Outcomes
Ø What is the fundamental trade-off between project cost
and project time?
Ø Expediting a project
Ø Resource leveling
Ø How to allocate limited resources to specific
activities/projects when there are competing demands for
the same limited resources?
Ø Case study review
1
2019-10-29
Expediting a Project
Ø Consider a “bridge construction” project.
Ø Digging the drainage ditch takes 10 days with a team of 3
workers.
Ø There are 3 other workers that can be added if they stop their
own task “concrete pouring”.
Ø They (i.e., 6 workers) can complete digging in 5 days at the cost
of delaying the task “concrete pouring”.
Ø The PM can also rent an excavator machine to get the job done
in 2 days.
Ø What is the key trade-off in making the decision?
Expediting a Project: Example
Relocation of a Hospital
Level 0
Organizing and Site Preparation
Physical Facilities and Infrastructure
Select administration staff
Purchase and deliver equipment
Site selection and survey
Construct hospital
Select medical equipment
Develop information system
Prepare final construction plans
Install medical equipment
Bring utilities to site
Train nurses and support staff
Level 1
Level 2
Interview applicants for
nursing and support staff
2
2019-10-29
Relocation of Hospital
Activity List, Duration, and Precedence Relationship
Activity
Immediate
Predecessors
Activity times
(wks)
0
start
start
A
B
B
A
12
9
10
10
24
10
C
D
A
E, G, H
F, I, J
K
35
40
15
4
6
0
START
Organizing and Site Preparation
A. Selecting administrative staff
B. Site selection and survey
C. Select medical equipment
D. Prepare final construction plans
E. Bring utilities to site
F. Interview applicants for nursing and support staff
Physical Facilities and Infrastructure
G. Purchase and deliver equipment
H. Construct hospital
I. Develop information system
J. Install medical equipment
K. Train nurses and support staff
FINISH
Developing the Schedule and Finding Critical Path
12
I 27
The Critical Path takes
69 weeks
15
0
A
12
12 F 22
12
Path
A-I-K
A-F-K
A-C-G-J-K
B-D-H-J-K
B-E-J-K
Time (wks)
33
28
67
69
43
63
6
10
12
Start
C
22
22
10
0
B
9
9
9
D 19
10
9
K 69
G
57
Finish
35
19
H
40
59
59
J
63
4
E 33
24
3
2019-10-29
Project Crashing
Project Crashing: Shortening (or expediting) some activities within a
project to reduce overall project completion time.
Ø Project Costs
ü Direct Costs: Labor, materials, and any other costs directly related to
activities
ü Indirect Costs: Administration, depreciation, financial, and other
overhead costs that can be avoided by reducing total project time
ü Penalty Costs: If project extends beyond some specific date.
Cost to Crash
Cost to Crash
Ø Normal time (NT) is the time necessary to complete an activity
under normal conditions.
Ø Normal cost (NC) is the activity cost associated with the normal
time.
Ø Crash time (CT) is the shortest possible time to complete an
activity.
Ø Crash cost (CC) is the activity cost associated with the crash
time.
𝐂𝐨𝐬𝐭 𝐭𝐨 𝐜𝐫𝐚𝐬𝐡 𝐩𝐞𝐫 𝐩𝐞𝐫𝐢𝐨𝐝 =
𝐂𝐂 − 𝐍𝐂
𝐍𝐓 − 𝐂𝐓
4
2019-10-29
Which Activities to Crash?
Ø Which activities and for how long should
be crashed?
0
A
12
12
15
12 F 22
12
Path
A-I-K
A-F-K
A-C-G-J-K
B-D-H-J-K
B-E-J-K
63
12
Start
C
22
22
10
0
B
9
9
9
D 19
G
57
Finish
35
19
10
9
K 69
6
10
Time (wks)
33
28
67
69
43
I 27
H
40
59
59
J
63
4
E 33
24
Which Activities to Crash?
Algorithm
Ø Which activities and for how long should be crashed?
Ø Determining the Minimum Cost Schedule:
1. Determine the project’s critical path(s).
2. Find the activity or activities on the critical path(s) with the lowest cost of
crashing per unit of time.
3. Reduce the time for this activity until…
a. It cannot be further reduced or
b. Until another path becomes critical, or
c. The increase in direct costs exceeds the savings that result from shortening
the project.
4. Repeat this procedure until the increase in direct costs is larger than the savings
generated by shortening the project.
5
2019-10-29
Hospital Example: Finding Cost to Crash
CC – NC
NT – CT
DIRECT COST AND TIME DATA FOR THE HOSPITAL PROJECT
Activity
Normal
Time (NT)
(weeks)
Normal
Cost
(NC)($)
Crash Time
(CT)(weeks)
Crash Cost
(CC)($)
Maximum
Time
Reduction
(week)
Cost of
Crashing per
Week ($)
A
12
$12,000
11
$13,000
1
1,000
B
9
50,000
7
64,000
2
7,000
C
10
4,000
5
7,000
5
600
D
10
16,000
8
20,000
2
2,000
E
24
120,000
14
200,000
10
8,000
F
10
10,000
6
16,000
4
1,500
G
35
500,000
25
530,000
10
3,000
H
40
1,200,000
35
1,260,000
5
12,000
I
15
40,000
10
52,500
5
2,500
J
4
10,000
1
13,000
3
1,000
K
6
30,000
5
34,000
1
4,000
Totals
$1,992,000
$2,209,500
Analyzing Cost – Time Trade-Offs
Hospital Example
Ø Project completion time was 69 weeks (Critical Path Analysis)
Ø Suppose that project indirect costs are $8,000 per week
Ø After 65 weeks, the Regional Hospital Board imposes cost of $20,000
per week if the hospital is not fully operational.
Ø Question: What is the saving per week shortening the project?
ü Saving per week (if completion time > 65) = $28,000
ü Saving per week (if completion time ≤ 65) = $8,000
6
2019-10-29
Analyzing Cost – Time Trade-Offs
Hospital Example
Ø Determine the minimum-cost schedule for the Hospital project.
Ø Project completion time = 69 weeks
Ø Project cost = $2,624,000
Direct
=
$1,992,000
Indirect
=
69($8,000) = $552,000
Penalty
=
(69 – 65)($20,000) = $80,000
A–I–K
A–F–K
A–C–G–J–K
B–D–H–J–K
B–E–J–K
33 weeks
28 weeks
67 weeks
69 weeks
43 weeks
Hospital Example
Which Activity To Crash?
DIRECT COST AND TIME DATA FOR THE HOSPITAL PROJECT
Activity
Normal
Time (NT)
(weeks)
Normal
Cost
(NC)($)
Crash Time
(CT)(weeks)
Crash Cost
(CC)($)
Maximum
Time
Reduction
(week)
Cost of
Crashing per
Week ($)
A
12
$12,000
11
$13,000
1
1,000
B
9
50,000
7
64,000
2
7,000
C
10
4,000
5
7,000
5
600
D
10
16,000
8
20,000
2
2,000
E
24
120,000
14
200,000
10
8,000
F
10
10,000
6
16,000
4
1,500
G
35
500,000
25
530,000
10
3,000
H
40
1,200,000
35
1,260,000
5
12,000
I
15
40,000
10
52,500
5
2,500
J
4
10,000
1
13,000
3
1,000
K
6
30,000
5
34,000
1
4,000
Totals
$1,992,000
$2,209,500
7
2019-10-29
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 1
Ø Step 1. The critical path is B–D–H–J–K.
Ø Step 2. The cheapest activity to crash per week is J at $1,000.
Ø Step 3. Crash activity J by its limit of three weeks because the critical
path remains unchanged. The new expected path times are
A–C–G–J–K: 64 weeks
Path
Time (wks)
B–D–H–J–K: 66 weeks
B–E–J–K: 40 weeks
A-I-K
33
A-F-K
28
Ø The net savings are 3($28,000) – 3($1,000) = $81,000. A-C-G-J-K
67
B-D-H-J-K
69
Ø The total project costs are now $2,624,000 - $81,000 = $2,543,000.
B-E-J-K
43
Analyzing Cost – Time Trade-Offs: Hospital Example
I
15
STAGE 1
A
12
Start
B
9
The critical path does not change
F
10
K
6
C
10
G
35
D
10
H
40
Finish
J
1
E
24
8
2019-10-29
Analyzing Cost – Time Trade-Offs: Hospital Example
DIRECT COST AND TIME DATA FOR THE HOSPITAL PROJECT
Activity
Normal
Time (NT)
(weeks)
Normal
Cost
(NC)($)
Crash Time
(CT)(weeks)
Crash Cost
(CC)($)
Maximum
Time
Reduction
(week)
Cost of
Crashing per
Week ($)
A
12
$12,000
11
$13,000
1
1,000
B
9
50,000
7
64,000
2
7,000
C
10
4,000
5
7,000
5
600
D
10
16,000
8
20,000
2
2,000
E
24
120,000
14
200,000
10
8,000
F
10
10,000
6
16,000
4
1,500
G
35
500,000
25
530,000
10
3,000
H
40
1,200,000
35
1,260,000
5
12,000
I
15
40,000
10
52,500
5
2,500
J
4
10,000
1
13,000
3
1,000
K
6
30,000
5
34,000
1
4,000
Totals
$1,992,000
$2,209,500
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 2
Ø Step 1. The critical path is B–D–H–J–K with 66 weeks.
Ø Step 2. The cheapest activity to crash per week is now D at $2,000.
Ø Step 3. Crash D by two weeks.
Ø The first week of reduction in activity D saves $28,000.
Ø Crashing D by a second week saves only $8,000 in indirect costs.
Ø Updated path times are
Path
A–C–G–J–K: 64 weeks and B–D–H–J–K: 64 weeks
A-I-K
A-F-K
Ø The net savings are $28,000 + $8,000 – 2($2,000) = $32,000.
A-C-G-J-K
Ø Total project costs are now $2,543,000 – $32,000 = $2,511,000.
B-D-H-J-K
B-E-J-K
Time (wks)
33
28
64
66
40
9
2019-10-29
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 2
I
15
A
12
F
10
Start
B
9
There are two critical paths
K
6
C
10
G
35
D
8
H
40
Finish
J
1
E
24
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 3
Step 1. The critical paths are B–D–H–J–K and A-C-G-J-K with 64 weeks
Step 2. Activities eligible to be crashed:
I
15
A
12
Start
B
9
F
10
K
6
C
10
G
35
D
8
H
40
E
24
Finish
J
1
A, B
A, H
$8,000
$13,000
C, B
$7,600
C, H
$12,600
G, B
G, H
$10,000
$15,000
K
$4,000
10
2019-10-29
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 3
Ø Step 1. The critical paths are B-D-H-J-K and A-C-G-J-K with 64 weeks
Ø Step 2. Activities eligible to be crashed:
(A, B); (C, B) or to crash Activity K
ü Candidates are those whose costs of crashing are less than the potential
savings; $8,000 per week.
Ø Step 3. We choose activity K to crash 1 week at $4,000 per week.
Updated path times are: A–C–G–J–K: and B–D–H–J–K with 63 weeks
Ø Net savings are $8,000 - $4,000 = $4,000
Ø Total project costs are $2,511,000 – $4,000 = $2,507,000
Analyzing Cost – Time Trade-Offs: Hospital Example
I
15
STAGE 3
A
12
Start
B
9
F
10
K
5
C
10
G
35
D
8
H
40
Finish
J
1
E
24
11
2019-10-29
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 4
Step 1. The critical paths are B–D–H–J–K and A-C-G-J-K with 63 weeks
Step 2. Activities eligible to be crashed:
I
15
A
12
Start
B
9
F
10
K
5
C
10
G
35
D
8
H
40
Finish
J
1
A, B
A, H
$8,000
$13,000
C, B
$7,600
C, H
$12,600
G, B
G, H
$10,000
$15,000
E
24
Analyzing Cost – Time Trade-Offs: Hospital Example
STAGE 4
Ø Step 1. The critical paths are still B-D-H-J-K and A-C-G-J-K with 63 weeks
Ø Step 2. Activities eligible to be crashed: (B,C) @ $7,600 per week.
Ø Step 3. Crash activities B and C by two weeks. Updated path times are
A–C–G–J–K: 61 weeks and B–D–H–J–K: 61 weeks
Ø The net savings are 2($8,000) – 2($7,600) = $800.
Ø Total project costs are now $2,507,000 – $800 = $2,506,200.
12
2019-10-29
Analyzing Cost-Time Trade-Offs: Hospital Example
Stage
Crash
Activity
Time
Redu
ction
Resulting
Critical
Path(s)
Project
Duration
Project
Direct
Costs
Crash
Cost
Added
Total
Indirect
Costs
Total
Penalty
Costs
Total
Project
Costs
0
—
—
B-D-H-J-K
69
1,992.0
—
552.0
80.0
2,624.0
1
J
3
B-D-H-J-K
66
1,992.0
3.0
528.0
20.0
2,543.0
2
D
2
64
1,995.0
4.0
512.0
0.0
2,511.0
3
K
1
63
1,999.0
4.0
504.0
0.0
2,507.0
4
B, C
2
61
2,003.0
15.2
488.0
0.0
2,506.2
B-D-H-J-K
A-C-G-J-K
B-D-H-J-K
A-C-G-J-K
B-D-H-J-K
A-C-G-J-K
Resource Leveling
Ø When the project is large and contains many resource over-allocations,
resource leveling must be accomplished:
Ø A technique in which start and finish dates are adjusted based on resource
constraints with the goal of balancing demand for resources with the
available supply.
Ø Purpose is to create a smoother distribution of resource usage.
Ø Resource leveling aims to minimize the period-by-period variations in
resource loading by shifting tasks within their slack allowances.
Ø Leveling is done by delaying or splitting tasks until the resources assigned to
them are no longer over-allocated.
13
2019-10-29
Activity Slack Allowance
S = 36
Question: How long each
activity can be delayed without
delaying the entire project?
12
48
I 27
15
S = LF – EF or LS – ES
63
S=0
S=2
S = 41
A
0
12
2 12 14
12 F 22
K 69
63
53 10 63
63
6 69
S=2
S=2
Start
C 22
12
14
10
S=0
S=0
22 G 57
24
59
24
Finish
35
S=0
0
B 9
D 19
9
0
9 9
9
10
19
19
19
H
40
59
59
59
59
J
63
4 63
S=0
S = 26 9 E 33
35 24 59
Resource Leveling Example
Ø Consider the following project.
Ø Find the project completion
time, all the paths from start to
end, and critical path.
Ø Find the number of required
resources (people) in every
week.
Ø Suggest a resource leveling
plan to smoothen the resource
allocation over project
schedule.
Activity
Immediate Duration # of resources
Predecessor (Week)
(People)
A
-
4
2
B
-
4
1
C
-
4
2
D
A
2
5
E
B
3
2
F
C
2
2
G
D
3
5
H
G
5
3
14
2019-10-29
Resource Leveling Example (Cont.)
Completion
Path
Time
ADGH
14
BE
7
CF
6
A
S
t
a
r
t
H
G
D
FI
N
IS
H
E
B
C
F
Resource Leveling Example (Cont.)
D (5)
G (5)
H (3)
A (2)
E (2)
B (1)
F (2)
C (2)
Week
1
2
3 4
5
6
7
8
9 10 11 12 13 14 15
Resources (people)
5
5
5 5
9
9
7
5
5
3
3
3
3
3
Over-allocated period
15
2019-10-29
Resource Leveling Example (Cont.)
D (5)
G (5)
H (3)
A (2)
E (2)
B (1)
Activity slack
F (2)
C (2)
Week
1
2
3 4
5
6
7
8
9 10 11 12 13 14 15
Resources (people)
5
5
5 5
9
9
7
5
5
3
3
3
3
3
Over-allocated period
Resource Leveling Example (Cont.)
D (5)
We can hire 5
workers at the
beginning without
laying off and
hiring over time!
G (5)
H (3)
A (2)
B (1)
E (2)
Activity slack
F (2)Activity slack
C (2)
Week
1
2
3 4
5
6
7
8
9 10 11 12 13 14 15
Resources (people)
5
5
5 5
9
7
5
9
7
5
7
5
5
5
3
5
3
5
3
5
3
5
3
5
16
2019-10-29
Allocating Scare Resources to Several Projects
Ø When the allocating scarce resources is extended to where several projects
are being carried out concurrently, the size and complexity of the problem
increase
Ø With several projects, we can link them together with pseudoactivities
Ø Pseudoactivities have duration but require no resources
Ø The use of pseudoactivities allows a set of projects to be linked and dealt
with as though it were a single project
§ The individual projects are interrelated by specifying predecessor/successor
relationships
§ They appear to be parts of one project
Multiple Projects Connected with Pseudoactivities
Ø Project manager faces the
problem of choosing
between different
outcomes that result from
different priority rules
Ø Must also deal with
different arrangements
and durations of
pseudoactivities (i.e.,
leveling rule)
17
2019-10-29
Criteria of Priority Rules
There are many measurable criteria to help select a priority rule
1. Schedule slippage
§ Amount project or set of projects delayed by application of a leveling rule
§ The PM must trade-off penalty costs or displeasure of clients against the
cost of adding resources
2. Resource utilization: extent to which resources are over or underworked
3. In-process inventory: amount of unfinished work in the system
Ø The minimum slack rule is probably the best overall priority rule according to
research.
Ø It gives the best combination of minimum project slippage, minimum
resource idle-time, and minimum in-process inventory.
Problems with Traditional Project Management
Ø When planning for a project, estimates for task durations are required.
Ø To increase the probability and high-confidence that the task completing on
time we consider additional safety time beyond the work content time
required to be embedded within the task duration.
Ø The more safety in a task the more there is a tendency to behave in the
following ways:
§ Not starting the task until the last moment (Student Syndrome)
§ Delaying (or pacing) completion of the task (Parkinson’s Law)
Ø As a result, the safety which was included at the planning stage is wasted
and tasks over-run.
18
2019-10-29
Goldratt’s Critical Chain
Ø Critical Chain Project Management was developed by Eli Goldratt in
response to many projects being dogged by poor performance
manifested in
§
§
§
§
Longer than expected durations
Frequently missed deadlines
Increased costs in excess of budget
Substantially less deliverables than originally promised
Ø Goldratt’s focus in the Critical Chain is on a single project with multiple
demands on a scarce resource
Ø The logic extends to the multiproject case without alteration
Goldratt’s Critical Chain (Cont.)
Ø Consider the three AOA
network diagrams.
Ø In scenario 1, there is only 1
path.
Ø In scenario 2, completion of
BCDE depends on three
activities.
Ø In scenario 3, there are two
completely independent paths
each consisting of 5 tasks.
19
2019-10-29
How Long?
Ø Each task takes 10 days
Ø What is the completion time
for each project?
§ All three would have the same
duration of 50 days
Ø Simple project with five
tasks takes the same time as
complex one with 11 tasks!
Part of Problem
Ø Part of the problem is the assumption that the activity times are
known with certainty
Ø Assume all activities are normally distributed
§ Mean of 10
§ Standard deviation of three
Ø Each is simulated 200 times
20
2019-10-29
Analysis
Ø This example clearly demonstrates how the commonly made
assumption of known activity times in practice can lead to quite
unrealistic project deadlines
Ø The results would have been even more dramatic had the activities
required some common resources
Ø Similarly, the results would have been more dramatic and realistic had
a nonsymmetrical distribution been used to model the activity times
Multitasking
Ø Multitasking is assigning team members to multiple projects and
having them allocate their time across these projects
Ø There is typically a penalty or cost associated with switching from
working on one project to another
21
2019-10-29
Multitasking (Cont.)
Ø Alternative Gantt Charts for Projects A and B
Switching from project to
project is likely to extend
activity times. Eliminating
such switching costs further
increases the benefits
associated with the Gantt
chart shown in Figure (b).
Resolving These Problems
Ø Goldratt suggests that the key to resolving this is to schedule the start
of new projects based on the availability of bottleneck resources
Ø He further suggests that time buffers be created between the
bottleneck resource and the resources that feed it
Ø He also suggests reducing the amount of safety time added to
individual tasks and then adding some fraction of the safety time
reduced back into the system as safety buffer for the entire project
22
2019-10-29
The Critical Chain
Ø Another limitation is the dependency
between resources and tasks is often
ignored
Ø Using traditional approaches, A1-C1 is the
critical path
Ø What if A1 and A2 are not independent
Ø Then path A1-C1 increases to 22 days, or
path A2-B1 increases to 18 days
If A1 is done first è A2-B1 will be finished in 7 + 5 + 6 = 18
If A2 is done first è A1-C1 will be finished in 5 + 7 + 10 = 22
Addressing Problem
Ø Need to consider both precedence relationships and resource
dependencies
Ø Goldratt proposes thinking in terms of the longest chain of
consecutively dependent tasks where such dependencies can arise
§ Referred to as critical chain
Ø There are two potential sources that can delay the project
§ Delay in the tasks that make up the critical chain
§ Delay in activity feeding the critical chain that results in delay of the
critical chain
23
2019-10-29
Project and Feeder Buffers
Project Buffer: A project buffer is inserted at the end of the project network
between the last task and the completion date.
§ Any delays on the longest chain of dependent tasks will consume some
of the buffer but will leave the completion date unchanged and so
protect the project.
Project and Feeder Buffers
Feeding Buffers: delays on paths of tasks feeding into the longest chain can
impact the project by delaying a subsequent task on the Critical Chain.
§ To protect against this, feeding buffers are inserted between the last
task on a feeding path and the Critical Chain.
24