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EMGT 501
HW #2
Answer
4.4-6 (b)Basis Z X1 X2 X3 X4 X5 X6 RHS
Z 1 -2 -4 -3 0 0 0 0X4 0 3 4 2 1 0 0 60X5 0 2 1 2 0 1 0 40X6 0 1 3 2 0 0 1 80Z 1 1 0 -1 1 0 0 60X2 0 3/4 1 1/2 1/4 0 0 15X5 0 5/4 0 3/2 -1/4 1 0 25X6 0 -5/4 0 1/2 -3/4 0 1 35Z 1 11/6 0 0 5/6 2/3 0 230/3X2 0 1/3 1 0 1/3 -1/3 0 20/3X3 0 5/6 0 1 -1/6 2/3 0 50/3X6 0 -5/3 0 0 -2/3 -1/3 1 80/3
(c)
.3/230with
)3/80,0,0,3/50,3/20,0(*)*,*,*,*,*,( solution Optimal 654321
Z
xxxxxx
6.1-4 (a) Dual formulation becomes
Min
s.t.
711232
411121
1021513
54321
54321
54321
yyyyy
yyyyy
yyyyy
54321 2090402525 yyyyy
.0,,,, 54321 yyyyy
# 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.
(b) Dual formulation becomes
Mins.t.
21 156 yy
1
473
352
563
24
21
21
21
21
21
yy
yy
yy
yy
yy
# 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.
6.1-5 (a)
Mins.t.
21 112 yy
12
2
1
21
21
21
yy
yy
yy
0,0 21 yy
(b)
It is clear that Z*=0, y1*=0, y2*=0.
Project Scheduling: PERT-CPM
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.
ActivityActivity
DescriptionImmediate
PredecessorsEstimated
TimeABCDEFGHIJKLMN
-ABCCED
E,GCF,IJJH
K,L
2 weeks4 weeks
10 weeks6 weeks4 weeks5 weeks7 weeks9 weeks7 weeks8 weeks4 weeks5 weeks2 weeks6 weeks
ExcavateLay the foundationPut up the rough wallPut up the roofInstall the exterior plumbingInstall the interior plumbingPut up the exterior sidingDo the exterior paintingDo the electrical workPut up the wallboardInstall the flooringDo the interior paintingInstall the exterior fixturesInstall the interior fixtures
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.
AON (Activity-on-Arc):
Each activity is represented by a node.
The arcs are used to show the precedence relationships between the activities.
AB
C
E
M N
START
FINISH
H
G
D
J
I
F
LK
410
4 76
7
9
8
54
62
nodearc
5
0
0(Estimated)Time
2
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
Path and Length
Critical Path
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.
ES :
Earliest Start time for a particular activity
EF :
Earliest Finish time for a particular activity
AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2ES=0EF=2
ES=2EF=6ES=6
EF=16
ES=16EF=20
ES=16EF=23
ES=16EF=22ES=22EF=29
ES=20EF=25
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.
AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2ES=0EF=2
ES=2EF=6
ES=6EF=1
ES=16EF=20
ES=16EF=23
ES=16EF=22ES=22EF=29
ES=20EF=25 ES=25
EF=33ES=33EF=37
ES=33EF=38
ES=38EF=44
ES=29EF=38
ES=38EF=40
ES=44EF=44
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.
LS:
Latest Start time for a particular activity
LF:
Latest Finish time for a particular activity
AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2
LS=0LF=0
LS=0LF=2
LS=2LF=6
LS=6LF=16
LS=16LF=20
LS=20LF=25
LS=25LF=33
LS=18LF=25
LS=34LF=38
LS=33LF=38
LS=38LF=44
LS=33LF=42
LS=42LF=44
LS=26LF=33
LS=20LF=26
LS=44LF=44
Latest
Start Time
Earliest
Start Time
S=( 2, 2 )F=( 6, 6 )
Latest
Finish Time
Earliest
Finish Time
S=(20,20)F=(25,25)
AB
C
E
MN
START
FINISH
H
G
D
J
I
LK
4
10
4 76
7
9
8
54
62
5
0
0
2
S=(0,0)F=(0,0)S=(0,0)
F=(2,2) S=(2,2)F=(6,6)
S=(16,16)F=(20,20)
S=(25,25)F=(33,33)
S=(16,18)F=(23,25)
S=(33,34)F=(37,38)
S=(33,33)F=(38,38)
S=(38,38)F=(44,44)
S=(29,33)F=(38,42)
S=(38,42)F=(40,44)
S=(22,26)F=(29,33)
S=(16,20)F=(22,26)
S=(44,44)F=(44,44)
F
S=(6,6)F=(16,16)
Critical Path
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.
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.
22
6
4
op
pmo
o pmo
Beta distribution
Mean :
Variance:
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
A
B
C
2
13
2
9
1
2
6
3
8
18
2
4
10
9
1
4
1
OE: Optimistic EstimateM : Most Likely EstimatePE: Pessimistic Estimate
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
91
Project Time 44p 92 p
94
94
Approximating Probability of Meeting Deadline
T = a project time has a normal distribution
with mean and ,
d = a deadline for the project = 47 weeks.
44p 92 p
13
4447
p
pdK
Assumption:A probability distribution of project time is a normal distribution.
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.
K
K
Time - Cost Trade - OffsCrashing an activity refers to taking special costly measures to reduce the time of an activity below its normal value.
Crash
Normal
Crashtime
Normaltime
Crash cost
Normal cost
Activitycost
Activitytime
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.
2
000,430$000,490$
Cost($1,000)
Crash Costper Week
Saved
A
B
J
$100
$ 50
$ 30
Activity
Time(week)
MaximumReduction
in Time(week)N NC C
1
2
2
$180
$320
$430
$280
$420
$490
2
4
8
1
2
6
N: Normal C: Crash
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.
= the reduction in the time of activity j
by crashing it
= the project time for the FINISH node
jx
.000,60000,50000,100 NBA xxxZ
40FINISHy
FINISHy
= 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:
iy
ix
Ex4
.4 EEF xyy
Immediate predecessor of activity J:
Activity F, which has time =
Activity I, which has time =
Relationship between these activities:
Fx5
Ix7
IIJ
FFJ
xyy
xyy
7
5
.000,60000,50000,100 NBA xxxZ
Minimize
.3,,2,1 NBA xxx
.0,0,,0,0
0,,0,0
FINISHNCB
NBA
yyyy
xxx
The Complete linear programming model
.40FINISHy
CCD
BBC
AB
xyy
xyy
xy
10
4
20
HHM xyy 9
One Immediate
Predecessor
Two ImmediatePredecessors
EEH
GGH
xyy
xyy
4
7
NNFINISH
MMFINISH
xyy
xyy
6
2
Finish Time = 40
Total Cost = $4,690,000
EMGT 501
HW #3
10.3-5
10.4-5
Due Day: Sep 27
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?
ActivityImmediate
PredecessorsEstimatedDuration
A - 5 monthsB - 1 monthsC B 2 monthsD A, C 4 monthsE A 6 monthsF D, E 3 months
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.
START FINISH
A C E F
B D
Using the PERT three-estimate approach, Sharon has obtained the following estimates of the duration of each activity.
Optimistic Most Likely PessimisticActivity 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.
(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)?
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