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EMGT 501
HW #2
SolutionsChapter 6 - SELF TEST 21
Chapter 6 - SELF TEST 22
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s.t.
u20u30u51Max 321
6u2uu
3uu2u0.5
4uu
321
321
31
0u ,u ,u 321
Ch. 6 – 21a.
0u ,5.0u ,4u
75z*3
*2
*1
*
1u
15
10
0
15
0
2u
30
01
0
30
0
3u
20
11/4
3/4
45/2
5/2
1s
0
1-1/4
-3/4
15/2
15/2
2s
0
01/2
-1/2
15
15
41/2
3/2
1530
0
2s
0
00
1
0
0
BCBasis
1u
2u
3s
75jz
jj cz
b.
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c. From the zj values for the surplus variables we see that the optimal primal solution is x1=15/2, x2=15, and x3=0.
d. The optimal value for the dual is shown in part b to equal 75. Substituting x1=15/2 and x2=15 into the primal objective function, we find that it gives the same value.
4(15/2)+3(15)=75
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Ch. 6 – 22a.
s.t.
x5x01Max 21
175xx3
100x
100x
20x
20x
21
2
1
2
1
0x ,x 21
s.t.
u175u100u100u20u20-Min 54321
5uuu-
10u3uu-
542
531
0u ,u ,u ,u ,u 54321
b.
The optimal solution to this problem is given by: u1=0, u2=0, u3=0, u4=5/3, and u5=10/3
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c. The optimal number of calls is given by the negative of the dual prices for the dual: x1=25 and x2=100.Commission=$750.
d. u4=5/3: $1.67 commission increase for an additional call for product 2.u5=10/3: $3.33 commission increase for an additional hour of selling time per month.
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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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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
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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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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
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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
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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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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
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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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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
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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
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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
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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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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
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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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22
6
op
6
pm4o
o pmo
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
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
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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
91
Project Time 44p 92 p
94
94
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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.
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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.
K
K
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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
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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.
2
000,430$000,490$
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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
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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
jx
.000,60000,50000,100 NBA xxxZ
40FINISHy
FINISHy
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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:
iy
ix
Ex4
.4 EEF xyy
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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
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.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
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.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
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EMGT 501
HW #3Chapter 10 - SELF TEST 7
Chapter 10 - SELF TEST 18
Due Day: Oct 3
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Ch. 10 – 7A project involving the installation of a computer system comprises eight activities. The following table lists immediate predecessors and activity times (in weeks).
ActivityImmediate
Predecessor Time
ABCDEFGH
--A
B,CDE
B,CF,G
36254393
a. Draw a project network.b. What are the critical activities?c. What is the expected project completion time?
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Ch. 10 – 18The manager of the Oak Hills Swimming Club is planning the club’s swimming team program. The first team practice is scheduled for May 1. The activities, their immediate predecessors, and the activity time estimates (in weeks) are as follows.
ActivityImmediatePredecessor
ABCDEFGHI
-AA
B,CBADG
E,H,F
Description
Meet with boardHire coachesReserve poolAnnounce programMeet with coachesOrder team suitsRegister swimmersCollect feesPlan first practice
Optimistic
142121111
Time (weeks)Most Probable
164232221
Pessimistic
286343331
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a. Draw a project network.b. Develop an activity schedule.c. What are the critical activities, and what is the expected
project completion time?d. If the club manager plans to start the project on
February1, what is the probability the swimming program will be ready by the scheduled May 1 date (13 weeks)? Should the manager begin planning the swimming program before February 1?