EMGT 501

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EMGT 501 HW 2 Answer
2
0
20/3
X
0
5/6
0
1
-1/6
2/3
0
50/3
3
X
0
-5/3
0
0
-2/3
-1/3
1
80/3
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(c)
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6.1-4 (a) Dual formulation becomes Min s.t.
of constraints of Dual 3 of constraints of
Primal 5 So, Dual is better than Primal because
the size of B-1 in Dual is smaller than that of
Primal.
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(b) Dual formulation becomes Min s.t.
of constraints of Dual 5 of constraints of
Primal 3 So, Primal is better than Dual because
the size of B-1 in Primal is smaller than that of
Dual.
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6.1-5 (a) Min s.t.
(b) It is clear that Z0, y10, y20.
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Project Scheduling PERT-CPM
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PERT (Program evaluation and review technique)
and CPM (Critical Path Method) makes a managerial
technique to help planning and displaying the
coordination of all the activities.
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Activity Description
Immediate Predecessors
Estimated Time
Activity
A B C D E F G H I J K L M N
Excavate Lay the foundation Put up the rough
wall Put up the roof Install the exterior
plumbing Install the interior plumbing Put up the
exterior siding Do the exterior painting Do the
electrical work Put up the wallboard Install the
flooring Do the interior painting Install the
exterior fixtures Install the interior fixtures
2 weeks 4 weeks 10 weeks 6 weeks 4 weeks 5
weeks 7 weeks 9 weeks 7 weeks 8 weeks 4 weeks 5
weeks 2 weeks 6 weeks
- A B C C E D E,G C F,I J J H K,L
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Immediate predecessors For any given activity,
its immediate predecessors are the activities
that must be completed by no later than the
starting time of the given activity.
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AON (Activity-on-Arc) Each activity is
represented by a node. The arcs are used to
show the precedence relationships between the
activities.
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START
0
(Estimated) Time
2
A
node
arc
B
4
10
C
D
I
7
6
4
E
5
7
F
G
8
J
L
K
H
5
4
9
N
M
2
6
FINISH
0
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Path and Length
START A B C D G H M
FINISH 2 4 10 6 7 9 2
40 weeks START A B C E F J
K N FINISH 2 4 10 4
5 8 4 6 43 weeks START A B C
E F J L N FINISH
2 4 10 4 5 8 5 6 44 weeks
Critical Path
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Critical Path A project time equals the length
of the longest path through a project network.
The longest path is called critical
path. Activities on a critical path are the
critical bottleneck activities where any delay in
their completion must be avoided to prevent
delaying project completion.
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ES Earliest Start time for a particular
activity EF Earliest Finish time for a
particular activity
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0
START
2
ES0 EF2
A
B
4
ES2 EF6
10
C
ES6 EF16
D
I
7
ES16 EF23
6
4
ES16 EF22
ES16 EF20
E
7
5
ES22 EF29
ES20 EF25
G
F
8
J
H
9
L
K
5
4
M
2
N
6
FINISH
0
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If an activity has only a single immediate
predecessor, then ES EF for the immediate
predecessor.
Earliest Start Time Rule The earliest start time
of an activity is equal to the largest of the
earliest finish times of its immediate
predecessors. ES largest EF of the
immediate predecessors.
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0
START
ES0 EF2
2
A
B
4
ES2 EF6
C
10
ES6 EF1
D
I
7
ES16 EF23
6
4
ES16 EF22
ES16 EF20
E
7
5
ES22 EF29
ES20 EF25
G
F
8
J
ES25 EF33
H
9
ES29 EF38
K
4
L
5
ES33 EF37
ES33 EF38
M
2
ES38 EF40
N
6
ES38 EF44
FINISH
0
ES44 EF44
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LS Latest Start time for a particular
activity LF Latest Finish time for a particular
activity
Latest Finish Time Rule The latest finish time
of an activity is equal to the smallest of the
latest finish times of its immediate successors.
LF the smallest LS of immediate successors.
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0
START
LS0 LF0
2
LS0 LF2
A
4
B
LS2 LF6
C
10
LS6 LF16
D
I
7
6
4
LS18 LF25
LS20 LF26
LS16 LF20
E
7
5
LS26 LF33
LS20 LF25
G
F
8
J
LS25 LF33
H
9
K
4
LS33 LF42
L
5
LS34 LF38
LS33 LF38
M
2
LS42 LF44
N
6
LS38 LF44
FINISH
0
LS44 LF44
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Latest Start Time
Earliest Start Time
S( 2, 2 ) F( 6, 6 )
Latest Finish Time
Earliest Finish Time
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START
0
S(0,0) F(0,0)
2
S(0,0) F(2,2)
A
Critical Path
4
S(2,2) F(6,6)
B
S(6,6) F(16,16)
C
10
7
S(16,18) F(23,25)
D
I
4
S(16,16) F(20,20)
6
S(16,20) F(22,26)
E
5
7
S(20,20) F(25,25)
S(22,26) F(29,33)
G
F
S(25,25) F(33,33)
8
J
H
S(29,33) F(38,42)
9
K
4
S(33,34) F(37,38)
L
5
S(33,33) F(38,38)
M
2
S(38,42) F(40,44)
N
6
S(38,38) F(44,44)
FINISH
0
S(44,44) F(44,44)
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Slack A difference between the latest finish
time and the earliest finish time.
Slack LF - EF Each activity with zero
slack is on a critical path. Any delay along
this path delays a whole project completion.
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Three-Estimates
Most likely Estimate (m) an estimate of the
most likely value of time. Optimistic Estimate
(o) an estimate of time under the most
favorable conditions. Pessimistic Estimate
(p) an estimate of time under the most
unfavorable conditions.
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Beta distribution
Mean
Variance
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Mean critical path A path through the project
network becomes the critical path if each
activity time equals its mean.
Activity
OE
M
PE
Mean
Variance
2
1
3
2
A B C
2
1
8
4
4
9
6
18
10
OE Optimistic Estimate M Most Likely
Estimate PE Pessimistic Estimate
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Activities on Mean Critical Path
Mean
Variance
A B C E F J L N
2 4 10 4 5 8 5 6
1 4 1 1 1
Project Time
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Approximating Probability of Meeting Deadline
Assumption A probability distribution of project
time is a normal distribution.
T a project time has a normal distribution
with mean and , d
a deadline for the project 47 weeks.
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Using a table for a standard normal
distribution, the probability of meeting the
deadline is
P ( T d ) P ( standard normal )
1 - P( standard normal )
1 - 0.1587
0.84.
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Time - Cost Trade - Offs
Crashing an activity refers to taking special
costly measures to reduce the time of an activity
below its normal value.
Activity cost
Crash
Crash cost
Normal
Normal cost
Crash time
Normal time
Activity time
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Activity J Normal point time 8 weeks, cost
430,000. Crash point time 6 weeks, cost
490,000. Maximum reduction in time 8 - 6 2
weeks. Crash cost per week saved
30,000.
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Maximum Reduction in Time (week)
Crash Cost per Week Saved
Cost (1,000)
Time (week)
Activity
N
N
C
C
1 2 2
A B J
100 50 30
180 320 430
280 420 490
2 4 8
1 2 6
N Normal C Crash
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Using LP to Make Crashing Decisions
Let Z be the total cost of crashing activities.
A problem is to minimize Z, subject to the
constraint that its project duration must be less
than or equal to the time desired by a project
manager.
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the reduction in the time of activity j
by crashing it the
project time for the FINISH node
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the start time of activity j Duration
of activity j its normal time Immediate
predecessor of activity F Activity E, which has
duration Relationship between these
activities
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Immediate predecessor of activity J Activity F,
which has time Activity I, which has time
Relationship between these activities
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The Complete linear programming model
Minimize
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One Immediate Predecessor
Two Immediate Predecessors
Finish Time 40 Total Cost 4,690,000
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EMGT 501 HW 3 10.3-5 10.4-5 Due Day Sep 27
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10.3-5. You are given the following information
about a project consisting of six activities
(a) Construct the project network for this
project. (b) Find the earliest times, latest
times, and slack for each activity. Which of the
paths is a critical path? (c) If all other
activities take the estimated amount of time,
what is the maximum duration of activity D
without delaying the completion of the project?
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10.4-5 Sharon Lowe, vice president for
marketing for the Electronic Toys Company, is
about to begin a project to design an advertising
campaign for a new line of toys. She wants the
project completed within 57 days in time to
launch the advertising campaign at the beginning
of the Christmas season. Sharon has identified
the six activities (labeled A, B, , F) needed to
execute this project. Considering the order in
which these activities need to occur, she also
has constructed the following project network.
A
C
E
F
START
FINISH
B
D
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Using the PERT three-estimate approach, Sharon
has obtained the following estimates of the
duration of each activity.
Optimistic
Most Likely Pessimistic Activity
Estimate Estimate
Estimate A 12
days 12 days
12 days B 15 days
21 days 39 days
C 12 days
15 days 18 days D
18 days 27 days
36 days E
12 days 18 days
24 days F 2
days 5 days
14 days
(a) Find the estimate of the mean and variance of
the duration of each
activity. (b) Find the mean critical path.
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(c) Use the mean critical path to find the
approximate probability that the
advertising campaign will be ready to launch
within 57 days. (d) Now consider the other
path through the project network. Find
the approximate probability that this path will
be completed within 57 days. (e) Since
these paths do not overlap, a better estimate of
the probability that the project will
finish within 57 days can be obtained as
follows. The project will finish within 57 days
if both paths are completed within 57
days. Therefore, the approximate
probability that the project will finish within
57 days is the product of the
probabilities found in parts (c) and
(d). Perform this calculation. What does this
answer say about the accuracy of the
standard procedure used in part (c)?