Transcript Slide 1
Lecture
5
Project Management
Chapter 17
1
Project Management
How is it different?
Limited time frame
Narrow focus, specific objectives
Why is it used?
Special needs
Pressures for new or improves products or services
Definition of a project
Unique, one-time sequence of activities designed
to accomplish a specific set of objectives in a
limited time frame
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Project Management
What are the Key Metrics
Time
Cost
Performance objectives
What are the Key Success Factors?
Top-down commitment
Having a capable project manager
Having time to plan
Careful tracking and control
Good communications
3
Project Management
What are the tools?
Work breakdown structure
Network diagram
Gantt charts
4
Project Manager
Responsible for:
Work
Human Resources
Communications
Quality
Time
Costs
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Key Decisions
Deciding which projects to implement
Selecting a project manager
Selecting a project team
Planning and designing the project
Managing and controlling project resources
Deciding if and when a project should be
terminated
6
Ethical Issues
Temptation to understate costs
Withhold information
Misleading status reports
Falsifying records
Compromising workers’ safety
Approving substandard work
http://www.pmi.org/
7
PERT and CPM
PERT: Program Evaluation and Review Technique
CPM: Critical Path Method
Graphically displays project activities
Estimates how long the project will take
Indicates most critical activities
Show where delays will not affect project
PERT and CPM have been used to plan, schedule, and control
a wide variety of projects:
R&D of new products and processes
Construction of buildings and highways
Maintenance of large and complex equipment
Design and installation of new systems
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PERT/CPM
PERT/CPM
used to plan the scheduling of individual
activities that make up a project.
Projects may have as many as several
thousand activities.
Complicating factor in carrying out the
activities
some activities depend on the completion of
other activities before they can be started.
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PERT/CPM
Project managers rely on PERT/CPM to help
them answer questions such as:
What is the total time to complete the project?
What are the scheduled start and finish dates for
each specific activity?
Which activities are critical?
must be completed exactly as scheduled to keep
the project on schedule?
How long can non-critical activities be delayed
before they cause an increase in the project
completion time?
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Planning and Scheduling
Activity
0
2
4
6
8
10 12
14
16
18
Locate new
facilities
Interview staff
Hire and train staff
Select and order
furniture
Remodel and install
phones
Furniture setup
Move in/startup
11
20
Project Network
Project network
constructed to model the precedence of the
activities.
Nodes represent activities
Arcs represent precedence relationships of the
activities
Critical path for the network
a path consisting of activities with zero slack
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Project Network – An Example
8 weeks
Locate
facilities
A
6 weeks
Order
furniture
B
F
11 weeks
Remodel
E
S
4 weeks
Interview
C
3 weeks
Furniture
setup
G
Move
in
1 week
9 weeks
Hire and
train
D
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Management Scientist Solution
Critical Path
Path
Length
Slack
(weeks)
A-B-F-G
A-E-G
C-D-G
18
20
14
2
0
6
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Uncertain Activity Times
Three-time estimate approach
the time to complete an activity assumed to
follow a Beta distribution
An activity’s mean completion time is:
t = (a + 4m + b)/6
a = the optimistic completion time estimate
b = the pessimistic completion time estimate
m = the most likely completion time estimate
An activity’s completion time variance is
2 = ((b-a)/6)2
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Uncertain Activity Times
In the three-time estimate approach, the critical
path is determined as if the mean times for the
activities were fixed times.
The overall project completion time is assumed to
have a normal distribution
with mean equal to the sum of the means of
activities along the critical path, and
variance equal to the sum of the variances of
activities along the critical path.
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Example
Immediate
Activity Predecessor
Optimistic
Time (a)
Most Likely
Time (m)
Pessimistic
Time (b)
A
--
4
6
8
B
--
1
4.5
5
C
A
3
3
3
D
A
4
5
6
E
A
0.5
1
1.5
F
B,C
3
4
5
G
B,C
1
1.5
5
H
E,F
5
6
7
I
E,F
2
5
8
J
D,H
2.5
2.75
4.5
K
G,I
3
5
7
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Management Scientist Solution
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Key Terminology
Network activities
ES: early start
EF: early finish
LS: late start
LF: late finish
Used to determine
Expected project duration
Slack time
Critical path
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The Network Diagram (cont’d)
Path
Critical path
The longest path; determines expected project duration
Critical activities
Sequence of activities that leads from the starting node
to the finishing node
AON path: S-1-2-6-7
Activities on the critical path
Slack
Allowable slippage for path; the difference the length
of path and the length of critical path
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Advantages of PERT
Forces managers to organize
Provides graphic display of activities
Identifies
Critical activities
Slack activities
4
2
1
5
6
3
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Limitations of PERT
Important activities may be omitted
Precedence relationships may not be correct
Estimates may include a fudge factor
May focus solely on critical path
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Linear Programming
George Dantzig – 1914 -2005
Concerned with optimal allocation of limited
resources such as
Materials
Budgets
Labor
Machine time
among competitive activities
under a set of constraints
George Dantzig – 1914 -2005
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Product Mix Example (from session 1)
Type 1
Type 2
Profit per unit
$60
$50
Assembly time per
unit
4 hrs
10 hrs
Inspection time per
unit
2 hrs
Storage space per
unit
3 cubic ft
Resource
Amount available
Assembly time
100 hours
Inspection time
22 hours
Storage space
39 cubic feet
1 hr
3 cubic ft
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Linear Programming Example
Variables
Maximize 60X1 + 50X2
Subject to
Objective function
4X1 + 10X2 <= 100
2X1 + 1X2 <= 22
Constraints
3X1 + 3X2 <= 39
Non-negativity Constraints
X1, X2 >= 0
What is a Linear Program?
• A LP is an optimization model that has
• continuous variables
• a single linear objective function, and
• (almost always) several constraints (linear equalities or inequalities)
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Linear Programming Model
Decision variables
Objective Function
unknowns, which is what model seeks to determine
for example, amounts of either inputs or outputs
goal, determines value of best (optimum) solution among all feasible (satisfy
constraints) values of the variables
either maximization or minimization
Constraints
restrictions, which limit variables of the model
limitations that restrict the available alternatives
Parameters: numerical values (for example, RHS of constraints)
Feasible solution: is one particular set of values of the decision
variables that satisfies the constraints
Feasible solution space: the set of all feasible solutions
Optimal solution: is a feasible solution that maximizes or minimizes
the objective function
There could be multiple optimal solutions
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Another Example of LP: Diet Problem
Energy requirement : 2000 kcal
Protein requirement : 55 g
Calcium requirement : 800 mg
Food
Energy (kcal)
Protein(g)
Calcium(mg)
Oatmeal
110
4
2
Price per
serving($)
3
Chicken
Eggs
Milk
Pie
205
160
160
420
32
13
8
4
12
54
285
22
24
13
9
24
Pork
260
14
80
13
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Example of LP : Diet Problem
oatmeal: at most 4 servings/day
chicken: at most 3 servings/day
eggs: at most 2 servings/day
milk: at most 8 servings/day
pie: at most 2 servings/day
pork: at most 2 servings/day
Design an optimal diet plan
which minimizes the cost per day
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Step 1: define decision variables
x1 = # of oatmeal servings
x2 = # of chicken servings
x3 = # of eggs servings
x4 = # of milk servings
x5 = # of pie servings
x6 = # of pork servings
Step 2: formulate objective function
• In this case, minimize total cost
minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
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Step 3: Constraints
Meet energy requirement
110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000
Meet protein requirement
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55
Meet calcium requirement
2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800
Restriction on number of servings
0x14, 0x23, 0x32, 0x48, 0x52, 0x62
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So, how does a LP look like?
minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
subject to
110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55
2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800
0x14
0x23
0x32
0x48
0x52
0x62
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Optimal Solution – Diet Problem
Using LINDO 6.1
Food
Oatmeal
# of servings
4
Chicken
Eggs
Milk
Pie
0
0
6.5
0
Pork
2
Cost of diet = $96.50 per day
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Optimal Solution – Diet Problem
Using Management Scientist
Food
Oatmeal
# of servings
4
Chicken
Eggs
Milk
Pie
0
0
6.5
0
Pork
2
Cost of diet = $96.50 per day
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Guidelines for Model Formulation
Understand the problem thoroughly.
Describe the objective.
Describe each constraint.
Define the decision variables.
Write the objective in terms of the decision
variables.
Write the constraints in terms of the decision
variables
Do not forget non-negativity constraints
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Product Mix Problem
•
•
•
•
•
•
Floataway Tours has $420,000 that can be used to
purchase new rental boats for hire during the summer.
The boats can be purchased from two different
manufacturers.
Floataway Tours would like to purchase at least 50 boats.
They would also like to purchase the same number from
Sleekboat as from Racer to maintain goodwill.
At the same time, Floataway Tours wishes to have a total
seating capacity of at least 200.
Formulate this problem as a linear program
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Product Mix Problem
Boat
Maximum
Builder
Cost
Speedhawk Sleekboat
Silverbird Sleekboat
Catman
Racer
Classy
Racer
$6000
$7000
$5000
$9000
Expected
Seating
3
5
2
6
Daily
Profit
$ 70
$ 80
$ 50
$110
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Product Mix Problem
Define the decision variables
x1 = number of Speedhawks ordered
x2 = number of Silverbirds ordered
x3 = number of Catmans ordered
x4 = number of Classys ordered
Define the objective function
Maximize total expected daily profit:
Max: (Expected daily profit per unit) x (Number of units)
Max: 70x1 + 80x2 + 50x3 + 110x4
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Product Mix Problem
Define the constraints
(1) Spend no more than $420,000:
6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000
(2) Purchase at least 50 boats:
x1 + x2 + x3 + x4 > 50
(3) Number of boats from Sleekboat equals number
of boats from Racer:
x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0
(4) Capacity at least 200:
3x1 + 5x2 + 2x3 + 6x4 > 200
Nonnegativity of variables:
xj > 0, for j = 1,2,3,4
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Product Mix Problem - Complete Formulation
Max 70x1 + 80x2 + 50x3 + 110x4
s.t.
6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000
x1 + x2 + x3 + x4 > 50
Boat
# purchased
x1 + x2 - x3 - x4 = 0
Speedhawk
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3x1 + 5x2 + 2x3 + 6x4 > 200
Silverbird
0
x1, x2, x3, x4 > 0
Catman
0
Classy
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Daily profit = $5040
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Marketing Application: Media Selection
Advertising Media
# of potential
customers reached
Cost ($) per
advertisement
Max times available
per month
Exposure Quality
Units
Day TV
1000
1500
15
65
Evening TV
2000
3000
10
90
Daily newspaper
1500
400
25
40
Sunday newspaper
2500
1000
4
60
Radio
300
100
30
20
Advertising budget for first month = $30000
At least 10 TV commercials must be used
At least 50000 customers must be reached
Spend no more than $18000 on TV adverts
Determine optimal media selection plan
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Media Selection Formulation
Step 1: Define decision variables
Step 2: Write the objective in terms of the decision variables
DTV = # of day time TV adverts
ETV = # of evening TV adverts
DN = # of daily newspaper adverts
SN = # of Sunday newspaper adverts
R = # of radio adverts
Maximize 65DTV+90ETV+40DN+60SN+20R
Step 3: Write the constraints in terms of the decision variables
DTV
ETV
DN
SN
R
+
25
<=
4
<=
30
<=
30000
0
DN
25
SN
2
R
30
Exposure = 2370 units
Availability of
Media
Budget
+
ETV
>=
10
1500DTV
+
3000ETV
<=
18000
TV Constraints
1000DTV
+
2000ETV
>=
50000
Customers reached
+
100R
<=
ETV
DTV
2500SN
+
10
10
3000ETV
+
1000SN
<=
DTV
+
1500DN
+
15
Value
1500DTV
+
400DN
<=
Variable
300R
DTV, ETV, DN, SN, R >= 0
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Applications of LP
Product mix planning
Distribution networks
Truck routing
Staff scheduling
Financial portfolios
Capacity planning
Media selection: marketing
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Possible Outcomes of a LP
A LP is either
Infeasible – there exists no solution which satisfies
all constraints and optimizes the objective function
or, Unbounded – increase/decrease objective
function as much as you like without violating any
constraint
or, Has an Optimal Solution
Optimal values of decision variables
Optimal objective function value
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Infeasible LP – An Example
minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16
x33+5x34
Subject to
x11+x12+x13+x14=100
x21+x22+x23+x24=200
x31+x32+x33+x34=150
x11+x21+x31=80
x12+x22+x32=90
x13+x23+x33=120
x14+x24+x34=170
xij>=0, i=1,2,3; j=1,2,3,4
Total demand exceeds total supply
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Unbounded LP – An Example
maximize 2x1 + x2
subject to
-x1 +
x2 1
x1 - 2x2 2
x1 , x 2 0
x2 can be increased indefinitely without violating any
constraint
=> Objective function value can be increased indefinitely
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Multiple Optima – An Example
maximize x1 + 0.5 x2
subject to
2x1 + x2 4
x1 + 2x2 3
x1 , x2 0
• x1= 2, x2= 0, objective function = 2
• x1= 5/3, x2= 2/3, objective function = 2
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